Working with lists in Python

Question

I have two non-unique lists of values, such as

["a", "b", "a", "c"]

and

 ["a", "b", "b", "f"]

I want to find which elements of the second list do not appear in the first.

I could code this by hand, but would prefer to use built-in functions. I can't figure out how because I keep bumping into the hashable / unhashable barrier.

If your barrier is that your values are unhashable, then tell us what your values are so we can help you work around it. — machine yearning, May 22 '16 at 21:00
I understand now. I thought the error message referred to the collective objects, not their members. Thanks. — Bernard Finucane, May 23 '16 at 19:06

Alex Hall · Answer 1 · 2016-05-22T21:09:37.060

In this case it would be

set(second_list) - set(first_list)

because strings are hashable and you can put them in sets. If you have non-hashable values then tell us what they are and we can probably come up with a workaround. For example:

You can convert a list to a tuple using tuple(the_list), can back using list(the_tuple).
You can convert a set to a tuple using tuple(the_set), can back using set(the_tuple).
You can convert a nested list to a nested tuple using tuple(map(tuple, the_list)) and back using list(map(list, the_tuple)).
You can convert a dictionary to a nested tuple using tuple(the_dict.items()) and back using dict(the_tuple).

Tried this and it worked. – Bernard Finucane May 23 '16 at 19:06 — Bernard Finucane, May 23 '16 at 19:06

score 1 · Answer 2 · answered May 22 '16 at 22:44

1

[x for x in ListB if x not in ListB]

I think this is a pretty fast way you can use.

answered May 22 '16 at 22:44

Rouzbeh

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Working with lists in Python

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