Java 8列表 <Map<String, Object> &gt;列表 <Map<String, Object> &gt;按键分组并按值计数

Question

I have the following list of maps

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

    Map<String, Object> m1 = new HashMap<String, Object>();
    m1.put("company", "LG");
    m1.put("billType", "A");
    m1.put("billPeriod", "09-2018");

    Map<String, Object> m2 = new HashMap<String, Object>();
    m2.put("company", "LG");
    m2.put("billType", "A");
    m2.put("billPeriod", "09-2018");

    Map<String, Object> m3 = new HashMap<String, Object>();
    m3.put("company", "LG");
    m3.put("billType", "A");
    m3.put("billPeriod", "09-2018");

    Map<String, Object> m4 = new HashMap<String, Object>();
    m4.put("company", "LG");
    m4.put("billType", "B");
    m4.put("billPeriod", "01-2019");

    Map<String, Object> m5 = new HashMap<String, Object>();
    m5.put("company", "LG");
    m5.put("billType", "B");
    m5.put("billPeriod", "01-2019");

    Map<String, Object> m6 = new HashMap<String, Object>();
    m6.put("company", "Samsung");
    m6.put("billType", "A");
    m6.put("billPeriod", "10-2018");

    Map<String, Object> m7 = new HashMap<String, Object>();
    m7.put("company", "Samsung");
    m7.put("billType", "A");
    m7.put("billPeriod", "10-2018");

    Map<String, Object> m8 = new HashMap<String, Object>();
    m8.put("company", "Samsung");
    m8.put("billType", "B");
    m8.put("billPeriod", "11-2018");

    listBeforeGroup.add(m1);listBeforeGroup.add(m2);
    listBeforeGroup.add(m3);listBeforeGroup.add(m4);
    listBeforeGroup.add(m5);listBeforeGroup.add(m6);

How do I get this output?

    //Desired Output - List<Map<String, Object>>
    //{company=LG, billType=A, billPeriod=09-2018, count=3}
    //{company=LG, billType=B, billPeriod=01-2019, count=2}
    //{company=Samsung, billType=A, billPeriod=10-2018, count=2}
    //{company=Samsung, billType=B, billPeriod=11-2018, count=1}

I tried this, using java 8 streams, but I can't get the desired output

List<Map<String, Object>> listAfterGroup = listBeforeGroup.stream().map(m -> m.entrySet().stream().collect(Collectors.toMap(p -> p.getKey(), p - >  p.getValue()))).collect(Collectors.toList());

And tried this, btw this solution gives a map but I don't want this

Map<Object, Long> listAfterGroup = listBeforeGroup.stream().flatMap(m -> m.entrySet().stream()).collect(Collectors.groupingBy(Map.Entry::getKey,Collectors.counting()));

I want to group the maps by the key "billPeriod" for example, and count items by the values, then generate a new list of maps.

Answer 1

You can create a class Company and then subsequent operations become much simpler.

class Company {
    String company;
    String billType;
    String billPeriod;

    public Company(String company, String billType, String billPeriod) {
        this.company = company;
        this.billType = billType;
        this.billPeriod = billPeriod;
    }

    // getters, setters, toString, etc
}

Initialize the list :

List<Company> list = new ArrayList<>();
list.add(new Company("LG", "A", "09-2018"));
list.add(new Company("LG", "A", "09-2018"));
list.add(new Company("LG", "A", "09-2018"));
list.add(new Company("LG", "B", "01-2019"));
list.add(new Company("LG", "B", "01-2019"));
list.add(new Company("Samsung", "A", "10-2018"));
list.add(new Company("Samsung", "A", "10-2018"));
list.add(new Company("Samsung", "B", "11-2018"));

Now for an example, you can group by company name :

Map<String, Long> map = list.stream().collect(
        Collectors.groupingBy(Company::getCompany, 
                              Collectors.mapping((Company c) -> c, Collectors.counting())));

Now it becomes much easier to perform other operations as you desire. Also, here instead of creating 8 maps you end up dealing with just 1 list .

Answer 2

It's really difficult to grouping and counting a map because your map data will be changed after you increase your counter value. Therefore, you must save the original data of the map, and save your counter value to the another map. Join 2 maps after your counting process is complete.

Map<Map<String, Object>, Long> counterData = listBeforeGroup.stream().collect(Collectors.groupingBy(m -> m, Collectors.counting()));

List<Map<String, Object>> listAfterGroup = new ArrayList<>();
for (Map<String, Object> m : counterData.keySet()) {
    Map<String, Object> newMap = new HashMap<>(m);
    newMap.put("count", counterData.get(m));
    listAfterGroup.add(newMap);
}

Update Java 8 approach, not easy to debug

List<Map<String, Object>> listAfterGroup = listBeforeGroup.stream().collect(Collectors.groupingBy(m -> m, Collectors.counting())).entrySet().stream().map(e -> {
    Map<String, Object> newMap = e.getKey();
    newMap.put("count", e.getValue());
    return newMap;
}).collect(Collectors.toList());

Answer 3

My approach with java 8:

    Function<Map<String, Object>, String> createHashKey = map ->
                     map.values().stream()
                    .map(val -> String.valueOf(val))
                    .collect(Collectors.joining());

    BinaryOperator<Map<String, Object>> mergeDuplicate = (map1, map2) -> {
        int incrementedCount = (int)map1.get("count") + 1;
        map1.put("count", incrementedCount);
        return map1;
    };

    List<Map<String, Object>> listAfterGroup = listBeforeGroup.stream()
            .collect(Collectors.toMap(createHashKey, el -> {
                        el.put("count", 1);
                        return el;
            },mergeDuplicate))
          .values().stream()
          .collect(toList());

Maybe not the most succinct solution but I think quite readable and easy to follow the logic of the code.