Java 8 流。除其他元素外的所有元素
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/35512258/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Java 8 stream. all elements EXCEPT other elements
提问by Beta033
I'm interested in identifying an approach that returns a list of elements excluding the elements in another list.
我有兴趣确定一种返回元素列表的方法，该列表不包括另一个列表中的元素。
for example
例如
List<Integer> multiplesOfThree = ... // 3,6,9,12 etc
List<Integer> evens = ... // 2,4,6,8 etc
List<Integer> others = multiplesOfThree.except(evens) // should return a list of elements that are not in the other list
how do you do this? i found an approach that's a bit clunky and difficult to read....
你怎么做到这一点？我发现了一种有点笨拙且难以阅读的方法......
multiplesOfThree.stream()
.filter(intval -> evens.stream().noneMatch(even -> even.intValue() == intval.intValue()))
回答by rgettman
You can use Stream's filtermethod, passing a Predicatethat ensures that the element doesn't exist in evens.
您可以使用Stream的filter方法，传递Predicate确保元素不存在于evens.
List<Integer> others = multiplesOfThree.stream()
.filter(i -> !evens.contains(i))
.collect(Collectors.toList());
But assuming you have a mutable List(e.g. ArrayList), you don't even need streams, just Collections's removeAllmethod.
但是假设你有一个可变的List（例如ArrayList），你甚至不需要流，只需要Collections's removeAllmethod。
multiplesOfThree.removeAll(evens);
回答by fabian
You could use
你可以用
multipleOfThree.stream()
.filter(((Predicate<Integer>) evens::contains).negate())
or more efficient for big evenlists
或者对于大even列表更有效
HashSet<Integer> evenSet = new HashSet<>(even);
multipleOfThree.stream()
.filter(((Predicate<Integer>) evenSet::contains).negate())
回答by Todd
There are a few solutions.
有几种解决方案。
First, without using streams, you can just create a new list and remove all elements from another collection from it...
首先，不使用流，您只需创建一个新列表并从中删除另一个集合中的所有元素...
final List<Integer> multiplesOfThree = Arrays.asList(3,6,9,12);
final List<Integer> evens = Arrays.asList(2,4,6,8,10,12);
final List<Integer> others1 = new ArrayList<>(multiplesOfThree);
others1.removeAll(evens);
Another solution would be to pass the stream through a filter():
另一种解决方案是通过过滤器（）传递流：
final List<Integer> others2 = multiplesOfThree
.stream()
.filter(x -> !evens.contains(x))
.collect(Collectors.toList());
(You may want to consider making evensa Setin this case).
（在这种情况下，您可能需要考虑制作evens一个Set）。
And finally, you could modify the logic above to represent the "evens" as a function rather than a collection of all even numbers. This is essentially the same as above, but you don't have to have a second collection.
最后，您可以修改上面的逻辑，将“偶数”表示为一个函数，而不是所有偶数的集合。这与上面的基本相同，但您不必有第二个集合。
final List<Integer> others3 = multiplesOfThree
.stream()
.filter(x -> x % 2 != 0)
.collect(Collectors.toList());
