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分享海量数据查询匹配，有哪些高效的实现方案？
拿数据二中的value值替换成数值一的key值
数据一数据二根据id可以对应
Java实现，有哪些比较好的高效率做法呢
有没有大侠给点思考或者方案
数据一数据二根据id可以对应
Java实现，有哪些比较好的高效率做法呢
有没有大侠给点思考或者方案
数据一：
[
{
"id": "1000001",
"a": "aValue",
"b": "bValue",
"c": "cValue"
},
{
"id": "1000002",
"e": "eValue",
"f": "fValue",
"g": "gValue",
"h": "hvalue"
}
]
---------------------------------------
数据二：
[
{
"id": "1000001",
"a": "A",
"b": "B",
"c": "C"
},
{
"id": "1000002",
"e": "E",
"f": "F",
"g": "G",
"h": "H"
}
]
---------------------------------------
结果：
[
{
"id": "1000001",
"A": "aValue",
"B": "bValue",
"C": "cValue"
},
{
"id": "1000002",
"E": "eValue",
"F": "fValue",
"G": "gValue",
"H": "hvalue"
}
]
...全文
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sinat_27054329 2020-03-12
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多谢大佬
戎码一生灬 2020-03-11
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算了，还是给你写完吧！
public static void main(String[] args) {
//准备数据
List<Map<String, String>> data1 = new ArrayList<>();
Map<String, String> data1Map = new HashMap<>();
data1Map.put("id", "1000001");
data1Map.put("a", "aValue");
data1Map.put("b", "bValue");
data1Map.put("c", "cValue");
data1.add(data1Map);
Map<String, String> data2Map = new HashMap<>();
data2Map.put("id", "1000002");
data2Map.put("e", "eValue");
data2Map.put("f", "fValue");
data2Map.put("g", "gValue");
data2Map.put("h", "hValue");
data1.add(data2Map);
List<Map<String, String>> data2 = new ArrayList<>();
Map<String, String> data3Map = new HashMap<>();
data3Map.put("id", "1000001");
data3Map.put("a", "A");
data3Map.put("b", "B");
data3Map.put("c", "C");
data2.add(data3Map);
Map<String, String> data4Map = new HashMap<>();
data4Map.put("id", "1000002");
data4Map.put("e", "E");
data4Map.put("f", "F");
data4Map.put("g", "G");
data4Map.put("h", "H");
data2.add(data4Map);
//开始操作
//分组数据2
Map<String, List<Map<String, String>>> groupIdMap = data2.stream().collect(Collectors.groupingBy(e -> e.get("id")));
//遍历数据1
List<Map<String, String>> result = new ArrayList<>();
data1.stream().forEach(e -> {
Map<String, String> item = groupIdMap.get(e.get("id")).get(0);//数据2的item
List<String> newKeys = new ArrayList<>();
item.values().stream().forEach(v -> {
newKeys.add(v);
});
List<String> newValues = new ArrayList<>();
e.forEach((k, v) -> newValues.add(v));
Map<String, String> resultMap = new HashMap<>();
for (int i = 0; i < newKeys.size(); i++) {
resultMap.put(newKeys.get(i), newValues.get(i));
}
result.add(resultMap);
});
System.out.println(result);
}
public static void main(String[] args) {
//准备数据
List<Map<String, String>> data1 = new ArrayList<>();
Map<String, String> data1Map = new HashMap<>();
data1Map.put("id", "1000001");
data1Map.put("a", "aValue");
data1Map.put("b", "bValue");
data1Map.put("c", "cValue");
data1.add(data1Map);
Map<String, String> data2Map = new HashMap<>();
data2Map.put("id", "1000002");
data2Map.put("e", "eValue");
data2Map.put("f", "fValue");
data2Map.put("g", "gValue");
data2Map.put("h", "hValue");
data1.add(data2Map);
List<Map<String, String>> data2 = new ArrayList<>();
Map<String, String> data3Map = new HashMap<>();
data3Map.put("id", "1000001");
data3Map.put("a", "A");
data3Map.put("b", "B");
data3Map.put("c", "C");
data2.add(data3Map);
Map<String, String> data4Map = new HashMap<>();
data4Map.put("id", "1000002");
data4Map.put("e", "E");
data4Map.put("f", "F");
data4Map.put("g", "G");
data4Map.put("h", "H");
data2.add(data4Map);
//开始操作
//分组数据2
Map<String, List<Map<String, String>>> groupIdMap = data2.stream().collect(Collectors.groupingBy(e -> e.get("id")));
//遍历数据1
List<Map<String, String>> result = new ArrayList<>();
data1.stream().forEach(e -> {
Map<String, String> item = groupIdMap.get(e.get("id")).get(0);//数据2的item
List<String> newKeys = new ArrayList<>();
item.values().stream().forEach(v -> {
newKeys.add(v);
});
List<String> newValues = new ArrayList<>();
e.forEach((k, v) -> newValues.add(v));
Map<String, String> resultMap = new HashMap<>();
for (int i = 0; i < newKeys.size(); i++) {
resultMap.put(newKeys.get(i), newValues.get(i));
}
result.add(resultMap);
});
System.out.println(result);
}
戎码一生灬 2020-03-11
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大佬这是伪代码么，结果不对
[/quote]写了一半，急于下班，就懒得写了，思路就是这个思路，你参照一下
sinat_27054329 2020-03-11
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大佬这是伪代码么，结果不对
戎码一生灬 2020-03-10
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public static void main(String[] args) {
//准备数据
List<Map<String, String>> data1 = new ArrayList<>();
Map<String, String> data1Map = new HashMap<>();
data1Map.put("id", "1000001");
data1Map.put("a", "aValue");
data1Map.put("b", "bValue");
data1Map.put("c", "cValue");
data1.add(data1Map);
Map<String, String> data2Map = new HashMap<>();
data2Map.put("id", "1000002");
data2Map.put("e", "eValue");
data2Map.put("f", "fValue");
data2Map.put("g", "gValue");
data2Map.put("h", "hValue");
data1.add(data2Map);
List<Map<String, String>> data2 = new ArrayList<>();
Map<String, String> data3Map = new HashMap<>();
data3Map.put("id", "1000001");
data3Map.put("a", "A");
data3Map.put("b", "B");
data3Map.put("c", "C");
data2.add(data3Map);
Map<String, String> data4Map = new HashMap<>();
data4Map.put("id", "1000002");
data4Map.put("e", "E");
data4Map.put("f", "F");
data4Map.put("g", "G");
data4Map.put("h", "H");
data2.add(data4Map);
//开始操作
//分组数据2
Map<String, List<Map<String, String>>> groupIdMap = data2.stream().collect(Collectors.groupingBy(e -> e.get("id")));
//遍历数据1
List<Map<String, String>> result = new ArrayList<>();
data1.stream().forEach(e -> {
Map<String, String> item = groupIdMap.get(e.get("id")).get(0);
Map<String, String> resultMap = new HashMap<>();
item.values().stream().forEach(v -> {
resultMap.put(v, "id");
});
});
}
//准备数据
List<Map<String, String>> data1 = new ArrayList<>();
Map<String, String> data1Map = new HashMap<>();
data1Map.put("id", "1000001");
data1Map.put("a", "aValue");
data1Map.put("b", "bValue");
data1Map.put("c", "cValue");
data1.add(data1Map);
Map<String, String> data2Map = new HashMap<>();
data2Map.put("id", "1000002");
data2Map.put("e", "eValue");
data2Map.put("f", "fValue");
data2Map.put("g", "gValue");
data2Map.put("h", "hValue");
data1.add(data2Map);
List<Map<String, String>> data2 = new ArrayList<>();
Map<String, String> data3Map = new HashMap<>();
data3Map.put("id", "1000001");
data3Map.put("a", "A");
data3Map.put("b", "B");
data3Map.put("c", "C");
data2.add(data3Map);
Map<String, String> data4Map = new HashMap<>();
data4Map.put("id", "1000002");
data4Map.put("e", "E");
data4Map.put("f", "F");
data4Map.put("g", "G");
data4Map.put("h", "H");
data2.add(data4Map);
//开始操作
//分组数据2
Map<String, List<Map<String, String>>> groupIdMap = data2.stream().collect(Collectors.groupingBy(e -> e.get("id")));
//遍历数据1
List<Map<String, String>> result = new ArrayList<>();
data1.stream().forEach(e -> {
Map<String, String> item = groupIdMap.get(e.get("id")).get(0);
Map<String, String> resultMap = new HashMap<>();
item.values().stream().forEach(v -> {
resultMap.put(v, "id");
});
});
}
似水流年风萧兮 2020-03-09
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关键两组数据读出来怎么组装？一条条循环遍历么？有没有高效的方法[/quote]
都是需要循环的，你要通过降低循环的时间复杂度来提高效率
sinat_27054329 2020-03-09
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有没有大佬知道
sinat_27054329 2020-03-08
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关键两组数据读出来怎么组装？一条条循环遍历么？有没有高效的方法
似水流年风萧兮 2020-03-08
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数据量大的时候，把这两个数据分别存入数据库，然后再读出来组装，数据量小则自己写一个解析类
