我有要求,我有地图列表
[{Men=1},{Men=2, Women=3},{Women=2,Boys=4}]
现在我需要将其设为 flatMap,使其看起来像
Gender=countOfValues
在上面的例子中,输出将是
{Men=3,Women=5,Boys=4}
目前我有以下代码:
private Map<String, Long> getGenderMap(
List<Map<String, Long>> genderToCountList) {
Map<String, Long> gendersMap = new HashMap<String, Long>();
Iterator<Map<String, Long>> genderToCountListIterator = genderToCountList
.iterator();
while (genderToCountListIterator.hasNext()) {
Map<String, Long> genderToCount = genderToCountListIterator.next();
Iterator<String> genderToCountIterator = genderToCount.keySet()
.iterator();
while (genderToCountIterator.hasNext()) {
String gender = genderToCountIterator.next();
if (gendersMap.containsKey(gender)) {
Long count = gendersMap.get(gender);
gendersMap.put(gender, count + genderToCount.get(gender));
} else {
gendersMap.put(gender, genderToCount.get(gender));
}
}
}
return gendersMap;
}
我们如何使用Java8使用lambda表达式来编写这段代码呢?
我不会为此使用任何 lambda,但我已经使用了Map.merge
和方法参考,两者都是在 Java 8 中引入的。
Map<String, Long> result = new HashMap<>();
for (Map<String, Long> map : genderToCountList)
for (Map.Entry<String, Long> entry : map.entrySet())
result.merge(entry.getKey(), entry.getValue(), Long::sum);
你也可以这样做Stream
s:
return genderToCountList.stream().flatMap(m -> m.entrySet().stream())
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, Long::sum));
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