list.remove() isn't removing all instances

Question

I'm facing rather bizarre situation with lists. It appears that Python is very selective about what zeros it will remove in this case:

count = 0
a = ["a",0,0,"b",None,"c","d",0,1,False,0,1,0,3,[],0,1,9,0,0,{},0,0,9]

for x in a:
    if x == 0:
        a.remove(x)
        count += 1

print(a, count)

Only 6 out of 10 zeros are removed. Why ?

please always use the generic [python] tag for all python questions — juanpa.arrivillaga, Oct 04 '19 at 13:53

score 3 · Accepted Answer · answered Oct 04 '19 at 13:56

3

A far better solution is to just create a new list that does not contain the 0s:

b = [x for x in a if x != 0]
count = len(a) - len(b)

answered Oct 04 '19 at 13:56

RemcoGerlich

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score 0 · Answer 2 · answered Oct 04 '19 at 13:54

0

To workaround the issue described by the guys in comments you can do the following:

count = 0
a = ["a",0,0,"b",None,"c","d",0,1,False,0,1,0,3,[],0,1,9,0,0,{},0,0,9]
for x in a.copy():
    if x == 0:
        a.remove(x)
        count += 1
print(a, count)

Then you will be iterating through the original a, while simultaneously reducing it by 0 whenever you encounter it.

Caveat:

>>> False == 0
True

answered Oct 04 '19 at 13:54

Grzegorz Skibinski

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So I need to make a duplicate list in order to modify one on the fly. – mr_incredible Oct 04 '19 at 13:58
Correct, if it's "in use" - and you still want to iterate through it, while simultaneously changing it - then yes - you need to duplicate it – Grzegorz Skibinski Oct 04 '19 at 13:58
@mr_incredible: another option is to go through it backwards. But duplicating or re-creating it is faster, more readable, less error prone. – RemcoGerlich Oct 04 '19 at 14:00

score 0 · Answer 3 · edited Oct 04 '19 at 18:19

0

x = ["a",0,0,"b",None,"c","d",0,1,False,0,1,0,3,[],0,1,9,0,0,{},0,0,9]

list(filter(lambda a: a != 0, x))

output

['a', 'b', None, 'c', 'd', 1, 1, 3, [], 1, 9, {}, 9]

edited Oct 04 '19 at 18:19

shahaf

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answered Oct 04 '19 at 14:02

Darjusch

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list.remove() isn't removing all instances

3 Answers3