java - 计算有多少 HashMap 条目具有给定值

Question

public final static HashMap<String, Integer> party = new HashMap<String, Integer>();
party.put("Jan",1);
party.put("John",1);
party.put("Brian",1);
party.put("Dave",1);
party.put("David",2);

我怎样才能返回有多少人的值1

Answer 1

我只是在 HashMap 值上使用 Collections.frequency() 方法，就像这样。

int count = Collections.frequency(party.values(), 1);
System.out.println(count);
===> 4

或通用解决方案，生成频率与数字的映射。

Map<Integer, Integer> counts = new HashMap<Integer, Integer>();
for (Integer c : party.values()) {
    int value = counts.get(c) == null ? 0 : counts.get(c);
    counts.put(c, value + 1);
}
System.out.println(counts);
==> {1=4, 2=1}

Answer 2

使用 Java 8：

party.values().stream().filter(v -> v == 1).count();

Answer 3

尝试这个：

int counter = 0;
Iterator it = party.entrySet().iterator();
while (it.hasNext()) {
  Map.Entry pairs = (Map.Entry)it.next();
  if(pairs.getValue() == 1){
    counter++; 
  }      
}
System.out.println("number of 1's: "+counter);

Answer 4

你可以用这个

HashMap<String, Integer> party = new HashMap<String, Integer>();
party.put("Jan",1);
party.put("John",1);
party.put("Brian",1);
party.put("Dave",1);
party.put("David",2);

Set<Entry<String, Integer>> set = party.entrySet();
for (Entry<String, Integer> me : set) {
    if(me.getValue()==1)
    System.out.println(me.getKey() + " : " + me.getValue());
}

Answer 5

按函数http://code.google.com/p/lambdaj/wiki/LambdajFeatures为很多这样的组尝试这个库

HashMap<String, Integer> party = new HashMap<String, Integer>();
    party.put("Jan",1);
    party.put("John",1);
    party.put("Brian",1);
    party.put("Dave",1);
    party.put("David",2);
    List<Integer> list = filter(equalTo(1),party.values());
    System.out.println(list.size());

您可能需要导入这些 Maven 依赖项

<dependency>
      <groupId>com.googlecode.lambdaj</groupId>
     <artifactId>lambdaj</artifactId>
    <version>2.3.3</version>

和 hamcrest 匹配器

equalTo(1)