[英]How to Convert a Map<String, List<String>> to Map<String, String> in java 8
[英]How to Convert a Map<String, List<String>> to Map<String, String> in java 8 functional APIs
我有一張下面的地圖,
[key = "car", value = ["bmw", "toyota"]]
[key = "bike", value = ["honda", "kawasaki"]]
我想使用以下Java 8功能性API將其轉換為另一張地圖,
[key = "bmw", value = "car"]
[key = "toyota", value = "car"]
[key = "honda", value = "bike"]
[key = "kawasaki", value = "bike"]
將地圖值展平為條目,然后收集它們:
Map<String, String> m2 = map
.entrySet()
.stream()
.flatMap(e -> e.getValue().stream().map(v -> new AbstractMap.SimpleEntry<>(v, e.getKey())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
可以通過導入AbstractMap.SimpleEntry
和Map.Entry
來縮短此時間。
不需要構造臨時Map.Entry
實例的解決方案是:
Map<String, String> result = source.entrySet().stream()
.collect(HashMap::new, (m,e)->e.getValue().forEach(k->m.put(k,e.getKey())), Map::putAll);
您可能會注意到與非流解決方案的相似之處
Map<String, String> result = new HashMap<>();
source.forEach((key, value) -> value.forEach(k -> result.put(k, key)));
或Java 8之前的解決方案
Map<String, String> result = new HashMap<>();
for(Map.Entry<String,List<String>> e: source.entrySet())
for(String key: e.getValue()) result.put(key, e.getKey());
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