Removing lists within a nested list based on condition

Question

I am have a nested list with the following:

[['X', 0],
['X', 0],
['X', ''],
['X', '']]

for i in myList:
    if i[1] == 0:
         myList.remove(i)

This will only remove the first list. Output would be:

[['X', 0],
['X', ''],
['X', '']]

How can I get it for both?

when you `remove(i)`, you change the list you're iterating over. So when you remove element 0, your loop will continue on with element 1, but what previously *was* element 1 is now element 0, because the previous element 0 was removed, so you skip a value. — Zinki, Apr 05 '18 at 12:16
As above but if you want to use the nethod you are using you will need to iterate over the list in reverse: `for i in reversed(range(len(myList))):` — AndrewK, Apr 05 '18 at 12:22
Or make a copy of the original list by using `for i in myList[:]` instead. — AndrewK, Apr 05 '18 at 12:27

score 3 · Answer 1 · answered Apr 05 '18 at 12:15

3

Using a list comprehension . Do not delete an element from a list while iterating over it.

Ex:

s = [['X', 0],
['X', 0],
['X', ''],
['X', '']]

print([i for i in s if i[1] != 0])

answered Apr 05 '18 at 12:15

Rakesh

1

I was writing the same answer, but you were faster. This is a better approach then iterating and changing the original list. – cezar Apr 05 '18 at 12:16
'Do not delete an element from a list while iterating over it.' Yea I knew it was bad practise and the source of why it was not working. Thanks for your input and help :) – arsenal88 Apr 05 '18 at 12:20

1 Answers1