Python3 - for loop remove list item

Question

I have code:

listA = [i for i in range(5)]

for x in listA :
    print(listA)
    print('prepare to remove %s'%x)
    listA.remove(x)
    listA = [i for i in range(x, 10)]
    print(listA)

I guess x in for loop will be: 0 1 2 3 4

but the result is 0 2 3 4

Why ? This is so weird...

I just want to know why the 1 are disappeared ,Why not print 0 2 4 or 01234 ? Just 1

someone can explain this?

but I just wanna know why the `x` print will be 0 2 3 4 , not 0 1 2 3 4 or 0 2 4 ? — Vic, Dec 02 '16 at 09:24
While you are at it: it is an horrible idea to use `list` as a variable name, as list is the built-in Python name for list constructors. — jsbueno, Dec 02 '16 at 09:29
@Jean-FrançoisFabre I don't think he wants to know how to do anything, just why the above does what it does. Otherwise, I would recommend never altering the container of a for loop — Tom Malkin, Dec 02 '16 at 09:29
Yes, I wouldn't bet on undefined behaviour, that's just like the old C : `a = b++ - --c;` — Jean-François Fabre, Dec 02 '16 at 09:31
@roganjosh I'm not saying it is. can we close this question now? — Jean-François Fabre, Dec 02 '16 at 09:32

score 1 · Answer 1 · answered Dec 02 '16 at 09:49

refer code below

during 1st iteration, lst.remove(x) effectively remove 1 item from lst with id xxxx, lst is now [1,2,3,4]
at 2nd iteration, since 0 already being remove, it will get 2 at index 1 when trying to get next item
from 2nd iteration onwards, like @Harlekuin mention, you are performing remove operation on a totally new created list, which won't impact original list with id xxxx, folowing iteration will continue on lst with id xxxx [1,2,3,4], so its 3 following by 4

code

lst = [0, 1, 2, 3, 4] # define raw list to make it clearer
# id(lst) = xxxx

# iterate over lst with id xxxx
for x in lst: 
    print('prepare to remove %s'%x)
    lst.remove(x) # 1st loop operate on list with id xxxx, 2nd loop onwards operate on list with id yyyy
    lst = [i for i in range(x, 10)] # lst being assign to another id, lets assume its yyyy now
    # id(lst) = yyyy

score 0 · Answer 2 · answered Dec 02 '16 at 09:27

It's because you remove x (being an integer), and then recreate the list with range(x, 10). The first time you do this it will remove the 1 from the original list, but the list is recreated before printing, so only the first loop will lose a number. In other words,

list.remove(x)

is useless once you:

list = [i for i in range(x, 10)]

score 0 · Answer 3 · answered Dec 02 '16 at 09:51

In first loop you remove 0 from listA which refers to oryginal [0, 1, 2, 3, 4] and you have [1, 2, 3, 4]. for uses reference to the same list so in next loop it use [1, 2, 3, 4] and it skips 1.

But meanwhile you assign new list to variable listA so in next loop you remove 1 from new list - but for still use reference to oryginal [1, 2, 3, 4] and it doesn't skip other elements.

score 0 · Answer 4 · answered Dec 02 '16 at 09:55

The problem is that you use the same list in your loop condition and inside.

# List to loop x over
listB = [i for i in range(5)]
# List to manipulate with remove
listA = [i for i in range(0, 10)]
for x in listB:
    print('loop: %s'%x)
    print('list before removing %s'%x)
    print(listA)
    listA.remove(x)
    print('list after removing %s'%x)
    print(listA)
    # Re-initialize listA
    listA = [i for i in range(x, 10)]

This produces the following output:

loop: 0
list before removing 0
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
list after removing 0
[1, 2, 3, 4, 5, 6, 7, 8, 9]
loop: 1
list before removing 1
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
list after removing 1
[0, 2, 3, 4, 5, 6, 7, 8, 9]
loop: 2
list before removing 2
[1, 2, 3, 4, 5, 6, 7, 8, 9]
list after removing 2
[1, 3, 4, 5, 6, 7, 8, 9]
loop: 3
list before removing 3
[2, 3, 4, 5, 6, 7, 8, 9]
list after removing 3
[2, 4, 5, 6, 7, 8, 9]
loop: 4
list before removing 4
[3, 4, 5, 6, 7, 8, 9]
list after removing 4
[3, 5, 6, 7, 8, 9]

Python3 - for loop remove list item

4 Answers4