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Java SortedSet添加全部

[英]Java SortedSet add all

The following code creates a sorted set that sorts by its values rather thank keys. 以下代码创建一个排序集,按其值排序,而不是感谢键。 vertexRank is an object responsible for getting the value. vertexRank是一个负责获取值的对象。 Everything works well except the code: vertexCentralities.addAll(vMap.entrySet()); 除了代码之外,一切都运行良好: vertexCentralities.addAll(vMap.entrySet()); What happens is that only the first entry from vMap is added to vertexCentralities rather than all entries. 会发生的是,只有vMap的第一个条目被添加到vertexCentralities而不是所有条目。

  1. How can I get all entries from vMap into vertexCentralities? 如何从vMap获取所有条目到vertexCentralities?

     SortedSet<Map.Entry<String, Double>> vertexCentralities = new TreeSet<Map.Entry<String, Double>>( new Comparator<Map.Entry<String, Double>>() { @Override public int compare(Map.Entry<String, Double> e1, Map.Entry<String, Double> e2) { return e2.getValue().compareTo(e1.getValue()); } }); SortedMap<String, Double> vMap = new TreeMap<String, Double>(); double curRank = 0; for(String vStr: g.getVertices()) { curRank = vertexRank.getVertexScore(vStr); vMap.put(vStr, curRank); } vertexCentralities.addAll(vMap.entrySet()); 

I tried running: 我试过跑:

public static final void main(final String[] args) {
    final String[] vStrs = new String[] { "A", "Z", "E", "R", "T", "Y" }; // init

    final SortedSet<Map.Entry<String, Double>> vertexCentralities = new TreeSet<Map.Entry<String, Double>>(new Comparator<Map.Entry<String, Double>>() {
        @Override
        public int compare(final Map.Entry<String, Double> e1, final Map.Entry<String, Double> e2) {
            return e2.getValue().compareTo(e1.getValue());
        }
    });
    final SortedMap<String, Double> vMap = new TreeMap<String, Double>();
    double curRank = 0;
    for (final String vStr : vStrs) {
        curRank = new Random().nextDouble() * 100.0; // replacing
                                                        // vertexRank.getVertexScore(vStr);
                                                        // for testing
        vMap.put(vStr, curRank);
    }
    vertexCentralities.addAll(vMap.entrySet());

    for (final Map.Entry<String, Double> entry : vertexCentralities) {
        System.out.println(entry.getKey() + ": " + entry.getValue());
    }

}

and the output was sorted by value: 并且输出按值排序:

A: 70.50008784770233
Z: 55.48252329485239
E: 37.31308600830347
Y: 32.534528844628255
T: 16.544965680467794
R: 12.258316023552872

Maybe your problem comes from somewhere else... like g.getVertices() or vertexRank.getVertexScore(vStr) 也许你的问题来自其他地方......比如g.getVertices()或者vertexRank.getVertexScore(vStr)

EDIT: I tried with duplicates values for the String and for the double : 编辑:我尝试使用Stringdouble重复值:

final String[] vStrs = new String[] { "A", "Z", "E", "R", "T", "Y", "A" };
curRank = new Random().nextInt(3);

and it looks like no duplicates are allowed. 并且看起来不允许重复。 Is this your problem? 这是你的问题吗?

EDIT: Found a solution if you want to allow multiple entry with the same Double : Replace your SortedSet vertexCentralities 's comparator condition to: 编辑:如果你想允许多个条目使用相同的Double找到一个解决方案:将SortedSet vertexCentralities的比较器条件替换为:

final int bValue = e2.getValue().compareTo(e1.getValue());
return bValue != 0 ? bValue : e2.getKey().compareTo(e1.getKey());

An alternative solution could be to use SimpleEntry<K, V> (inner public static class of java.util.AbstractMap) and avoid the SortedMap: 另一种解决方案可能是使用SimpleEntry<K, V> (java.util.AbstractMap的内部公共静态类)并避免使用SortedMap:

final SortedSet<Map.Entry<String, Double>> vertexCentralities = new TreeSet<Map.Entry<String, Double>>(new Comparator<Map.Entry<String, Double>>() {
    @Override
    public int compare(final Map.Entry<String, Double> e1, final Map.Entry<String, Double> e2) {
        final int bValue = e2.getValue().compareTo(e1.getValue());
        return bValue != 0 ? bValue : e2.getKey().compareTo(e1.getKey());
    }
});
double curRank = 0;
for (final String vStr : g.getVertices()) {
    curRank = vertexRank.getVertexScore(vStr);
    vertexCentralities.add(new SimpleEntry<String, Double>(vStr, curRank));
}

You should be able to have duplicate Key or Value but not both. 您应该能够拥有重复的KeyValue但不能同时拥有两个。

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