从指定列中减去pandas列

Question

如何动态地从指定列中减去多个Pandas DataFrame列的值。在这种情况下，如何从存款中减去A、B和C列，并将该值放置在相应的A、B和C列中。

   date         deposit       A                 B           C
0  2017-01-15   12            5                 10          12
1  2017-01-16   20            10                4           32
2  2017-01-17   5             50                10          18
3  2017-01-18   22            15                20          12

应该产生：

   date         deposit       A                 B           C
0  2017-01-15   12            7                 2           0
1  2017-01-16   20            10                16         -12
2  2017-01-17   5            -45               -5          -13
3  2017-01-18   22            7                 2           10

- DougKruger

Answer 1

In [226]: df[['A','B','C']] = df.deposit.values[:, None] - df[['A','B','C']]

In [227]: df
Out[227]:
         date  deposit   A   B   C
0  2017-01-15       12   7   2   0
1  2017-01-16       20  10  16 -12
2  2017-01-17        5 -45  -5 -13
3  2017-01-18       22   7   2  10

- MaxU - stand with Ukraine

Answer 2

loc + rsub

cols = ['A', 'B', 'C']
df.loc[:, cols] = df[cols].rsub(df.deposit, 0)
df


         date  deposit   A   B   C
0  2017-01-15       12   7   2   0
1  2017-01-16       20  10  16 -12
2  2017-01-17        5 -45  -5 -13
3  2017-01-18       22   7   2  10

inplace
我更喜欢在原地进行操作inplace

df.update(df[['A', 'B', 'C']].rsub(df.deposit, 0))

df

         date  deposit   A   B   C
0  2017-01-15       12   7   2   0
1  2017-01-16       20  10  16 -12
2  2017-01-17        5 -45  -5 -13
3  2017-01-18       22   7   2  10

复制
总体而言，我更偏爱

df.assign(**df[['A', 'B', 'C']].rsub(df.deposit, 0).to_dict('list'))

         date  deposit   A   B   C
0  2017-01-15       12   7   2   0
1  2017-01-16       20  10  16 -12
2  2017-01-17        5 -45  -5 -13
3  2017-01-18       22   7   2  10

- piRSquared

Answer 3

for c in ['A','B','C']:
    df[c]=df['deposit']-df[c]

- alex314159