Remove all the elements that occur in one list from another

Solution 1:

Python has a language feature called List Comprehensions that is perfectly suited to making this sort of thing extremely easy. The following statement does exactly what you want and stores the result in l3:

l3 = [x for x in l1 if x not in l2]

l3 will contain [1, 6].

Solution 2:

One way is to use sets:

>>> set([1,2,6,8]) - set([2,3,5,8])
set([1, 6])

Note, however, that sets do not preserve the order of elements, and cause any duplicated elements to be removed. The elements also need to be hashable. If these restrictions are tolerable, this may often be the simplest and highest performance option.

Solution 3:

Performance Comparisons

Comparing the performance of all the answers mentioned here on Python 3.9.1 and Python 2.7.16.

Python 3.9.1

Answers are mentioned in order of performance:

Arkku's set difference using subtraction "-" operation - (91.3 nsec per loop)

mquadri$ python3 -m timeit -s "l1 = set([1,2,6,8]); l2 = set([2,3,5,8]);" "l1 - l2"
5000000 loops, best of 5: 91.3 nsec per loop

Moinuddin Quadri's using set().difference()- (133 nsec per loop)

mquadri$ python3 -m timeit -s "l1 = set([1,2,6,8]); l2 = set([2,3,5,8]);" "l1.difference(l2)"
2000000 loops, best of 5: 133 nsec per loop

Moinuddin Quadri's list comprehension with set based lookup- (366 nsec per loop)

 mquadri$ python3 -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "[x for x in l1 if x not in l2]"
 1000000 loops, best of 5: 366 nsec per loop

Donut's list comprehension on plain list - (489 nsec per loop)

 mquadri$ python3 -m timeit -s "l1 = [1,2,6,8]; l2 = [2,3,5,8];" "[x for x in l1 if x not in l2]"
 500000 loops, best of 5: 489 nsec per loop

Daniel Pryden's generator expression with set based lookup and type-casting to list - (583 nsec per loop) : Explicitly type-casting to list to get the final object as list, as requested by OP. If generator expression is replaced with list comprehension, it'll become same as Moinuddin Quadri's list comprehension with set based lookup.
```
 mquadri$ mquadri$ python3 -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "list(x for x in l1 if x not in l2)"
 500000 loops, best of 5: 583 nsec per loop
```

Moinuddin Quadri's using filter() and explicitly type-casting to list (need to explicitly type-cast as in Python 3.x, it returns iterator) - (681 nsec per loop)

 mquadri$ python3 -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "list(filter(lambda x: x not in l2, l1))"
 500000 loops, best of 5: 681 nsec per loop

Akshay Hazari's using combination of functools.reduce + filter -(3.36 usec per loop) : Explicitly type-casting to list as from Python 3.x it started returned returning iterator. Also we need to import functools to use reduce in Python 3.x
```
 mquadri$ python3 -m timeit "from functools import reduce; l1 = [1,2,6,8]; l2 = [2,3,5,8];" "list(reduce(lambda x,y : filter(lambda z: z!=y,x) ,l1,l2))"
 100000 loops, best of 5: 3.36 usec per loop
```

Python 2.7.16

Answers are mentioned in order of performance:

Arkku's set difference using subtraction "-" operation - (0.0783 usec per loop)

mquadri$ python -m timeit -s "l1 = set([1,2,6,8]); l2 = set([2,3,5,8]);" "l1 - l2"
10000000 loops, best of 3: 0.0783 usec per loop

Moinuddin Quadri's using set().difference()- (0.117 usec per loop)

mquadri$ mquadri$ python -m timeit -s "l1 = set([1,2,6,8]); l2 = set([2,3,5,8]);" "l1.difference(l2)"
10000000 loops, best of 3: 0.117 usec per loop

Moinuddin Quadri's list comprehension with set based lookup- (0.246 usec per loop)

 mquadri$ python -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "[x for x in l1 if x not in l2]"
 1000000 loops, best of 3: 0.246 usec per loop

Donut's list comprehension on plain list - (0.372 usec per loop)

 mquadri$ python -m timeit -s "l1 = [1,2,6,8]; l2 = [2,3,5,8];" "[x for x in l1 if x not in l2]"
 1000000 loops, best of 3: 0.372 usec per loop

Moinuddin Quadri's using filter() - (0.593 usec per loop)

 mquadri$ python -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "filter(lambda x: x not in l2, l1)"
 1000000 loops, best of 3: 0.593 usec per loop

Daniel Pryden's generator expression with set based lookup and type-casting to list - (0.964 per loop) : Explicitly type-casting to list to get the final object as list, as requested by OP. If generator expression is replaced with list comprehension, it'll become same as Moinuddin Quadri's list comprehension with set based lookup.
```
 mquadri$ python -m timeit -s "l1 = [1,2,6,8]; l2 = set([2,3,5,8]);" "list(x for x in l1 if x not in l2)"
 1000000 loops, best of 3: 0.964 usec per loop
```

Akshay Hazari's using combination of functools.reduce + filter -(2.78 usec per loop)

 mquadri$ python -m timeit "l1 = [1,2,6,8]; l2 = [2,3,5,8];" "reduce(lambda x,y : filter(lambda z: z!=y,x) ,l1,l2)"
 100000 loops, best of 3: 2.78 usec per loop

Solution 4:

Expanding on Donut's answer and the other answers here, you can get even better results by using a generator comprehension instead of a list comprehension, and by using a set data structure (since the in operator is O(n) on a list but O(1) on a set).

So here's a function that would work for you:

def filter_list(full_list, excludes):
    s = set(excludes)
    return (x for x in full_list if x not in s)

The result will be an iterable that will lazily fetch the filtered list. If you need a real list object (e.g. if you need to do a len() on the result), then you can easily build a list like so:

filtered_list = list(filter_list(full_list, excludes))

Solution 5:

Use the Python set type. That would be the most Pythonic. :)

Also, since it's native, it should be the most optimized method too.

See:

http://docs.python.org/library/stdtypes.html#set

http://docs.python.org/library/sets.htm (for older python)

# Using Python 2.7 set literal format.
# Otherwise, use: l1 = set([1,2,6,8])
#
l1 = {1,2,6,8}
l2 = {2,3,5,8}
l3 = l1 - l2