Question

我有多个TreeMap，我只想在同一个键的一个TreeMap求和值和有效中求和。像：

   TreeMap<String,Long> sum(TreeMap<String,Long> tm1,TreeMap<String,Long> tm2);

我试图做到这一点，但是1.我无法将列表再次转换为TreeMap，如果等于，则密钥重复：

    TreeMap<String,Long> tm1=new TreeMap<String, Long>();
    ...
    TreeMap<String,Long> tm2=new TreeMap<String, Long>();
    ...       
    List<Map.Entry<String,Long>> first = new ArrayList<Map.Entry<String,Long>>(tm1.entrySet());
    List<Map.Entry<String,Long>> second = new ArrayList<Map.Entry<String,Long>>(tm2.entrySet());
    Iterable<Map.Entry<String,Long>> all = Iterables.mergeSorted(
            ImmutableList.of(first, second), new Ordering<Map.Entry<String, Long>>() {
        @Override
        public int compare(java.util.Map.Entry<String, Long> stringLongEntry, java.util.Map.Entry<String, Long> stringLongEntry2) {
            return stringLongEntry.getKey().compareTo(stringLongEntry2.getKey());
        }
    });
    TreeMap<String,Long> mappedMovies = Maps.uniqueIndex(... ??)

编辑：我无法使用Java 8，因为该程序在Amazon Web Services中仅支持Java 1.7的Haddoop程序中运行。

Answer 1

您可以使用Java 8 Streams来实现此目的。请考虑以下代码：

// Testdata - The first map
Map<String, Long> m1 = new TreeMap<>();
m1.put("A", 1L);
m1.put("B", 1L);
m1.put("C", 1L);

// Testdata - The second map
Map<String, Long> m2 = new TreeMap<>();
m2.put("C", 2L);
m2.put("D", 2L);
m2.put("E", 2L);

// Summarize using streams
final Map<String, Long> summarized =
        Stream.concat(m1.entrySet().stream(), m2.entrySet().stream())     // Stream both maps
              .collect(Collectors.groupingBy(                             // Collect the map
                          Map.Entry::getKey,                              // Group by key
                          Collectors.summingLong(Map.Entry::getValue)));  // Value is the sum

System.out.println("Summarized: " + summarized);                          // Print the output

汇总的Map按键分组，并在值上汇总。输出是：

总结：{A = 1，B = 1，C = 3，D = 2，E = 2}

如果你想把它放在一个函数中，只需这样做：

public Map<String, Long> summarize(
        final Map<String, Long> m1, 
        final Map<String, Long> m2) {

    return Stream.concat(m1.entrySet().stream(), m2.entrySet().stream())
                 .collect(groupingBy(
                          Map.Entry::getKey,
                          summingLong(Map.Entry::getValue)));
}

要阅读有关Java 8流的更多信息，请查看Oracle docs。

Answer 2

以下函数计算总和：

public static TreeMap<String, Long> sum(TreeMap<String, Long> first,
        TreeMap<String, Long> second) {
    TreeMap<String, Long> result = new TreeMap<String, Long>(first);

    for (Entry<String, Long> e : second.entrySet()) {
        Long l = result.get(e.getKey());
        result.put(e.getKey(), e.getValue() + (l == null ? 0 : l));
    }

    return result;
}

测试代码：

TreeMap<String, Long> first = new TreeMap<String, Long>();
TreeMap<String, Long> second = new TreeMap<String, Long>();

first.put("x", 1L);
first.put("y", 5L);

second.put("x", 2L);
second.put("y", 3L);
second.put("z", 5L);

System.out.println(sum(first, second));

输出：

{x=3, y=8, z=5}

修改

一个小的优化是复制最大的TreeMap并迭代最小的public static TreeMap<String, Long> sum(TreeMap<String, Long> first, TreeMap<String, Long> second) { // optimization (copy the largest tree map and iterate over the // smallest) if (first.size() < second.size()) { TreeMap<String, Long> t = first; first = second; second = t; } TreeMap<String, Long> result = new TreeMap<String, Long>(first); for (Entry<String, Long> e : second.entrySet()) { Long l = result.get(e.getKey()); result.put(e.getKey(), e.getValue() + (l == null ? 0 : l)); } return result; }。这减少了查找/插入的次数。

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如何有效地将Java中的两个TreeMap相加？

2 个答案: