javascript sort and remap array

Question

I'm using the following code (courtesy of JQuery Javascript Sort Array by Highest Count) to sort a list of strings from highest to lowest count:

var items = {}, sortableItems = [], i, len, element, listOfStrings; 

    listOfStrings = JSON.parse(the_chems);


for (i = 0, len = listOfStrings.length; i < len; i += 1) 
    {
        if (items.hasOwnProperty(listOfStrings[i])) {
            items[listOfStrings[i]] += 1;
        } else {
            items[listOfStrings[i]] = 1;
        }
    }

for (element in items) 
    {
        if (items.hasOwnProperty(element)) {
            sortableItems.push([element, items[element]]);
        }
    }

sortableItems.sort(function (first, second) 
    {
        return second[1] - first[1];
    });

Instead of this type of array input

["red", "red", "red", "blue", "blue"]

Which returns

[ [ "red", 3 ], [ "blue", 2 ] ]

I would like to use an array like

[["red","apple"], ["red","chilli"], ["red","melon"], ["blue","ocean"], ["blue","eyes"]]

And return

[["red", 3, ["apple","chilli","melon"]], ["blue", 2, ["blue","ocean"]]

A note that the sort (provided the count is the 2nd property) can be done by result.sort((count_a, count_b) => count_b[1] - count_a[1]); — haz
– haz, Commented Aug 22, 2017 at 11:06

Nina Scholz · Accepted Answer · 2017-08-21 16:16:34Z

3

You could use a hash table and collect the result in an array.

var array = [["red", "apple"], ["red", "chilli"], ["red", "melon"], ["blue", "ocean"], ["blue", "eyes"]],
    hash = Object.create(null),
    result = [];

array.forEach(function (a) {
    if (!hash[a[0]]) {
        hash[a[0]] = [a[0], 0, []];
        result.push(hash[a[0]]);
    }
    hash[a[0]][1]++;
    hash[a[0]][2].push(a[1]);
});

console.log(result);

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

edited Aug 21, 2017 at 16:16

answered Aug 21, 2017 at 16:10

Nina Scholz

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Comments

Jonas Wilms · Accepted Answer · 2017-08-21 16:16:11Z

2

You could use a Map and some cool destructuring:

var hash = new Map(), result = [];

input.forEach(([color,fruit]) => {
 if(hash.has(color)){
  var arr = hash.get(color);
  arr[1]++;
  arr[2].push(fruit);
 }else{
  var tmp = [color,1,[fruit]];
  hash.set(color,tmp);
  result.push(tmp);
 }
});

Try it out (Hint: that would be a good usecase for objects...)

Slower but maybe a bit more readable:

var hash = new Map();

input.forEach(([color,fruit])=>{
 if(hash.has(color)){
  hash.get(color).push(fruit);
 }else{
  hash.set(color,[fruit]);
 }
})

var result = [...hash.entries()].map(([color,fruits])=>[color,fruits.length,fruits]);

edited Aug 21, 2017 at 16:16

answered Aug 21, 2017 at 16:09

Jonas Wilms

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Comments

Adam Jenkins · Accepted Answer · 2017-08-21 16:19:25Z

1

Why not create a useful data structure like this:

{
  red:["apple","chili","melon"],
  blue:["ocean","eyes"]
}

Done with one loop and no extraneous hash table:

var arr = [["red","apple"], ["red","chilli"], ["red","melon"], ["blue","ocean"], ["blue","eyes"]];
arr.reduce(function(c,element) {
    if(!c[element[0]]) {
        c[element[0]] = [];
    }
    c[element[0]].push(element[1]);
    return c;
},{});

For those who don't care about readable code:

arr.reduce(function(c,element) {
    return (c[element[0]] ? c[element[0]].push(element[1]) : c[element[0]] = [element[1]]) && c;
},{});

answered Aug 21, 2017 at 16:19

Adam Jenkins

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Comments

Manuel Otto · Accepted Answer · 2017-08-21 16:20:41Z

1

Also possible without map...

function groupAndSort(items){
    var output = []
    for(var i=0;i<items.length;i++){
        var exists = false;
        for(var u=0;u<output.length;u++){
            if(output[u][0]==items[i][0]){
                output[u][1]++
                output[u][2].push(items[i][1])
                exists = true
                break
            }
        }
        if(!exists)
            output.push([items[i][0],1,[items[i][1]]])
    }
    output.sort(function (first, second){
        return second[1] - first[1];
    });
    return output
}

edited Aug 21, 2017 at 16:20

answered Aug 21, 2017 at 16:10

Manuel Otto

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Comments

Anurag Singh Bisht · Accepted Answer · 2017-08-21 16:20:56Z

1

You can use Array.reduce to restructure the array.

const arr = [["red","apple"], ["red","chilli"], ["red","melon"], ["blue","ocean"], ["blue","eyes"]];

const newArr = arr.reduce((acc , curr) => {
  let flag = false;
  acc.forEach(a => {
    if(a[0] === curr[0]) {
      a[1]++;
      a[2].push(curr[1]);
      flag = true;
    }
  });
  if(!flag) {
    let newArr = [curr[0], 1, [curr[1]]];
    acc.push(newArr);
  }
  return acc;
}, []);

console.log(newArr);

answered Aug 21, 2017 at 16:20

Anurag Singh Bisht

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Comments

Christopher Messer · Accepted Answer · 2017-08-21 16:21:47Z

Here's another solution. First the function creates a simple count of each color. From there, we go through the object and filter / map through the relevant items, and push our final desired output.

const testArray = [["red","apple"], ["red","chilli"], ["red","melon"], ["blue","ocean"], ["blue","eyes"]]

function modifyArray(arr) {
  let colorCount = arr.reduce((allColors, color) => {
    if (color[0] in allColors) {
      allColors[color[0]]++
	}
	else {
	  allColors[color[0]] = 1
	}
	return allColors
  }, {})

  let newArr = [];

  for (var key in colorCount) {
    let coloredItems = arr.filter(val => val[0] === key).map(val => val[1])
    newArr.push([key, colorCount[key], coloredItems])
  }

  return newArr;
}

console.log(modifyArray(testArray))

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