Group by Length javascript [closed]

Question

I have a array a = ["one", "two", "three", "four", "five"] and I want to count length of each string in this array and return an object:

{
    3: ["one", "two"],
    4: ["four", "five"],
    5: ["three"]
}

How can I solve that? Thanks in advance

Answer 1

you can do it with reduce

const strings = ["one", "two", "three", "four", "five"]


const result = strings.reduce((res, s) => {
  const strLength = s.length
  const existing = res[strLength] || []

  return {
    ...res,
    [strLength]: [...existing, s]
  }

}, {})


console.log(result)

Answer 2

var a = ["one", "two", "three", "four", "five"]
const result = {}

for (let i = 0; i < a.length; i++) {
  const item = a[i]

  const length = item.length

  if (!result[length]) {
    result[length] = []
  }

  result[length].push(item)
}
console.log(result)

Answer 3

You can use Array.prototype.reduce to group the strings by their length.

const 
  strs = ["one", "two", "three", "four", "five"],
  result = strs.reduce((r, s) => {
    if (!r[s.length]) {
      r[s.length] = [];
    }
    r[s.length].push(s);
    return r;
  }, {});

console.log(result);

You can also do it more concisely, as shown below:

const 
  strs = ["one", "two", "three", "four", "five"],
  result = strs.reduce((r, s) => ((r[s.length] ??= []).push(s), r), {});

console.log(result);

Answer 4

const a = ["one", "two", "three", "four", "five"]

const result = a.reduce((acc, item) => (
  Object.keys(acc).find(key => key == item.length) 
    ?? (acc[item.length] = []), 
  acc[item.length].push(item), 
  acc
), {})

console.log(result)

Answer 5

const a = ["one", "two", "three", "four", "five"]
const b = {}
a.forEach(str => {
  if (b[str.length]) {
    b[str.length].push(str);
  }
  else {
    b[str.length] = [str];
  }
})
console.log(b)

output:

{
  3: ["one", "two"],
  4: ["four", "five"],
  5: ["three"]
}