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[英]Elegant way to count how many pieces of each type are present in a position
[英]the quickest way to Implement an algorithm to count how many times each string is present in an array of strings using Java
只是想知道实现一种算法以使用Java计算字符串数组中每个字符串存在多少次的最快方法是什么?
这是我尝试过的方法,并且有效,但是我担心它可能会“作弊”,因为它偏离了这个问题:
{
String[] stringArray = {"Random", "Words", "Here","Random", "Words", "Here","Random", "Words", "Here","Random", "Words", "Here"};
List asList = Arrays.asList(stringArray);
Set<String> mySet = new HashSet<>(asList);
mySet.stream().forEach((s) -> {
System.out.println(s + " " +Collections.frequency(asList,s));
});
}
最简单的方法是使用Map#merge()
:
Map<String, Integer> m = new HashMap<>();
for (String s : array)
m.merge(s, 1, Integer::sum);
之后, m
将把字符串作为键,将出现的值作为值:
m.forEach((k, v) -> System.out.format("%s occured %s time(s)\n", k, v));
通过在流中使用收集器:
Arrays.stream(list).collect(Collectors.groupingBy(e -> e, Collectors.counting()))
因此,如果您有这样的事情:
String[] list = new String[4];
list[0] = "something";
list[1] = "gfddfgdfg";
list[2] = "something";
list[3] = "somet444hing";
System.out.println(Arrays.stream(list).collect(Collectors.groupingBy(e -> e, Collectors.counting())));
输出将是:
{gfddfgdfg=1, something=2, somet444hing=1}
我将使用groupingBy返回计数图。
Map<String, Long> counts = Stream.of(array)
.collect(Collectors.groupingBy(w -> w, Collectors.counting()));
也可以打印这些
Stream.of(array)
.collect(Collectors.groupingBy(w -> w, Collectors.counting()))
.forEach((k, v) -> System.out.println(k + " occurred " + v " times));
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