Question

我正在尝试创建一个函数，以查看数组中的任何字符是否在字符串中，如果是，则多少个字符。

我尝试计算每种模式，但是太多了。我尝试使用Python中“ in”运算符的替代方法，但效果不佳

function calc_fit(element) {
  var fitness_let = ["e", "l", "m", "n", "t"]

  }
}

element是字符串，而Fitness_let数组是我需要检查的东西数组，以查看它们是否在字符串中，如果是，则检查多少。

Answer 1

您可以使用 map 和 filter 来计数具有相同数组值的出现次数：

let str="I love JavaScript and Node.js ";
let arr=str.replace(/[^a-zA-Z]/g, '').split('');

const mapped = [...new Set(arr)].map(a => `${a} occurs ${arr.filter(a1 => a1 === a).length  } time(s)`);
console.log(mapped);

Answer 2

首先，为了概括该方法，最好将calc_fit()也采用字母数组作为参数。然后，您可以从数组创建一个Map，每个字母的计数器从0开始。最后，遍历字符串并在需要时增加每个字母的相应计数器。

function calc_fit(element, fitness_let)
{
    // Create a Map from the array of letters to search.
    let map = new Map(fitness_let.map(l => ([l, 0])));

    // Traverse the string and increment counter of letters.    
    for (const c of element)
    {
        if (map.has(c))
            map.set(c, map.get(c) + 1);
    }
    
    return map;
}

let res = calc_fit("This is a string with some letters", ["e","l","m","n","t"]);
res.forEach((counter, letter) => console.log(`${letter} => ${counter}`));

.as-console {background-color:black !important; color:lime;}
.as-console-wrapper {max-height:100% !important; top:0;}

Answer 3

一种方法是迭代数组，并对每个字母进行全局正则表达式删除。然后，将替换后的字符串长度与原始输入长度进行比较，以确定出现的次数。

function calc_fit(element) {
var fitness_let = ["e", "l", "m", "n", "t"];

for (var i=0; i < fitness_let.length; i++) {
    var letter = fitness_let[i];
    var numtimes = element.length - element.replace(new RegExp(letter, 'g'), '').length;
    console.log(fitness_let[i] + " occurs: " + numtimes + " times.");
}

}

var input = "elements are elemental";
calc_fit(input);

Answer 4

如果您想获得否。每个角色出现的次数比您可以使用reduce和Map

let getTotal = (element) => {
  let fitness = ["e", "l", "m", "n", "t"]
  let newMap = new Map(fitness.map(v=>[v,v]))
  return element.split('').reduce((op,inp)=>{
    if(newMap.has(inp)){
      op[inp] = op[inp] || 0
      op[inp]++
    }
    return op
  },{})
}

console.log(getTotal('element'))
console.log(getTotal('eleabc'))

您可以使用join来构建带有交替|和单词边界的正则表达式，以获取总数

let getTotal = (element) =>{
  let fitness = ["e", "l", "m", "n", "t"]
  let reg = '\\b' + fitness.join('|') + '\\b'
  let pattern = new RegExp(reg,'gi')
  return (element.match(pattern) || []).length
}

console.log(getTotal('element'))
console.log(getTotal('eleabc'))

Answer 5

您可以创建一个哈希映射并利用reduce来计数找到的所有实例

示例：

const counts = {};
["e", "l", "m", "n", "t"].forEach( e => counts[e] = 0 );
const letters = "hey look! a string!".split("");
const results = letters.reduce( (acc, curr) => {
	if (acc.hasOwnProperty(curr)) { acc[curr] += 1; }
  return acc;
}, counts);


console.log(results);

Answer 6

这里有一个稍微不同的方法，它依赖于函数生成器。

在其他解决方案上使用此功能没有相关的原因，但是它可以在整个周期内提供额外的控制。

请注意，字符串仅迭代一次，因此整个“迭代”周期应该非常快。

说明直接在下面的代码中。

输出是一个对象，其中每个键都是一个字符，并保留出现的次数，并保留所有搜索到的指针。

如果未传递搜索的字符数组，它将自动在calc_fit函数内部构建，这将分别返回大写和小写的出现，包括标点和符号。不过，这很容易自定义。

// Iterates a string and yield the current looped char if it exists in the chars list.
function* matchChars(s, chars) {
  for (var char of s) {
    if (chars.indexOf(char) > -1) yield { char: char };
  }
}

// not sure why the function was named in this way in the original code, but hey, it's the OP's function name.
function calc_fit(element, chars) {
  chars = chars || [...new Set(element)];
  // builds an object from the above array, where the structure has the key which is the char, and the value which is initially 0.
  const matchList = chars.reduce((acc,next) => (acc[next] = 0, acc), {});
  // Iterates all the matches. For each match, it increments the amount of matches of matchList.
  for (var match of matchChars(element, chars)) matchList[match.char]++;
  // finally, returns matchList.
  return matchList;
}

// assertions: should match all the characters.
console.log(calc_fit('element', ["e", "l", "m", "n", "t"]));
// assertions: should return all zeros.
console.log(calc_fit('', ["e", "l", "m", "n", "t"]));
// assertions: should automatically detect chars, even upper case and lower case.
console.log(calc_fit('hello, world. ThIs is beatiful!'));

如何计算字符串中特定字符的次数

6 个答案: