how to convert list of lists to a set in python so

2019-05-06 16:03发布

问题:

I have a list users_with_invites_ids_list, formed by loop where I append values to the list, in python that looks like this:

...[ObjectId('55119e14bf2e4e010d8b48f2')], [ObjectId('54624128bf2e4e5e558b5a52')], [ObjectId('53a6e7bc763f4aa0308b4569')], [ObjectId('55241823bf2e4e59508b494c')]]

when I try:

    users_with_invites_ids_set = set(users_with_invites_ids_list)

I get:

TypeError: unhashable type: 'list'

How do I convert this list of lists to a set?

EDIT

based on answer I've done the following:

#convert list to set
first_tuple_list = [tuple(lst) for lst in users_with_invites_ids_list]
users_with_invites_ids_set = set(first_tuple_list)

Which yields the following:

 (ObjectId('542ac5a6763f4a82188b4a51'),), (ObjectId('54496fe6bf2e4efe348bd344'),), (ObjectId('54c96339bf2e4ee62c8b48e0'),)])

How do I get each ObjectId without the () around each one. It's keeping me from comparing this set to other set's of ids.

回答1:

You would need to convert the inner lists to tuples, assuming each of those ObjectId('55119e14bf2e4e010d8b48f2') is hashable:

users_with_invites_ids_set = set(tuple(x) for x in users_with_invites_ids_list)

Working example:

>>> class ObjectId(object):
...   def __init__(self, v):
...     self.v = v
... 
>>> list_of_lists = [[ObjectId('55119e14bf2e4e010d8b48f2')], [ObjectId('54624128bf2e4e5e558b5a52')], [ObjectId('53a6e7bc763f4aa0308b4569')], [ObjectId('55241823bf2e4e59508b494c')]]
>>> set(tuple(x) for x in list_of_lists)
set([(<__main__.ObjectId object at 0x7f71483cfc50>,), (<__main__.ObjectId object at 0x7f71483cfd10>,), (<__main__.ObjectId object at 0x7f71483cfcd0>,), (<__main__.ObjectId object at 0x7f71483cfc90>,)])

In case if you are looking to create the set of the ObjectId's alone, you could do:

>>> set(x for lst in list_of_lists for x in lst)
set([<__main__.ObjectId object at 0x7f71483cfb10>, <__main__.ObjectId object at 0x7f71483db050>, <__main__.ObjectId object at 0x7f71483cfad0>, <__main__.ObjectId object at 0x7f71483cfd50>])


回答2:

While the accepted answer is good, if you'd like something simpler for comparison and don't want to deal with the immutability of tuples, you could also try it using good old-fashioned string comparison:

list1 = [[], [60], [95], [60, 95]]
list2 = [[], [95], [60], [60, 95]]
print(set(str(x) for x in list1) == set(str(x) for x in list2))


标签: python list set