Swapping list elements in python where the expressions contain function calls

Question

if arr = [4,3,2,1] and i want to swap the first value with the minimum of the array , if am using this on python

arr[0] , arr[arr.index(min(arr))] = min(arr) , arr[0] 
#or   
arr[0] , arr[arr.index(min(arr))] = arr[arr.index(min(arr))] , arr[0]

they are not working but if i do this

b = arr.index(min(arr))  
#and then  
arr[0] , arr[b] = arr[b] , arr[0]

this works fine. Can anyone explain why ?

Answer 1

It has to do with the order of the operations.

We can subclass lists to instrument them to show what they're doing.

class TracingList(list):
    def __getitem__(self, key):
        value = super().__getitem__(key)
        print(self, "reading", key, "=", value)
        return value

    def __setitem__(self, key, value):
        print(self, "writing", key, "=", value)
        super().__setitem__(key, value)

    def index(self, value):
        index = super().index(value)
        print(self, "finding index of", value, "=", index)
        return index


arr = TracingList([4, 3, 2, 1])
arr[0], arr[arr.index(min(arr))] = min(arr), arr[0]
print(arr)
print("===")

arr = TracingList([4, 3, 2, 1])
arr[0], arr[arr.index(min(arr))] = arr[arr.index(min(arr))], arr[0]
print(arr)
print("===")

arr = TracingList([4, 3, 2, 1])
b = arr.index(min(arr))
arr[0], arr[b] = arr[b], arr[0]
print(arr)

prints out

[4, 3, 2, 1] reading 0 = 4
[4, 3, 2, 1] writing 0 = 1
[1, 3, 2, 1] finding index of 1 = 0
[1, 3, 2, 1] writing 0 = 4
[4, 3, 2, 1]
===
[4, 3, 2, 1] finding index of 1 = 3
[4, 3, 2, 1] reading 3 = 1
[4, 3, 2, 1] reading 0 = 4
[4, 3, 2, 1] writing 0 = 1
[1, 3, 2, 1] finding index of 1 = 0
[1, 3, 2, 1] writing 0 = 4
[4, 3, 2, 1]
===
[4, 3, 2, 1] finding index of 1 = 3
[4, 3, 2, 1] reading 3 = 1
[4, 3, 2, 1] reading 0 = 4
[4, 3, 2, 1] writing 0 = 1
[1, 3, 2, 1] writing 3 = 4
[1, 3, 2, 4]