blocks|key|446311|text|一种简单的方法是使用列的列表重新分配数据帧，并根据需要进行重新排列。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|446312|这就是你现在所拥有的：|446313|In+[6]:+df
Out[6]:
++++++++++0+++++++++1+++++++++2+++++++++3+++++++++4++++++mean
0++0.445598++0.173835++0.343415++0.682252++0.582616++0.445543
1++0.881592++0.696942++0.702232++0.696724++0.373551++0.670208
2++0.662527++0.955193++0.131016++0.609548++0.804694++0.632596
3++0.260919++0.783467++0.593433++0.033426++0.512019++0.436653
4++0.131842++0.799367++0.182828++0.683330++0.019485++0.363371
5++0.498784++0.873495++0.383811++0.699289++0.480447++0.587165
6++0.388771++0.395757++0.745237++0.628406++0.784473++0.588529
7++0.147986++0.459451++0.310961++0.706435++0.100914++0.345149
8++0.394947++0.863494++0.585030++0.565944++0.356561++0.553195
9++0.689260++0.865243++0.136481++0.386582++0.730399++0.561593

In+[7]:+cols+=+df.columns.tolist()

In+[8]:+cols
Out[8]:+[0L,+1L,+2L,+3L,+4L,+'mean']|code-block|syntax|javascript|446314|以您想要的任何方式重新排列cols。这就是我如何将最后一个元素移动到第一个位置：|offset|length|style|CODE|446315|In+[12]:+cols+=+cols[-1:]+%2B+cols[:-1]

In+[13]:+cols
Out[13]:+['mean',+0L,+1L,+2L,+3L,+4L]|446316|然后对数据帧重新排序，如下所示：|446317|In+[16]:+df+=+df[cols]++#++++OR++++df+=+df.ix[:,+cols]

In+[17]:+df
Out[17]:
+++++++mean+++++++++0+++++++++1+++++++++2+++++++++3+++++++++4
0++0.445543++0.445598++0.173835++0.343415++0.682252++0.582616
1++0.670208++0.881592++0.696942++0.702232++0.696724++0.373551
2++0.632596++0.662527++0.955193++0.131016++0.609548++0.804694
3++0.436653++0.260919++0.783467++0.593433++0.033426++0.512019
4++0.363371++0.131842++0.799367++0.182828++0.683330++0.019485
5++0.587165++0.498784++0.873495++0.383811++0.699289++0.480447
6++0.588529++0.388771++0.395757++0.745237++0.628406++0.784473
7++0.345149++0.147986++0.459451++0.310961++0.706435++0.100914
8++0.553195++0.394947++0.863494++0.585030++0.565944++0.356561
9++0.561593++0.689260++0.865243++0.136481++0.386582++0.730399|446318|entityMap^0|0|0|0|D|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@$K|10|L|11|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

One easy way would be to reassign the dataframe with a list of the columns, rearranged as needed. 

This is what you have now: 

<pre><code>In [6]: df
Out[6]:
 0 1 2 3 4 mean
0 0.445598 0.173835 0.343415 0.682252 0.582616 0.445543
1 0.881592 0.696942 0.702232 0.696724 0.373551 0.670208
2 0.662527 0.955193 0.131016 0.609548 0.804694 0.632596
3 0.260919 0.783467 0.593433 0.033426 0.512019 0.436653
4 0.131842 0.799367 0.182828 0.683330 0.019485 0.363371
5 0.498784 0.873495 0.383811 0.699289 0.480447 0.587165
6 0.388771 0.395757 0.745237 0.628406 0.784473 0.588529
7 0.147986 0.459451 0.310961 0.706435 0.100914 0.345149
8 0.394947 0.863494 0.585030 0.565944 0.356561 0.553195
9 0.689260 0.865243 0.136481 0.386582 0.730399 0.561593

In [7]: cols = df.columns.tolist()

In [8]: cols
Out[8]: [0L, 1L, 2L, 3L, 4L, 'mean']
</code></pre>

Rearrange <code>cols</code> in any way you want. This is how I moved the last element to the first position: 

<pre><code>In [12]: cols = cols[-1:] + cols[:-1]

In [13]: cols
Out[13]: ['mean', 0L, 1L, 2L, 3L, 4L]
</code></pre>

Then reorder the dataframe like this: 

<pre><code>In [16]: df = df[cols] # OR df = df.ix[:, cols]

In [17]: df
Out[17]:
 mean 0 1 2 3 4
0 0.445543 0.445598 0.173835 0.343415 0.682252 0.582616
1 0.670208 0.881592 0.696942 0.702232 0.696724 0.373551
2 0.632596 0.662527 0.955193 0.131016 0.609548 0.804694
3 0.436653 0.260919 0.783467 0.593433 0.033426 0.512019
4 0.363371 0.131842 0.799367 0.182828 0.683330 0.019485
5 0.587165 0.498784 0.873495 0.383811 0.699289 0.480447
6 0.588529 0.388771 0.395757 0.745237 0.628406 0.784473
7 0.345149 0.147986 0.459451 0.310961 0.706435 0.100914
8 0.553195 0.394947 0.863494 0.585030 0.565944 0.356561
9 0.561593 0.689260 0.865243 0.136481 0.386582 0.730399
</code></pre>

blocks|key|446330|text|这样如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|446331|df.insert(0,+'mean',+df['mean'])|code-block|syntax|javascript|446332|http://pandas.pydata.org/pandas-docs/stable/dsintro.html#column-selection-addition-deletion|offset|length|446333|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|2J|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|T|8|@]|9|@$I|U|J|V|1|W]]|A|$]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|H]]]]

How about:
<pre><code>df.insert(0, 'mean', df['mean'])
</code></pre>
<a href="http://pandas.pydata.org/pandas-docs/stable/dsintro.html#column-selection-addition-deletion" rel="noreferrer">http://pandas.pydata.org/pandas-docs/stable/dsintro.html#column-selection-addition-deletion</a>

blocks|key|228315|text|这个问题已经回答了before，但是reindex_axis现在被弃用了，所以我建议使用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|228316|df+=+df.reindex(sorted(df.columns),+axis=1)|code-block|syntax|javascript|228317|对于那些想要指定他们想要的顺序而不仅仅是对它们进行排序的人，以下是详细说明的解决方案：|228318|df+=+df.reindex(['the','order','you','want'],+axis=1)|228319|现在，您希望如何对列名列表进行排序实际上不是一个pandas问题，而是一个Python列表操作问题。有很多方法可以做到这一点，我认为this+answer有一种非常巧妙的方式。|228320|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/11067027/python-pandas-re-ordering-columns-in-a-dataframe-based-on-column-name|1|https://stackoverflow.com/a/51935892/1479269^0|I|C|9|6|0|0|0|0|0|O|6|1U|B|1|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@$9|1C|A|1D|1|1E]]|E|$]]|$1|Q|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]]]

This question has been answered <a href="https://stackoverflow.com/questions/11067027/python-pandas-re-ordering-columns-in-a-dataframe-based-on-column-name">before</a> but <code>reindex_axis</code> is deprecated now so I would suggest to use:
<pre><code>df = df.reindex(sorted(df.columns), axis=1)
</code></pre>
For those who want to specify the order they want instead of just sorting them, here's the solution spelled out:
<pre><code>df = df.reindex(['the','order','you','want'], axis=1)
</code></pre>
Now, how you want to sort the list of column names is really not a <code>pandas</code> question, that's a Python list manipulation question. There are many ways of doing that, and I think <a href="https://stackoverflow.com/a/51935892/1479269">this answer</a> has a very neat way of doing it.

blocks|key|228352|text|你也可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|228353|df+=+df[['mean',+'0',+'1',+'2',+'3']]|code-block|syntax|javascript|228354|您可以使用以下命令获取列的列表：|228355|cols+=+list(df.columns.values)|228356|输出将产生：|228357|['0',+'1',+'2',+'3',+'mean']|228358|然后，在将...which放入第一个函数之前，可以很容易地手动重新排列它|228359|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

You could also do something like this:

<pre><code>df = df[['mean', '0', '1', '2', '3']]
</code></pre>

You can get the list of columns with:

<pre><code>cols = list(df.columns.values)
</code></pre>

The output will produce:

<pre><code>['0', '1', '2', '3', 'mean']
</code></pre>

...which is then easy to rearrange manually before dropping it into the first function

blocks|key|446487|text|此函数使您不必列出数据集中的每个变量，而只需对其中的几个变量进行排序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|446488|def+order(frame,var):
++++if+type(var)+is+str:
++++++++var+=+[var]+#let+the+command+take+a+string+or+list
++++varlist+=[w+for+w+in+frame.columns+if+w+not+in+var]
++++frame+=+frame[var%2Bvarlist]
++++return+frame+|code-block|syntax|javascript|446489|它有两个参数，第一个是数据集，第二个是要放在前面的数据集中的列。|446490|因此，在我的示例中，我有一个名为Frame的数据集，其中包含变量A1、A2、B1、B2、Total和Date。如果我想把Total带到前面，那么我要做的就是：|446491|frame+=+order(frame,['Total'])|446492|如果我想把Total和Date放在前面，那么我会这样做：|446493|frame+=+order(frame,['Total','Date'])|446494|编辑：|446495|另一种有用的用法是，如果您有一个不熟悉的表，并且正在查找其中包含特定术语的变量，如VAR1、VAR2等。您可以执行如下代码：|446496|frame+=+order(frame,[v+for+v+in+frame.columns+if+"VAR"+in+v])|446497|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

This function avoids you having to list out every variable in your dataset just to order a few of them. 

<pre><code>def order(frame,var):
 if type(var) is str:
 var = [var] #let the command take a string or list
 varlist =[w for w in frame.columns if w not in var]
 frame = frame[var+varlist]
 return frame 
</code></pre>

It takes two arguments, the first is the dataset, the second are the columns in the data set that you want to bring to the front. 

So in my case I have a data set called Frame with variables A1, A2, B1, B2, Total and Date. If I want to bring Total to the front then all I have to do is: 

<pre><code>frame = order(frame,['Total'])
</code></pre>

If I want to bring Total and Date to the front then I do:

<pre><code>frame = order(frame,['Total','Date'])
</code></pre>

EDIT:

Another useful way to use this is, if you have an unfamiliar table and you're looking with variables with a particular term in them, like VAR1, VAR2,... you may execute something like: 

<pre><code>frame = order(frame,[v for v in frame.columns if "VAR" in v])
</code></pre>

blocks|key|446533|text|我自己也遇到了类似的问题，只是想补充一下我想要的东西。我喜欢用于更改列顺序的reindex_axis()+method。这是可行的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|446534|df+=+df.reindex_axis(['mean']+%2B+list(df.columns[:-1]),+axis=1)|code-block|syntax|javascript|446535|基于@Jorge的注释的另一种方法：|446536|df+=+df.reindex(columns=['mean']+%2B+list(df.columns[:-1]))|446537|虽然在微基准测试中，reindex_axis似乎比reindex稍微快一些，但我认为我更喜欢后者，因为它直接。|446538|entityMap^0|12|L|0|0|0|0|A|C|P|7|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

I ran into a similar question myself, and just wanted to add what I settled on. I liked the <code>reindex_axis() method</code> for changing column order. This worked:

<pre><code>df = df.reindex_axis(['mean'] + list(df.columns[:-1]), axis=1)
</code></pre>

An alternate method based on the comment from @Jorge:

<pre><code>df = df.reindex(columns=['mean'] + list(df.columns[:-1]))
</code></pre>

Although <code>reindex_axis</code> seems to be slightly faster in micro benchmarks than <code>reindex</code>, I think I prefer the latter for its directness.

blocks|key|228511|text|如果你知道另一列的位置，我相信@Aman's+answer是最好的。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|228512|如果你不知道mean的位置，只知道它的名字，你就不能直接求助于cols+=+cols[-1:]+%2B+cols[:-1]。下面是我能想到的下一个最好的办法：|style|CODE|228513|meanDf+=+pd.DataFrame(df.pop('mean'))
#+now+df+doesn't+contain+"mean"+anymore.+Order+of+join+will+move+it+to+left+or+right:
meanDf.join(df)+#+has+mean+as+first+column
df.join(meanDf)+#+has+mean+as+last+column|code-block|syntax|javascript|228514|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/13148611/1082349^0|F|E|0|0|6|4|V|S|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@$A|Z|B|10|F|G]|$A|11|B|12|F|G]]|9|@]|C|$]]|$1|H|3|I|5|J|7|13|8|@]|9|@]|C|$K|L]]|$1|M|3|-4|5|6|7|14|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

I believe <a href="https://stackoverflow.com/a/13148611/1082349">@Aman's answer</a> is the best if you know the location of the other column.

If you don't know the location of <code>mean</code>, but only have its name, you cannot resort directly to <code>cols = cols[-1:] + cols[:-1]</code>. Following is the next-best thing I could come up with:

<pre><code>meanDf = pd.DataFrame(df.pop('mean'))
# now df doesn't contain "mean" anymore. Order of join will move it to left or right:
meanDf.join(df) # has mean as first column
df.join(meanDf) # has mean as last column
</code></pre>

blocks|key|446588|text|简单地这样做，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|446589|df+=+df[['mean']+%2B+df.columns[:-1].tolist()]|code-block|syntax|javascript|446590|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Simply do,

<pre><code>df = df[['mean'] + df.columns[:-1].tolist()]
</code></pre>

blocks|key|228570|text|只需按所需的顺序分配列名：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|228571|In+[39]:+df
Out[39]:+
++++++++++0+++++++++1+++++++++2+++++++++3+++++++++4++mean
0++0.172742++0.915661++0.043387++0.712833++0.190717+++++1
1++0.128186++0.424771++0.590779++0.771080++0.617472+++++1
2++0.125709++0.085894++0.989798++0.829491++0.155563+++++1
3++0.742578++0.104061++0.299708++0.616751++0.951802+++++1
4++0.721118++0.528156++0.421360++0.105886++0.322311+++++1
5++0.900878++0.082047++0.224656++0.195162++0.736652+++++1
6++0.897832++0.558108++0.318016++0.586563++0.507564+++++1
7++0.027178++0.375183++0.930248++0.921786++0.337060+++++1
8++0.763028++0.182905++0.931756++0.110675++0.423398+++++1
9++0.848996++0.310562++0.140873++0.304561++0.417808+++++1

In+[40]:+df+=+df[['mean',+4,3,2,1]]|code-block|syntax|javascript|228572|现在，“mean”列出现在前面：|228573|In+[41]:+df
Out[41]:+
+++mean+++++++++4+++++++++3+++++++++2+++++++++1
0+++++1++0.190717++0.712833++0.043387++0.915661
1+++++1++0.617472++0.771080++0.590779++0.424771
2+++++1++0.155563++0.829491++0.989798++0.085894
3+++++1++0.951802++0.616751++0.299708++0.104061
4+++++1++0.322311++0.105886++0.421360++0.528156
5+++++1++0.736652++0.195162++0.224656++0.082047
6+++++1++0.507564++0.586563++0.318016++0.558108
7+++++1++0.337060++0.921786++0.930248++0.375183
8+++++1++0.423398++0.110675++0.931756++0.182905
9+++++1++0.417808++0.304561++0.140873++0.310562|228574|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Just assign the column names in the order you want them:

<pre><code>In [39]: df
Out[39]: 
 0 1 2 3 4 mean
0 0.172742 0.915661 0.043387 0.712833 0.190717 1
1 0.128186 0.424771 0.590779 0.771080 0.617472 1
2 0.125709 0.085894 0.989798 0.829491 0.155563 1
3 0.742578 0.104061 0.299708 0.616751 0.951802 1
4 0.721118 0.528156 0.421360 0.105886 0.322311 1
5 0.900878 0.082047 0.224656 0.195162 0.736652 1
6 0.897832 0.558108 0.318016 0.586563 0.507564 1
7 0.027178 0.375183 0.930248 0.921786 0.337060 1
8 0.763028 0.182905 0.931756 0.110675 0.423398 1
9 0.848996 0.310562 0.140873 0.304561 0.417808 1

In [40]: df = df[['mean', 4,3,2,1]]
</code></pre>

Now, 'mean' column comes out in the front:

<pre><code>In [41]: df
Out[41]: 
 mean 4 3 2 1
0 1 0.190717 0.712833 0.043387 0.915661
1 1 0.617472 0.771080 0.590779 0.424771
2 1 0.155563 0.829491 0.989798 0.085894
3 1 0.951802 0.616751 0.299708 0.104061
4 1 0.322311 0.105886 0.421360 0.528156
5 1 0.736652 0.195162 0.224656 0.082047
6 1 0.507564 0.586563 0.318016 0.558108
7 1 0.337060 0.921786 0.930248 0.375183
8 1 0.423398 0.110675 0.931756 0.182905
9 1 0.417808 0.304561 0.140873 0.310562
</code></pre>

blocks|key|228670|text|您需要按所需的顺序创建列的新列表，然后使用df+=+df[cols]以此新顺序重新排列列。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|228671|cols+=+['mean']++%2B+[col+for+col+in+df+if+col+!=+'mean']
df+=+df[cols]|code-block|syntax|javascript|228672|您还可以使用更通用的方法。在本例中，插入最后一列(用-1表示)作为第一列。|228673|cols+=+[df.columns[-1]]+%2B+[col+for+col+in+df+if+col+!=+df.columns[-1]]
df+=+df[cols]|228674|如果列存在于DataFrame中，您还可以使用这种方法以所需的顺序重新排序列。|228675|inserted_cols+=+['a',+'b',+'c']
cols+=+([col+for+col+in+inserted_cols+if+col+in+df]+
++++++++%2B+[col+for+col+in+df+if+col+not+in+inserted_cols])
df+=+df[cols]|228676|entityMap^0|L|D|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|T|$]]

You need to create a new list of your columns in the desired order, then use <code>df = df[cols]</code> to rearrange the columns in this new order.

<pre><code>cols = ['mean'] + [col for col in df if col != 'mean']
df = df[cols]
</code></pre>

You can also use a more general approach. In this example, the last column (indicated by -1) is inserted as the first column.

<pre><code>cols = [df.columns[-1]] + [col for col in df if col != df.columns[-1]]
df = df[cols]
</code></pre>

You can also use this approach for reordering columns in a desired order if they are present in the DataFrame.

<pre><code>inserted_cols = ['a', 'b', 'c']
cols = ([col for col in inserted_cols if col in df] 
 + [col for col in df if col not in inserted_cols])
df = df[cols]
</code></pre>

blocks|key|446712|text|只需键入要更改的列名，并为新位置设置索引。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|446713|def+change_column_order(df,+col_name,+index):
++++cols+=+df.columns.tolist()
++++cols.remove(col_name)
++++cols.insert(index,+col_name)
++++return+df[cols]|code-block|syntax|javascript|446714|对于您的情况，这将类似于：|446715|df+=+change_column_order(df,+'mean',+0)|446716|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Just type the column name you want to change, and set the index for the new location.

<pre><code>def change_column_order(df, col_name, index):
 cols = df.columns.tolist()
 cols.remove(col_name)
 cols.insert(index, col_name)
 return df[cols]
</code></pre>

For your case, this would be like:

<pre><code>df = change_column_order(df, 'mean', 0)
</code></pre>

blocks|key|446727|text|使用T怎么样？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|446728|df+=+df.T.reindex(['mean',+0,+1,+2,+3,+4]).T|code-block|syntax|javascript|446729|entityMap^0|2|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

How about using <code>T</code>?
<pre><code>df = df.T.reindex(['mean', 0, 1, 2, 3, 4]).T
</code></pre>

blocks|key|446748|text|您可以执行以下操作(借用Aman答案中的部分内容)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|446749|cols+=+df.columns.tolist()
cols.insert(0,+cols.pop(-1))

cols
>>>['mean',+0L,+1L,+2L,+3L,+4L]

df+=+df[cols]|code-block|syntax|javascript|446750|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You could do the following (borrowing parts from Aman's answer):

<pre><code>cols = df.columns.tolist()
cols.insert(0, cols.pop(-1))

cols
&gt;&gt;&gt;['mean', 0L, 1L, 2L, 3L, 4L]

df = df[cols]
</code></pre>

blocks|key|228817|text|我想从一个数据帧中将两列放在前面，而我并不确切知道所有列的名称，因为它们是从之前的pivot语句生成的。因此，如果您处于同样的情况:将您知道名称的列放在前面，然后让它们跟在“所有其他列”后面，我想出了以下一般解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|228818|df+=+df.reindex_axis(['Col1','Col2']+%2B+list(df.columns.drop(['Col1','Col2'])),+axis=1)|code-block|syntax|javascript|228819|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I wanted to bring two columns in front from a dataframe where I do not know exactly the names of all columns, because they are generated from a pivot statement before.
So, if you are in the same situation: To bring columns in front that you know the name of and then let them follow by &quot;all the other columns&quot;, I came up with the following general solution:
<pre><code>df = df.reindex_axis(['Col1','Col2'] + list(df.columns.drop(['Col1','Col2'])), axis=1)
</code></pre>

blocks|key|228861|text|Check+doc+here.+DataFrame.sort_index(axis=1)是非常干净的。然后是concat|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|228862|entityMap|0|LINK|mutability|MUTABLE|url|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.sort_index.html^0|G|S|1I|6|0|E|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@$9|S|A|T|1|U]]|E|$]]|$1|F|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

<code>DataFrame.sort_index(axis=1)</code> is quite clean.<a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.sort_index.html" rel="nofollow noreferrer">Check doc here</a>.
And then <code>concat</code>

blocks|key|446921|text|我喜欢Shoresh's+answer在您不知道列的位置时使用set功能来删除列，但是这对我的目的不起作用，因为我需要保持原始的列顺序(具有任意的列标签)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|446922|不过，通过使用boltons包中的IndexedSet，我实现了这一点。|446923|我还需要重新添加多个列标签，因此对于更一般的情况，我使用了以下代码：|446924|from+boltons.setutils+import+IndexedSet
cols+=+list(IndexedSet(df.columns.tolist())+-+set(['mean',+'std']))
cols[0:0]+=['mean',+'std']
df+=+df[cols]|code-block|syntax|javascript|446925|希望这对任何人在这个帖子中寻找一般的解决方案都是有用的。|446926|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/46166838/9075180|1|http://boltons.readthedocs.io/en/latest/setutils.html#^0|3|G|0|0|H|A|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|F|3|G|5|6|7|16|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|17|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|19|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

I liked <a href="https://stackoverflow.com/a/46166838/9075180">Shoresh's answer</a> to use set functionality to remove columns when you don't know the location, however this didn't work for my purpose as I need to keep the original column order (which has arbitrary column labels). 

I got this to work though by using <a href="http://boltons.readthedocs.io/en/latest/setutils.html#" rel="nofollow noreferrer">IndexedSet</a> from the boltons package.

I also needed to re-add multiple column labels, so for a more general case I used the following code:

<pre><code>from boltons.setutils import IndexedSet
cols = list(IndexedSet(df.columns.tolist()) - set(['mean', 'std']))
cols[0:0] =['mean', 'std']
df = df[cols]
</code></pre>

Hope this is useful to anyone searching this thread for a general solution.

blocks|key|446961|text|您可以使用可用于两个轴的reindex：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|446962|df
#+++++++++++0+++++++++1+++++++++2+++++++++3+++++++++4++++++mean
#+0++0.943825++0.202490++0.071908++0.452985++0.678397++0.469921
#+1++0.745569++0.103029++0.268984++0.663710++0.037813++0.363821
#+2++0.693016++0.621525++0.031589++0.956703++0.118434++0.484254
#+3++0.284922++0.527293++0.791596++0.243768++0.629102++0.495336
#+4++0.354870++0.113014++0.326395++0.656415++0.172445++0.324628
#+5++0.815584++0.532382++0.195437++0.829670++0.019001++0.478415
#+6++0.944587++0.068690++0.811771++0.006846++0.698785++0.506136
#+7++0.595077++0.437571++0.023520++0.772187++0.862554++0.538182
#+8++0.700771++0.413958++0.097996++0.355228++0.656919++0.444974
#+9++0.263138++0.906283++0.121386++0.624336++0.859904++0.555009

df.reindex(['mean',+*range(5)],+axis=1)

#++++++++mean+++++++++0+++++++++1+++++++++2+++++++++3+++++++++4
#+0++0.469921++0.943825++0.202490++0.071908++0.452985++0.678397
#+1++0.363821++0.745569++0.103029++0.268984++0.663710++0.037813
#+2++0.484254++0.693016++0.621525++0.031589++0.956703++0.118434
#+3++0.495336++0.284922++0.527293++0.791596++0.243768++0.629102
#+4++0.324628++0.354870++0.113014++0.326395++0.656415++0.172445
#+5++0.478415++0.815584++0.532382++0.195437++0.829670++0.019001
#+6++0.506136++0.944587++0.068690++0.811771++0.006846++0.698785
#+7++0.538182++0.595077++0.437571++0.023520++0.772187++0.862554
#+8++0.444974++0.700771++0.413958++0.097996++0.355228++0.656919
#+9++0.555009++0.263138++0.906283++0.121386++0.624336++0.859904|code-block|syntax|javascript|446963|entityMap^0|C|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can use <code>reindex</code> which can be used for both axis:

<pre><code>df
# 0 1 2 3 4 mean
# 0 0.943825 0.202490 0.071908 0.452985 0.678397 0.469921
# 1 0.745569 0.103029 0.268984 0.663710 0.037813 0.363821
# 2 0.693016 0.621525 0.031589 0.956703 0.118434 0.484254
# 3 0.284922 0.527293 0.791596 0.243768 0.629102 0.495336
# 4 0.354870 0.113014 0.326395 0.656415 0.172445 0.324628
# 5 0.815584 0.532382 0.195437 0.829670 0.019001 0.478415
# 6 0.944587 0.068690 0.811771 0.006846 0.698785 0.506136
# 7 0.595077 0.437571 0.023520 0.772187 0.862554 0.538182
# 8 0.700771 0.413958 0.097996 0.355228 0.656919 0.444974
# 9 0.263138 0.906283 0.121386 0.624336 0.859904 0.555009

df.reindex(['mean', *range(5)], axis=1)

# mean 0 1 2 3 4
# 0 0.469921 0.943825 0.202490 0.071908 0.452985 0.678397
# 1 0.363821 0.745569 0.103029 0.268984 0.663710 0.037813
# 2 0.484254 0.693016 0.621525 0.031589 0.956703 0.118434
# 3 0.495336 0.284922 0.527293 0.791596 0.243768 0.629102
# 4 0.324628 0.354870 0.113014 0.326395 0.656415 0.172445
# 5 0.478415 0.815584 0.532382 0.195437 0.829670 0.019001
# 6 0.506136 0.944587 0.068690 0.811771 0.006846 0.698785
# 7 0.538182 0.595077 0.437571 0.023520 0.772187 0.862554
# 8 0.444974 0.700771 0.413958 0.097996 0.355228 0.656919
# 9 0.555009 0.263138 0.906283 0.121386 0.624336 0.859904
</code></pre>

blocks|key|446984|text|这里有一种移动现有列的方法，它将修改现有的数据帧。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|446985|my_column+=+df.pop('column+name')
df.insert(3,+my_column.name,+my_column)++#+Is+in-place|code-block|syntax|javascript|446986|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's a way to move one existing column that will modify the existing dataframe in place.
<pre><code>my_column = df.pop('column name')
df.insert(3, my_column.name, my_column) # Is in-place
</code></pre>

blocks|key|447003|text|下面是一个函数，可以对任意数量的列执行此操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|447004|def+mean_first(df):
++++ncols+=+df.shape[1]++++++++#+Get+the+number+of+columns
++++index+=+list(range(ncols))+#+Create+an+index+to+reorder+the+columns
++++index.insert(0,ncols)++++++#+This+puts+the+last+column+at+the+front
++++return(df.assign(mean=df.mean(1)).iloc[:,index])+#+new+df+with+last+column+(mean)+first|code-block|syntax|javascript|447005|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is a function to do this for any number of columns. 

<pre><code>def mean_first(df):
 ncols = df.shape[1] # Get the number of columns
 index = list(range(ncols)) # Create an index to reorder the columns
 index.insert(0,ncols) # This puts the last column at the front
 return(df.assign(mean=df.mean(1)).iloc[:,index]) # new df with last column (mean) first
</code></pre>

blocks|key|228986|text|将任意列移动到任意位置：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|228987|import+pandas+as+pd
df+=+pd.DataFrame({"A":+[1,2,3],+
+++++++++++++++++++"B":+[2,4,8],+
+++++++++++++++++++"C":+[5,5,5]})

cols+=+df.columns.tolist()
column_to_move+=+"C"
new_position+=+1

cols.insert(new_position,+cols.pop(cols.index(column_to_move)))
df+=+df[cols]|code-block|syntax|javascript|228988|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Moving any column to any position: 

<pre><code>import pandas as pd
df = pd.DataFrame({"A": [1,2,3], 
 "B": [2,4,8], 
 "C": [5,5,5]})

cols = df.columns.tolist()
column_to_move = "C"
new_position = 1

cols.insert(new_position, cols.pop(cols.index(column_to_move)))
df = df[cols]
</code></pre>

blocks|key|229018|text|如果您的列名太长而无法键入，那么您可以通过包含以下位置的整数列表指定新的顺序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|229019|数据：|229020|++++++++++0+++++++++1+++++++++2+++++++++3+++++++++4++++++mean
0++0.397312++0.361846++0.719802++0.575223++0.449205++0.500678
1++0.287256++0.522337++0.992154++0.584221++0.042739++0.485741
2++0.884812++0.464172++0.149296++0.167698++0.793634++0.491923
3++0.656891++0.500179++0.046006++0.862769++0.651065++0.543382
4++0.673702++0.223489++0.438760++0.468954++0.308509++0.422683
5++0.764020++0.093050++0.100932++0.572475++0.416471++0.389390
6++0.259181++0.248186++0.626101++0.556980++0.559413++0.449972
7++0.400591++0.075461++0.096072++0.308755++0.157078++0.207592
8++0.639745++0.368987++0.340573++0.997547++0.011892++0.471749
9++0.050582++0.714160++0.168839++0.899230++0.359690++0.438500|code-block|syntax|javascript|229021|通用示例：|229022|new_order+=+[3,2,1,4,5,0]
print(df[df.columns[new_order]])++

++++++++++3+++++++++2+++++++++1+++++++++4++++++mean+++++++++0
0++0.575223++0.719802++0.361846++0.449205++0.500678++0.397312
1++0.584221++0.992154++0.522337++0.042739++0.485741++0.287256
2++0.167698++0.149296++0.464172++0.793634++0.491923++0.884812
3++0.862769++0.046006++0.500179++0.651065++0.543382++0.656891
4++0.468954++0.438760++0.223489++0.308509++0.422683++0.673702
5++0.572475++0.100932++0.093050++0.416471++0.389390++0.764020
6++0.556980++0.626101++0.248186++0.559413++0.449972++0.259181
7++0.308755++0.096072++0.075461++0.157078++0.207592++0.400591
8++0.997547++0.340573++0.368987++0.011892++0.471749++0.639745
9++0.899230++0.168839++0.714160++0.359690++0.438500++0.050582|229023|尽管看起来我只是显式地以不同的顺序输入列名，但有一列“mean”这一事实应该清楚地表明，new_order与实际位置相关，而不是列名。|offset|length|style|CODE|229024|对于OP问题的具体情况：|229025|new_order+=+[-1,0,1,2,3,4]
df+=+df[df.columns[new_order]]
print(df)

+++++++mean+++++++++0+++++++++1+++++++++2+++++++++3+++++++++4
0++0.500678++0.397312++0.361846++0.719802++0.575223++0.449205
1++0.485741++0.287256++0.522337++0.992154++0.584221++0.042739
2++0.491923++0.884812++0.464172++0.149296++0.167698++0.793634
3++0.543382++0.656891++0.500179++0.046006++0.862769++0.651065
4++0.422683++0.673702++0.223489++0.438760++0.468954++0.308509
5++0.389390++0.764020++0.093050++0.100932++0.572475++0.416471
6++0.449972++0.259181++0.248186++0.626101++0.556980++0.559413
7++0.207592++0.400591++0.075461++0.096072++0.308755++0.157078
8++0.471749++0.639745++0.368987++0.340573++0.997547++0.011892
9++0.438500++0.050582++0.714160++0.168839++0.899230++0.359690|229026|这种方法的主要问题是多次调用相同的代码会产生不同的结果，因此需要小心:)|229027|entityMap^0|0|0|0|0|0|18|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|15|8|@$O|16|P|17|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|18|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|19|8|@]|9|@]|A|$G|H]]|$1|W|3|X|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

If your column names are too-long-to-type then you could specify the new order through a list of integers with the positions:
Data:
<pre><code> 0 1 2 3 4 mean
0 0.397312 0.361846 0.719802 0.575223 0.449205 0.500678
1 0.287256 0.522337 0.992154 0.584221 0.042739 0.485741
2 0.884812 0.464172 0.149296 0.167698 0.793634 0.491923
3 0.656891 0.500179 0.046006 0.862769 0.651065 0.543382
4 0.673702 0.223489 0.438760 0.468954 0.308509 0.422683
5 0.764020 0.093050 0.100932 0.572475 0.416471 0.389390
6 0.259181 0.248186 0.626101 0.556980 0.559413 0.449972
7 0.400591 0.075461 0.096072 0.308755 0.157078 0.207592
8 0.639745 0.368987 0.340573 0.997547 0.011892 0.471749
9 0.050582 0.714160 0.168839 0.899230 0.359690 0.438500
</code></pre>
Generic example:
<pre><code>new_order = [3,2,1,4,5,0]
print(df[df.columns[new_order]]) 

 3 2 1 4 mean 0
0 0.575223 0.719802 0.361846 0.449205 0.500678 0.397312
1 0.584221 0.992154 0.522337 0.042739 0.485741 0.287256
2 0.167698 0.149296 0.464172 0.793634 0.491923 0.884812
3 0.862769 0.046006 0.500179 0.651065 0.543382 0.656891
4 0.468954 0.438760 0.223489 0.308509 0.422683 0.673702
5 0.572475 0.100932 0.093050 0.416471 0.389390 0.764020
6 0.556980 0.626101 0.248186 0.559413 0.449972 0.259181
7 0.308755 0.096072 0.075461 0.157078 0.207592 0.400591
8 0.997547 0.340573 0.368987 0.011892 0.471749 0.639745
9 0.899230 0.168839 0.714160 0.359690 0.438500 0.050582
</code></pre>
Although it might seem like I'm just explicitly typing the column names in a different order, the fact that there's a column 'mean' should make it clear that <code>new_order</code> relates to actual positions and not column names.
For the specific case of OP's question:
<pre><code>new_order = [-1,0,1,2,3,4]
df = df[df.columns[new_order]]
print(df)

 mean 0 1 2 3 4
0 0.500678 0.397312 0.361846 0.719802 0.575223 0.449205
1 0.485741 0.287256 0.522337 0.992154 0.584221 0.042739
2 0.491923 0.884812 0.464172 0.149296 0.167698 0.793634
3 0.543382 0.656891 0.500179 0.046006 0.862769 0.651065
4 0.422683 0.673702 0.223489 0.438760 0.468954 0.308509
5 0.389390 0.764020 0.093050 0.100932 0.572475 0.416471
6 0.449972 0.259181 0.248186 0.626101 0.556980 0.559413
7 0.207592 0.400591 0.075461 0.096072 0.308755 0.157078
8 0.471749 0.639745 0.368987 0.340573 0.997547 0.011892
9 0.438500 0.050582 0.714160 0.168839 0.899230 0.359690
</code></pre>
The main problem with this approach is that calling the same code multiple times will create different results each time, so one needs to be careful :)

blocks|key|229046|text|书中最Hackiest的方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|229047|df.insert(0,+"test",+df["mean"])
df+=+df.drop(columns=["mean"]).rename(columns={"test":+"mean"})|code-block|syntax|javascript|229048|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Hackiest method in the book
<pre><code>df.insert(0, &quot;test&quot;, df[&quot;mean&quot;])
df = df.drop(columns=[&quot;mean&quot;]).rename(columns={&quot;test&quot;: &quot;mean&quot;})
</code></pre>

blocks|key|447122|text|大多数答案都不够通用，而且pandas的reindex_axis方法有点单调乏味，因此我提供了一个简单的函数，使用字典将任意数量的列移动到任意位置，其中key+=列名，value+=要移动到的位置。如果您的数据帧很大，将True传递给'big_data‘，则函数将返回有序列列表。你可以使用这个列表来分割你的数据。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|447123|def+order_column(df,+columns,+big_data+=+False):

++++"""Re-Orders+dataFrame+column(s)
+++++++Parameters+:+
+++++++df++++++--+dataframe
+++++++columns+--+a+dictionary:
++++++++++++++++++key+++=+current+column+position/index+or+column+name
++++++++++++++++++value+=+position+to+move+it+to++
+++++++big_data+--+boolean+
++++++++++++++++++True+=+returns+only+the+ordered+columns+as+a+list
++++++++++++++++++++++++++the+user+user+can+then+slice+the+data+using+this
++++++++++++++++++++++++++ordered+column
++++++++++++++++++False+=+default+-+return+a+copy+of+the+dataframe
++++"""
++++ordered_col+=+df.columns.tolist()

++++for+key,+value+in+columns.items():

++++++++ordered_col.remove(key)
++++++++ordered_col.insert(value,+key)

++++if+big_data:

++++++++return+ordered_col

++++return+df[ordered_col]

#+e.g.
df+=+pd.DataFrame({'chicken+wings':+np.random.rand(10,+1).flatten(),+'taco':+np.random.rand(10,1).flatten(),
++++++++++++++++++++++++++'coffee':+np.random.rand(10,+1).flatten()})
df['mean']+=+df.mean(1)

df+=+order_column(df,+{'mean':+0,+'coffee':1+})

>>>|code-block|syntax|javascript|447124|​|447125|📷|atomic|offset|length|447126|447127|col+=+order_column(df,+{'mean':+0,+'coffee':1+},+True)

col
>>>
['mean',+'coffee',+'chicken+wings',+'taco']

#+you+could+grab+it+by+doing+this

df+=+df[col]|447128|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/9597f088d4ed74965c77aaa530ca776e.png|imageAlt^0|0|0|0|0|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|11|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|12|8|@]|9|@$L|13|M|14|1|15]]|A|$]]|$1|N|3|H|5|6|7|16|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X|Y|-4]]]]

Most of the answers did not generalize enough and pandas reindex_axis method is a little tedious, hence I offer a simple function to move an arbitrary number of columns to any position using a dictionary where key = column name and value = position to move to. If your dataframe is large pass True to 'big_data' then the function will return the ordered columns list. And you could use this list to slice your data.

<pre><code>def order_column(df, columns, big_data = False):

 """Re-Orders dataFrame column(s)
 Parameters : 
 df -- dataframe
 columns -- a dictionary:
 key = current column position/index or column name
 value = position to move it to 
 big_data -- boolean 
 True = returns only the ordered columns as a list
 the user user can then slice the data using this
 ordered column
 False = default - return a copy of the dataframe
 """
 ordered_col = df.columns.tolist()

 for key, value in columns.items():

 ordered_col.remove(key)
 ordered_col.insert(value, key)

 if big_data:

 return ordered_col

 return df[ordered_col]

# e.g.
df = pd.DataFrame({'chicken wings': np.random.rand(10, 1).flatten(), 'taco': np.random.rand(10,1).flatten(),
 'coffee': np.random.rand(10, 1).flatten()})
df['mean'] = df.mean(1)

df = order_column(df, {'mean': 0, 'coffee':1 })

&gt;&gt;&gt;
</code></pre>

<a href="https://i.stack.imgur.com/uqZWm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uqZWm.png" alt="output"></a>

<pre><code>col = order_column(df, {'mean': 0, 'coffee':1 }, True)

col
&gt;&gt;&gt;
['mean', 'coffee', 'chicken wings', 'taco']

# you could grab it by doing this

df = df[col]

</code></pre>

blocks|key|229138|text|我认为这是一个稍微更简洁的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|229139|df.insert(0,+'mean',+df.pop("mean"))|code-block|syntax|javascript|229140|这个解决方案有点类似于@JoeHeffer的解决方案，但这只是一个线性函数。|229141|在这里，我们从数据帧中删除列"mean"，并使用相同的列名将其附加到索引0。|offset|length|style|CODE|229142|entityMap^0|0|0|0|E|6|10|1|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]|$K|W|L|X|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

I think this is a slightly neater solution:
<pre class="lang-py prettyprint-override"><code>df.insert(0, 'mean', df.pop(&quot;mean&quot;))
</code></pre>
This solution is somewhat similar to @JoeHeffer 's solution but this is one liner.
Here we remove the column <code>&quot;mean&quot;</code> from the dataframe and attach it to index <code>0</code> with the same column name.

blocks|key|447247|text|我有一个在pandas中重新排序列名的非常具体的用例。有时，我会在基于现有列的数据框架中创建一个新列。默认情况下，pandas会在末尾插入我的新列，但我希望新列插入到派生它的现有列的旁边。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|447248|​|447249|📷|atomic|offset|length|447250|447251|def+rearrange_list(input_list,+input_item_to_move,+input_item_insert_here):
++++'''
++++Helper+function+to+re-arrange+the+order+of+items+in+a+list.
++++Useful+for+moving+column+in+pandas+dataframe.

++++Inputs:
++++++++input_list+-+list
++++++++input_item_to_move+-+item+in+list+to+move
++++++++input_item_insert_here+-+item+in+list,+insert+before+

++++returns:
++++++++output_list
++++'''
++++#+make+copy+for+output,+make+sure+it's+a+list
++++output_list+=+list(input_list)

++++#+index+of+item+to+move
++++idx_move+=+output_list.index(input_item_to_move)

++++#+pop+off+the+item+to+move
++++itm_move+=+output_list.pop(idx_move)

++++#+index+of+item+to+insert+here
++++idx_insert+=+output_list.index(input_item_insert_here)

++++#+insert+item+to+move+into+here
++++output_list.insert(idx_insert,+itm_move)

++++return+output_list


import+pandas+as+pd

#+step+1:+create+sample+dataframe
df+=+pd.DataFrame({
++++'motorcycle':+['motorcycle1',+'motorcycle2',+'motorcycle3'],
++++'initial_odometer':+[101,+500,+322],
++++'final_odometer':+[201,+515,+463],
++++'other_col_1':+['blah',+'blah',+'blah'],
++++'other_col_2':+['blah',+'blah',+'blah']
})
print('Step+1:+create+sample+dataframe')
display(df)
print()

#+step+2:+add+new+column+that+is+difference+between+final+and+initial
df['change_odometer']+=+df['final_odometer']-df['initial_odometer']
print('Step+2:+add+new+column')
display(df)
print()

#+step+3:+rearrange+columns
ls_cols+=+df.columns
ls_cols+=+rearrange_list(ls_cols,+'change_odometer',+'final_odometer')
df=df[ls_cols]
print('Step+3:+rearrange+columns')
display(df)|code-block|syntax|javascript|447252|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/cac938dca4f733f1c9602b278167df05.png|imageAlt^0|0|0|0|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@$G|10|H|11|1|12]]|A|$]]|$1|I|3|C|5|6|7|13|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|14|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V|W|-4]]]]

I have a very specific use case for re-ordering column names in pandas. Sometimes I am creating a new column in a dataframe that is based on an existing column. By default pandas will insert my new column at the end, but I want the new column to be inserted next to the existing column it's derived from. 

<a href="https://i.stack.imgur.com/EenQB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EenQB.png" alt="enter image description here"></a>

<pre class="lang-py prettyprint-override"><code>def rearrange_list(input_list, input_item_to_move, input_item_insert_here):
 '''
 Helper function to re-arrange the order of items in a list.
 Useful for moving column in pandas dataframe.

 Inputs:
 input_list - list
 input_item_to_move - item in list to move
 input_item_insert_here - item in list, insert before 

 returns:
 output_list
 '''
 # make copy for output, make sure it's a list
 output_list = list(input_list)

 # index of item to move
 idx_move = output_list.index(input_item_to_move)

 # pop off the item to move
 itm_move = output_list.pop(idx_move)

 # index of item to insert here
 idx_insert = output_list.index(input_item_insert_here)

 # insert item to move into here
 output_list.insert(idx_insert, itm_move)

 return output_list


import pandas as pd

# step 1: create sample dataframe
df = pd.DataFrame({
 'motorcycle': ['motorcycle1', 'motorcycle2', 'motorcycle3'],
 'initial_odometer': [101, 500, 322],
 'final_odometer': [201, 515, 463],
 'other_col_1': ['blah', 'blah', 'blah'],
 'other_col_2': ['blah', 'blah', 'blah']
})
print('Step 1: create sample dataframe')
display(df)
print()

# step 2: add new column that is difference between final and initial
df['change_odometer'] = df['final_odometer']-df['initial_odometer']
print('Step 2: add new column')
display(df)
print()

# step 3: rearrange columns
ls_cols = df.columns
ls_cols = rearrange_list(ls_cols, 'change_odometer', 'final_odometer')
df=df[ls_cols]
print('Step 3: rearrange columns')
display(df)
</code></pre>

blocks|key|229240|text|我认为这个函数更简单。您只需在开头和/或结尾指定列的子集：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|229241|def+reorder_df_columns(df,+start=None,+end=None):
++++"""
++++++++This+function+reorder+columns+of+a+DataFrame.
++++++++It+takes+columns+given+in+the+list+`start`+and+move+them+to+the+left.
++++++++Its+also+takes+columns+in+`end`+and+move+them+to+the+right.
++++"""
++++if+start+is+None:
++++++++start+=+[]
++++if+end+is+None:
++++++++end+=+[]
++++assert+isinstance(start,+list)+and+isinstance(end,+list)
++++cols+=+list(df.columns)
++++for+c+in+start:
++++++++if+c+not+in+cols:
++++++++++++start.remove(c)
++++for+c+in+end:
++++++++if+c+not+in+cols+or+c+in+start:
++++++++++++end.remove(c)
++++for+c+in+start+%2B+end:
++++++++cols.remove(c)
++++cols+=+start+%2B+cols+%2B+end
++++return+df[cols]|code-block|syntax|javascript|229242|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I think this function is more straightforward. You Just need to specify a subset of columns at the start or the end or both:

<pre><code>def reorder_df_columns(df, start=None, end=None):
 """
 This function reorder columns of a DataFrame.
 It takes columns given in the list `start` and move them to the left.
 Its also takes columns in `end` and move them to the right.
 """
 if start is None:
 start = []
 if end is None:
 end = []
 assert isinstance(start, list) and isinstance(end, list)
 cols = list(df.columns)
 for c in start:
 if c not in cols:
 start.remove(c)
 for c in end:
 if c not in cols or c in start:
 end.remove(c)
 for c in start + end:
 cols.remove(c)
 cols = start + cols + end
 return df[cols]
</code></pre>

blocks|key|229261|text|只是翻转经常会有帮助。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|229262|df[df.columns[::-1]]|code-block|syntax|javascript|229263|或者只是拖着脚步看一看。|229264|import+random
cols+=+list(df.columns)
random.shuffle(cols)
df[cols]|229265|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Just flipping helps often.

<pre><code>df[df.columns[::-1]]
</code></pre>

Or just shuffle for a look.

<pre><code>import random
cols = list(df.columns)
random.shuffle(cols)
df[cols]
</code></pre>

blocks|key|447319|text|对我有效的一个非常简单的解决方案是在df.columns上使用.reindex|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|447320|df+=+df[df.columns.reindex(['mean',+0,+1,+2,+3,+4])[0]]|code-block|syntax|javascript|447321|entityMap^0|I|A|V|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

A pretty straightforward solution that worked for me is to use <code>.reindex</code> on <code>df.columns</code>:
<pre><code>df = df[df.columns.reindex(['mean', 0, 1, 2, 3, 4])[0]]
</code></pre>

blocks|key|229285|text|假设您有df+with+columns+A+B+C。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|229286|最简单的方法是：|229287|df+=+df.reindex(['B','C','A'],+axis=1)|code-block|syntax|javascript|229288|entityMap^0|4|2|K|1|M|1|O|1|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

Suppose you have <code>df</code> with columns <code>A</code> <code>B</code> <code>C</code>.
The most simple way is:
<pre><code>df = df.reindex(['B','C','A'], axis=1)
</code></pre>

blocks|key|447369|text|这是一个非常简单的答案(只有一行)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|447370|在将'n‘列添加到df中之后，可以这样做，如下所示。|447371|import+numpy+as+np
import+pandas+as+pd

df+=+pd.DataFrame(np.random.rand(10,+5))
df['mean']+=+df.mean(1)
df
+++++++++++0+++++++++++1+++++++++++2+++++++++++3+++++++++++4++++++++mean
0+++0.929616++++0.316376++++0.183919++++0.204560++++0.567725++++0.440439
1+++0.595545++++0.964515++++0.653177++++0.748907++++0.653570++++0.723143
2+++0.747715++++0.961307++++0.008388++++0.106444++++0.298704++++0.424512
3+++0.656411++++0.809813++++0.872176++++0.964648++++0.723685++++0.805347
4+++0.642475++++0.717454++++0.467599++++0.325585++++0.439645++++0.518551
5+++0.729689++++0.994015++++0.676874++++0.790823++++0.170914++++0.672463
6+++0.026849++++0.800370++++0.903723++++0.024676++++0.491747++++0.449473
7+++0.526255++++0.596366++++0.051958++++0.895090++++0.728266++++0.559587
8+++0.818350++++0.500223++++0.810189++++0.095969++++0.218950++++0.488736
9+++0.258719++++0.468106++++0.459373++++0.709510++++0.178053++++0.414752


###+here+you+can+add+below+line+and+it+should+work+
#+Don't+forget+the+two+(())+'brackets'+around+columns+names.Otherwise,+it'll+give+you+an+error.

df+=+df[list(('mean',0,+1,+2,3,4))]
df

++++++++mean+++++++++++0+++++++++++1+++++++++++2+++++++++++3+++++++++++4
0+++0.440439++++0.929616++++0.316376++++0.183919++++0.204560++++0.567725
1+++0.723143++++0.595545++++0.964515++++0.653177++++0.748907++++0.653570
2+++0.424512++++0.747715++++0.961307++++0.008388++++0.106444++++0.298704
3+++0.805347++++0.656411++++0.809813++++0.872176++++0.964648++++0.723685
4+++0.518551++++0.642475++++0.717454++++0.467599++++0.325585++++0.439645
5+++0.672463++++0.729689++++0.994015++++0.676874++++0.790823++++0.170914
6+++0.449473++++0.026849++++0.800370++++0.903723++++0.024676++++0.491747
7+++0.559587++++0.526255++++0.596366++++0.051958++++0.895090++++0.728266
8+++0.488736++++0.818350++++0.500223++++0.810189++++0.095969++++0.218950
9+++0.414752++++0.258719++++0.468106++++0.459373++++0.709510++++0.178053|code-block|syntax|javascript|447372|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

<h3>Here is a very simple answer to this(only one line).</h3>

You can do that after you added the 'n' column into your df as follows.

<pre><code>import numpy as np
import pandas as pd

df = pd.DataFrame(np.random.rand(10, 5))
df['mean'] = df.mean(1)
df
 0 1 2 3 4 mean
0 0.929616 0.316376 0.183919 0.204560 0.567725 0.440439
1 0.595545 0.964515 0.653177 0.748907 0.653570 0.723143
2 0.747715 0.961307 0.008388 0.106444 0.298704 0.424512
3 0.656411 0.809813 0.872176 0.964648 0.723685 0.805347
4 0.642475 0.717454 0.467599 0.325585 0.439645 0.518551
5 0.729689 0.994015 0.676874 0.790823 0.170914 0.672463
6 0.026849 0.800370 0.903723 0.024676 0.491747 0.449473
7 0.526255 0.596366 0.051958 0.895090 0.728266 0.559587
8 0.818350 0.500223 0.810189 0.095969 0.218950 0.488736
9 0.258719 0.468106 0.459373 0.709510 0.178053 0.414752


### here you can add below line and it should work 
# Don't forget the two (()) 'brackets' around columns names.Otherwise, it'll give you an error.

df = df[list(('mean',0, 1, 2,3,4))]
df

 mean 0 1 2 3 4
0 0.440439 0.929616 0.316376 0.183919 0.204560 0.567725
1 0.723143 0.595545 0.964515 0.653177 0.748907 0.653570
2 0.424512 0.747715 0.961307 0.008388 0.106444 0.298704
3 0.805347 0.656411 0.809813 0.872176 0.964648 0.723685
4 0.518551 0.642475 0.717454 0.467599 0.325585 0.439645
5 0.672463 0.729689 0.994015 0.676874 0.790823 0.170914
6 0.449473 0.026849 0.800370 0.903723 0.024676 0.491747
7 0.559587 0.526255 0.596366 0.051958 0.895090 0.728266
8 0.488736 0.818350 0.500223 0.810189 0.095969 0.218950
9 0.414752 0.258719 0.468106 0.459373 0.709510 0.178053

</code></pre>

blocks|key|229377|text|要根据列的名称将现有列设置为另一列的右侧/左侧，请执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|229378|def+df_move_column(df,+col_to_move,+col_left_of_destiny="",+right_of_col_bool=True):
++++cols+=+list(df.columns.values)
++++index_max+=+len(cols)+-+1

++++if+not+right_of_col_bool:
++++++++#+set+left+of+a+column+"c",+is+like+putting+right+of+column+previous+to+"c"
++++++++#+...+except+if+left+of+1st+column,+then+recursive+call+to+set+rest+right+to+it
++++++++aux+=+cols.index(col_left_of_destiny)
++++++++if+not+aux:
++++++++++++for+g+in+[x+for+x+in+cols[::-1]+if+x+!=+col_to_move]:
++++++++++++++++df+=+df_move_column(
++++++++++++++++++++++++df,+
++++++++++++++++++++++++col_to_move=g,+
++++++++++++++++++++++++col_left_of_destiny=col_to_move
++++++++++++++++++++++++)
++++++++++++return+df
++++++++col_left_of_destiny+=+cols[aux+-+1]

++++index_old+=+cols.index(col_to_move)
++++index_new+=+0
++++if+len(col_left_of_destiny):
++++++++index_new+=+cols.index(col_left_of_destiny)+%2B+1

++++if+index_old+==+index_new:
++++++++return+df

++++if+index_new+<+index_old:
++++++++index_new+=+np.min([index_new,+index_max])
++++++++cols+=+(
++++++++++++cols[:index_new]
++++++++++++%2B+[cols[index_old]]
++++++++++++%2B+cols[index_new:index_old]
++++++++++++%2B+cols[index_old+%2B+1+:]
++++++++)
++++else:
++++++++cols+=+(
++++++++++++cols[:index_old]
++++++++++++%2B+cols[index_old+%2B+1+:+index_new]
++++++++++++%2B+[cols[index_old]]
++++++++++++%2B+cols[index_new:]
++++++++)

++++df+=+df[cols]
++++return+df|code-block|syntax|javascript|229379|例如。|229380|cols+=+list("ABCD")
df2+=+pd.DataFrame(np.arange(4)[np.newaxis,+:],+columns=cols)
for+k+in+cols:
++++print(30+*+"-")
++++for+g+in+[x+for+x+in+cols+if+x+!=+k]:
++++++++df_new+=+df_move_column(df2,+k,+g)
++++++++print(f"{k}+after+{g}:++{df_new.columns.values}")
for+k+in+cols:
++++print(30+*+"-")
++++for+g+in+[x+for+x+in+cols+if+x+!=+k]:
++++++++df_new+=+df_move_column(df2,+k,+g,+right_of_col_bool=False)
++++++++print(f"{k}+before+{g}:++{df_new.columns.values}")|229381|输出：|229382|​|229383|📷|atomic|offset|length|229384|229385|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/17808cfe19d5ff690ede0e005b794e6f.png|imageAlt^0|0|0|0|0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|15|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|17|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|19|8|@]|9|@$R|1A|S|1B|1|1C]]|A|$]]|$1|T|3|N|5|6|7|1D|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11|12|-4]]]]

To set an existing column right/left of another, based on their names:
<pre><code>def df_move_column(df, col_to_move, col_left_of_destiny=&quot;&quot;, right_of_col_bool=True):
 cols = list(df.columns.values)
 index_max = len(cols) - 1

 if not right_of_col_bool:
 # set left of a column &quot;c&quot;, is like putting right of column previous to &quot;c&quot;
 # ... except if left of 1st column, then recursive call to set rest right to it
 aux = cols.index(col_left_of_destiny)
 if not aux:
 for g in [x for x in cols[::-1] if x != col_to_move]:
 df = df_move_column(
 df, 
 col_to_move=g, 
 col_left_of_destiny=col_to_move
 )
 return df
 col_left_of_destiny = cols[aux - 1]

 index_old = cols.index(col_to_move)
 index_new = 0
 if len(col_left_of_destiny):
 index_new = cols.index(col_left_of_destiny) + 1

 if index_old == index_new:
 return df

 if index_new &lt; index_old:
 index_new = np.min([index_new, index_max])
 cols = (
 cols[:index_new]
 + [cols[index_old]]
 + cols[index_new:index_old]
 + cols[index_old + 1 :]
 )
 else:
 cols = (
 cols[:index_old]
 + cols[index_old + 1 : index_new]
 + [cols[index_old]]
 + cols[index_new:]
 )

 df = df[cols]
 return df
</code></pre>
E.g.
<pre><code>cols = list(&quot;ABCD&quot;)
df2 = pd.DataFrame(np.arange(4)[np.newaxis, :], columns=cols)
for k in cols:
 print(30 * &quot;-&quot;)
 for g in [x for x in cols if x != k]:
 df_new = df_move_column(df2, k, g)
 print(f&quot;{k} after {g}: {df_new.columns.values}&quot;)
for k in cols:
 print(30 * &quot;-&quot;)
 for g in [x for x in cols if x != k]:
 df_new = df_move_column(df2, k, g, right_of_col_bool=False)
 print(f&quot;{k} before {g}: {df_new.columns.values}&quot;)
</code></pre>
Output:
<a href="https://i.stack.imgur.com/LieHi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LieHi.png" alt="enter image description here" /></a>

blocks|key|447433|text|与上面的答案类似，还有一种替代方法，使用deque()及其rotate()方法。rotate方法获取列表中的最后一个元素，并将其插入到开头：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|447434|from+collections+import+deque

columns+=+deque(df.columns.tolist())
columns.rotate()

df+=+df[columns]|code-block|syntax|javascript|447435|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Similar to the top answer, there is an alternative using deque() and its rotate() method. The rotate method takes the last element in the list and inserts it to the beginning:
<pre><code>from collections import deque

columns = deque(df.columns.tolist())
columns.rotate()

df = df[columns]
</code></pre>

blocks|key|229444|text|您可以使用名称列表对数据框列进行重新排序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|229445|df+=+df.filter(list_of_col_names)|offset|length|style|CODE|229446|entityMap^0|0|0|X|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

You can reorder the dataframe columns using a list of names with:
<code>df = df.filter(list_of_col_names)</code>

blocks|key|447495|text|我认为这与Dmitriy+Work一样，显然是最简单的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|447496|df["mean"]+=+df.mean(1)
l+=++list(np.arange(0,len(df.columns)+-1+))
l.insert(0,-1)
df.iloc[:,l]|code-block|syntax|javascript|447497|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I thought of the same as Dmitriy Work, clearly easiest answer:
<pre><code>df[&quot;mean&quot;] = df.mean(1)
l = list(np.arange(0,len(df.columns) -1 ))
l.insert(0,-1)
df.iloc[:,l]
</code></pre>

blocks|key|229500|text|我试着做了一个排序函数，你可以参考Stata的order命令对列进行重新排序/移动。最好是创建一个py文件(其名称可以是order.py)，并将其保存在一个目录中并调用它的函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|229501|def+order(dataframe,cols,f_or_l=None,before=None,+after=None):

#만든이:+김완석,+Stata로+뚝딱뚝딱+저자,+blog.naver.com/sanzo213+운영
#+갖다+쓰시거나+수정을+하셔도+되지만+출처는+꼭+밝혀주세요
#+cols옵션+및+befor/after옵션에+튜플이+가능하게끔+수정했으며,+오류문구+수정함(2021.07.12,1)
#+칼럼이+멀티인덱스인+상태에서+reset_index()메소드+사용했을+시+적용안되는+걸+수정함(2021.07.12,2)+

import+pandas+as+pd
if+(type(cols)==str)+or+(type(cols)==int)+or+(type(cols)==float)+or+(type(cols)==bool)+or+type(cols)==tuple:++++
++++cols=[cols]
++++
dd=list(dataframe.columns)
for+i+in+cols:
++++i
++++dd.remove(i)+#cols요소를+제거함
++++
if+(f_or_l==None)+&+((before==None)+&+(after==None)):
++++print('f_or_l옵션을+쓰시거나+아니면+before옵션/after옵션+쓰셔야되요')
++++
if+((f_or_l=='first')+or+(f_or_l=='last'))+&+~((before==None)+&+(after==None)):
++++print('f_or_l옵션+사용시+before+after+옵션+사용불가입니다.')
++++
if+(f_or_l=='first')+&+(before==None)+&+(after==None):
++++new_order=cols%2Bdd
++++dataframe=dataframe[new_order]
++++return+dataframe

if+(f_or_l=='last')+&+(before==None)+&+(after==None):+++
++++new_order=dd%2Bcols
++++dataframe=dataframe[new_order]
++++return+dataframe
++++
if+(before!=None)+&+(after!=None):
++++print('before옵션+after옵션+둘다+쓸+수+없습니다.')
++++

if+(before!=None)+&+(after==None)+&+(f_or_l==None):

++++if+not((type(before)==str)+or+(type(before)==int)+or+(type(before)==float)+or
+++++++(type(before)==bool)+or+((type(before)!=list))+or+
+++++++((type(before)==tuple))):
++++++++print('before옵션은+칼럼+하나만+입력가능하며+리스트+형태로도+입력하지+마세요.')
++++
++++else:
++++++++b=dd[:dd.index(before)]
++++++++a=dd[dd.index(before):]
++++++++
++++++++new_order=b%2Bcols%2Ba
++++++++dataframe=dataframe[new_order]++
++++++++return+dataframe
++++
if+(after!=None)+&+(before==None)+&+(f_or_l==None):

++++if+not((type(after)==str)+or+(type(after)==int)+or+(type(after)==float)+or
+++++++(type(after)==bool)+or+((type(after)!=list))+or+
+++++++((type(after)==tuple))):
++++++++++++
++++++++print('after옵션은+칼럼+하나만+입력가능하며+리스트+형태로도+입력하지+마세요.')++

++++else:
++++++++b=dd[:dd.index(after)%2B1]
++++++++a=dd[dd.index(after)%2B1:]
++++++++
++++++++new_order=b%2Bcols%2Ba
++++++++dataframe=dataframe[new_order]
++++++++return+dataframe|code-block|syntax|javascript|229502|下面的python代码是我创建的order函数的一个示例。我希望您可以使用我的order函数轻松地重新排序列:)|229503|#+module

import+pandas+as+pd
import+numpy+as+np
from+order+import+order+#+call+order+function+from+order.py+file

#+make+a+dataset

columns='a+b+c+d+e+f+g+h+i+j+k'.split()
dic={}

n=-1
for+i+in+columns:
++++
++++n%2B=1
++++dic[i]=list(range(1%2Bn,10%2B1%2Bn))
data=pd.DataFrame(dic)
print(data)

#+use+order+function+(1)+:+order+column+e+in+the+first

data2=order(data,'e',f_or_l='first')
print(data2)

#+use+order+function+(2):+order+column+e+in+the+last+,+"data"+dataframe

print(order(data,'e',f_or_l='last'))


#+use+order+function+(3)+:+order+column+i+before+column+c+in+"data"+dataframe

print(order(data,'i',before='c'))


#+use+order+function+(4)+:+order+column+g+after+column+b+in+"data"+dataframe

print(order(data,'g',after='b'))

#+use+order+function+(4)+:+order+columns+['c',+'d',+'e']+after+column+i+in+"data"+dataframe

print(order(data,['c',+'d',+'e'],after='i'))|229504|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I tried making a order function which you can reorder/move column(s) with reference of order command of Stata. it would be better to make a py file( the name of which may be order.py) and save it in a directory and call it function
<pre><code>def order(dataframe,cols,f_or_l=None,before=None, after=None):

#만든이: 김완석, Stata로 뚝딱뚝딱 저자, blog.naver.com/sanzo213 운영
# 갖다 쓰시거나 수정을 하셔도 되지만 출처는 꼭 밝혀주세요
# cols옵션 및 befor/after옵션에 튜플이 가능하게끔 수정했으며, 오류문구 수정함(2021.07.12,1)
# 칼럼이 멀티인덱스인 상태에서 reset_index()메소드 사용했을 시 적용안되는 걸 수정함(2021.07.12,2) 

import pandas as pd
if (type(cols)==str) or (type(cols)==int) or (type(cols)==float) or (type(cols)==bool) or type(cols)==tuple: 
 cols=[cols]
 
dd=list(dataframe.columns)
for i in cols:
 i
 dd.remove(i) #cols요소를 제거함
 
if (f_or_l==None) &amp; ((before==None) &amp; (after==None)):
 print('f_or_l옵션을 쓰시거나 아니면 before옵션/after옵션 쓰셔야되요')
 
if ((f_or_l=='first') or (f_or_l=='last')) &amp; ~((before==None) &amp; (after==None)):
 print('f_or_l옵션 사용시 before after 옵션 사용불가입니다.')
 
if (f_or_l=='first') &amp; (before==None) &amp; (after==None):
 new_order=cols+dd
 dataframe=dataframe[new_order]
 return dataframe

if (f_or_l=='last') &amp; (before==None) &amp; (after==None): 
 new_order=dd+cols
 dataframe=dataframe[new_order]
 return dataframe
 
if (before!=None) &amp; (after!=None):
 print('before옵션 after옵션 둘다 쓸 수 없습니다.')
 

if (before!=None) &amp; (after==None) &amp; (f_or_l==None):

 if not((type(before)==str) or (type(before)==int) or (type(before)==float) or
 (type(before)==bool) or ((type(before)!=list)) or 
 ((type(before)==tuple))):
 print('before옵션은 칼럼 하나만 입력가능하며 리스트 형태로도 입력하지 마세요.')
 
 else:
 b=dd[:dd.index(before)]
 a=dd[dd.index(before):]
 
 new_order=b+cols+a
 dataframe=dataframe[new_order] 
 return dataframe
 
if (after!=None) &amp; (before==None) &amp; (f_or_l==None):

 if not((type(after)==str) or (type(after)==int) or (type(after)==float) or
 (type(after)==bool) or ((type(after)!=list)) or 
 ((type(after)==tuple))):
 
 print('after옵션은 칼럼 하나만 입력가능하며 리스트 형태로도 입력하지 마세요.') 

 else:
 b=dd[:dd.index(after)+1]
 a=dd[dd.index(after)+1:]
 
 new_order=b+cols+a
 dataframe=dataframe[new_order]
 return dataframe
</code></pre>
python code below is an example of order function I made. I hope you can reorder column(s) so easily with my order function :)
<pre><code># module

import pandas as pd
import numpy as np
from order import order # call order function from order.py file

# make a dataset

columns='a b c d e f g h i j k'.split()
dic={}

n=-1
for i in columns:
 
 n+=1
 dic[i]=list(range(1+n,10+1+n))
data=pd.DataFrame(dic)
print(data)

# use order function (1) : order column e in the first

data2=order(data,'e',f_or_l='first')
print(data2)

# use order function (2): order column e in the last , &quot;data&quot; dataframe

print(order(data,'e',f_or_l='last'))


# use order function (3) : order column i before column c in &quot;data&quot; dataframe

print(order(data,'i',before='c'))


# use order function (4) : order column g after column b in &quot;data&quot; dataframe

print(order(data,'g',after='b'))

# use order function (4) : order columns ['c', 'd', 'e'] after column i in &quot;data&quot; dataframe

print(order(data,['c', 'd', 'e'],after='i'))
</code></pre>

blocks|key|447601|text|另一种选择是使用后跟reset_index()的set_index()方法。请注意，我们首先对要移到数据帧前面的列执行pop()操作，以便在重置索引时避免名称冲突：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|447602|df.set_index(df.pop('column_name'),+inplace=True)
df.reset_index(inplace=True)|code-block|syntax|javascript|447603|447604|有关详细信息，请参阅How+to+change+the+order+of+dataframe+columns+in+pandas。|447605|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.reset_index.html|1|https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.set_index.html|2|https://towardsdatascience.com/how-to-change-the-order-of-dataframe-columns-in-pandas-c91370326053^0|A|D|O|B|1N|5|A|D|0|O|B|1|0|0|0|A|1I|2|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@$9|16|A|17|1|18]|$9|19|A|1A|1|1B]]|E|$]]|$1|F|3|G|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|1E|8|@]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|N|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]|V|$5|Q|R|S|E|$T|W]]|X|$5|Q|R|S|E|$T|Y]]]]

Another option would be to use <a href="https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.set_index.html" rel="nofollow noreferrer"><code>set_index()</code></a> method followed by a <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.reset_index.html" rel="nofollow noreferrer"><code>reset_index()</code></a>. Note that we first <code>pop()</code> the column we intend to move to the front of the dataframe, so that we avoid name collision when resetting the index:
<pre><code>df.set_index(df.pop('column_name'), inplace=True)
df.reset_index(inplace=True)
</code></pre>
<hr />
For more details see <a href="https://towardsdatascience.com/how-to-change-the-order-of-dataframe-columns-in-pandas-c91370326053" rel="nofollow noreferrer">How to change the order of dataframe columns in pandas</a>.

blocks|key|229591|text|这里有一个非常简单的方法的例子。如果要从excel复制标题，请使用.split('\t')|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|229592|df+=+df['FILE_NAME+DISPLAY_PATH+SHAREPOINT_PATH+RETAILER+LAST_UPDATE'.split()]|code-block|syntax|javascript|229593|entityMap^0|X|C|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Here's an example of a super easy way to do it. If you're copying the headers from excel use <code>.split('\t')</code>
<pre><code>df = df['FILE_NAME DISPLAY_PATH SHAREPOINT_PATH RETAILER LAST_UPDATE'.split()]
</code></pre>

blocks|key|447671|text|排序并不能确保保留正确的顺序。通过将“mean”与列列表连接起来，它将是。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|447672|cols_list+=+['mean']+%2B+df.columns.tolist()
df['mean']+=+df.mean(1)
df+=+df[cols_list]|code-block|syntax|javascript|447673|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Sorting doesn't ensure the correct order preserved. By concatenating ['mean'] with the columns list it will be.
<pre class="lang-py prettyprint-override"><code>cols_list = ['mean'] + df.columns.tolist()
df['mean'] = df.mean(1)
df = df[cols_list]
</code></pre>

I have the following <code>DataFrame</code> (<code>df</code>):

<pre><code>import numpy as np
import pandas as pd

df = pd.DataFrame(np.random.rand(10, 5))
</code></pre>

I add more column(s) by assignment:

<pre><code>df['mean'] = df.mean(1)
</code></pre>

How can I move the column <code>mean</code> to the front, i.e. set it as first column leaving the order of the other columns untouched?

How to change the order of DataFrame columns?

Python

我有以下DataFrame (df)：import numpy as npimport pandas as pddf = pd.DataFrame(np.random.rand(10, 5))我通过赋值添加更多的列：df['mean'] = df.mean(1)如何将列mean移到最前面，即将其设置为第一列，保持其他列...

问如何更改DataFrame列的顺序？
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