Javascript中的经典字数算法

Question

请..。伙计们，我哪里出问题了？

经典的字数算法:给定一个字符串数组，为每个不同的字符串返回一个带键的Map，值为该字符串在数组中出现的次数。

wordCount("a“、"b”、"a“、"c”、"b")→{"a"：2、"b"：2、"c"：1}

wordCount("c“、"b”、"a")→{"a"：1、"b"：1、"c"：1}

wordCount("c"，"c")→{"c"：4}

到目前为止我的代码

function wordCount(arrayOfStrings) {
    const map = {};

    const arr = arrayOfStrings;

    for (let i = 0; i < arr.length; i++) {

        let arr2 = arr.charAt(i);

        if (arr.indexOf(arr2) === arr.lastIndexOf(arr2)) {
            map.push({
                arr: arr2
            });
        }
    }
}

wordCount(["a", "b", "a", "c", "b"])

及以下是我要通过的考试

test(`Expect the wordCount of ["one", "fish", "two", "fish", "red", "fish", "blue", "fish"] to equal {one: 1, fish: 4, two: 1, red: 1, blue: 1}`, () => {
expect(wordCount([ 'one', 'fish', 'two', 'fish', 'red', 'fish', 'blue', 'fish' ])).toEqual({ one: 1, fish: 4, two: 1, red: 1, blue: 1 });
});

test(`Expect the wordCount of ["str", "hell", "str", "str"] to equal {str: 3, hell: 1}`, () => {
expect(wordCount([ 'str', 'hell', 'str', 'str' ])).toEqual({ str: 3, hell: 1 });
});

test(`Expect the wordCount of ["a", "b", "a", "c", "b"] to equal {"a": 2, "b": 2, "c": 1}`, () => {
expect(wordCount([ 'a', 'b', 'a', 'c', 'b' ])).toEqual({ a: 2, b: 2, c: 1 });
});

test(`Expect the wordCount of [1, "chair", "cane", "chair"] to equal {1: 1, chair: 2, cane: 1}`, () => {
expect(wordCount([ 1, 'chair', 'cane', 'chair' ])).toEqual({ 1: 1, chair: 2, cane: 1 });
});

test(`Expect the wordCount of ["ch", "chair", "cane", "chair", "ai", "ir"] to equal { ch: 1, chair: 2, cane: 1, ai: 1, ir: 1 }`, () => {
expect(wordCount([ 'ch', 'chair', 'cane', 'chair', 'ai', 'ir' ])).toEqual({ ch: 1, chair: 2, cane: 1, ai: 1, ir: 1 });
});

Answer 1

从目前来看，你的做法是完全错误的。您所需要做的就是将数组中的每个字符串添加为一个属性(如果它还不是一个属性)，如果是的话，则增加它的值。

function wordCount(arrayOfStrings) {
    const map = {};
    for (let i = 0; i < arrayOfStrings.length; ++i) {
      if (arrayOfStrings[i] in map)
        map[arrayOfStrings[i]]++;
      else
        map[arrayOfStrings[i]] = 1;
    }

    return map;
}

该代码检查数组中的每个字符串，看看它是否已经是正在构造的映射(一个普通对象)的属性。如果是，则值将递增；如果不是，则创建一个新属性并将其初始化为1。

使用.reduce()会更整洁一些

function wordCount(arr) {
  return arr.reduce(function(map, word) {
    if (word in map)
      map[word]++;
    else
      map[word] = 1;
    return map;
  }, {});
}

Answer 2

最简洁和最简单的方法是reduce

const wordCount = arr => arr.reduce((a, c) => ((a[c] = (a[c] || 0) + 1), a), {});

Answer 3

尝试这个(基于您的代码)：

function wordCount(arrayOfStrings) {
    const map = {};
    const arr = arrayOfStrings;

    for (let i = 0; i < arr.length; i++) {
        map[arr[i]] = (map[arr[i]] || 0) +1;
    }
    return map;
}