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そのような宣言でネストされたマップをどのように反復処理する必要がありますか?

  • Map<String, Multiset<String>>

このハッシュ生成タスクを行うためのより効果的な方法である他のハッシュマップ/リストがあるかどうかを提案してください。

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

String[] foobarness = {"foo" , "bar", "ness", "foo", 
    "bar", "foo", "ness", "bar", "foo", "ness", "foo", 
    "bar", "foo", "ness", "bar", "ness", "foo", "bar", 
    "foo", "ness"};
String[] types = {"type::1", "type::2", "type::3", 
    "type::4",};

Map<String, Multiset<String>> typeTextCount = 
new HashMap<String, Multiset<String>>();

Multiset<String> textAndCount 
    = TreeMultiset.create();

for (int i=0; i<types.length; i++) {
    // I know it's kinda weird but in my task, 
    //  i want to keep adding only 1 to the count for each entry.
    // Please suggest if there is a better hashmap/list for such task.
    if ((types[i]== "type::1") or (types[i]== "type::3")) {
        for (String text : foobarness) {
            // I don't worry too much about how i 
            //  populate the Map, it is iterating through 
            //  the Map that I have problem with.           
            textAndCount.put(text, 1); 
        }
    }

    if ((types[i]== "type::2") or (types[i]== "type::4")) {
        for (String text : foobarness) 
            textAndCount.put(text, 1);
    }
}

これで、ハッシュマップが作成されました。その複雑なネストされたマップをどのように反復処理すればよいでしょうか? 以下のコードを試してみましたが、マルチセットから最初の getValue() しか得られませんでした:

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();
}
4

1 に答える 1

2

あなたのコードでは、あなたのMultisetをあなたに追加していませんMap。そのため、出力が表示されません。

あなたのコードで私はこれをしました：

typeTextCount.put(types[i], textAndCount);

ループ内で、同じ反復子を使用して、次のようなすべての出力を確認できます。

type::3 :   [bar x 24, foo x 32, ness x 24]
type::2 :   [bar x 24, foo x 32, ness x 24]
type::4 :   [bar x 24, foo x 32, ness x 24]
type::1 :   [bar x 24, foo x 32, ness x 24]

編集:参照用の完全なコード:

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[] types      =
                                   {
            "type::1", "type::2", "type::3", "type::4",
                                   };
    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++) {
            // I know it's kinda weird but in my task,
            // I want to keep adding only 1 to the count for each entry.
            // Please suggest if there is a better hashmap/list for such task.
            if (("type::1".equals(types[i])) || ("type::3".equals(types[i]))) {
                for (String text : foobarness) {
                    // I don't worry too much about how i
                    // populate the Map, it is iterating through
                    // the Map that I have problem with.
                    textAndCount.add(text, 1);
                }
            }

            if (("type::2".equals(types[i])) || ("type::4".equals(types[i]))) {
                for (String text : foobarness)
                    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();
        }
    }
}
于 2011-12-15T15:25:51.530 に答える