Sort the array and organize it by frequency

Question

Sort the array and organize it by frequency

Given the input array:

 array = [ 1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20 ]

The expected output is:

 [ [ 1, 1, 1, 1 ], [ 2, 2, 2 ], 4, 5, 10, [ 20, 20 ], 391, 392, 591 ]

Using the following function to generate the result:

var array = [ 1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]

function answer(ArrayFromAbove) {
  var length = array.length;
  for (var i = 0; i < length; i++) {
    for (var j = 0; j < (length - i - 1); j++) {
      if (array[j] > array[j + 1]) {
        var tmp = array[j];
        array[j] = array[j + 1];
        array[j + 1] = tmp;

      }
    }
  }
}
answer(array);
console.log(array);

Expected return value:

[ [ 1, 1, 1, 1 ], [ 2, 2, 2 ], 4, 5, 10, [ 20, 20 ], 391, 392, 591 ]

javascript arrays

Answer 1

Answer №1

You may want to consider utilizing the reduce method instead. This involves tallying up the occurrences of each number, then cycling through the sorted entries and adding the value to the result array (either as an array if there are multiple values, or as a single number if there's only one):

const input = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20];
/* create an object like:
{
  "1": 4,
  "2": 3,
  "4": 1,
  "5": 1,
  "10": 1,
  "20": 2,
  "391": 1,
  "392": 1,
  "591": 1
} */
const inputCounts = input.reduce((a, num) => {
  a[num] = (a[num] || 0) + 1;
  return a;
}, {});

const output = Object.entries(inputCounts)
  // turn (string) key to number:
  .map(([key, val]) => [Number(key), val])
  .sort((a, b) => a[0] - b[0])
  .reduce((a, [num, count]) => {
    a.push(
      count === 1
      ? num
      : new Array(count).fill(num)
    );
    return a;
  }, []);
console.log(output);

Answer 2

You may want to consider utilizing the reduce method instead. This involves tallying up the occurrences of each number, then cycling through the sorted entries and adding the value to the result array (either as an array if there are multiple values, or as a single number if there's only one):

const input = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20];
/* create an object like:
{
  "1": 4,
  "2": 3,
  "4": 1,
  "5": 1,
  "10": 1,
  "20": 2,
  "391": 1,
  "392": 1,
  "591": 1
} */
const inputCounts = input.reduce((a, num) => {
  a[num] = (a[num] || 0) + 1;
  return a;
}, {});

const output = Object.entries(inputCounts)
  // turn (string) key to number:
  .map(([key, val]) => [Number(key), val])
  .sort((a, b) => a[0] - b[0])
  .reduce((a, [num, count]) => {
    a.push(
      count === 1
      ? num
      : new Array(count).fill(num)
    );
    return a;
  }, []);
console.log(output);

Answer 3

Answer №2

To start with, eliminate duplicates in an array using the Set method and then sort the array using

<a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort" rel="nofollow noreferrer">Array.prototype.sort()</a>

. After that, determine the count of each number in the array and populate a new Array with it before appending it to the output.

const array=[1,3,5,672,457,456,3,6,11,3,1,1,1,29,29]
function countNumbers(arr,num){
  return Array(arr.filter(x => x === num).length).fill(num);
}
const uniqueNums = [...new Set(array)].sort((a,b) => a - b);
const finalOutput = uniqueNums.map(uniqueNum => {
  let arr = countNumbers(array,uniqueNum);
  return (arr.length > 1) ? arr : uniqueNum;
})
console.log(finalOutput)

The above method involves looping through the array multiple times. You can achieve the same result by looping through the array only twice as demonstrated below:

let numbers = [1,3,5,672,457,456,3,6,11,3,1,1,1,29,29]
numbers = numbers.sort((a,b) => a-b);
const finalResult = [];
let tempArr = [];
for(let j = 0; j < numbers.length + 1; ++j){
  if(numbers[j - 1] === numbers[j] || j === 0){
    tempArr.push(numbers[j]);
  }
  else{
    finalResult.push((tempArr.length === 1) ? tempArr[0] : tempArr);
    tempArr = [numbers[j]];
  }
}
//finalResult.push(tempArr)
console.log(finalResult);

Answer 4

To start with, eliminate duplicates in an array using the Set method and then sort the array using

<a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort" rel="nofollow noreferrer">Array.prototype.sort()</a>

. After that, determine the count of each number in the array and populate a new Array with it before appending it to the output.

const array=[1,3,5,672,457,456,3,6,11,3,1,1,1,29,29]
function countNumbers(arr,num){
  return Array(arr.filter(x => x === num).length).fill(num);
}
const uniqueNums = [...new Set(array)].sort((a,b) => a - b);
const finalOutput = uniqueNums.map(uniqueNum => {
  let arr = countNumbers(array,uniqueNum);
  return (arr.length > 1) ? arr : uniqueNum;
})
console.log(finalOutput)

The above method involves looping through the array multiple times. You can achieve the same result by looping through the array only twice as demonstrated below:

let numbers = [1,3,5,672,457,456,3,6,11,3,1,1,1,29,29]
numbers = numbers.sort((a,b) => a-b);
const finalResult = [];
let tempArr = [];
for(let j = 0; j < numbers.length + 1; ++j){
  if(numbers[j - 1] === numbers[j] || j === 0){
    tempArr.push(numbers[j]);
  }
  else{
    finalResult.push((tempArr.length === 1) ? tempArr[0] : tempArr);
    tempArr = [numbers[j]];
  }
}
//finalResult.push(tempArr)
console.log(finalResult);

Answer 5

Answer №3

To determine the number of times each element appears, you can tally up the occurrences and organize them into arrays:

 const frequency = new Map();

 for(const item of elements.sort())
   frequency.set(item, (frequency.get(item) || 0) + 1);

const output = [];

for(const [count, item] of frequency.entries())
  output.push(new Array(count).fill(item));

Answer 6

To determine the number of times each element appears, you can tally up the occurrences and organize them into arrays:

 const frequency = new Map();

 for(const item of elements.sort())
   frequency.set(item, (frequency.get(item) || 0) + 1);

const output = [];

for(const [count, item] of frequency.entries())
  output.push(new Array(count).fill(item));

Answer 7

Answer №4

To populate an array based on the number of occurrences of each unique number, you can utilize the reduce method and then fill the array accordingly.

var array = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]

let reduced = array.reduce((op,inp) => {
    op[inp] ? op[inp]++ : (op[inp]=1);
    return op;
}, {})

let op = Object.keys(reduced).map(e => 
    (reduced[e] === 1 ? e : new Array(reduced[e]).fill(e)))
        .sort((a,b) => 
            (Array.isArray(a) ? a[0] : a) - (Array.isArray(b) ? b[0] : b))

console.log(op)

Answer 8

To populate an array based on the number of occurrences of each unique number, you can utilize the reduce method and then fill the array accordingly.

var array = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]

let reduced = array.reduce((op,inp) => {
    op[inp] ? op[inp]++ : (op[inp]=1);
    return op;
}, {})

let op = Object.keys(reduced).map(e => 
    (reduced[e] === 1 ? e : new Array(reduced[e]).fill(e)))
        .sort((a,b) => 
            (Array.isArray(a) ? a[0] : a) - (Array.isArray(b) ? b[0] : b))

console.log(op)

Answer 9

Answer №5

First, organize the array by grouping all numbers into sub-arrays. Then, replace those sub-arrays with a single element representing each.

var nums = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]
function reorganize(arr){
arr.sort((a,b) => a - b);
var tempResult = [];
while(arr.length > 0){
var val = arr.shift();
if(tempResult.length == 0 || tempResult[tempResult.length-1][0] !== val){
tempResult.push([val]);
        }else{
tempResult[tempResult.length-1].push(val);
        }
    }

while(tempResult.length > 0){
var subArr = tempResult.shift();
if(subArr.length == 1) arr.push(subArr[0]);
else arr.push(subArr);
    }

}
reorganize(nums);
console.log(JSON.stringify(nums));

Answer 10

First, organize the array by grouping all numbers into sub-arrays. Then, replace those sub-arrays with a single element representing each.

var nums = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]
function reorganize(arr){
arr.sort((a,b) => a - b);
var tempResult = [];
while(arr.length > 0){
var val = arr.shift();
if(tempResult.length == 0 || tempResult[tempResult.length-1][0] !== val){
tempResult.push([val]);
        }else{
tempResult[tempResult.length-1].push(val);
        }
    }

while(tempResult.length > 0){
var subArr = tempResult.shift();
if(subArr.length == 1) arr.push(subArr[0]);
else arr.push(subArr);
    }

}
reorganize(nums);
console.log(JSON.stringify(nums));

Answer 11

Answer №6

If you want to achieve this, here is what you can do:

Start by sorting the array using Array.sort
Next, utilize Array.reduce to generate the final array
- For the initial value, simply add it to the resulting array
- For all subsequent values, check if the last value in the resulting array is an array
- If the last value is an array, compare the first value with the current one
  - If they match, append it to the existing array
  - If not, add it as a new entry in the main resulting array
- If the last value is not an array, compare it with the current value
  - If they match, replace the element in the resulting array with an array of 2 identical values
  - If not, add it as a new entry in the main resulting array

let array = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];
let result = array.sort((a,b) => a-b).reduce((a,c) => {
  if(a.length) {
    let lastVal = a[a.length-1];
    if(Array.isArray(lastVal)) {
      if(lastVal[0] == c) lastVal.push(c);
      else a.push(c);
    } else if(lastVal == c) a[a.length-1] = [c,c];
    else a.push(c);
  }
  else a.push(c);
  return a;
}, []);

console.log(result);

Answer 12

If you want to achieve this, here is what you can do:

Start by sorting the array using Array.sort
Next, utilize Array.reduce to generate the final array
- For the initial value, simply add it to the resulting array
- For all subsequent values, check if the last value in the resulting array is an array
- If the last value is an array, compare the first value with the current one
  - If they match, append it to the existing array
  - If not, add it as a new entry in the main resulting array
- If the last value is not an array, compare it with the current value
  - If they match, replace the element in the resulting array with an array of 2 identical values
  - If not, add it as a new entry in the main resulting array

let array = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];
let result = array.sort((a,b) => a-b).reduce((a,c) => {
  if(a.length) {
    let lastVal = a[a.length-1];
    if(Array.isArray(lastVal)) {
      if(lastVal[0] == c) lastVal.push(c);
      else a.push(c);
    } else if(lastVal == c) a[a.length-1] = [c,c];
    else a.push(c);
  }
  else a.push(c);
  return a;
}, []);

console.log(result);

Answer 13

Answer №7

Below is my own take on the problem:

var data = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]

function rearrangeArray(input) {
  return input.sort((a,b) => a - b)
    .reduce((collection, nextItem) => {
      // First item, simply add it
      if (!collection.length) {
        collection.push(nextItem)
        return collection
      }
      let previous = collection.pop()
      if (Array.isArray(previous)) {
        let [ number ] = previous
        if (number === nextItem) {
          previous.push(nextItem)
          collection.push(previous)
          return collection
        } else {
          collection.push(previous, nextItem)
          return collection
        }
      } else if (previous === nextItem ) {
        collection.push([previous, nextItem])
        return collection
      } else {
        collection.push(previous, nextItem)
        return collection
      }
    }, []);
}

console.log(rearrangeArray(data));

Answer 14

Below is my own take on the problem:

var data = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]

function rearrangeArray(input) {
  return input.sort((a,b) => a - b)
    .reduce((collection, nextItem) => {
      // First item, simply add it
      if (!collection.length) {
        collection.push(nextItem)
        return collection
      }
      let previous = collection.pop()
      if (Array.isArray(previous)) {
        let [ number ] = previous
        if (number === nextItem) {
          previous.push(nextItem)
          collection.push(previous)
          return collection
        } else {
          collection.push(previous, nextItem)
          return collection
        }
      } else if (previous === nextItem ) {
        collection.push([previous, nextItem])
        return collection
      } else {
        collection.push(previous, nextItem)
        return collection
      }
    }, []);
}

console.log(rearrangeArray(data));

Answer 15

Answer №8

const numbers = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]
const sortedNumbers = []
const subArray = []
const length = numbers.length;
numbers.sort(function sortArray(a,b) {
    return a - b;
});
for(let i = 0; i < length; i++)
{
    if(subArray.length == 0 || subArray[0] == numbers[i])
    {
        subArray.push(numbers[i]);
    }
    else if(subArray.length > 1 && numbers[i] != subArray[0])
    {
        sortedNumbers.push(subArray);
        subArray = [numbers[i]];     
    }
    else if(subArray.length == 1 && numbers[i] != subArray[0])
    {
        sortedNumbers.push(subArray[0]);
        subArray = [numbers[i]];
    } 
}
if(subArray.length == 1)
{
    sortedNumbers.push(subArray[0]);
}
else
{
    sortedNumbers.push(subArray);
}
console.log(sortedNumbers);
//return(sortedNumbers);

Answer 16

const numbers = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20]
const sortedNumbers = []
const subArray = []
const length = numbers.length;
numbers.sort(function sortArray(a,b) {
    return a - b;
});
for(let i = 0; i < length; i++)
{
    if(subArray.length == 0 || subArray[0] == numbers[i])
    {
        subArray.push(numbers[i]);
    }
    else if(subArray.length > 1 && numbers[i] != subArray[0])
    {
        sortedNumbers.push(subArray);
        subArray = [numbers[i]];     
    }
    else if(subArray.length == 1 && numbers[i] != subArray[0])
    {
        sortedNumbers.push(subArray[0]);
        subArray = [numbers[i]];
    } 
}
if(subArray.length == 1)
{
    sortedNumbers.push(subArray[0]);
}
else
{
    sortedNumbers.push(subArray);
}
console.log(sortedNumbers);
//return(sortedNumbers);

Answer 17

Answer №9

Utilizing Array.reduce() and Array.sort() to aggregate the duplicates into an object, then sorting and mapping the elements into a fresh array based on the key of the duplicate objects.

const newArray=[1,2,4,591,392,391,2,5,10,2,1,1,1,20,20].sort((a,b)=> a - b);

let accumulator = newArray.reduce((accumulator, element, index, array) => {
   if(element === array[index+1] && accumulator[element]){
     accumulator[element] = accumulator[element].concat(element) ;
   } else if(element === array[index+1]){
    accumulator[element] = [element].concat(element);
   }
   return accumulator;
}, {});
let resultArray = newArray.filter((element, index, array) => element !== array[index+1]).map((element) => accumulator[element]? accumulator[element]: element);
console.log(resultArray);

Answer 18

Utilizing Array.reduce() and Array.sort() to aggregate the duplicates into an object, then sorting and mapping the elements into a fresh array based on the key of the duplicate objects.

const newArray=[1,2,4,591,392,391,2,5,10,2,1,1,1,20,20].sort((a,b)=> a - b);

let accumulator = newArray.reduce((accumulator, element, index, array) => {
   if(element === array[index+1] && accumulator[element]){
     accumulator[element] = accumulator[element].concat(element) ;
   } else if(element === array[index+1]){
    accumulator[element] = [element].concat(element);
   }
   return accumulator;
}, {});
let resultArray = newArray.filter((element, index, array) => element !== array[index+1]).map((element) => accumulator[element]? accumulator[element]: element);
console.log(resultArray);

Sort the array and organize it by frequency

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

Answer №9

Similar questions

Utilizing React Developer Tools to easily click a button

The message "Missing property 'type' in type 'typeof WithXXX'" indicates an error

Revise Script to Duplicate Alt Attribute onto Miniatures

Display icon within ng-bind-html when the value is not true

Here's a method for using a for loop in ReactJS to update the state when a certain condition is met. Loops can efficiently update the state without having to iterate

Having difficulty redirecting to a dynamic URL with Next.js

Locate all the elements in one array that also exist in another - PHP

Is it possible for an HTML5 video element to stream m3u files?

Utilizing Angular Filters to Assign Values to an Object within a Controller

Transferring information from JSON to HTML

The Dialog feature offered by jQuery-UI

When a frameset is clicked on

How to make Sails.js package accessible across all modules

"Transferring a variable from the parent Layout component to its children components in Next

What are the steps to decode a JSON response using jQuery?

AngularJS is failing to recognize the onload event when loading a URL

Transforming ANSI to UTF-8 in Java

What is the process for installing Angular version 1.1.5 using npm?

Leveraging NgRx for Managing Arrays

Evaluation of outgoing HTTP requests through acceptance testing was the primary focus