Question

我有以下地图列表

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);

如何获得此输出？

    //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}

我使用Java 8流尝试了此操作，但无法获得所需的输出

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

并尝试了此方法，顺便说一句，此解决方案给出了一张地图，但我不希望这样做

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

例如，我想按“ billPeriod”键对地图进行分组，并按值对项目计数，然后生成新的地图列表。

Answer 1

您可以创建一个类sms:<phone number>, e.g. sms:5550101234 Send an SMS message to <phone number> using the default messaging app，随后的操作将变得更加简单。

Company

初始化列表：

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
}

现在举个例子，您可以按公司名称分组：

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"));

现在，执行所需的其他操作变得容易得多。此外，在这里您没有创建 8张地图，而只处理了 1个列表。

Answer 2

对地图进行分组和计数真的很困难，因为增加计数器值后，地图数据将被更改。因此，您必须保存地图的原始数据，并将计数器值保存到另一个地图。计数过程完成后，加入2张地图。

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);
}

更新Java 8方法，不容易调试

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

我使用Java 8的方法：

    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());

也许不是最简洁的解决方案，但我认为它非常易读并且易于遵循代码的逻辑。

Java 8 List <map <string，object =“” >>到List <map <string，object =“” >>按键分组并按值计数

3 个答案: