基於密鑰的 Map 過濾列表使用 Java 流

Question

有List<Map<String,String>>，如下：

[
 {"A": "ABC","B": "ABCD","C": "0.945"},
 {"A": "ABC","B": "AB","C": "0.871"},
 {"A": "XYZ","B": "XYZA","C": "0.765"},
 {"A": "ABC","B": "ABD","C": "0.945"},
 {"A": "XYZ","B": "XYZB","C": "0.563"}
]

問題

在列表上方過濾以獲得 Map，其中鍵為“A”，該“A”的最高值為“C”。

{"ABC": "0.945", "XYZ": "0.765"}

初始代碼

Map<String, String> m0 = new HashMap<>();
m0.put("A", "ABC");
m0.put("B", "ABCD");
m0.put("C", "0.945");

Map<String, String> m1 = new HashMap<>();
m1.put("A", "ABC");
m1.put("B", "AB");
m1.put("C", "0.871");

Map<String, String> m2 = new HashMap<>();
m2.put("A", "XYZ");
m2.put("B", "XYZA");
m2.put("C", "0.765");

Map<String, String> m3 = new HashMap<>();
m3.put("A", "ABC");
m3.put("B", "ABD");
m3.put("C", "0.945");

Map<String, String> m4 = new HashMap<>();
m4.put("A", "XYZ");
m4.put("B", "XYZB");
m4.put("C", "0.563");

List<Map<String,String>> res = Arrays.asList(m0,m1,m2,m3,m4);

我能夠得到：

[{A=ABC, B=ABCD, C=0.945}, {A=XYZ, B=XYZA, C=0.765}]

使用下面的代碼：

List<Map<String, String>> result = res.stream().collect(Collectors.groupingBy(map->map.get("A")))
                .entrySet().stream().map(m -> m.getValue().stream()
                        .max(Comparator.comparingDouble(s -> Double.parseDouble(s.get("C")))).get())
                .collect(Collectors.toList());

Answer 1

當您開始丟棄不需要的條目並僅保留A和C的值作為鍵/值對時，會更容易。

Map<String, Double> result = res.stream()
        .map(m -> Map.entry(m.get("A"), 
                            Double.parseDouble(m.get("C"))))
        .collect(Collectors.groupingBy(Entry::getKey,
                Collectors.mapping(Entry::getValue, 
                                   Collectors.maxBy(Double::compare))))
        .entrySet().stream()
        .collect(Collectors.toMap(Entry::getKey, e -> e.getValue().get()));

根據需要產生{ABC=0.945, XYZ=0.765}作為result ； 但使用兩個流來擺脫Optional<Double>類型的值。