Problem with how pop() and enumerate() interact

Question

When I use List.pop inside a for loop that uses enumerate, the process terminates before getting to the end of the list. How do I get around having pop cut short the process?

I've checked that if I write the loop using pop but without enumerate, then it works as expected. Likewise, if I remove pop and use enumerate to do something else, it also works as expected.

Here is the code:

x=[0,2,4,6,8,10] 
for i in enumerate(x):
x.pop(0)
print(x)

I expect to have the following printed:

[2,4,6,8,10]
[4,6,8,10]
[6,8,10]
[8,10]
[10]
[]

Instead, I receive

[2,4,6,8,10]
[4,6,8,10]
[6,8,10]

If I run it again then I receive

[8,10]
[10]

and if I run it again I receive

[]

Rakesh · Accepted Answer · 2019-07-19 17:29:42Z

2

Use range

Ex:

x=[0,2,4,6,8,10] 
for i in range(len(x)):
    x.pop(0)
    print(x)

Output:

[2, 4, 6, 8, 10]
[4, 6, 8, 10]
[6, 8, 10]
[8, 10]
[10]
[]

Note: Modifying a list while iterating it is not good practice.

edited Jul 19, 2019 at 17:29

answered Jul 19, 2019 at 17:24

Rakesh

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1 Comment

fox-daniel Over a year ago

Thanks for answering. Can you explain why len(x) isn't also problematic sense x is getting modified? I'm Ok with using the version you suggest, but I'd like to understand why the enumerate version doesn't work.

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