Finding elements not in a list

Question

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<

Answer 1

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)

Answer 2

No, z is undefined. item contains a list of integers.

I think what you're trying to do is this:

#z defined elsewhere
item = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

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

As has been stated in other answers, you may want to try using sets.

Answer 3

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]

Answer 4

If you run a loop taking items from z, how do you expect them not to be in z? IMHO it would make more sense comparing items from a different list to z.

Answer 5

In the case where item and z are sorted iterators, we can reduce the complexity from O(n^2) to O(n+m) by doing this

def iexclude(sorted_iterator, exclude_sorted_iterator):
    next_val = next(exclude_sorted_iterator)
    for item in sorted_iterator:
        try:
            while next_val < item:
                next_val = next(exclude_sorted_iterator)
                continue
            if item == next_val:
                continue
        except StopIteration:
            pass
        yield item

If the two are iterators, we also have the opportunity to reduce the memory footprint not storing z (exclude_sorted_iterator) as a list.

Answer 6

>>> item = set([0,1,2,3,4,5,6,7,8,9])
>>> z = set([2,3,4])
>>> print item - z
set([0, 1, 5, 6, 7, 8, 9])