Python list subtraction operation

if duplicate and ordering items are problem :

[i for i in a if not i in b or b.remove(i)]

a = [1,2,3,3,3,3,4]
b = [1,3]
result: [2, 3, 3, 3, 4]

If the lists allow duplicate elements, you can use Counter from collections:

from collections import Counter
result = list((Counter(x)-Counter(y)).elements())

If you need to preserve the order of elements from x:

result = [ v for c in [Counter(y)] for v in x if not c[v] or c.subtract([v]) ]

I think the easiest way to achieve this is by using set().

>>> x = [1,2,3,4,5,6,7,8,9,0]  
>>> y = [1,3,5,7,9]  
>>> list(set(x)- set(y))
[0, 2, 4, 6, 8]

The other solutions have one of a few problems:

1. They don't preserve order, or
2. They don't remove a precise count of elements, e.g. for x = [1, 2, 2, 2] and y = [2, 2] they convert y to a set, and either remove all matching elements (leaving [1] only) or remove one of each unique element (leaving [1, 2, 2]), when the proper behavior would be to remove 2 twice, leaving [1, 2], or
3. They do O(m * n) work, where an optimal solution can do O(m + n) work

Alain was on the right track with Counter to solve #2 and #3, but that solution will lose ordering. The solution that preserves order (removing the first n copies of each value for n repetitions in the list of values to remove) is:

from collections import Counter

x = [1,2,3,4,3,2,1]  
y = [1,2,2]  
remaining = Counter(y)

out = []
for val in x:
    if remaining[val]:
        remaining[val] -= 1
    else:
        out.append(val)
# out is now [3, 4, 3, 1], having removed the first 1 and both 2s.

Try it online!

To make it remove the last copies of each element, just change the for loop to for val in reversed(x): and add out.reverse() immediately after exiting the for loop.

Constructing the Counter is O(n) in terms of y's length, iterating x is O(n) in terms of x's length, and Counter membership testing and mutation are O(1), while list.append is amortized O(1) (a given append can be O(n), but for many appends, the overall big-O averages O(1) since fewer and fewer of them require a reallocation), so the overall work done is O(m + n).

You can also test for to determine if there were any elements in y that were not removed from x by testing:

remaining = +remaining  # Removes all keys with zero counts from Counter
if remaining:
    # remaining contained elements with non-zero counts

You could also try this if the values are unique anyway:

list(set(x) - set(y))

We can use set methods as well to find the difference between two list

x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]
y = [1, 3, 5, 7, 9]
list(set(x).difference(y))
[0, 2, 4, 6, 8]

The answer provided by @aaronasterling looks good, however, it is not compatible with the default interface of list: x = MyList(1, 2, 3, 4) vs x = MyList([1, 2, 3, 4]). Thus, the below code can be used as a more python-list friendly:

class MyList(list):
    def __init__(self, *args):
        super(MyList, self).__init__(*args)

    def __sub__(self, other):
        return self.__class__([item for item in self if item not in other])

Example:

x = MyList([1, 2, 3, 4])
y = MyList([2, 5, 2])
z = x - y

from collections import Counter

y = Counter(y)
x = Counter(x)

print(list(x-y))

list1 = ['a', 'c', 'a', 'b', 'k'] 
list2 = ['a', 'a', 'a', 'a', 'b', 'c', 'c', 'd', 'e', 'f'] 
for e in list1: 
    try: 
        list2.remove(e) 
    except ValueError: 
        print(f'{e} not in list') 
list2 
# ['a', 'a', 'c', 'd', 'e', 'f']

This will change list2. if you want to protect list2 just copy it and use the copy of list2 in this code.