合并两个流以创建一个新的[重复]

Question

我想同时执行两个流，并生成作为前两个流的组合的第三个流

Stream<String> a = Stream.of("f","b","z");
Stream<String> b = Stream.of("foo","bar","baz");

我想从这两个中生成一个新的，它将与这一个等价

Stream<String> c = Stream.of("f:foo","b:bar","z:baz");

我目前找到的是stream.concat或stream.of

 Stream<String> result = Stream.concat(a, b); or
 Stream<String> result = Stream.of(a, b).flatMap(s -> s);

Stream.of("f","b","z","foo","bar","baz");

List<String> aList = a.collect(Collectors.toList());
List<String> bList = b.collect(Collectors.toList());
List<String> result = new ArrayList<>();
for(int i = 0; i< aList.size(); i++){
    result.add(aList.get(i)+":"+bList.get(i));
}
Stream<String> cc = result.stream();

Answer 1

这个操作通常称为zip，它并不作为streamAPI中的函数存在。自己实现它有点困难，您可以从以下答案中看到:https://stackoverflow.com/a/23529010/4137489

该函数在Google Guava库中实现,但：

https://guava.dev/releases/23.0/api/docs/com/google/common/collect/streams.html#zip-java.util.stream.stream-java.util.stream.stream-java.util.function.bifunction-

Streams.zip(a, b, (aElem, bElem) -> aElem + ":" + bElem));