累积Java流，然后才对其进行处理

问题描述

我的文档如下所示：

data.txt

100, "some text"
101, "more text"
102, "even more text"

我使用正则表达式处理它并返回一个新的处理过的文档，如下所示：

I processed it using regex and returned a new processed documents as the follow:

Stream<String> lines = Files.lines(Paths.get(data.txt);
Pattern regex = Pattern.compile("([\\d{1,3}]),(.*)");

List<MyClass> result =
  lines.map(regex::matcher)
       .filter(Matcher::find)
       .map(m -> new MyClass(m.group(1), m.group(2)) //MyClass(int id, String text)
       .collect(Collectors.toList());

这将返回已处理的MyClass列表。可以并行运行，一切正常。

This returns a list of MyClass processed. Can run in parallel and everything is ok.

问题是我现在有这个：

data2.txt

data2.txt

101, "some text
the text continues in the next line
and maybe in the next"
102, "for a random
number
of lines"
103, "until the new pattern of new id comma appears"

所以，我不知何故需要连接从流中读取的行，直到出现新的匹配。（像缓冲区？）

So, I somehow need to join lines that are being read from the stream until a new match appear. (Something like an buffer?)

我试图收集字符串然后收集MyCla ss（），但没有成功，因为我实际上无法拆分流。

I tried to Collect strings and then collect MyClass(), but with no success, because I cannot actually split streams.

降低连接线的想法，但我会连接线，我无法减少并生成一个新的行流。

Reduce comes to mind to concatenate lines, but I'll concatenate just lines and I cannot reduce and generate a new stream of lines.

如何用java 8流解决这个问题？

Any ideas how to solve this with java 8 streams?

推荐答案

这是 java.util.Scanner 的工作。随着即将推出的Java 9，你会写：

This is a job for java.util.Scanner. With the upcoming Java 9, you would write:

List<MyClass> result;
try(Scanner s=new Scanner(Paths.get("data.txt"))) {
    result = s.findAll("(\\d{1,3}),\\s*\"([^\"]*)\"")
                //MyClass(int id, String text)
    .map(m -> new MyClass(Integer.parseInt(m.group(1)), m.group(2)))
    .collect(Collectors.toList());
}
result.forEach(System.out::println);

但是因为 Stream 生成 findAll 在Java 8下不存在，我们需要一个辅助方法：

but since the Stream producing findAll does not exist under Java 8, we’ll need a helper method:

private static Stream<MatchResult> matches(Scanner s, String pattern) {
    Pattern compiled=Pattern.compile(pattern);
    return StreamSupport.stream(
        new Spliterators.AbstractSpliterator<MatchResult>(1000,
                         Spliterator.ORDERED|Spliterator.NONNULL) {
        @Override
        public boolean tryAdvance(Consumer<? super MatchResult> action) {
            if(s.findWithinHorizon(compiled, 0)==null) return false;
            action.accept(s.match());
            return true;
        }
    }, false);
}

更换 findAll 这个帮助方法，我们得到

Replacing findAll with this helper method, we get

List<MyClass> result;
try(Scanner s=new Scanner(Paths.get("data.txt"))) {

    result = matches(s, "(\\d{1,3}),\\s*\"([^\"]*)\"")
               // MyClass(int id, String text)
    .map(m -> new MyClass(Integer.parseInt(m.group(1)), m.group(2)))
    .collect(Collectors.toList());
}

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