I want to take all the elements in the list except the current element. Here is the List:

nums = [1, 1, 3, 5, 2, 3, 4, 4]

I tried using list comprehension:

[(nums[:x:],nums[x 1:]) for x in range(len(nums)) ]

This is the output I am getting:

[([], [1, 3, 5, 2, 3, 4, 4]),
 ([1], [3, 5, 2, 3, 4, 4]),
 ([1, 1], [5, 2, 3, 4, 4]),
 ([1, 1, 3], [2, 3, 4, 4]),
 ([1, 1, 3, 5], [3, 4, 4]),
 ([1, 1, 3, 5, 2], [4, 4]),
 ([1, 1, 3, 5, 2, 3], [4]),
 ([1, 1, 3, 5, 2, 3, 4], [])]

I am getting two blank lists one in the beginning and one at the end which is correct. I want to know if there is a way where I can edit these 2 blank lists to [1] by putting any validation?

In essence I need to calculate the product of all the numbers except the current number and I am not allowed to use division.

Also, there is a time constraint that I cannot use nested/multiple for loops as test cases will fail on exceeding time limit

CodePudding user response：

Based on your final goal, it seems like you don't need two separate lists for before/after the current number. Instead, concatenate them and apply prod. That way you don't even need to worry about the empty list issue.

from math import prod

nums = [1, 1, 3, 5, 2, 3, 4, 4]

prods = [prod(nums[:i]   nums[i 1:]) for i in range(len(nums))]
print(prods) # [1440, 1440, 480, 288, 720, 480, 360, 360]

CodePudding user response：

Not very pretty, but this works:

nums = [1, 1, 3, 5, 2, 3, 4, 4]

n = [(nums[:x:],nums[x 1:]) if x!=0 and x!=len(nums)-1 else ((nums[:x:],[1]) if x==len(nums)-1 else ([1],nums[x 1:])) for x in range(len(nums)) ]

print(n)

Output:

[([1], [1, 3, 5, 2, 3, 4, 4]), ([1], [3, 5, 2, 3, 4, 4]), ([1, 1], [5, 2, 3, 4, 4]), ([1, 1, 3], [2, 3, 4, 4]), ([1, 1, 3, 5], [3, 4, 4]), ([1, 1, 3, 5, 2], [4, 4]), ([1, 1, 3, 5, 2, 3], [4]), ([1, 1, 3, 5, 2, 3, 4], [1])]