Most efficient way to multiply every column of a large pandas dataframe with every other column of the same dataframe

Question

Suppose I have a dataset that looks something like:

INDEX   A   B   C
    1   1   1   0.75
    2   1   1   1
    3   1   0   0.35
    4   0   0   1
    5   1   1   0

I want to get a dataframe that looks like the following, with the original columns, and all possible interactions between columns:

INDEX   A   B   C       A_B     A_C     B_C
    1   1   1   0.75    1       0.75    0.75
    2   1   1   1       1       1       1
    3   1   0   0.35    0       0.35    0
    4   0   0   1       0       0       0
    5   1   1   0       1       0       0

My actual datasets are pretty large (~100 columns). What is the fastest way to achieve this?

I could, of course, do a nested loop, or similar, to achieve this but I was hoping there is a more efficient way.

Answer 1

You could use itertools.combinations for this:

>>> import pandas as pd
>>> from itertools import combinations
>>> df = pd.DataFrame({
...     "A": [1,1,1,0,1],
...     "B": [1,1,0,0,1],
...     "C": [.75,1,.35,1,0]
... })
>>> df.head()
   A  B     C
0  1  1  0.75
1  1  1  1.00
2  1  0  0.35
3  0  0  1.00
4  1  1  0.00
>>> for col1, col2 in combinations(df.columns, 2):
...     df[f"{col1}_{col2}"] = df[col1] * df[col2]
...
>>> df.head()
   A  B     C  A_B   A_C   B_C
0  1  1  0.75    1  0.75  0.75
1  1  1  1.00    1  1.00  1.00
2  1  0  0.35    0  0.35  0.00
3  0  0  1.00    0  0.00  0.00
4  1  1  0.00    1  0.00  0.00

If you need to vectorize an arbitrary function on the pairs of columns you could use:

import numpy as np

def fx(x, y):
    return np.multiply(x, y)

for col1, col2 in combinations(df.columns, 2):
    df[f"{col1}_{col2}"] = np.vectorize(fx)(df[col1], df[col2])

Answer 2

I am not aware of a native pandas function to solve this, but itertools.combinations would be an improvement over a nested loop.

You could do something like:

from itertools import combinations

df = pd.DataFrame(data={"A": [1,1,1,0,1], 
                        "B": [1,1,0,0,1], 
                        "C": [0.75, 1, 0.35, 1, 0]})

for comb in combinations(df.columns, 2): 
    col_name = comb[0] + "_" + comb[1]
    result[col_name] = df[comb[0]] * df[comb[1]]