Get difference between two lists

Question

I have two lists in Python, like these:

temp1 = ['One', 'Two', 'Three', 'Four'] temp2 = ['One', 'Two'] 

I need to create a third list with items from the first list which aren't present in the second one. From the example I have to get

temp3 = ['Three', 'Four'] 

Are there any fast ways without cycles and checking?

Answer 1

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 

Answer 2

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])).