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[英]What would be the best way to define a Map<String, Map<String, String>> map in Java?
[英]What's the best way to sum two Map<String,String>?
我有以下地图。
Map<String,String> map1= new HashMap<String, String>(){{
put("no1","123"); put("no2","5434"); put("no5","234");}};
Map<String,String> map1 = new HashMap<String, String>(){{
put("no1","523"); put("no2","234"); put("no3","234");}};
sum(map1, map2);
我想将它们加入到一个中,将类似的键值一起总结。 什么;使用java 7或guava库我能做到的最好的方法是什么?
预期产出
Map<String, String> output = { { "no1" ,"646"}, { "no2", "5668"}, {"no5","234"}, {"no3","234" } }
private static Map<String, String> sum(Map<String, String> map1, Map<String, String> map2) {
Map<String, String> result = new HashMap<String, String>();
result.putAll(map1);
for (String key : map2.keySet()) {
String value = result.get(key);
if (value != null) {
Integer newValue = Integer.valueOf(value) + Integer.valueOf(map2.get(key));
result.put(key, newValue.toString());
} else {
result.put(key, map2.get(key));
}
}
return result;
}
尝试这个
Map<String, List<String>> map3 = new HashMap<String, List<String>>();
for (Entry<String, String> e : map1.entrySet()) {
List<String> list = new ArrayList<String>();
list.add(e.getValue());
String v2 = map2.remove(e.getKey());
if (v2 != null) {
list.add(v2);
}
map3.put(e.getKey(), list);
}
for (Entry<String, String> e : map2.entrySet()) {
map3.put(e.getKey(), new ArrayList<String>(Arrays.asList(e.getValue())));
}
Java 8引入了Map.merge(K,V,BiFunction) ,这使得这很容易,如果不是特别简洁:
Map<String, String> result = new HashMap<>(map1);
//or just merge into map1 if mutating it is okay
map2.forEach((k, v) -> result.merge(k, v, (a, b) ->
Integer.toString(Integer.parseInt(a) + Integer.parseInt(b))));
如果你反复这样做,你将要解析并创建很多字符串。 如果你一次只生成一个地图,你最好保持一个字符串列表,只解析和求和一次。
Map<String, List<String>> deferredSum = new HashMap<>();
//for each map
mapN.forEach((k, v) ->
deferredSum.computeIfAbsent(k, x -> new ArrayList<String>()).add(v));
//when you're done
Map<String, String> result = new HashMap<>();
deferredSum.forEach((k, v) -> result.put(k,
Long.toString(v.stream().mapToInt(Integer::parseInt).sum())));
如果这个求和是一个常见的操作,请考虑使用Integer作为您的值类型是否更有意义; 在这种情况下,您可以使用Integer::sum
作为合并函数,并且不再需要维护延迟总和列表。
尝试这个
Map<String,String> map1= new HashMap<String, String>(){{
put("no1","123"); put("no2","5434"); put("no5","234");}};
Map<String,String> map2 = new HashMap<String, String>(){{
put("no1","523"); put("no2","234"); put("no3","234");}};
Map<String,String> newMap=map1;
for(String a:map2.keySet()){
if(newMap.keySet().contains(a)){
newMap.put(a,""+(Integer.parseInt(newMap.get(a))+Integer.parseInt(map2.get(a))));
}
else{
newMap.put(a,map2.get(a));
}
}
for(String k : newMap.keySet()){
System.out.println("key : "+ k + " value : " + newMap.get(k));
}
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