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- Check if two unordered lists are equal [duplicate] 8 answers
- 检查两个无序列表是否相等[重复] 8个答案
I'm trying to write tests for my Django app and I need to check many times if 2 lists have the same objects (i.e. that every object in A is also in B and vice versa).
我正在尝试为我的Django app编写测试，我需要多次检查2个列表是否具有相同的对象（即A中的每个对象也在B中，反之亦然）。
I read on the assertLists/Sequence/Equal etc but for what I saw if the lists have the same objects but in a different order (A = [a,b,c], B = [b,c,a]) then it returns an error, which I don't want it to be an error because they both have the same objects.
我读了assertLists / Sequence / Equal等但是对于我所看到的，如果列表具有相同的对象但是顺序不同（A = [a，b，c]，B = [b，c，a]）那么它返回一个错误，我不希望它是一个错误，因为它们都有相同的对象。
Is there a way to check this without looping the lists?
有没有办法在不循环列表的情况下检查这个？
2 个解决方案
#1
9
You can use assertCountEqual in Python 3, or assertItemsEqual in Python 2.
您可以在Python 3中使用assertCountEqual，或在Python 2中使用assertItemsEqual。
From the Python 3 docs for assertCountEqual:
来自assertCountEqual的Python 3文档：
Test that sequence first contains the same elements as second, regardless of their order. When they don’t, an error message listing the differences between the sequences will be generated.
测试该序列首先包含与第二个相同的元素，而不管它们的顺序如何。如果没有，将生成列出序列之间差异的错误消息。
Duplicate elements are not ignored when comparing first and second. It verifies whether each element has the same count in both sequences. Equivalent to:
assertEqual(Counter(list(first)), Counter(list(second)))but works with sequences of unhashable objects as well.比较第一个和第二个时，不会忽略重复的元素。它验证每个元素在两个序列中是否具有相同的计数。相当于：assertEqual（Counter（list（first）），Counter（list（second））），但也适用于不可用对象的序列。
#2
2
If you are staying with list() datatype then the cleanest way if your lists are not too large would be :
如果您使用list（）数据类型，那么如果列表不是太大，最简洁的方法是：
sorted(list_1) == sorted(list_2)
Have a look at this Question (which is the same): Check if two unordered lists are equal
看看这个问题（相同）：检查两个无序列表是否相等
#1
9
You can use assertCountEqual in Python 3, or assertItemsEqual in Python 2.
您可以在Python 3中使用assertCountEqual，或在Python 2中使用assertItemsEqual。
From the Python 3 docs for assertCountEqual:
来自assertCountEqual的Python 3文档：
Test that sequence first contains the same elements as second, regardless of their order. When they don’t, an error message listing the differences between the sequences will be generated.
测试该序列首先包含与第二个相同的元素，而不管它们的顺序如何。如果没有，将生成列出序列之间差异的错误消息。
Duplicate elements are not ignored when comparing first and second. It verifies whether each element has the same count in both sequences. Equivalent to:
assertEqual(Counter(list(first)), Counter(list(second)))but works with sequences of unhashable objects as well.比较第一个和第二个时，不会忽略重复的元素。它验证每个元素在两个序列中是否具有相同的计数。相当于：assertEqual（Counter（list（first）），Counter（list（second））），但也适用于不可用对象的序列。
#2
2
If you are staying with list() datatype then the cleanest way if your lists are not too large would be :
如果您使用list（）数据类型，那么如果列表不是太大，最简洁的方法是：
sorted(list_1) == sorted(list_2)
Have a look at this Question (which is the same): Check if two unordered lists are equal
看看这个问题（相同）：检查两个无序列表是否相等