如何在创建附加列时将 function 应用于整个 pandas df？

Question

So my question is a follow up from a previous post which has had some updates:

Problem

Given a (400 * 18) pandas dataframe I would like to produce a (400*153) size dataframe which includes the unique pairs of the 18 existing columns (18./2,(18-2).=153). I have so far managed this with one row, however I am not able to apply it correctly to all 400 observations.

Attempt so far

My code:

#df2 = (400 * 18) df
#list(combinations(df2.iloc[i], 2)): yields 152 unique pairs of the 18 variables

rows = {} #storing data in a dict as 400 keys which have  a list of 153 pairs
vars = {} #final dictionary for the computed pairs to be stored in

#For each pair I need to compute 400 values because n=400
for j in range(0,153): 
  for i in range(0,400):
      #grabbing all unique pairs of the variables and storing it as a list for each observation
      rows[i] = list(combinations(df2.iloc[i], 2)) 
      for x in rows.keys():
        #Performing the computation of each pair in each row
        vars[x] = rows[x][j][0] * rows[x][j][1] 

#vars yields a dictionary containing 400 keys each 
#only having one value within them rather than 153 values

My approach was to create a dictionary containing 400 keys which would then have 153 values each, where each of the 153 values is the product of the pairs that were acquired previously. So far I am receiving a dictionary of 400 keys each which only have 1 value within it, as opposed to the 153 I am wanting

Answer 1

You can simply apply across each row (axis=1) and create a new series based on the index of the combination and the results.

Here's a sample with 5 columns producing 10 ( 5!/(2!(5-2)!) ) results.

Currently, column names are being generated from the index provided by enumerate , but you could also modify the keys to give more meaningful column names.

from itertools import combinations

import pandas as pd

df2 = pd.DataFrame({'A': [0, 1, 2],
                    'B': [3, 4, 5],
                    'C': [6, 7, 8],
                    'D': [9, 10, 11],
                    'E': [12, 13, 14]})

new_df = df2.apply(lambda s:
                   pd.Series(
                       {i: c for i, c in enumerate(combinations(s.values, 2))}
                   ),
                   axis=1)

# For Display
print(new_df.to_string(index=False))

Output:

     0      1       2       3      4       5       6       7       8        9
(0, 3) (0, 6)  (0, 9) (0, 12) (3, 6)  (3, 9) (3, 12)  (6, 9) (6, 12)  (9, 12)
(1, 4) (1, 7) (1, 10) (1, 13) (4, 7) (4, 10) (4, 13) (7, 10) (7, 13) (10, 13)
(2, 5) (2, 8) (2, 11) (2, 14) (5, 8) (5, 11) (5, 14) (8, 11) (8, 14) (11, 14)