熊猫: 添加一个列到一个多索引列数据框架

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最佳答案

It's actually pretty simple (FWIW, I originally thought to do it your way):

df['bar', 'three'] = [0, 1, 2]
df = df.sort_index(axis=1)
print(df)


bar                        baz
one       two  three       one       two
A -0.212901  0.503615      0 -1.660945  0.446778
B -0.803926 -0.417570      1 -0.336827  0.989343
C  3.400885 -0.214245      2  0.895745  1.011671

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If we want to add a multi-level column:

Source DF:

In [221]: df
Out[221]:
first        bar                 baz
second       one       two       one       two
A      -1.089798  2.053026  0.470218  1.440740
B       0.488875  0.428836  1.413451 -0.683677
C      -0.243064 -0.069446 -0.911166  0.478370

Option 1: adding result of division: bar / baz as a new foo column

In [222]: df = df.join(df[['bar']].div(df['baz']).rename(columns={'bar':'foo'}))


In [223]: df
Out[223]:
first        bar                 baz                 foo
second       one       two       one       two       one       two
A      -1.089798  2.053026  0.470218  1.440740 -2.317647  1.424980
B       0.488875  0.428836  1.413451 -0.683677  0.345873 -0.627250
C      -0.243064 -0.069446 -0.911166  0.478370  0.266761 -0.145172

Option 2: adding multi-level column with three "sub-columns":

In [235]: df = df.join(pd.DataFrame(np.random.rand(3,3),
...:                           columns=pd.MultiIndex.from_product([['new'], ['one','two','three']]),
...:                             index=df.index))


In [236]: df
Out[236]:
first        bar                 baz                 new
second       one       two       one       two       one       two     three
A      -1.089798  2.053026  0.470218  1.440740  0.274291  0.636257  0.091048
B       0.488875  0.428836  1.413451 -0.683677  0.668157  0.456931  0.227568
C      -0.243064 -0.069446 -0.911166  0.478370  0.333824  0.363060  0.949672

小开

If you want to add multiple columns to a multiindex column dataframe, you can try

All same value for columns

df[[("foo", "bar1"), ("foo", "bar2")]] = 2

        bar                 baz            foo
one       two       one       two bar1 bar2
0  0.487880 -0.487661 -1.030176  0.100813    2    2
1  0.267913  1.918923  0.132791  0.178503    2    2
2  1.550526 -0.312235 -1.177689 -0.081596    2    2

Same value for each column

df[[("foo", "bar1"), ("foo", "bar2")]] = [2, 3]

        bar                 baz            foo
one       two       one       two bar1 bar2
0  0.487880 -0.487661 -1.030176  0.100813    2    3
1  0.267913  1.918923  0.132791  0.178503    2    3
2  1.550526 -0.312235 -1.177689 -0.081596    2    3

Different value for each cell

df[[("foo", "bar1"), ("foo", "bar2")]] = [[1,2], [3,4], [5,6]] # shape is (3, 2) where 3 is index length and 2 is new added column length

        bar                 baz            foo
one       two       one       two bar1 bar2
0  0.487880 -0.487661 -1.030176  0.100813    1    2
1  0.267913  1.918923  0.132791  0.178503    3    4
2  1.550526 -0.312235 -1.177689 -0.081596    5    6

Another usecase is that we have a single index dataframe, and we want to concat it to the multi index dataframe

        bar                 baz
one       two       one       two     concat to      bar1  bar2
0  0.487880 -0.487661 -1.030176  0.100813   <---------  0     1     2
1  0.267913  1.918923  0.132791  0.178503               1     3     4
2  1.550526 -0.312235 -1.177689 -0.081596               2     5     6

Generate a list of tuples for columns

df[[("foo", col) for col in single_index_df.columns]] = single_index_df

        bar                 baz            foo
one       two       one       two bar1 bar2
0  0.487880 -0.487661 -1.030176  0.100813    1    2
1  0.267913  1.918923  0.132791  0.178503    3    4
2  1.550526 -0.312235 -1.177689 -0.081596    5    6

Create a new multi index columns dataframe from the single index dataframe as Option 2 of MaxU - stop genocide of UA

df = df.join(pd.DataFrame(single_index_df.values,
columns=pd.MultiIndex.from_product([['foo'], single_index_df.columns]),
index=single_index_df.index))

Create a multi index dataframe from single index dataframe with pd.concat({'foo': single_index_df}, axis=1)

df = pd.concat([df, pd.concat({'foo': single_index_df}, axis=1)], axis=1)