Python , array wont change after once in iteration(swapping)

you can use:

for i in range(5):  
    b, c = map(int, input().split())
    b_index = a.index(b)
    c_index = a.index(c)
    a[b_index], a[c_index] = a[c_index], a[b_index]

here is a piece of code to understand the behavior:

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

def get_index(i, s):
    idx = a.index(i)
    print(s, i, idx)
    return idx

a[get_index(1, 'left')], a[get_index(2, 'left')] = a[get_index(2, 'right')], a[get_index(1, 'right')]
print(a)

output:

right 2 1
[1, 2, 3, 4, 5]
right 1 0
[1, 2, 3, 4, 5]
left 1 0
[1, 2, 3, 4, 5]
left 2 0
[2, 2, 3, 4, 5]
[1, 2, 3, 4, 5]

as you can see get_index(2, 'left') is not the same with get_index(2, 'right'), this shows that a.index(c) doesn't have the same value in the left and in the right side of the = operator

back to your code:

a[a.index(b)], a[a.index(c)] = a[a.index(c)], a[a.index(b)]

first, it is executed the code from the right side of = operator so you have (c, b), then a[a.index(b)] will take the value c, at this step you have 2 of c in your list a, when a.index(c) it is evaluated will return the first c found from the left to right, in the final step you want to set value b to where c was in the beginning, but now you have 2 of c, if c < b you will end with the desired output because the method list.index will return the position of actual c; if b < c then b will take the place where b was in the beginning and this will result in no change in your list a

example:

case b < c
a = [1, 2, 3, 4, 5] --> swap 2 with 5; b = 2, c = 5
a[a.index(b)], a[a.index(c)] =  5, 2
a[a.index(b)] = 5 --> a = [1, 5, 3, 4, 5]
      -->  a.index(5) --> --> ↑
      ↑
a[a.index(c)] = 2 --> a =  [1, 2, 3, 4, 5]


case b > c
a = [1, 2, 3, 4, 5] --> swap 2 with 5; b = 5, c = 2
a[a.index(b)], a[a.index(c)] =  2, 5
a[a.index(b)] = 2 --> a = [1, 2, 3, 4, 2]
      -->  a.index(2) --> --> ↑
      ↑
a[a.index(c)] = 5 --> a =  [1, 5, 3, 4, 2]