有没有办法检查两个集合是否包含相同的元素，独立于顺序？ [英] Is there a way to check if two Collections contain the same elements, independent of order?

问题描述

我一直在寻找一个操作像 Arrays.equals（a1，a2），但忽略元素顺序的方法。我无法在Google Collections中找到它（类似 Iterables.elementsEqual（），但它确实是排序）和JUnit（ assertEquals（）显然只是调用集合上的 equals（），这取决于集合实现，这不是我想要的）
这将是最好的，如果这样的方法将 Iterable s，但我也很好，简单地采取集合 s
这样的方法当然会考虑集合中的任何重复元素（因此不能简单地测试 containsAll（））。 p>

注意，我不是在问如何实现这样的事情，我只是想知道是否有任何标准的集合库有它。

解决方案

Apache commons-collections有 CollectionUtils＃isEqualCollection ：

< blockquote>

如果给定的Collections包含完全相同的基数的相同元素，则返回true。

也就是说，如果e的基数

我认为，这是完全相同的你以后的。

I've been looking for a method that operates like Arrays.equals(a1, a2), but ignoring the element order. I haven't been able to find it in either Google Collections (something like Iterables.elementsEqual(), but that does account for ordering) and JUnit (assertEquals() obviously just calls equals() on the Collection, which depends on the Collection implementation, and that's not what I want) It would be best if such a method would take Iterables, but I'm also fine with simply taking Collections Such a method would of course take into account any duplicate elements in the collection (so it can't simply test for containsAll()).

Note that I'm not asking how to implement such a thing, I was just wondering if any of the standard Collections libraries have it.

解决方案

Apache commons-collections has CollectionUtils#isEqualCollection:

Returns true if the given Collections contain exactly the same elements with exactly the same cardinality.

That is, if the cardinality of e in a is equal to the cardinality of e in b, for each element e in a or b.

Which is, I think, exactly what you're after.

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