To get elements which are in temp1 but not in temp2 :

In [5]: list(set(temp1) – set(temp2))

Out[5]: [‘Four’, ‘Three’]

Beware that it is asymmetric :

In [5]: set([1, 2]) – set([2, 3])

Out[5]: set([1])

where you might expect/want it to equal set([1, 3]). If you do want set([1, 3]) as your answer, you can use set([1, 2]).symmetric_difference(set([2, 3])).

The existing solutions all offer either one or the other of:

Faster than O(n*m) performance.

Preserve order of input list.

But so far no solution has both. If you want both, try this:

s = set(temp2)

temp3 = [x for x in temp1 if x not in s]

Performance test

import timeit

init = ‘temp1 = list(range(100)); temp2 = [i * 2 for i in range(50)]’

print timeit.timeit(‘list(set(temp1) – set(temp2))’, init, number = 100000)

print timeit.timeit(‘s = set(temp2);[x for x in temp1 if x not in s]’, init, number = 100000)

print timeit.timeit(‘[item for item in temp1 if item not in temp2]’, init, number = 100000)

Results:

4.34620224079 # ars’ answer

4.2770634955 # This answer

30.7715615392 # matt b’s answer

The method I presented as well as preserving order is also (slightly) faster than the set subtraction because it doesn’t require construction of an unnecessary set. The performance difference would be more noticable if the first list is considerably longer than the second and if hashing is expensive. Here’s a second test demonstrating this:

init = ”’

temp1 = [str(i) for i in range(100000)]

temp2 = [str(i * 2) for i in range(50)]

”’

Results:

11.3836875916 # ars’ answer

3.63890368748 # this answer (3 times faster!)

37.7445402279 # matt b’s answer