Java8:使用字符串的字符数创建 HashMap

问题描述

想知道有没有比如下计算给定字符串的字符数更简单的方法?

Wondering is there more simple way than computing the character count of a given string as below?

String word = "AAABBB";
    Map<String, Integer> charCount = new HashMap();
    for(String charr: word.split("")){
        Integer added = charCount.putIfAbsent(charr, 1);
        if(added != null)
            charCount.computeIfPresent(charr,(k,v) -> v+1);
    }

    System.out.println(charCount);

推荐答案

计算字符串中每个字符出现次数的最简单方法，完全支持 Unicode (Java 11+)¹:

Simplest way to count occurrence of each character in a string, with full Unicode support (Java 11+)¹:

String word = "AAABBB";
Map<String, Long> charCount = word.codePoints().mapToObj(Character::toString)
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
System.out.println(charCount);

^{1) 完全支持 Unicode 的 Java 8 版本在答案的末尾.}

输出

{A=3, B=3}

<小时>

更新:对于 Java 8+(不支持来自补充平面的字符，例如表情符号):

---

UPDATE: For Java 8+ (doesn't support characters from supplemental planes, e.g. emoji):

Map<String, Long> charCount = IntStream.range(0, word.length())
        .mapToObj(i -> word.substring(i, i + 1))
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

<小时>

更新 2: 也适用于 Java 8+.

---

UPDATE 2: Also for Java 8+.

我错了，以为 codePoints() 直到 Java 9 才添加.它是在 Java 8 中添加到 CharSequence 接口，因此 String 在 Java 8 中，并显示为 在 Java 9 中添加 用于更高版本的 javadoc.

I was mistaken, thinking that codePoints() wasn't added until Java 9. It was added in Java 8 to the CharSequence interface, so it doesn't show in javadoc for String in Java 8, and shows as added in Java 9 for later versions of the javadoc.

但是，Character.toString (int codePoint) 方法直到 Java 11 才被添加，所以要使用 Character.toString (char c) 方法，我们可以使用 chars() 在 Java 8 中:

However, the Character.toString​(int codePoint) method wasn't added until Java 11, so to use the Character.toString​(char c) method, we can use chars() in Java 8:

Map<String, Long> charCount = word.chars().mapToObj(c -> Character.toString((char) c))
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

或者对于完整的 Unicode 支持，包括.补充平面，我们可以使用 codePoints() 和 String(int[] codePoints, int offset, int count) 构造函数，在 Java 8 中:

Or for full Unicode support, incl. supplemental planes, we can use codePoints() and the String(int[] codePoints, int offset, int count) constructor, in Java 8:

Map<String, Long> charCount = word.codePoints()
        .mapToObj(cp -> new String(new int[] { cp }, 0, 1))
        .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

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