python multiply all elements except ith element in list and return list [closed]-CodePudding

Suppose I have a list .I want to multiply its all elements except with the i'th element.

Example:

input: [1,2,3,4]
output: [24,12,8,6]

Here, the output

 24 is 2*3*4
 12 is 1*3*4
 8 is 1*2*4
 6 is 1*2*3

Any help would be appreciated

CodePudding user response：

A simple solution using two for loops:

l=[1,2,3,4]
out=[]
for i in range(len(l)):
    prod=1 
    for j in range(len(l)):
        if(j!=i): #Iterate through list once more and compare the indices
            prod=prod*l[j]
    out.append(prod) 
print(out)

Output is: [24, 12, 8, 6]

CodePudding user response：

A solution using itertools.combinations (and reversed to get the right order)

from itertools import combinations
from operator import mul
from functools import reduce

values = [1, 2, 3, 4]
result = [reduce(mul, c) for c in combinations(reversed(values), r=len(values) - 1)]
print(result)  # [6, 8, 12, 24]

Or using your logic directly : iterate and for each index don't use the corresponding value

values = [1, 2, 3, 4]
result = []
for i in range(len(values)):
    x = 1
    for idx, value in enumerate(values):
        if i != idx:
            x *= value
    result.append(x)

print(result)  # [24, 12, 8, 6]

CodePudding user response：

You can use numpy, like below:

import numpy as np
arr = np.array([1,2,3,4])
[np.prod(np.delete(arr, idx)) for idx in range(len(arr))]
# [24, 12, 8, 6]

If you want to not install numpy you can do like below: ([::-1] reverse liest)

from itertools import combinations
import math

arr = [1, 2, 3, 4]
[math.prod(a) for a in combinations(arr, r=len(arr) - 1)][::-1]
# [24, 12, 8, 6]