作为新列附加到 Pandas 中的 DataFrame

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时间:2020-09-13 23:45:02  来源:igfitidea点击:

Append to a DataFrame in Pandas as new column

pythonpandas

提问by TheStrangeQuark

I have two DataFrames with the same indexing and want to append the second to the first. Lets say I have:

我有两个具有相同索引的 DataFrame,并希望将第二个附加到第一个。可以说我有:

df1 = pd.DataFrame([1,2,3], index = [2,3,4])
df2 = pd.DataFrame([3,5,3], index = [2,3,4])
df1 = df1.append(df2)

which returns

返回

   0
2  1
3  2
4  3
2  3
3  5
4  3

But I want it to append a new column where the indexes match:

但我希望它附加一个索引匹配的新列:

2  1  3
3  2  5
4  3  3

Is there a way to do this?

有没有办法做到这一点?

采纳答案by EdChum

Use concatand pass param axis=1to concatenate the list of dfs column-wise:

使用concat并传递 paramaxis=1来按列连接 dfs 列表:

In [3]:

df1 = pd.DataFrame([1,2,3], index = [2,3,4])
df2 = pd.DataFrame([3,5,3], index = [2,3,4])
pd.concat([df1,df2], axis=1)
Out[3]:
   0  0
2  1  3
3  2  5
4  3  3

You can also use joinbut you have to rename the column first:

您也可以使用,join但您必须先重命名该列:

In [6]:

df1.join(df2.rename(columns={0:'x'}))
Out[6]:
   0  x
2  1  3
3  2  5
4  3  3

Or mergespecifying that you wish to match on indices:

或者merge指定您希望匹配索引:

In [8]:

df1.merge(df2.rename(columns={0:'x'}), left_index=True, right_index=True )
Out[8]:
   0  x
2  1  3
3  2  5
4  3  3

回答by vk1011

If the indexes match exactly and there's only one column in the other DataFrame (like your question has), then you could even just add the other DataFrame as a new column.

如果索引完全匹配并且另一个 DataFrame 中只有一列(就像您的问题一样),那么您甚至可以将另一个 DataFrame 添加为新列。

>>> df1['new_column'] = df2
>>> df1
   0  new_column
2  1           3
3  2           5
4  3           3

In general, the concatapproach is better. If you have different indexes, you can choose to do an inner joinor outer join.

一般来说,这种concat方法更好。如果你有不同的索引,你可以选择做一个inner joinouter join

>>> df2 = pd.DataFrame([3,5,3], index = [2,3,5])
>>> df2
   0
2  3
3  5
5  3

>>> pd.concat([df1, df2], axis=1, join='inner')
   0  0
2  1  3
3  2  5

>>> pd.concat([df1, df2], axis=1, join='outer')
    0   0
2   1   3
3   2   5
4   3 NaN
5 NaN   3