HashMap是put还是putAll？ -Java

Question

通过引用放置哈希图，并通过副本放置哈希图。 我该怎么办？ 另一个问题是String[] types数量并不是真正已知的，因此创建Multiset<String> textAndCount = TreeMultiset.create();多个实例Multiset<String> textAndCount = TreeMultiset.create(); 不是很有帮助。 我有以下代码，但是两种类型的输出都相同。

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;

import com.google.common.collect.Multiset;
import com.google.common.collect.TreeMultiset;


public class TestIterator {

private static String[] foobarness  = {"foo", "bar", "ness", "foo", "bar", "foo", "ness", "bar", "foo", "ness", "foo", "bar", "foo", "ness", "bar", "ness", "foo", "bar", "foo", "ness"};

private static String[] foobarness2  = {"bar", "ness", "foo", "bar", "foo", "ness", "bar", "foo", "ness", "foo", "foo", "bar", "foo", "ness", "bar", "foo", "ness", "bar", "foo", "ness", "ness", "bar", "foo", "ness"};

private static String[] types = {"type::1", "type::2"};


public static void main(String[] args) {
    Map<String, Multiset<String>> typeTextCount = 
        new HashMap<String, Multiset<String>>();

    Multiset<String> textAndCount = TreeMultiset.create();
    for (int i = 0; i < types.length; i++) {
        if ("type::1".equals(types[i])) {
            for (String text : foobarness)
                textAndCount.add(text, 1);
        }
        if ("type::2".equals(types[i])) {
            for (String text : foobarness2) 
                textAndCount.add(text, 1);
        }
        typeTextCount.put(types[i], textAndCount);
    }

    Iterator<Entry<String, Multiset<String>>> itTTC = 
        typeTextCount.entrySet().iterator();

    while (itTTC.hasNext()) {
        Map.Entry textCt = (Map.Entry) itTTC.next();
        System.out.println(textCt.getKey() + " :\t" + textCt.getValue());
        itTTC.remove();
    }
}

我的输出来自上面的代码：

type::2 :   [bar x 13, foo x 17, ness x 14]
type::1 :   [bar x 13, foo x 17, ness x 14]

正确的输出应为：

type::1 :   [bar x 6, foo x 8, ness x 6]
type::2 :   [bar x 7, foo x 9, ness x 8]

Answer 1

在您的for循环中移动Multiset<String> textAndCount = TreeMultiset.create() 。 两个“类型”都共享相同的多集，因此您的计数增加了一倍。

您的for循环可能如下所示：

    for (int i = 0; i < types.length; i++) {
        Multiset<String> textAndCount = TreeMultiset.create();
        if ("type::1".equals(types[i])) {
            for (String text : foobarness)
                textAndCount.add(text, 1);
        }
        if ("type::2".equals(types[i])) {
            for (String text : foobarness2)
                textAndCount.add(text, 1);
        }
        typeTextCount.put(types[i], textAndCount);
    }

在使用它的同时，还可以通过使用for-each样式循环来改进地图的迭代。 如果你遍历这是热衷于消除每个条目，你可以换你的entrySet一个consumingIterable为相同的功能。

    for (Entry<String, Multiset<String>> textCt : Iterables.consumingIterable(typeTextCount
            .entrySet())) {
        System.out.println(textCt.getKey() + " :\t" + textCt.getValue());
    }

这产生输出：

type::2 :   [bar x 7, foo x 9, ness x 8]
type::1 :   [bar x 6, foo x 8, ness x 6]

如果您不喜欢该命令，建议您使用“ Ordering来获取条目的排序列表。

Answer 2

Multiset<String> textAndCount = TreeMultiset.create(); 应该在循环内。 如果您放入该设置的副本，则输出为

type::1 :   [bar x 6, foo x 8, ness x 6]
type::2 :   [bar x 13, foo x 17, ness x 14]