Finding elements not in a list

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?

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)

>> items = [1,2,3,4]
>> Z = [3,4,5,6]

>> print list(set(items)-set(Z))
[1, 2]

Using list comprehension:

print [x for x in item if x not in Z]

or using filter function :

filter(lambda x: x not in Z, item)

Using set in any form may create a bug if the list being checked contains non-unique elements, e.g.:

print item

Out[39]: [0, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9]

print Z

Out[40]: [3, 4, 5, 6]

set(item) - set(Z)

Out[41]: {0, 1, 2, 7, 8, 9}

vs list comprehension as above

print [x for x in item if x not in Z]

Out[38]: [0, 1, 1, 2, 7, 8, 9]

or filter function:

filter(lambda x: x not in Z, item)

Out[38]: [0, 1, 1, 2, 7, 8, 9]

list1 = [1,2,3,4]; list2 = [0,3,3,6]

print set(list2) - set(list1)