按天数拆分java.util.Date集合 [英] Split java.util.Date collection by Days

问题描述

请你帮我一下。我无法弄清楚算法。

Could you help me, please. I can't figure out the algorithm.

我有一组日期排序的集合，例如ArrayList这样的：

I have a sorted collection of Dates, for example ArrayList like this:

Wed Jan 22 00:00:00 MSK 2014
Wed Jan 22 00:30:00 MSK 2014
Wed Jan 23 01:00:00 MSK 2014
Wed Jan 23 01:30:00 MSK 2014
Wed Jan 23 02:00:00 MSK 2014
Wed Jan 24 02:30:00 MSK 2014
Wed Jan 24 03:00:00 MSK 2014
Wed Jan 24 03:30:00 MSK 2014
Wed Jan 24 04:00:00 MSK 2014
Wed Jan 28 04:30:00 MSK 2014
Wed Jan 28 05:00:00 MSK 2014

我需要该列表的另一个版本，分组（从00:00:00到23:59:59），因此它可以是列表数组，例如 List< List< Date>> 。在示例之后，它应该是大小为4的列表，第二个对象==

I need another version of this list, grouped by day (from 00:00:00 till 23:59:59), so it could be an array of lists, like List<List<Date>>. Following the example it should be the list with size 4, with second Object ==

List<Date> {Wed Jan 23 01:00:00 MSK 2014;
Wed Jan 23 01:30:00 MSK 2014;
Wed Jan 23 02:00:00 MSK 2014; }

似乎很简单。但我找不到解决问题的简便方法。

Seems like an easy task. but I can't find convenient way to solve it. Thanks in advance.

更新
根据TEXHIK的回答，宽度为JDK 7，可以这样完成：

UPDATE According to TEXHIK's answer, width JDK 7 it could be done like this:

 public List<List<Date>> split(List<Date> value) {
        List<List<Date>> result = new ArrayList<>();

        int day = value.iterator().next().getDate();
        List<Date> newListEntry = new ArrayList<>();

        for (Date date : value) {
            if (date.getDate() == day) {
                newListEntry.add(date);
            }
            else {
                day = date.getDate();
                result.add(newListEntry);
                newListEntry = new ArrayList<>();
                newListEntry.add(date);
            }
        }
        result.add(newListEntry);//because the last sublist was not added

        return result;
    }

没关系，除非已弃用java.util.Date方法。会很高兴看到使用Java 8或Joda Time的答案。

It's OK, unless it's deprecated methods of java.util.Date. Will be glad to see the answers using Java 8 or Joda Time.

推荐答案

如果您对UTC的日子感到满意，生活就会变得愉快更简单：

If you're happy with UTC days, life becomes simpler:

• 由于unix纪元的毫秒数开始于一天的边界，因此您可以除以


• 每天长达24小时，这在很多情况下都很方便（例如，如果您要对数据进行分析）

在这种情况下，您可以使用类似以下的命令：

In that case, you can just use something like:

long millisPerDay = TimeUnit.DAYS.toMillis(1);
Map<Long, List<Date>> datesByDay = new HashMap<>();
for (Date date : dates) {
    long day = date.getTime() / millisPerDay;
    List<Dates> dayDates = datesByDay.get(day);
    if (dayDates == null) {
        dayDates = new ArrayList<>();
        datesByDay.put(day, dayDates);
    }
    dayDates.add(date);
}

当然，多幅地图，例如从番石榴中，将使此过程变得更简单：

Of course, a Multimap, e.g. from Guava, will make this simpler:

long millisPerDay = TimeUnit.DAYS.toMillis(1);
Multimap<Long, Date> datesByDay = ArrayListMultimap.create();
for (Date date : dates) {
    long day = date.getTime() / millisPerDay;
    datesByDay.put(day, date);
}

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