java - java获取字符串中每个字符的相对频率

Question

我想获得字符串中每个唯一字符的平均值。我这里有一个例子，因为它会更容易说明。

字符串:

The big brown fox

每个字符(包括空格)的平均值:

T = 1/17 = .058
h = 1/17 = .058
e = 1/17 = .058
' '= 3/17 = .176
b = 2/17 = .117
i = 1/17 = .058
g = 1/17 = .058
r = 1/17 = .058
o = 2/17 = .117
w = 1/17 = .058
n = 1/17 = .058
f = 1/17 = .058
x = 1/17 = .058

到目前为止我所有的尝试都失败了，我想我的大脑现在不工作了。我该如何编码？任何帮助或意见将不胜感激。

我有这个代码作为解决方案。当我将代码复制粘贴到堆栈中时，我突然想到了这一点。我希望这不是重新发帖，因为我几分钟前刚刚回答了同样的问题，但它没有出现。

 Map<String, Integer> storeCharCount = new HashMap<String, Integer>();

        String a = "The big brown fox";


        for (int x=0; x<a.length(); x++){
            char getChar = a.charAt(x);
            String convGetChar = Character.toString(getChar);

            Integer countChar = storeCharCount.get(convGetChar);
            storeCharCount.put(convGetChar, (countChar==null?countChar=1:countChar+1));

        }
        System.out.println("Map: "+ storeCharCount);
        double RelFrequency = 0;
        for (Map.Entry<String, Integer> getValue: storeCharCount.entrySet()){

            RelFrequency = (double)(getValue.getValue())/(a.length());
            System.out.println("Character "+getValue.getKey() +"  Relative Frequency: "+RelFrequency);

        }

这是输出

Map: {f=1, g=1,  =3, e=1, b=2, n=1, o=2, h=1, i=1, w=1, T=1, r=1, x=1}
Character f  Relative Frequency: 0.058823529411764705
Character g  Relative Frequency: 0.058823529411764705
Character    Relative Frequency: 0.17647058823529413
Character e  Relative Frequency: 0.058823529411764705
Character b  Relative Frequency: 0.11764705882352941
Character n  Relative Frequency: 0.058823529411764705
Character o  Relative Frequency: 0.11764705882352941
Character h  Relative Frequency: 0.058823529411764705
Character i  Relative Frequency: 0.058823529411764705
Character w  Relative Frequency: 0.058823529411764705
Character T  Relative Frequency: 0.058823529411764705
Character r  Relative Frequency: 0.058823529411764705
Character x  Relative Frequency: 0.058823529411764705   

Answer 1

这是我的建议。首先迭代字符串中的每个字符，如果找到则更新频率，否则添加 1。然后再次迭代以打印结果:

        String s = "The big brown fox";
    Map<Character, Float> m = new TreeMap<Character, Float>();
    for (char c : s.toCharArray()) {
        if (m.containsKey(c))
            m.put(c, m.get(c) + 1);
        else
            m.put(c, 1f);
    }

    for (char c : s.toCharArray()) {
        float freq = m.get(c) / s.length();
        System.out.println(c + " " + freq);
    }