Let say I have A=[2,4, 3 ,1], I want to compute the sum of element inside by skipping one element in each step. What I try is:

s=[ ]
a=0
for i in range(len(A)):
    for j in range(len(A)):
        if i==j:
            continue
        else :
            a  =A[j]
    s.append(a)

When I print the result s I am getting

print(s)

s=[8, 14, 21, 30]

What I want to have is:

s=[ 8, 6, 7, 9]

Where

8=4 3 1 we skip A[0]
6=2 3 1 we skip A[1]
7=2 4 1 we skip A[2]
9=2 4 3 we skip A[3]

CodePudding user response：

How about computing the sum and then returning the sum minus each item, using list comprehension?

sum_a = sum(A)
output = [sum_a - a for a in A]
print(output) # [8, 6, 7, 9]

CodePudding user response：

The if statement is already skipping an iteration, so all you need is to repeat the initialization. You can do this by re-initializing the sum variable (a) inside the outer loop instead of outside.

A = [2, 4, 3, 1]
s = []

for i in range(len(A)):
    a = 0
    for j in range(len(A)):
        if i == j:
            continue
        else:
            a  = A[j]
    s.append(a)

print(s)

Output:

[8, 6, 7, 9]

CodePudding user response：

I like @j1-lee answer, but here's a nifty one-liner variation on that:

output = [sum(A) - a for a in A]

CodePudding user response：

You could build a new list from slices, excluding the unwanted value, then sum

[sum(A[:i]   A[i 1:]) for i in range(len(A))]

Or, since sum results can themselves be added, use 2 slices

[sum(A[:i])   sum(A[i 1:]) for i in range(len(A))]