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Home/ Questions/Q 569669
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T13:19:12+00:00 2026-05-13T13:19:12+00:00

So heres my code: item = [0,1,2,3,4,5,6,7,8,9] z = [] # list of integers

  • 0

So heres my code:

item = [0,1,2,3,4,5,6,7,8,9]
z = []  # list of integers

for item in z:
    if item not in z:
        print item

z contains a list of integers. I want to compare item to z and print out the numbers that are not in z when compared to item.

I can print the elements that are in z when compared not item, but when I try and do the opposite using the code above nothing prints.

Any help?

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  1. Editorial Team
    Editorial Team
    2026-05-13T13:19:12+00:00Added an answer on May 13, 2026 at 1:19 pm

    Your code is not doing what I think you think it is doing. The line for item in z: will iterate through z, each time making item equal to one single element of z. The original item list is therefore overwritten before you’ve done anything with it.

    I think you want something like this:

    item = [0,1,2,3,4,5,6,7,8,9]
    
    for element in item:
        if element not in z:
            print(element)
    

    But you could easily do this like:

    [x for x in item if x not in z]
    

    or (if you don’t mind losing duplicates of non-unique elements):

    set(item) - set(z)
    
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