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

Answer 1

If you run into TypeError: unhashable type: 'list' you need to turn lists or sets into tuples, e.g.

set(map(tuple, list_of_lists1)).symmetric_difference(set(map(tuple, list_of_lists2)))

See also How to compare a list of lists/sets in python?

Answer 2

In [5]: list(set(temp1) - set(temp2))
Out[5]: ['Four', 'Three']

Beware that

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'll need to use set([1, 2]).symmetric_difference(set([2, 3])).

Answer 3

i'll toss in since none of the present solutions yield a tuple:

temp3 = tuple(set(temp1) - set(temp2))

alternatively:

#edited using @Mark Byers idea. If you accept this one as answer, just accept his instead.
temp3 = tuple(x for x in temp1 if x not in set(temp2))

Like the other non-tuple yielding answers in this direction, it preserves order

Answer 4

Here are a few simple, order-preserving ways of diffing two lists of strings.

Code

An unusual approach using pathlib:

import pathlib


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

p = pathlib.Path(*temp1)
r = p.relative_to(*temp2)
list(r.parts)
# ['Three', 'Four']

This assumes both lists contain strings with equivalent beginnings. See the docs for more details. Note, it is not particularly fast compared to set operations.

---

A straight-forward implementation using itertools.zip_longest:

import itertools as it


[x for x, y in it.zip_longest(temp1, temp2) if x != y]
# ['Three', 'Four']

Answer 5

Try this:

temp3 = set(temp1) - set(temp2)

Answer 6

(list(set(a)-set(b))+list(set(b)-set(a)))