计算 Javascript 中字符串中每个字母的出现次数

Question

I have seen similar questions like this asked before, such as counting characters in a given string. However when it comes to comparing given string to the letters of the alphabet and returning an object with occurences such as:

const letters = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

const sampleString = "a bee";

const results = {
 a: 1,
 b: 1,
 c: 0,
 d: 0,
 e: 2,
 f: 0,
 ...
}

Answer 1

I like using Object.fromEntries for this:

 const sampleString = "a bee"; const result = Object.fromEntries(Array.from("abcdefghijklmnopqrstuvwxyz", ch => [ch, 0])); for (let ch of sampleString) if (ch in result) result[ch]++; console.log(result);

Answer 2

We can use Array.reduce() , to count letters in the sampleString.

 const letters = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"] const sampleString = "a bee"; const all = letters.reduce((acc, c) => (acc[c] = 0)); const result = [...sampleString].reduce((acc, c) => { if (c in acc) acc[c]++; return acc; }, letters.reduce((acc, c) => { acc[c] = 0; return acc; }, {}) ) console.log('Result:', result)

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Answer 3

Using Array.reduce and String.match may be an idea. So, for each letter of letters , use match (length) to determine the frequency of the letter in the given sample.

 const letters = `abcdefghijklmnopqrstuvwxyz`.split(``); const freq = (chr, sample) => (sample.match(RegExp(chr, `g`)) || []).length; const result = letters.reduce( (acc, chr) => ({...acc, [chr]: freq(chr, acc.sample)}), {sample: "a bee"}); console.log(result);

Answer 4

Or a more undestandable way

const letters = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
const sampleString = "a bee";
results = {};
sampleString.forEach(letter=>{
   if(letters.includes(letter)) {
      if(results[letter] === undefined) results[letter]=0;
      results[letter]++;
   }
});

Answer 5

making the value as a key is a bad practice, because it will be hard to read, I recommend to call it what it is:

const letters = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
const tempString = "This is a string.";
let results = []
letters.forEach( char =>{
  const count = countCharacterOccurences(tempString, char)
  results.push({ alphabet: char, count })
})

function countCharacterOccurences(string, char) {
   return string.split(char).length - 1;
}

console.log(results)

//filter alphabet with value
const alphabetWithCount = results.filter( result => {
   return result.count > 0
})
console.log(alphabetWithCount)

//log alphabet only with counts
const alphabetOnlyWithCounts = alphabetWithCounts.map( alphabetWithCount => {
    return alphabetWithCount.alphabet
})

console.log(alphabetOnlyWithCounts)

//

Answer 6

You can apply .reduce() on the letters directly and in each iteration use RegExp() to replace all letters that do not match the current letter and the count the remaining letters in the string, if any, with .length :

 const letters = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]; const sampleString = "a bee"; const results = letters.reduce((result,letter) => ({...result, [letter]: sampleString.replace(new RegExp(`[^${letter}]`,'g'), "").length }), {}); console.log( results );

NOTE :

Please note that:

sr.replace(new RegExp('[^a]','g'), "").length

for example, is equivalent to:

sr.replace(/[^a]/g,"").length

See: JavaScript: How many times a character occurs in a string?