blocks|key|4403231|text|是的，使用set_index，您可以将Locality作为行索引。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4403232|data.set_index('Locality',+inplace=True)|code-block|syntax|javascript|4403233|如果未提供inplace=True，set_index将返回修改后的数据帧作为结果。|4403234|示例：|4403235|>+import+pandas+as+pd
>+df+=+pd.DataFrame([['ABBOTSFORD',+427000,+448000],
+++++++++++++++++++++['ABERFELDIE',+534000,+600000]],
++++++++++++++++++++columns=['Locality',+2005,+2006])

>+df
+++++Locality++++2005++++2006
0++ABBOTSFORD++427000++448000
1++ABERFELDIE++534000++600000

>+df.set_index('Locality',+inplace=True)
>+df
++++++++++++++2005++++2006
Locality++++++++++++++++++
ABBOTSFORD++427000++448000
ABERFELDIE++534000++600000

>+df.loc['ABBOTSFORD']
2005++++427000
2006++++448000
Name:+ABBOTSFORD,+dtype:+int64

>+df.loc['ABBOTSFORD'][2005]
427000

>+df.loc['ABBOTSFORD'].values
array([427000,+448000])

>+df.loc['ABBOTSFORD'].tolist()
[427000,+448000]|4403236|entityMap|0|LINK|mutability|MUTABLE|url|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.set_index.html^0|J|8|5|9|0|0|0|5|C|I|9|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1A|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

Yes, with <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.set_index.html" rel="noreferrer">set_index</a> you can make <code>Locality</code> your row index.

<pre><code>data.set_index('Locality', inplace=True)
</code></pre>

If <code>inplace=True</code> is not provided, <code>set_index</code> returns the modified dataframe as a result.

Example:

<pre><code>&gt; import pandas as pd
&gt; df = pd.DataFrame([['ABBOTSFORD', 427000, 448000],
 ['ABERFELDIE', 534000, 600000]],
 columns=['Locality', 2005, 2006])

&gt; df
 Locality 2005 2006
0 ABBOTSFORD 427000 448000
1 ABERFELDIE 534000 600000

&gt; df.set_index('Locality', inplace=True)
&gt; df
 2005 2006
Locality 
ABBOTSFORD 427000 448000
ABERFELDIE 534000 600000

&gt; df.loc['ABBOTSFORD']
2005 427000
2006 448000
Name: ABBOTSFORD, dtype: int64

&gt; df.loc['ABBOTSFORD'][2005]
427000

&gt; df.loc['ABBOTSFORD'].values
array([427000, 448000])

&gt; df.loc['ABBOTSFORD'].tolist()
[427000, 448000]
</code></pre>

blocks|key|4403252|text|您可以使用set_index更改索引，如前所述。您不需要手动交换行和列，pandas中有一个转置(data.T)方法可以为您完成此操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4403253|>+df+=+pd.DataFrame([['ABBOTSFORD',+427000,+448000],
++++++++++++++++++++['ABERFELDIE',+534000,+600000]],
++++++++++++++++++++columns=['Locality',+2005,+2006])

>+newdf+=+df.set_index('Locality').T
>+newdf

Locality++++ABBOTSFORD++ABERFELDIE
2005++++++++427000++++++534000
2006++++++++448000++++++600000|code-block|syntax|javascript|4403254|然后，您可以获取dataframe列值并将其转换为列表：|4403255|>+newdf['ABBOTSFORD'].values.tolist()

[427000,+448000]|4403256|entityMap^0|5|9|1D|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$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]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

You can change the index as explained already using <code>set_index</code>.
You don't need to manually swap rows with columns, there is a transpose (<code>data.T</code>) method in pandas that does it for you:

<pre><code>&gt; df = pd.DataFrame([['ABBOTSFORD', 427000, 448000],
 ['ABERFELDIE', 534000, 600000]],
 columns=['Locality', 2005, 2006])

&gt; newdf = df.set_index('Locality').T
&gt; newdf

Locality ABBOTSFORD ABERFELDIE
2005 427000 534000
2006 448000 600000
</code></pre>

then you can fetch the dataframe column values and transform them to a list:

<pre><code>&gt; newdf['ABBOTSFORD'].values.tolist()

[427000, 448000]
</code></pre>

blocks|key|4403422|text|另一种简单的方法是将列分配给数据框索引|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4403423|data+=+{
++'Locality':+['ABBOTSFORD',+'ABERFELDIE',+'AIREYS+INLET'],
++'2005':+[427000,+534000,+459000+],
++'2006':+[448000,+448000,+448000],
++'2007':+[602500,+602500,+602500],
++'2008':+[600000,+710000,+517500],
++'2009':+[638500,+775000,+512500]
}

df+=+pd.DataFrame(data)

#+set+the+locality+column+as+the+index
df.index+=+df['Locality']|code-block|syntax|javascript|4403424|如果不再需要Locality列作为列，只需删除它即可|4403425|df.drop('Locality',+axis=1)|4403426|你最终会得到|4403427|++++++++++++++%7C+2005+++++%7C+2006+++%7C+2007+++%7C+2008+++%7C+2009
Locality++++++%7C-------------------------------------------++++++++++++++
ABBOTSFORD++++%7C+427000+++%7C+448000+%7C+602500+%7C+600000+%7C+638500
ABERFELDIE++++%7C+534000+++%7C+448000+%7C+602500+%7C+710000+%7C+775000
AIREYS+INLET++%7C+459000+++%7C+448000+%7C+602500+%7C+517500+%7C+512500|4403428|entityMap^0|0|0|0|0|0|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]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Another simple approach is to assign the column to the data frame index
<pre class="lang-py prettyprint-override"><code>data = {
 'Locality': ['ABBOTSFORD', 'ABERFELDIE', 'AIREYS INLET'],
 '2005': [427000, 534000, 459000 ],
 '2006': [448000, 448000, 448000],
 '2007': [602500, 602500, 602500],
 '2008': [600000, 710000, 517500],
 '2009': [638500, 775000, 512500]
}

df = pd.DataFrame(data)

# set the locality column as the index
df.index = df['Locality']
</code></pre>
And if you no longer want the Locality column as a column, you can just drop it
<pre class="lang-py prettyprint-override"><code>df.drop('Locality', axis=1)
</code></pre>
You'll end up with
<pre class="lang-sh prettyprint-override"><code>
 | 2005 | 2006 | 2007 | 2008 | 2009
Locality |------------------------------------------- 
ABBOTSFORD | 427000 | 448000 | 602500 | 600000 | 638500
ABERFELDIE | 534000 | 448000 | 602500 | 710000 | 775000
AIREYS INLET | 459000 | 448000 | 602500 | 517500 | 512500
</code></pre>

I have a spreadsheet like this:

<pre><code>Locality 2005 2006 2007 2008 2009

ABBOTSFORD 427000 448000 602500 600000 638500
ABERFELDIE 534000 600000 735000 710000 775000
AIREYS INLET459000 440000 430000 517500 512500
</code></pre>

I don't want to manually swap the column with the row. Could it be possible to use pandas reading data to a list as this:

<pre><code>data['ABBOTSFORD']=[427000,448000,602500,600000,638500]
data['ABERFELDIE']=[534000,600000,735000,710000,775000]
data['AIREYS INLET']=[459000,440000,430000,517500,512500]
</code></pre>

Could pandas use column as index?

我有一个这样的电子表格：Locality    2005    2006    2007    2008    2009ABBOTSFORD  427000  448000  602500  600000  638500ABERFELDIE  534000  600000  735000  710000  775000...

问pandas可以使用列作为索引吗？
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问pandas可以使用列作为索引吗？
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