我有一个列表users_with_invites_ids_list
,由循环形成,我将值附加到列表中,在python中看起来像这样:
...[ObjectId('55119e14bf2e4e010d8b48f2')], [ObjectId('54624128bf2e4e5e558b5a52')], [ObjectId('53a6e7bc763f4aa0308b4569')], [ObjectId('55241823bf2e4e59508b494c')]]
当我尝试:
users_with_invites_ids_set = set(users_with_invites_ids_list)
我明白了:
TypeError: unhashable type: 'list'
如何将此列表列表转换为一组?
修改
基于回答我做了以下事情:
#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)
产生以下结果:
(ObjectId('542ac5a6763f4a82188b4a51'),), (ObjectId('54496fe6bf2e4efe348bd344'),), (ObjectId('54c96339bf2e4ee62c8b48e0'),)])
如何在每个ObjectId
周围找不到()
。它阻止我将这个集合与其他集合的ID进行比较。
答案 0 :(得分:5)
您需要将内部列表转换为元组,假设每个ObjectId('55119e14bf2e4e010d8b48f2')
都是可清除的:
users_with_invites_ids_set = set(tuple(x) for x in users_with_invites_ids_list)
工作示例:
>>> 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>,)])
如果你想单独创建ObjectId
的集合,你可以这样做:
>>> 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>])
答案 1 :(得分:1)
虽然可接受的答案是好的,但是如果您希望进行更简单的比较并且不想处理元组的不变性,则也可以使用老式的字符串比较来尝试它:
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))
答案 2 :(得分:0)
如果marker1.getGeometry()
的{{1}}中有list of lists
,则可以使用此函数将其转换为一个lists
中的integer
:
set
示例:
list
输出:
outer_list = []
def lists_to_list(nested_lists):
for el in nested_lists:
if type(el) == list:
lists_to_list(el)
else:
outer_list.append(el)
return set(outer_list)