计数出现的HashMap

Question

我正在尝试使用 hashmap 来统计每天的疾病数量：

public static main(String[] args){
 Disease cholera=new Disease("cholera");
 Disease dengue=new Disease("dengue");

 List<Diagnosis> diagnoses = Arrays.asList(
            new Diagnosis(cholera, 0), // registered cholera on day 0
            new Diagnosis(cholera, 0),
            new Diagnosis(cholera, 1),
            new Diagnosis(cholera, 1),
            new Diagnosis(cholera, 2),
            new Diagnosis(cholera, 2)
    );
   
   printFreq(diagnosis);

}
  
public static void printFreq(List<Diagnosis> diagnoses) {
    Map<Diagnosis, Integer> hm = new HashMap();
    for (Diagnosis x : diagnoses) {

        if (!hm.containsKey(x)) {
            hm.put(x, 1);
        } else {
            hm.put(x, hm.get(x) + 1);
        }
    }

但如果我调用 printFreq(diagnoses) 我会得到：{{cholera, 0}=1, {cholera, 1}=1, {cholera, 2}=1, {cholera, 0}=1, {dengue, 0}=1，{霍乱，1}=1，{霍乱，2}=1}。我该如何解决这个问题 {{cholera,0}=2,{dengue0}=1,{cholera,1}=2,{cholera,2}=2}}

我不能更改诊断或疾病类，但它们看起来是这样的：

public class Disease {

private final String name;

public Disease(String name) {
    this.name = name;
}

public String getName() {
    return name;
}

@Override
public boolean equals(Object o) {
    if (this == o) return true;
    if (o == null || getClass() != o.getClass()) return false;

    Disease disease = (Disease) o;

    return name.equals(disease.name);
}

@Override
public int hashCode() {
    return name.hashCode();
}

@Override
public String toString() {
    return name;
}

诊断：

 public class Diagnosis {

private final Disease disease;
private final int day;

public Diagnosis(Disease disease, int day) {
    this.disease = disease;
    this.day = day;
}

public Disease getDisease() {
    return disease;
}

public int getDay() {
    return day;
}

@Override
public String toString() {
    return "{" + disease + ", " + day + "}";
}

Answer 1

您可以添加一个包装类以使生活更轻松。在这段代码中，我有一个包装类DiagnosisMetric，它包装了Diagnosis 类。 以下是printFreq函数的新实现。

    public static void printFreq(List<Diagnosis> diagnoses) {
        Map<DiagnosisMetric, Long> collect = diagnoses.stream().
             collect(Collectors.groupingBy(DiagnosisMetric::new, counting()));
        System.out.println(collect);
    }

而且，以下是包装类。请注意，我已经按照要求实现了equals 和hashCode。

    public static class DiagnosisMetric {
      private Diagnosis diagnosis;

        public DiagnosisMetric(Diagnosis s) {
            this.diagnosis = s;
        }

        public Diagnosis getDiagnosis() {
            return diagnosis;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            DiagnosisMetric that = (DiagnosisMetric) o;
            return diagnosis.getDisease().getName().
              equals(that.getDiagnosis().getDisease().getName())
                    && diagnosis.getDay() == that.getDiagnosis().getDay();
        }

        @Override
        public int hashCode() {
            return Objects.hash(diagnosis.getDay(), 
                   diagnosis.getDisease().getName());
        }

        @Override
        public String toString() {
            String disease = diagnosis.getDisease().getName();
            int day = diagnosis.getDay();
            return "{" + disease + ", " + day + "}";
        }
    }

Answer 2

不需要使用Diagnosis 键存储中间映射的临时解决方案（如前所述，如果没有正确实现hashCode 和equals，它不能用作映射键）是这样的：

• 使用包含diagnosis.day 和disease.name 的原始列表作为Collectors.groupingBy 中的密钥包装器
• 计算频率（例如使用Collectors.summingInt）
• 使用Supplier<Map>提供按列表内容排序的树状图
• 打印统计数据

diagnoses.stream()
    .collect(Collectors.groupingBy(
        d -> Arrays.asList(d.getDay(), d.getDisease().getName()),
        () -> new TreeMap<List<?>, Integer> (Comparator
            .comparingInt((List k) -> (Integer) k.get(0))
            .thenComparing((List k) -> (String) k.get(1))
        ),
        Collectors.summingInt(d -> 1)
    )) // Map<List, Integer> created here
    .forEach((k, v) -> System.out.println(k + " = " + v));

输出：

[0, cholera] = 2
[0, dengue] = 1
[1, cholera] = 2
[2, cholera] = 2

Answer 3

要解决这个问题，我们需要做的是重写 Diagnosis 类的 hashCode 和 equals 方法，如下所示：

@Override
public boolean equals(Object obj) {
    Diagnosis diagnosis = (Diagnosis) obj;
    return this.day == diagnosis.day && this.disease == diagnosis.disease;
}
    
@Override
public int hashCode() {
    char[] charArr = this.disease.toString().toCharArray();
    int sum = 0;
    for (int i = 0; i < charArr.length; i++) {
        sum += charArr[i];
    }
    return sum;
}

您还需要覆盖Disease 类的equals 方法，下面是代码：

@Override
public boolean equals(Object obj) {
    Disease disease = (Disease) obj;
    return this.disease.equalsIgnoreCase(disease.toString());
}