# Get difference between two lists

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 timeitinit = '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' answer4.2770634955 # This answer30.7715615392 # 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' answer3.63890368748 # this answer (3 times faster!)37.7445402279 # matt b's answer`

Can be done using python XOR operator.

- This will remove the duplicates in each list
- This will show difference of temp1 from temp2 and temp2 from temp1.

`set(temp1) ^ set(temp2)`