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

Question

ArrayList<Map<String, String>> result1

result1就好像

(1, a)

(2, a)

(3, b)

(4, e)

(5, e)

ArrayList<Map<String, String>> result2

result2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())&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .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 dataList<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 dataList<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 dataprivate static Map<String, String> getMapData(String k, String v) {&nbsp; &nbsp; Map<String, String> m = new HashMap<>();&nbsp; &nbsp; m.put(k, v);&nbsp; &nbsp; 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 listfor (Map<String, String> m : list1) {&nbsp; &nbsp; for (Map.Entry<String, String> entry : m.entrySet()) {&nbsp; &nbsp; &nbsp; &nbsp; List<String> valueList = new ArrayList<>();&nbsp; &nbsp; &nbsp; &nbsp; valueList.add(entry.getValue());&nbsp; &nbsp; &nbsp; &nbsp; result.put(entry.getKey(), valueList);&nbsp; &nbsp; }}// process the second list; merge with the firstfor (Map<String, String> m : list2) {&nbsp; &nbsp; for (Map.Entry<String, String> entry : m.entrySet()) {&nbsp; &nbsp; &nbsp; &nbsp; String k = entry.getKey();&nbsp; &nbsp; &nbsp; &nbsp; List<String> valueList = result.get(k);&nbsp; &nbsp; &nbsp; &nbsp; if (valueList == null) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; valueList = new ArrayList<>();&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; valueList.add(entry.getValue());&nbsp; &nbsp; &nbsp; &nbsp; result.put(k, valueList);&nbsp; &nbsp; }}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 8result1.forEach((k1, v1) ->&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; mergedResult.put(k1, new HashMap<String, String>() {{&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; put(v1, result2.get(k1));&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }}));Java 9 或更高版本result1.forEach((k1, v1) -> mergedResult.put(k1,&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Map.of(v1, result2.get(k1))));