熊猫连接忽略_索引不工作

我正在尝试列绑定数据框架，而且遇到了熊猫 concat的问题，因为 ignore_index=True似乎不起作用:

df1 = pd.DataFrame({'A': ['A0', 'A1', 'A2', 'A3'],
'B': ['B0', 'B1', 'B2', 'B3'],
'D': ['D0', 'D1', 'D2', 'D3']},
index=[0, 2, 3,4])


df2 = pd.DataFrame({'A1': ['A4', 'A5', 'A6', 'A7'],
'C': ['C4', 'C5', 'C6', 'C7'],
'D2': ['D4', 'D5', 'D6', 'D7']},
index=[ 5, 6, 7,3])
df1
#     A   B   D
# 0  A0  B0  D0
# 2  A1  B1  D1
# 3  A2  B2  D2
# 4  A3  B3  D3


df2
#    A1   C  D2
# 5  A4  C4  D4
# 6  A5  C5  D5
# 7  A6  C6  D6
# 3  A7  C7  D7


dfs = [df1,df2]
df = pd.concat( dfs,axis=1,ignore_index=True)
print df

结果就是

     0    1    2    3    4    5
0   A0   B0   D0  NaN  NaN  NaN
2   A1   B1   D1  NaN  NaN  NaN
3   A2   B2   D2   A7   C7   D7
4   A3   B3   D3  NaN  NaN  NaN
5  NaN  NaN  NaN   A4   C4   D4
6  NaN  NaN  NaN   A5   C5   D5
7  NaN  NaN  NaN   A6   C6   D6

即使我使用

 df1.reset_index()
df2.reset_index()

然后再试试

pd.concat([df1,df2],axis=1)

结果还是一样！

144153 次浏览
小开

The ignore_index option is working in your example, you just need to know that it is ignoring the axis of concatenation which in your case is the columns. (Perhaps a better name would be ignore_labels.) If you want the concatenation to ignore the index labels, then your axis variable has to be set to 0 (the default).

小开

Agree with the comments, always best to post expected output.

Is this what you are seeking?

df1 = pd.DataFrame({'A': ['A0', 'A1', 'A2', 'A3'],
'B': ['B0', 'B1', 'B2', 'B3'],
'D': ['D0', 'D1', 'D2', 'D3']},
index=[0, 2, 3,4])


df2 = pd.DataFrame({'A1': ['A4', 'A5', 'A6', 'A7'],
'C': ['C4', 'C5', 'C6', 'C7'],
'D2': ['D4', 'D5', 'D6', 'D7']},
index=[ 5, 6, 7,3])




df1 = df1.transpose().reset_index(drop=True).transpose()
df2 = df2.transpose().reset_index(drop=True).transpose()




dfs = [df1,df2]
df = pd.concat( dfs,axis=0,ignore_index=True)


print df






0   1   2
0  A0  B0  D0
1  A1  B1  D1
2  A2  B2  D2
3  A3  B3  D3
4  A4  C4  D4
5  A5  C5  D5
6  A6  C6  D6
7  A7  C7  D7
小开
最佳答案

If I understood you correctly, this is what you would like to do.

import pandas as pd


df1 = pd.DataFrame({'A': ['A0', 'A1', 'A2', 'A3'],
'B': ['B0', 'B1', 'B2', 'B3'],
'D': ['D0', 'D1', 'D2', 'D3']},
index=[0, 2, 3,4])


df2 = pd.DataFrame({'A1': ['A4', 'A5', 'A6', 'A7'],
'C': ['C4', 'C5', 'C6', 'C7'],
'D2': ['D4', 'D5', 'D6', 'D7']},
index=[ 4, 5, 6 ,7])




df1.reset_index(drop=True, inplace=True)
df2.reset_index(drop=True, inplace=True)


df = pd.concat( [df1, df2], axis=1)

Which gives:

    A   B   D   A1  C   D2
0   A0  B0  D0  A4  C4  D4
1   A1  B1  D1  A5  C5  D5
2   A2  B2  D2  A6  C6  D6
3   A3  B3  D3  A7  C7  D7

Actually, I would have expected that df = pd.concat(dfs,axis=1,ignore_index=True) gives the same result.

This is the excellent explanation from jreback:

ignore_index=True ‘ignores’, meaning doesn’t align on the joining axis. it simply pastes them together in the order that they are passed, then reassigns a range for the actual index (e.g. range(len(index))) so the difference between joining on non-overlapping indexes (assume axis=1 in the example), is that with ignore_index=False (the default), you get the concat of the indexes, and with ignore_index=True you get a range.

小开

Thanks for asking. I had the same issue. For some reason "ignore_index=True" doesn't help in my case. I wanted to keep index from the first dataset and ignore the second index a this worked for me

X_train=pd.concat([train_sp, X_train.reset_index(drop=True, inplace=True)], axis=1)
小开

You can use numpy's concatenate to achieve the result.

cols = df1.columns.to_list() + df2.columns.to_list()
dfs = [df1,df2]
df = np.concatenate(dfs, axis=1)
df = pd.DataFrame(df, columns=cols)


Out[1]:
A   B   D  A1   C  D2
0  A0  B0  D0  A4  C4  D4
1  A1  B1  D1  A5  C5  D5
2  A2  B2  D2  A6  C6  D6
3  A3  B3  D3  A7  C7  D7
小开

In case you want to retain the index of the left data frame, set the index of df2 to be df1 using set_index:

pd.concat([df1, df2.set_index(df1.index)], axis=1)

提交答案