拆分Java的ArrayList中所包含对象的属性 [英] Split Java ArrayList by properties of contained Objects

问题描述

我有有个约会值的的ArrayList 包含对象。
现在我想设法创建一个新的的ArrayList 为包含所有从主的ArrayList 具有对象每年同年的日期值。

I have an ArrayList containing Objects that have a date value. Now I want to manage to create a new ArrayList for each year that contains all the Objects from the main ArrayList that have the same year in their date Value.

因此​​，所有的另一个从2010年去的对象在一个列表中，全部来自1999年。

So all Objects from 2010 go in one List, all from 1999 in another.

推荐答案

您需要一个<一个href=\"http://java.sun.com/javase/6/docs/api/java/util/Map.html\"><$c$c>Map<Year,List<DatedObject>>,甚至 的SortedMap 像 TreeMap的 。

You need a Map<Year,List<DatedObject>>, maybe even SortedMap like a TreeMap.

您也可以使用<一个href=\"http://guava-libraries.google$c$c.com/svn-history/r8/trunk/javadoc/com/google/common/collect/Multimap.html\">multimap番石榴。

从本质上讲，你每年从地图（也许只是一个整数）的列表＆LT; D​​atedObject＆GT; ，在属于那年。

Essentially you map from each year (perhaps just an Integer), the List<DatedObject> that belong in that year.

而自从几年有一种天然的排序顺序，您可能想看看的SortedMap 为您提供您需要的功能。最有可能的答案是肯定的。

And since years have a natural sorting order, you may want to see if SortedMap provides you with functionalities that you need. Most likely the answer is yes.

下面是在Java中的一个片段，显示您如何填充地图。还需要注意的是<一个href=\"http://java.sun.com/javase/6/docs/api/java/util/NavigableMap.html\"><$c$c>NavigableMap是用来代替的SortedMap ; 的NavigableMap 允许包含的范围查询（见相关的问题）。

Here's a snippet in Java that shows you how you can populate the map. Note also that NavigableMap is used instead of SortedMap; NavigableMap allows inclusive range queries (see related question).

    class DatedObject {
        final String name; final int year;
        DatedObject(String name, int year) {
            this.name = name; this.year = year;
        }
        @Override public String toString() {
            return String.format("%s (%d)", name, year);
        }
    }
    List<DatedObject> masterList = Arrays.asList(
        new DatedObject("A", 2010),
        new DatedObject("B", 2009),
        new DatedObject("C", 2006),
        new DatedObject("D", 2010),
        new DatedObject("E", 2009),     
        new DatedObject("F", 2011)          
    );
    NavigableMap<Integer,List<DatedObject>> objectsByYear =
        new TreeMap<Integer,List<DatedObject>>();
    for (DatedObject obj : masterList) {
        List<DatedObject> yearList = objectsByYear.get(obj.year);
        if (yearList == null) {
            objectsByYear.put(obj.year, yearList = new ArrayList<DatedObject>());
        }
        yearList.add(obj);          
    }
    System.out.println(objectsByYear);
    // prints "{2006=[C (2006)], 2009=[B (2009), E (2009)],
    //          2010=[A (2010), D (2010)], 2011=[F (2011)]}"
    System.out.println(objectsByYear.get(2011));
    // prints "[F (2011)]"
    System.out.println(objectsByYear.subMap(2007, true, 2010, true));
    // prints "{2009=[B (2009), E (2009)], 2010=[A (2010), D (2010)]}"

相关问题

• How当仅支持半开区间做包括的范围查询（ALA SortedMap.subMap）



Related questions

• How to do inclusive range queries when only half-open range is supported (ala SortedMap.subMap)

如果你绝对在列表与LT坚持;列表与LT; D​​atedObject＆GT;＆GT; partitionedList ，然后建立映射同上，只需用遵循它：

If you absolutely insists on a List<List<DatedObject>> partitionedList, then build the map as above, and simply follow it with:

List<List<DatedObject>> partitionedList =
    new ArrayList<List<DatedObject>>(objectsByYear.values());
System.out.println(partitionedList);
// prints "[[C (2006)], [B (2009), E (2009)], [A (2010), D (2010)], [F (2011)]]"

---

的multimap中示例

您也可以使用 Multimap之番石榴和<一个href=\"http://guava-libraries.google$c$c.com/svn-history/r8/trunk/javadoc/com/google/common/collect/Multimaps.html#index%28java.lang.Iterable,%20com.google.common.base.Function%29\"><$c$c>Multimaps.index实用方法如下：

---

The MultiMap Example

You can also use Multimap from Guava, and Multimaps.index utility method as follows:

    Multimap<Integer,DatedObject> mmap = Multimaps.index(
        masterList,
        new Function<DatedObject, Integer>(){
            @Override public Integer apply(DatedObject from) {
                return from.year;
            }
        }
    );
    System.out.println(mmap);
    // prints "{2010=[A (2010), D (2010)], 2009=[B (2009), E (2009)],
    //          2006=[C (2006)], 2011=[F (2011)]}"

API链接



• <$c$c>com.google.common.collect.Multimap


• <$c$c>com.google.common.collect.Multimaps


• <$c$c>com.google.common.base.Function



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