使用Java中的流拆分单词列表

Question

I am having this method that takes a number of lists, which contain lines of books. I am combing them to a stream to then iterate over them to split on all non-letter's \\\\P{L} .

Is there a way to avoid the for-each loop and process this within a stream?

private List<String> getWordList(List<String>... lists) {
        List<String> wordList = new ArrayList<>();

        Stream<String> combinedStream = Stream.of(lists)
                .flatMap(Collection::stream);
        List<String> combinedLists = combinedStream.collect(Collectors.toList());

        for (String line: combinedLists) {
            wordList.addAll(Arrays.asList(line.split("\\P{L}")));
        }

        return wordList;
}

Answer 1

Having stream, you can simply " flatMap " further and return the result:

return combinedStream
        .flatMap(str -> Arrays.stream(str.split("\\P{L}")))
        .collect(Collectors.toList());

To put it altogether:

private List<String> getWordList(List<String>... lists) {
    return Stream.of(lists)
        .flatMap(Collection::stream)
        .flatMap(str -> Arrays.stream(str.split("\\P{L}")))
        .collect(Collectors.toList());
}

Answer 2

You don't need to introduce so many variables :

private List<String> getWordList(List<String>... lists) {

    return Stream.of(lists) // Stream<Stream<String>>
                 .flatMap(Collection::stream) // Stream<String> 
                 .flatMap(Pattern.compile("\\P{L}")::splitAsStream) //Stream<String>     
                 .collect(toList()); // List<String>
}

As underlined by Holger, .flatMap(Pattern.compile("\\\\P{L}")::splitAsStream)
should be favored over .flatMap(s -> Arrays.stream(s.split("\\\\P{L}"))) to spare array allocation and pattern compilation performed for each element of the stream.

Answer 3

You can combine all the list and flatMap for result

private List<String> getWordList(List<String>... lists) {
    return Stream.of(lists)
    .flatMap(Collection::stream)
    .flatMap(str -> Arrays.stream(str.split("\\P{L}")))
    .collect(Collectors.toList());
}