Pandas 申请创建多列，使用多列作为输入

Question

I am trying to use a function to create multiple outputs, using multiple columns as inputs. Here's my attempt:

df = pd.DataFrame(np.random.randint(0,10,size=(6, 4)), columns=list('ABCD'))
df.head()

    A   B   C   D
0   8   2   5   0
1   9   9   8   6
2   4   0   1   7
3   8   4   0   3
4   5   6   9   9

def some_func(a, b, c):
    return a+b, a+b+c

df['dd'], df['ee'] = df.apply(lambda x: some_func(a = x['A'], b = x['B'], c = x['C']), axis=1, result_type='expand')

df.head()

   A    B   C   D   dd  ee
0   8   2   5   0   0   1
1   9   9   8   6   0   1
2   4   0   1   7   0   1
3   8   4   0   3   0   1
4   5   6   9   9   0   1

The outputs are all 0 for the first new column, and all 1 for the next new column. I am interested in the correct solution, but I am also curious about why my code resulted this way.

Answer 1

You can assign to subset ['dd','ee'] :

def some_func(a, b, c):
    return a+b, a+b+c

df[['dd','ee']] = df.apply(lambda x: some_func(a = x['A'], 
                                               b = x['B'], 
                                               c = x['C']), axis=1, result_type='expand')
print (df)
   A  B  C  D  dd  ee
0  4  7  7  3  11  18
1  2  1  3  4   3   6
2  4  7  6  0  11  17
3  0  9  1  1   9  10
4  5  6  5  9  11  16
5  3  2  4  9   5   9

If possible, better/ fatser is use vectorized solution:

df = df.assign(dd = df.A + df.B, ee = df.A + df.B + df.C)

Answer 2

Just to explain the 0, 1 part. 0 and 1 are actually the column names of

df.apply(lambda x: some_func(a = x['A'], b = x['B'], c = x['C']), axis=1, result_type='expand')

That is

x = df.apply(lambda x: some_func(a = x['A'], b = x['B'], c = x['C']), axis=1, result_type='expand')
a, b = x
print(a)    # first column name
print(b)    # second column name

output:
0
1

Finally, you assign

df['dd'], df['ee'] = 0, 1

results in

   A    B   C   D   dd  ee
0   8   2   5   0   0   1
1   9   9   8   6   0   1
2   4   0   1   7   0   1
3   8   4   0   3   0   1
4   5   6   9   9   0   1

Answer 3

Alternative way:

df['dd'], df['ee'] = zip(*df.apply(lambda x: some_func(x['A'], x['B'], x['C]) )