Apply function in pandas with multiple args and multiple outputs

Question

I have a function foo() which takes two arguments as two columns of pandas named year and month and return a list of four numbers.

  df['A'],df['B'],df['C'],df['D']=  df.apply(lambda x: foo(x.year, x.month), axis=1,result_type="expand")

It just gives me four columns named A,B,C,D with 0,1,2,3 respectively filled in them. What am I doing wrong?

Kindly don't answer for single variable outputs or single variable arguments. There are plenty of examples out there for that. Thankyou so much for the help.

Can you add more part of function and data examples so that we can recreate the environment you are working in? — Rupesh
– Rupesh, Commented Jun 6, 2020 at 12:10

Shubham Sharma · Accepted Answer · 2020-06-06 13:30:59Z

1

You are incorrectly assigning the results of df.apply with optional parameter result_type='expand', instead, Use:

df[['A', 'B', 'C', 'D']] =  df.apply(lambda x: foo(x.year, x.month), axis=1, result_type="expand")

Consider the example demonstrating this,

df = pd.DataFrame({'col1': np.arange(5), 'col2': np.arange(5) * 2})

#print(df)
   col1  col2
0     0     0
1     1     2
2     2     4
3     3     6
4     4     8

Reproducing the problem situation,

df['A'], df['B'] = df.apply(lambda s: (s['col1']**2, s['col2']**2), axis=1, result_type='expand')

#print(df)
   col1  col2  A  B
0     0     0  0  1
1     1     2  0  1
2     2     4  0  1
3     3     6  0  1
4     4     8  0  1

The solution is,

df[['A', 'B']] = df.apply(lambda s: (s['col1']**2, s['col2']**2), axis=1, result_type='expand')

#print(df)
   col1  col2   A   B
0     0     0   0   0
1     1     2   1   4
2     2     4   4  16
3     3     6   9  36
4     4     8  16  64

OR:

df['A'], df['B'] = df.apply(
lambda s: (s['col1']**2, s['col2']**2), axis=1, result_type='expand').T.to_numpy()

#print(df)
   col1  col2   A   B
0     0     0   0   0
1     1     2   1   4
2     2     4   4  16
3     3     6   9  36
4     4     8  16  64

edited Jun 6, 2020 at 13:30

answered Jun 6, 2020 at 12:13

Shubham Sharma

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6 Comments

blastoise Over a year ago

Can you explain a little what the difference is? And why was mine giving 0,1,2,3?

Shubham Sharma Over a year ago

Well, I guess if you try to unpack the results of .apply with result_type='expand it just returns the headers of newly created columns (I guess its internal implementation behaviour).

Shubham Sharma Over a year ago

@jazrael @anky Can you explain about the behaviour of result_type=expand used here? Why its returning headers when we try to unpack it?

anky Over a year ago

When you unpack a dataframe , it by default chooses to return column headers example: print(list(df)) would return column names, for accessing the values you need to access the underlying array you need to use .to_numpy() : df['col3'],df['col4'] = df.apply(lambda s: (s['col1']**2, s['col2']**2), axis=1, result_type='expand').T.to_numpy()

Shubham Sharma Over a year ago

@anky Makes sense.:), btw, Thanks for taking time out for explaining.

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