在Java中查找重复值和相同值的出现次数

Question

我正在使用Java。我有一张地图如下所示：

List<Map<String, String>> listMap = new ArrayList<Map<String, String>>();

我正在将带有一些键值对的Map对象插入到上面的List<Map<String, String>> listMap中。

Map<String, String> map1 = new HashMap<String, String>();

map1.put("jobDescription", "Java Developer-SpringBoot");
map1.put("interviewType", "L2");
map1.put("hired", "yes");
listMap.add(map1);

Map<String, String> map2 = new HashMap<String, String>();

map2.put("jobDescription", "Java Developer-SpringBoot");
map2.put("interviewType", "L2");
map2.put("hired", "yes");
listMap.add(map2);

Map<String, String> map3 = new HashMap<String, String>();
map3.put("jobDescription", "Java Developer-SpringBoot");
map3.put("interviewType", "L1");
map3.put("hired", "no");
listMap.add(map3);

现在，我想迭代

listMap(`List<Map<String, String>> listMap`) 

然后查找映射中的键jobDescription是否有任何重复/相同的值，然后检查interviewType键值的值，并查看该值出现的次数。

在上面的示例中，键jobDescription的值在所有映射对象(即Java Developer-SpringBoot)中都是相同的。然后验证键interviewType的值，并查看每个值的出现次数(在上述情况下，L2重复两次，L1重复一次)。最后，我需要再构造一个包含我的观察结果的Map。

例如(为了说明起见，描述了这些数据，但是这实际上应该进入一个新的Map中)。

"jobDescription"             - "Count of L2" -  "Count of L1"
-------------------------------------------------------------------
"Java Developer-SpringBoot"      2                 1

有人能帮我吗？

下面给出了我正在尝试的代码：

package com.test;
    
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Sample {

    public static void main(String[] args) {
        
        List<Map<String, String>> listMap = new ArrayList<Map<String, String>>();
        Map<String, String> map1 = new HashMap<String, String>();

        map1.put("jobDescription", "Java Developer-SpringBoot");
        map1.put("interviewType", "L2");
        map1.put("hired", "yes");
        listMap.add(map1);
        
        Map<String, String> map2 = new HashMap<String, String>();
        
        map2.put("jobDescription", "Java Developer-SpringBoot");
        map2.put("interviewType", "L2");
        map2.put("hired", "yes");
        listMap.add(map2);
       
        Map<String, String> map3 = new HashMap<String, String>();
        map3.put("jobDescription", "Java Developer-SpringBoot");
        map3.put("interviewType", "L1");
        map3.put("hired", "no");
        listMap.add(map3);
        Map<String, Map<String, String>> requiredMap = new HashMap<String, Map<String, String>>();
        for (Map<String, String> someMap : listMap) {
            int count = Collections.frequency(someMap.values(), "L2");
            
        }
    }

}

Answer 1

使用map存储数据看起来很奇怪，如下所示：

所有的值都被限制在相同的type(String).

• There上，不限制键。

对类Job进行建模是一种更合适的方法。然后遵循@Joe的建议，Group by multiple field names in java 8

下面的程序将输出

Java开发人员-SpringBootL1:1，L2:2

import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

public class Sample {

    public static void main(String[] args) {
        List<Job> jobs = new ArrayList<>();
        jobs.add(new Job("Java Developer-SpringBoot", "L2", "yes"));
        jobs.add(new Job("Java Developer-SpringBoot", "L2", "yes"));
        jobs.add(new Job("Java Developer-SpringBoot", "L1", "no"));
        Map<String, Map<String, Long>> jobDescriptionToInterviewTypeToCountMapMap = jobs.stream().collect(Collectors.groupingBy(Job::getJobDescription,
                Collectors.groupingBy(Job::getInterviewType, Collectors.counting())));
        for (Map.Entry<String, Map<String, Long>> entry : jobDescriptionToInterviewTypeToCountMapMap.entrySet()) {
            System.out.println(entry.getKey() + " " + entry.getValue().entrySet().stream().map((e) ->
                    e.getKey() + ":" + e.getValue()).collect(Collectors.joining(", ")));
        }
    }

    public static class Job {
        public Job(String jobDescription, String interviewType, String hired) {
            this.jobDescription = jobDescription;
            this.interviewType = interviewType;
            this.hired = hired;
        }

        private String jobDescription;
        private String interviewType;
        private String hired;

        public String getJobDescription() {
            return jobDescription;
        }

        public void setJobDescription(String jobDescription) {
            this.jobDescription = jobDescription;
        }

        public String getInterviewType() {
            return interviewType;
        }

        public void setInterviewType(String interviewType) {
            this.interviewType = interviewType;
        }

        public String getHired() {
            return hired;
        }

        public void setHired(String hired) {
            this.hired = hired;
        }
    }
}

Answer 2

流在列表和筛选映射上，其中包含一个带有键的条目：jobDescription和value：Java Developer-SpringBoot，平面映射以获取所有映射的所有条目，过滤器条目以interviewType作为键，将每个条目映射到其值，收集到使用Function.identity()映射并映射到频率：

Map<String,Long> result =
listMap.stream()
        .filter(m -> m.entrySet().contains(Map.entry("jobDescription","Java Developer-SpringBoot")))
        .flatMap(m -> m.entrySet().stream())
        .filter(e -> e.getKey().equals("interviewType"))
        .map(Map.Entry::getValue)
        .collect(Collectors.groupingBy(Function.identity(),Collectors.counting()));

System.out.println(result);

//output: {L1=1, L2=2}