如何将两个 ArrayList<Map<String, String>> 合并为一个？

Question

ArrayList<Map<String, String>> result1result1就好像(1, a)(2, a)(3, b)(4, e)(5, e)ArrayList<Map<String, String>> result2result2就好像(1,android)(2,ios)(3,android)(4,android)(5,ios)我想合并两张地图来构建一张这样的地图(1, ( a, android))(2, ( a, ios))(3, ( b, android))(4, (e, android))(5, (e, ios))如何做到这一点？

Answer 1

您可以使用and合并两个流Stream.concat()并将它们分组：Collectors.groupingBy()Collectors.mapping()

Map<String, String> first = Map.of("1", "a", "2", "a");

Map<String, String> second = Map.of("1", "android", "2", "ios");

Map<String, List<String>> result = Stream.concat(first.entrySet().stream(), second.entrySet().stream())

            .collect(groupingBy(Entry::getKey, mapping(Entry::getValue, toList())));

System.out.println(result);

将输出：

{1=[a, android], 2=[a, ios]}

Answer 2

这是得出结果的一种方法：

输入数据：

// The first list of data

List<Map<String, String>> list1 = new ArrayList<>();

list1.add(getMapData("1", "a"));

list1.add(getMapData("2", "a"));

list1.add(getMapData("3", "b"));

list1.add(getMapData("4", "e"));

list1.add(getMapData("5", "e"));

list1.add(getMapData("999", "x"));

System.out.println(list1);

资料一：[{1=a}, {2=a}, {3=b}, {4=e}, {5=e}, {999=x}]

// The second list of data

List<Map<String, String>> list2 = new ArrayList<>();

list2.add(getMapData("1", "android"));

list2.add(getMapData("2", "ios"));

list2.add(getMapData("3", "android"));

list2.add(getMapData("4", "android"));

list2.add(getMapData("5", "ios"));

list2.add(getMapData("888", "zzzzz"));

System.out.println(list2);

数据2：[{1=android}, {2=ios}, {3=android}, {4=android}, {5=ios}, {888=zzzzz}]

// utility method for creating test data

private static Map<String, String> getMapData(String k, String v) {

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

    m.put(k, v);

    return m;

}

结果过程：

输出存储到Map<String, List<String>>：

Map<String, List<String>> result = new HashMap<>();

// process the first list

for (Map<String, String> m : list1) {

    for (Map.Entry<String, String> entry : m.entrySet()) {

        List<String> valueList = new ArrayList<>();

        valueList.add(entry.getValue());

        result.put(entry.getKey(), valueList);

    }

}

// process the second list; merge with the first

for (Map<String, String> m : list2) {

    for (Map.Entry<String, String> entry : m.entrySet()) {

        String k = entry.getKey();

        List<String> valueList = result.get(k);

        if (valueList == null) {

            valueList = new ArrayList<>();

        }

        valueList.add(entry.getValue());

        result.put(k, valueList);

    }

}

System.out.println(result);

结果：

{1=[a, android], 2=[a, ios], 3=[b, android], 4=[e, android], 5=[e, ios], 888=[zzzzz], 999=[x]}

Answer 3

您也可以尝试这种方法：

Map<String, String> result1 = new HashMap<>();

// initialize result1 ...

Map<String, String> result2 = new HashMap<>();

// initialize result2 ...

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

高达 Java 8

result1.forEach((k1, v1) ->

            mergedResult.put(k1, new HashMap<String, String>() {{

                put(v1, result2.get(k1));

            }}));

Java 9 或更高版本

result1.forEach((k1, v1) -> mergedResult.put(k1, 

                                 Map.of(v1, result2.get(k1))));