需要帮助使字符串输入返回计算字母对频率的直方图

Question

我需要编写一个函数，它将字符串作为输入并按降序输出字母对频率的直方图。

我设法编写了一个返回单个字符计数的函数，但我不知道如何进行配对。

这是我目前所拥有的：

var string = "etc";
var histogram = {};

for (var i = 0, len = string.length; i < len; i++) {
    var char = string[i];
    if ((string[i] !== " ") && (string[i] !== " ")) {
        histogram[char] = (histogram[char] || 0) + 1;
    }
};

console.log(histogram);

该功能有效，我能够让它将所有空白区域都留在直方图中。不过，我不知道从这里去哪里。如何让它评估对，以及如何遗漏任何单个字符（那些后面没有另一个字符的字符）......非常感谢任何帮助。

Answer 1

另一种方法，将问题分解成小块：

const inPairs = (xs) => 
  [...xs].reduce((a, x, i) => i == 0 ? a : [...a, xs[i - 1] + x], [])

const pairFreq = str => str         //   "this is a good thing"
  .split (/\s+/)                    //=> ["this","is","a","good","thing"]
  .filter (s => s.length > 1)       //=> ["this","is","good","thing"]
  .flatMap (inPairs)                //=> ["th","hi","is","is","go","oo","od","th","hi","in","ng"]
  .reduce ( (a, s) => ({...a, [s]: (a[s] || 0) + 1}), {})
                                    //=> {"th":2,"hi":2,"is":2,"go":1,"oo":1,"od":1,"in":1,"ng":1}

console .log (
  pairFreq('this is a good thing')
)

如果您愿意，显然您可以内联inPairs。我喜欢这种转变方式，简单地将步骤串联起来，使我朝着最终目标前进。

Answer 2

您需要使用string[i+1] 来获取配对中的下一个字母。

并且为避免在使用string[i+1]时访问数组外，循环的限制应为string.length-1。

var string = "this is a good thing";
var histogram = {};

for (var i = 0, len = string.length - 1; i < len; i++) {
  if ((string[i] !== " ") && (string[i + 1] !== " ")) {
    let pair = string.substr(i, 2);
    histogram[pair] = (histogram[pair] || 0) + 1;
  }
};

console.log(histogram);