在Java8流上执行操作时如何保持状态?
我需要解析一个字符串,该字符串由不同的整数组成,表示某个用户正在或不在看电视时的时段 我首先拆分字符串并将其收集到ArrayList中:在Java8流上执行操作时如何保持状态?,java,string,java-8,java-stream,Java,String,Java 8,Java Stream,我需要解析一个字符串,该字符串由不同的整数组成,表示某个用户正在或不在看电视时的时段 我首先拆分字符串并将其收集到ArrayList中: final List<String> separated = stream(split(s, "A")) .map(str -> "A" + str) .flatMap(str -> stream(split(str, "B")).map(s2 -> s2.startsWith("A") ? s2 : "B" + s2))
final List<String> separated = stream(split(s, "A"))
.map(str -> "A" + str)
.flatMap(str -> stream(split(str, "B")).map(s2 -> s2.startsWith("A") ? s2 : "B" + s2))
collect(Collectors.toList());
我觉得这是一个巨大的倒退,因为我打破了整个河流管道只是为了回到老学校的每一个
有什么方法可以在一条流管道中完成整个处理吗?我正在考虑创建一个自定义收集器,它将查看集合中的最后一个元素,并在此基础上计算正确的LocalDateTime对象
示例:
输入字符串:A60B80A60,这意味着某人正在观看某个内容60分钟,然后停止观看80分钟,然后再次观看60分钟
因此,我想得到一个包含对象的列表:
1从:0:00到:1:00,观看:真实
2从:1:00到:2:20,观看:错误
3从:2:20到:3:20,观看:正确
每个对象的计算都需要了解前一个对象的相关知识这不是关于连续对,而是关于收集累积前缀。在函数式编程中,这种操作通常被称为scanLeft,它出现在许多函数式语言中,如。不幸的是,在当前的Java8流API实现中没有它,所以我们只能用forEachOrdered来模拟它。让我们创建一个模型对象:
static class WatchPeriod {
static final DateTimeFormatter FORMATTER = DateTimeFormatter.ofPattern("HH:mm");
final LocalTime start;
final Duration duration;
final boolean watched;
WatchPeriod(LocalTime start, Duration duration, boolean watched) {
this.start = start;
this.duration = duration;
this.watched = watched;
}
// Takes string like "A60" and creates WatchPeriod starting from 00:00
static WatchPeriod forString(String watchPeriod) {
return new WatchPeriod(LocalTime.of(0, 0),
Duration.ofMinutes(Integer.parseInt(watchPeriod.substring(1))),
watchPeriod.startsWith("A"));
}
// Returns new WatchPeriod which start time is adjusted to start
// right after the supplied previous period
WatchPeriod after(WatchPeriod previous) {
return new WatchPeriod(previous.start.plus(previous.duration), duration, watched);
}
@Override
public String toString() {
return "from: "+start.format(FORMATTER)+", to: "+
start.plus(duration).format(FORMATTER)+", watched: "+watched;
}
}
现在,我们可以将输入字符串(如A60B80A60)拆分为标记A60、B80、A60,将这些标记映射到WatchPeriod对象,然后将它们存储到结果列表中:
String input = "A60B80A60";
List<WatchPeriod> result = new ArrayList<>();
Pattern.compile("(?=[AB])").splitAsStream(input)
.map(WatchPeriod::forString)
.forEachOrdered(wp -> result.add(result.isEmpty() ? wp :
wp.after(result.get(result.size()-1))));
result.forEach(System.out::println);
如果您不介意使用第三方库,“我的免费”增强了Stream API,在其他功能中添加了缺少的操作:
String input = "A60B80A60";
List<WatchPeriod> result = StreamEx.split(input, "(?=[AB])")
.map(WatchPeriod::forString).scanLeft((prev, next) -> next.after(prev));
result.forEach(System.out::println);
结果是一样的。您能提供输入字符串的示例吗?以及所需的输出?@RobAu,我刚刚编辑了我的答案。重复的问题的答案应该包含足够的信息:
from: 00:00, to: 01:00, watched: true
from: 01:00, to: 02:20, watched: false
from: 02:20, to: 03:20, watched: true
String input = "A60B80A60";
List<WatchPeriod> result = StreamEx.split(input, "(?=[AB])")
.map(WatchPeriod::forString).scanLeft((prev, next) -> next.after(prev));
result.forEach(System.out::println);