使用函数从使用熊猫的特定列输入返回多列输出

Question

I would like to add two new columns to my dataframe by applying a function that takes inputs from multiple, specific pre-existing columns.

Here is my approach which works for returning one column, but not multiple:

Here is my DataFrame:

d = {'a': [3,0,2,2],
    'b': [0,1,2,3],
    'c': [1,1,2,3],
    'd': [2,2,1,3]}

df = pd.DataFrame(d)

I'm trying to apply this function:

def myfunc(a,b,c):
    if a > 2 and b > 2:
        print('condition 1',a,b)
        return pd.Series((a,b))
    elif a < 2 and c < 2:
        print('condition 2',a,c)
        return pd.Series((b,c))
    else:
        print('no condition')
        return pd.Series((None,None))

Like this:

df['e'],df['f'] = df.apply(lambda x: myfunc(x['a'],x['b'],x['c']),axis=1)

Output:

no condition
no condition
condition 2 0 1
no condition
no condition

DataFrame result:

How can I input multiple columns and get multiple columns out?

Answer 1

Your function is going to return one series with either NAs or with a 2-tuple when my_funct matched.

One way to fix it is to return Series instead, that will be automatically expanded by apply:

def myfunc(col1,col2,col3):
    if col1 == 'x' and col2 == 'y':
        return pd.Series((col1,col2))
    if col2 == 'a' and col3 == 'b':
        return pd.Series(('yes','no'))

Note the double brackets to pass one argument as a tuple. A list would be fine too.

Answer 2

The issue is with the assignment, not myfunc

When you try to unpack a dataframe as a tuple, it returns the column lables. That's why you get (0, 1) for everything

df['e'], df['f'] = pd.DataFrame([[8, 9]] * 1000000, columns=['Told', 'You'])
print(df)

   a  b  c  d     e    f
0  3  0  1  2  Told  You
1  0  1  1  2  Told  You
2  2  2  2  1  Told  You
3  2  3  3  3  Told  You

Use join

df.join(df.apply(lambda x: myfunc(x['a'],x['b'],x['c']),axis=1))

Or pd.concat

pd.concat([df, df.apply(lambda x: myfunc(x['a'],x['b'],x['c']),axis=1)], axis=1)

both give

   a  b  c  d    e    f
0  3  0  1  2  NaN  NaN
1  0  1  1  2  1.0  1.0
2  2  2  2  1  NaN  NaN
3  2  3  3  3  NaN  NaN