Difference Between Two Lists with Duplicates in Python

You didn't specify if the order matters. If it does not, you can do this in >= Python 2.7:

l1 = ['a', 'b', 'c', 'b', 'c']
l2 = ['a', 'b', 'c', 'b']

from collections import Counter

c1 = Counter(l1)
c2 = Counter(l2)

diff = c1-c2
print list(diff.elements())

Create Counters for both lists, then subtract one from the other.

from collections import Counter

a = [1,2,3,1,2]
b = [1,2,3,1]

c = Counter(a)
c.subtract(Counter(b))

To take into account both duplicates and the order of elements:

from collections import Counter

def list_difference(a, b):
    count = Counter(a) # count items in a
    count.subtract(b)  # subtract items that are in b
    diff = []
    for x in a:
        if count[x] > 0:
           count[x] -= 1
           diff.append(x)
    return diff

Example

print(list_difference("z y z x v x y x u".split(), "x y z w z".split()))
# -> ['y', 'x', 'v', 'x', 'u']

Python 2.5 version:

from collections import defaultdict 

def list_difference25(a, b):
    # count items in a
    count = defaultdict(int) # item -> number of occurrences
    for x in a:
        count[x] += 1

    # subtract items that are in b
    for x in b: 
        count[x] -= 1

    diff = []
    for x in a:
        if count[x] > 0:
           count[x] -= 1
           diff.append(x)
    return diff