Subtracting two lists in Python

Question

In Python, How can one subtract two non-unique, unordered lists? Say we have a = [0,1,2,1,0] and b = [0, 1, 1] I\'d like to do something like

Answer 1

I attempted to find a more elegant solution, but the best I could do was basically the same thing that Dyno Fu said:

from copy import copy

def subtract_lists(a, b):
    """
    >>> a = [0, 1, 2, 1, 0]
    >>> b = [0, 1, 1]
    >>> subtract_lists(a, b)
    [2, 0]

    >>> import random
    >>> size = 10000
    >>> a = [random.randrange(100) for _ in range(size)]
    >>> b = [random.randrange(100) for _ in range(size)]
    >>> c = subtract_lists(a, b)
    >>> assert all((x in a) for x in c)
    """
    a = copy(a)
    for x in b:
        if x in a:
            a.remove(x)
    return a

Answer 2

I know "for" is not what you want, but it's simple and clear:

for x in b:
  a.remove(x)

Or if members of b might not be in a then use:

for x in b:
  if x in a:
    a.remove(x)

Answer 3

to use list comprehension:

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

would do the trick. It would change b in the process though. But I agree with jkp and Dyno Fu that using a for loop would be better.

Perhaps someone can create a better example that uses list comprehension but still is KISS?

Answer 4

You can use the map construct to do this. It looks quite ok, but beware that the map line itself will return a list of Nones.

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

map(lambda x:a.remove(x), b)
a

Answer 5

Here's a relatively long but efficient and readable solution. It's O(n).

def list_diff(list1, list2):
    counts = {}
    for x in list1:
        try:
            counts[x] += 1
        except:
            counts[x] = 1
    for x in list2:
        try:
            counts[x] -= 1
            if counts[x] < 0:
                raise ValueError('All elements of list2 not in list2')
        except:
            raise ValueError('All elements of list2 not in list1') 
    result = []
    for k, v in counts.iteritems():
        result += v*[k] 
    return result

a = [0, 1, 1, 2, 0]
b = [0, 1, 1]
%timeit list_diff(a, b)
%timeit list_diff(1000*a, 1000*b)
%timeit list_diff(1000000*a, 1000000*b)
100000 loops, best of 3: 4.8 µs per loop
1000 loops, best of 3: 1.18 ms per loop
1 loops, best of 3: 1.21 s per loop

Answer 6

c = [i for i in b if i not in a]