Given a data frame of numeric values, I would like to perform plus, minus, multiply & divide on all combinations of columns.
What would be the fastest approach to do this for combinations of 3 and above?
A minimal reproducible example is given below with combinations of 2.
import numpy as np
import pandas as pd
from itertools import combinations
from itertools import permutations
from sklearn.datasets import load_boston
# the dataset
X, y = load_boston(return_X_y=True)
X = pd.DataFrame(X)
combos2 = list(combinations(X.columns,2))
perm3 = list(permutations(X.columns,3)) # how would i do this with out typing out all the permutations
for i in combos2:
X[f'{i[0]}_X_{i[1]}'] = X.iloc[:,i[0]]*X.iloc[:,i[1]] # Multiply
X[f'{i[0]}_+_{i[1]}'] = X.iloc[:,i[0]]+X.iloc[:,i[1]] # Add
X[f'{i[0]}_-_{i[1]}'] = X.iloc[:,i[0]]-X.iloc[:,i[1]] # Subtract
X[f'{i[0]}_/_{i[1]}'] = X.iloc[:,i[0]]/(X.iloc[:,i[1]]+1e-20) # Divide
I was thinking of a way to add the "operators + * - / into the combinations so it can be written in fewer lines than manually typing out all the combinations, but I don't know where to begin?
I would like all orders: i.e (a * b + c) , (a * b - c) , (a * b / c) etc
Ideally leaving no duplicate columns. i.e (a + b + c) and (c + b + a)
For example if I had 3 columns a b c. I want a new column (a * b + c).
Here's a naive solution that outputs the combinations of 2 & 3 of all the columns.
from sklearn.datasets import load_boston
from itertools import combinations
import operator as op
X, y = load_boston(return_X_y=True)
X = pd.DataFrame(X)
comb= list(combinations(X.columns,3))
def operations(x,a,b):
if (x == '+'):
d = op.add(a,b)
if (x == '-'):
d = op.sub(a,b)
if (x == '*'):
d = op.mul(a,b)
if (x == '/'): # divide by 0 error
d = op.truediv(a,(b + 1e-20))
return d
for x in ['*','/','+','-']:
for y in ['*','/','+','-']:
for i in comb:
a = X.iloc[:,i[0]].values
b = X.iloc[:,i[1]].values
c = X.iloc[:,i[2]].values
d = operations(x,a,b)
e = operations(y,d,c)
X[f'{i[0]}{x}{i[1]}{y}{i[2]}'] = e
X[f'{i[0]}{x}{i[1]}'] = d
X = X.loc[:,~X.columns.duplicated()]