Souvik Ray Souvik Ray - 1 year ago 49
Python Question

Swaping two elements in a list shows unexpected behaviour

I have a list:

lis = [12,45,15,67,89]

I want to swap 12 and 89 such that the list should look

lis = [89,45,15,67,12]

When I do it this way

lis[0], lis[lis.index(89)] = lis[lis.index(89)], lis[0]

Nothing is changed

lis = [12,45,15,67,89]

But when I do it this way

lis5[0], lis5[4] = lis5[4], lis5[0]

It works perfect

lis = [89,45,15,67,12]

So why it isn't working the first way? PS-The whole reason I want to do the first way is because I want to find the max element in a list and then swap it with the first element of the list.

Something like:

max1 = max(lis)
lis[0], lis[lis.index(max1)] = lis[lis.index(max1)], lis[0]

Answer Source

In short, lis.index(max1) in left/right sides are evaluated at different times; then evaluated to different values (0 for the left side, 4 for the right side).

a, b = b, a

Above statement:

  • creates (b, a) tuple
  • unpacks the above tuple to a and b
    • a = old-value-of-b
    • b = old-value-of-a

lis[0], lis[lis.index(89)] = lis[lis.index(89)], lis[0]


lis[0], lis[lis.index(89)] = lis[4], lis[0]


lis[0], lis[lis.index(89)] = 89, 0


lis[0] = 89
lis[lis.index(89)] = 0


lis[0] = 89
lis[0] = 0    # NOTE lis.index(89) == 0  at this point

So lis[0] assigned 89, then assigned original value 0 back. => (only the first element changed, and re-assigned original value)

To avoid this problem, assign the index before the swap statement:

idx = lis.index(89)
lis[0], lis[idx] = lis[idx], lis[0]
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