Reverse list where indexes are even

January 9, 2022

I would like to reverse a list but only where the indexes are even.
So my goal is that a list that starts like :

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

Should look like :

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

So to check where indexes are even I write :

for j in range(0,len(A)):
     if j % 2 == 0:
         print(A[j])        #prints all the values at equal indexes.

How would I only reverse A where the above condition is met to for B?
So to explain more , I would like indexes of array A to swap where they are even . Indexes that are even are : 0 , 2 , 4 where the values of A are
1 , 3 , 5 . I want these values to switch to 5,3,1 but indexes that are not even should stay the same.

Is there a quick method like :

A[::-1]

>Solution :

Others like Michael Szczesny have suggested more concise solutions like:

A[::2] = reversed(A[::2])

already in the comments, but here is one that is more verbose in case it helps you see what is going on.

We’ll assume that the list starting out is already sorted.

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


#We isolate the even indices
even_indices = []
for i in range(len(A)):
    if i%2==0:
        even_indices.append(A[i])

#We reverse the even indices
even_indices = list(reversed(even_indices))
print(even_indices)

B = []
#We put everything back together, if the index is odd, we use the value from A, if not, we use the value from even_indices, which has been reversed.
for i in range(len(A)):
    if i%2==1:
        B.append(A[i])
    else:
        B.append(even_indices[i//2])


print(B)

Output: