blocks|key|264494|text|示例代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|264495|a+=+pd.Series([1,2,3,4],+index=[7,2,8,9])
b+=+pd.Series([5,6,7,8],+index=[7,2,8,9])
data+=+pd.DataFrame({'a':+a,'b':b,+'idx_col':a.index})|code-block|syntax|javascript|264496|Pandas允许您使用Series作为值和列名作为键，从dict创建DataFrame。当它找到一个Series作为值时，它使用Series索引作为DataFrame索引的一部分。这种数据对齐是Pandas的主要好处之一。因此，除非您有其他需求，否则新创建的DataFrame具有重复值。在上面的示例中，data['idx_col']具有与data.index相同的数据。|offset|length|style|CODE|264497|entityMap^0|0|0|B|6|S|4|Y|9|1E|6|1S|6|22|9|3L|9|48|F|4Q|A|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]|$I|17|J|18|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|N|$]]

Example code:

<pre><code>a = pd.Series([1,2,3,4], index=[7,2,8,9])
b = pd.Series([5,6,7,8], index=[7,2,8,9])
data = pd.DataFrame({'a': a,'b':b, 'idx_col':a.index})
</code></pre>

Pandas allows you to create a <code>DataFrame</code> from a <code>dict</code> with <code>Series</code> as the values and the column names as the keys. When it finds a <code>Series</code> as a value, it uses the <code>Series</code> index as part of the <code>DataFrame</code> index. This data alignment is one of the main perks of Pandas. Consequently, unless you have other needs, the freshly created <code>DataFrame</code> has duplicated value. In the above example, <code>data['idx_col']</code> has the same data as <code>data.index</code>.

blocks|key|47949|text|我不确定我是否完全理解你的问题，但这是你想要做的吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|47950|pd.DataFrame(data=dict(s1=s1,+s2=s2),+index=s1.index)|code-block|syntax|javascript|47951|(这里甚至不需要index=s1.index)|offset|length|style|CODE|47952|entityMap^0|0|0|8|E|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Not sure I fully understand your question, but is this what you want to do?

<pre><code>pd.DataFrame(data=dict(s1=s1, s2=s2), index=s1.index)
</code></pre>

(<code>index=s1.index</code> is not even necessary here)

blocks|key|47968|text|Pandas将自动对齐这些连续传递的参数，并创建联合索引，它们在这里恰好是相同的。reset_index将索引移动到列。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|47969|In+[2]:+s1+=+Series(randn(5),index=[1,2,4,5,6])

In+[4]:+s2+=+Series(randn(5),index=[1,2,4,5,6])

In+[8]:+DataFrame(dict(s1+=+s1,+s2+=+s2)).reset_index()
Out[8]:+
+++index++++++++s1++++++++s2
0++++++1+-0.176143++0.128635
1++++++2+-1.286470++0.908497
2++++++4+-0.995881++0.528050
3++++++5++0.402241++0.458870
4++++++6++0.380457++0.072251|code-block|syntax|javascript|47970|entityMap^0|15|B|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|$]]

Pandas will automatically align these passed in series and create the joint index
They happen to be the same here. <code>reset_index</code> moves the index to a column.

<pre><code>In [2]: s1 = Series(randn(5),index=[1,2,4,5,6])

In [4]: s2 = Series(randn(5),index=[1,2,4,5,6])

In [8]: DataFrame(dict(s1 = s1, s2 = s2)).reset_index()
Out[8]: 
 index s1 s2
0 1 -0.176143 0.128635
1 2 -1.286470 0.908497
2 4 -0.995881 0.528050
3 5 0.402241 0.458870
4 6 0.380457 0.072251
</code></pre>

blocks|key|47979|text|基于join()的解决方案的简化|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|47980|df+=+a.to_frame().join(b)|code-block|syntax|javascript|47981|entityMap^0|2|6|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|$]]

A simplification of the solution based on <code>join()</code>:

<pre><code>df = a.to_frame().join(b)
</code></pre>

blocks|key|264555|text|我使用pandas将我的numpy数组或iseries转换成一个数据帧，然后按键添加和添加额外的列，作为‘预测’。如果需要将数据帧转换回列表，则使用values.tolist()|type|unstyled|depth|inlineStyleRanges|entityRanges|data|264556|output=pd.DataFrame(X_test)
output['prediction']=y_pred

list=output.values.tolist()+++++|code-block|syntax|javascript|264557|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 used pandas to convert my numpy array or iseries to an dataframe then added and additional the additional column by key as 'prediction'. If you need dataframe converted back to a list then use values.tolist()

<pre><code>output=pd.DataFrame(X_test)
output['prediction']=y_pred

list=output.values.tolist() 
</code></pre>

blocks|key|48005|text|如果您正在尝试连接长度相等的Series，但是它们的索引不匹配(这是一种常见的场景)，那么将它们连接起来将在它们不匹配的任何地方生成NAs。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|48006|x+=+pd.Series({'a':1,'b':2,})
y+=+pd.Series({'d':4,'e':5})
pd.concat([x,y],axis=1)

#Output+(I've+added+column+names+for+clarity)
Index+++x++++y
a++++++1.0++NaN
b++++++2.0++NaN
d++++++NaN++4.0
e++++++NaN++5.0|code-block|syntax|javascript|48007|假设您并不关心索引是否匹配，那么解决方案就是在连接两个Series之前对它们重新编制索引。如果为drop=False+(这是默认值)，那么Pandas会将旧索引保存在新数据帧的列中(为简单起见，将索引放在此处)。|offset|length|style|CODE|48008|pd.concat([x.reset_index(drop=True),y.reset_index(drop=True)],axis=1)

#Output+(column+names+added):
Index+++x+++y
0+++++++1+++4
1+++++++2+++5|48009|entityMap^0|0|0|1C|A|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|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|$]]

If you are trying to join Series of equal length but their indexes don't match (which is a common scenario), then concatenating them will generate NAs wherever they don't match.
<pre class="lang-py prettyprint-override"><code>x = pd.Series({'a':1,'b':2,})
y = pd.Series({'d':4,'e':5})
pd.concat([x,y],axis=1)

#Output (I've added column names for clarity)
Index x y
a 1.0 NaN
b 2.0 NaN
d NaN 4.0
e NaN 5.0
</code></pre>
Assuming that you don't care if the indexes match, the solution is to reindex both Series before concatenating them. If <code>drop=False</code>, which is the default, then Pandas will save the old index in a column of the new dataframe (the indexes are dropped here for simplicity).
<pre class="lang-py prettyprint-override"><code>pd.concat([x.reset_index(drop=True),y.reset_index(drop=True)],axis=1)

#Output (column names added):
Index x y
0 1 4
1 2 5
</code></pre>

I have two Series <code>s1</code> and <code>s2</code> with the same (non-consecutive) indices. How do I combine <code>s1</code> and <code>s2</code> to being two columns in a DataFrame and keep one of the indices as a third column?

Combining two Series into a DataFrame in pandas

我有两个具有相同(非连续)索引的系列s1和s2。如何将s1和s2组合为DataFrame中的两列，并将其中一个索引保留为第三列？

问在熊猫中将两个序列合并为一个DataFrame
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