获得两组之间的差异

So if I have two sets:

所以，如果我有两套：

Set<int> test1 = new HashSet<Integer>();
test1.add(1);
test1.add(2);
test1.add(3);

Set<int> test2 = new HashSet<Integer>();
test2.add(1);
test2.add(2);
test2.add(3);
test2.add(4);
test2.add(5);

Is there a way to compare them and only have a set of 4 and 5 returned?

有没有办法比较它们，只有一组4和5返回？

4 个解决方案

#1

125

Try this

尝试这个

test2.removeAll(test1);

Set#removeAll

集＃的removeAll

Removes from this set all of its elements that are contained in the specified collection (optional operation). If the specified collection is also a set, this operation effectively modifies this set so that its value is the asymmetric set difference of the two sets.

从此集合中删除指定集合中包含的所有元素（可选操作）。如果指定的集合也是一个集合，则此操作会有效地修改此集合，以使其值为两个集合的不对称集合差异。

#2

If you use Guava (former Google Collections) library there is a solution:

如果您使用Guava（以前的Google Collections）库，则有一个解决方案：

SetView<Number> difference = com.google.common.collect.Sets.difference(test2, test1);

The returned SetView is a Set, it is a live representation you can either make immutable or copy to another set. test1 and test2 are left intact.

返回的SetView是一个Set，它是一个实时表示，您可以使其不可变或复制到另一个集合。 test1和test2保持不变。

#3

Yes:

是：

test2.removeAll(test1)

Although this will mutate test2, so create a copy if you need to preserve it.

虽然这会改变test2，但如果需要保留它，请创建一个副本。

Also, you probably meant <Integer> instead of <int>.

此外，您可能需要而不是。

#4

If you are using Java 8, you could try something like this:

如果您使用的是Java 8，可以尝试这样的方法：

public Set<Number> difference(final Set<Number> set1, final Set<Number> set2){
    final Set<Number> larger = set1.size() > set2.size() ? set1 : set2;
    final Set<Number> smaller = larger.equals(set1) ? set2 : set1;
    return larger.stream().filter(n -> !smaller.contains(n)).collect(Collectors.toSet());
}

#1

125