Python loop using zip with assignment

Question

I'm trying to use a for loop in order to modify values in a list, based on values on a second list. Hence I though that the "zip" function would come in handy, however I'm not getting the result I'm expecting. See an example here:

list_a = [0,0,0,0]
list_b = [1,2,3,4]

for a, b in zip(list_a, list_b):
a = b*2

I though that this modifies list_a, however it does not. My understanding was that, inside the loop, "a" was first equal to list_a[0], then list_a[1], etc, in other words views of the list.

I have actually 2 questions:

a) what are "a" and "b" inside the loop, if they are not views of the two lists?

b) is there a pythonic way of implementing this loop (i.e. something different from looping over i in range(len(list_a))?

pythonic way ? probably do your operations in a comprehension and create a new list, assigning to list_a at the end. — Paritosh Singh
– Paritosh Singh, Commented Apr 7, 2019 at 18:12
a=.. does not modify a[0]. It just assigns a new value to the variable. view is a numpy array concept, not a list one. — hpaulj
– hpaulj, Commented Apr 7, 2019 at 18:14
"was first equal to list_a[0], then list_a[1], etc, in other words views of the list." No, they are not views. list_a[i] returns the object contained by that index in the list. a = list_a[0] makes the name a reference some object, that happens to be the first element of list_a, but that object has no knowledge of that — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Apr 7, 2019 at 18:15

dcg · Accepted Answer · 2019-04-07 18:19:03Z

2

You can try something like:

list_a = [0,0,0,0]
list_b = [1,2,3,4]

for i, b in enumerate(list_b):
    list_a[i] = b*2

You can get the same result with:

[b*2 for b in list_b]

edited Apr 7, 2019 at 18:19

answered Apr 7, 2019 at 18:13

dcg

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