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 ?

@smci: Except that it's not actually left-to-right here. It could have been designed to be left-to-right, but it wasn't. (I really wish Python multiple assignment had been designed to evaluate all expressions left-to-right, then start assigning things.) — user2357112
– user2357112, Commented Aug 16, 2018 at 21:31
@user2357112: then please explain what it is. It is parsed left-to-right, then each expression is dynamically evaluated, yes? For a tuple assignment L1, L2 = R1, R2, L1 is evaluated first, then R1, then L2, then R2? — smci
– smci, Commented Aug 16, 2018 at 22:07
@smci: Nope. R1, R2, L1, (first assignment happens here), L2, (second assignment happens here). — user2357112
– user2357112, Commented Aug 16, 2018 at 22:09
@smci: The older answer mentions the exception for assignments, but in a way that's easy to miss. I've edited it to be a bit more clear. — user2357112
– user2357112, Commented Aug 16, 2018 at 22:22

AKX · Accepted Answer · 2018-08-16 20:16:51Z

8

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]

answered Aug 16, 2018 at 20:16

AKX

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bjschoenfeld Over a year ago

To explain, arr[0] gets set before arr[arr.index(min(arr))] gets evaluated and set. Thus once arr[0] is set to 1, the index of the min becomes 0. So arr[arr.index(min(arr))] assigns 4 to index 0.

Michael H. Over a year ago

This means that if you change the order of the assignments, i.e. arr = [4,3,2,1]; arr[arr.index(min(arr))], arr[0] = arr[0], min(arr); print(arr), you get the "correct" result. :o

smci Over a year ago

Ok but there's no explanation of what the order of evaluation actually is, or why it is so, or what principle is being followed. This is just like saying "The order is whatever it is, because it is so". Python evaluates expressions L-to-R, except for assignments, where the RHS is evaluated before the LHS

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