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Javascript: Split an array according to a pattern: items 1, 5, 10, then 2, 6, 11, then 3, 7, 12

I am trying to split an array which has a repeating pattern of elements 1, 2, 3, and 4. I want to turn my array [1,2,3,4,5,6,7,8,9,10] into four arrays: [1,5,10], [2,6,11], [3,7,12], and [4,8,13]. I tried using multiples, but the result creates the new arrays in a wrong order. Here is my attempt:
var upload_names_and_ids = [
"Certificat de salaire", //first line is the upload's visible title
"certificat-de-salaire", //second line is the upload's id
"no-info-circle", //third line is the info-circle class
"", //fourth line is the info-circle text
"Allocations Familiales",
"alloc-familiales",
"no-info-circle",
"",
"Courrier Impot (déclaration précédente)",
"courrier-impot",
"info-circle right",
""
];
//Seperate our first array into 4
var upload_names = [];
var upload_ids = [];
var upload_info_circle_class = [];
var upload_info_circle_content = [];
for (var i=0; i<upload_names_and_ids.length; i++){
if (i%4==0) {
upload_info_circle_content.push(upload_names_and_ids[i]);
} else if (i%3==0) {
upload_info_circle_class.push(upload_names_and_ids[i]);
} else if (i%2==0) {
upload_names.push(upload_names_and_ids[i]);
} else {
upload_ids.push(upload_names_and_ids[i]);
}
}
Any help is much appreciated, thank you!

You could take a remainder with index and wanted length.
const
array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16],
length = 4,
result = array.reduce(
(r, v, i) => (r[i % length].push(v), r),
Array.from({ length }, _ => [])
);
console.log(result);
If you like to use predeclared array directly, you could replace this line
Array.from({ length }, _ => [])
with
[upload_names, upload_ids, upload_info_circle_class, upload_info_circle_content]
where the accumulator of Array#reduce keeps the object references.

It's not i%3==0 (which matches 0, 3, 6, …) but i%4==1 (to match 1, 5, 10, …). Same for i%2==0.

I would add a helper sliceN that takes an array and a positive integer. Then returns an array of arrays where the inner arrays are of length n.
sliceN([1,2,3,4,5,6,7,8,9], 3) //=> [[1,2,3], [4,5,6], [7,8,9]]
sliceN([1,2,3,4,5,6], 2) //=> [[1,2], [3,4], [5,6]]
Then also add a helper transpose that transposes a matrix.
transpose([[1,2,3], [4,5,6], [7,8,9]]) //=> [[1,4,7], [2,5,8], [3,6,9]]
transpose([[1,2], [3,4], [5,6]]) //=> [[1,3,5], [2,4,6]]
With these two helpers you can create the wanted result with ease.
const upload_names_and_ids = [
"Certificat de salaire", //first line is the upload's visible title
"certificat-de-salaire", //second line is the upload's id
"no-info-circle", //third line is the info-circle class
"", //fourth line is the info-circle text
"Allocations Familiales",
"alloc-familiales",
"no-info-circle",
"",
"Courrier Impot (déclaration précédente)",
"courrier-impot",
"info-circle right",
""
];
const [
upload_names,
upload_ids,
upload_info_circle_class,
upload_info_circle_content,
] = transpose(sliceN(upload_names_and_ids, 4));
console.log(upload_names);
console.log(upload_ids);
console.log(upload_info_circle_class);
console.log(upload_info_circle_content);
function sliceN(array, n) {
const slices = [];
for (let i = 0; i < array.length; i += n) {
slices.push(array.slice(i, i + n));
}
return slices;
}
function transpose(rows) {
if (rows.length == 0) return [];
const columns = rows[0].map(cell => Array.of(cell));
for (let iRow = 1; iRow < rows.length; iRow += 1) {
for (let iCol = 0; iCol < columns.length; iCol += 1) {
columns[iCol].push(rows[iRow][iCol]);
}
}
return columns;
}
If you are already use a library with helper functions chances are that one or both of these data transforming methods are present. sliceN can often be found as something with split, slice or chunk in the name. transpose is very specific and if present will probably be present under the same name.
As an example Ramda offers both these methods.
R.transpose(R.splitEvery(4, upload_names_and_ids))

Javascript - Create dynamic index of numeric array index

I have an array like this
var a= [[1, 2, [6, 7, 8]], 4, 5];
and another array that indicates a specific element by describing the index:
var index= [0, 2, 2 ]; // = 8 in a
What i need is use variable "index" to create the index of "a" and replace the correspondent element (8) with another value, for example "hello".
How can i do?

Use Array.reduce() to iterate the index array without the last element (using Array.slice()), and get the sub array. Splice the text to last index inside the sub array:
const index = [0, 2, 2 ]; // = 8 in a
const a = [[1, 2, [6, 7, 8]], 4, 5];
const updateIndex = (arr, index, replacement) => {
index
.slice(0, -1)
.reduce((c, i) => c[i], arr)
.splice(index[index.length -1], 1, replacement);
}
updateIndex(a, index, 'hello');
console.log(a);

Basically each index of index array represents a level of nesting and the value at index represents the index of array in which target element lies or the index of target element itself if it is last element of array index. The problem can be easily solved recursively
For simplicity, I am calling index array, path
var path = [0, 2, 2]
var values = [[1, 2, [6, 7, 8]], 4, 5];
function findValues(path, values) {
//base case, if length = 1 we have the index of target element
if(path.length == 1) {
var targetIndex = path[0]
//read it
console.log(values[path]) //8
// modify it
values[path] = 'hello'
} else {
// pick the current nesting level
var currentLevel = path.shift()
// go one level down
findValues(path, values[currentLevel])
}
}
findValues(path, values)
console.log(values) // [[1, 2, [6, 7, "hello"]], 4, 5]
Ofcouse the function assumes that path and values array are in valid state i.e. target element exists on the given path, you might want to modify this function to validate input and handle edge cases

You can create a function that accesses the desired array and changes the value.
Take a look:
var index = [0, 2, 2 ]; // = 8 in a
var a = [[1, 2, [6, 7, 8]], 4, 5];
function updateIndex(arr, index, value) {
var elementIndex = index.pop();
var tempArray = a;
index.forEach(item => tempArray = tempArray[item]);
tempArray[elementIndex] = value;
}
updateIndex(a, index, 'hello');
console.log(a);

You could take a function which uses a vopy of indices and save the last index for accessing the reduced array for assignment of the value.
This proposal creates missing arrays as well.
function setValue(array, [...indices], value) {
var last = indices.pop();
indices.reduce((a, i) => a[i] = a[i] || [], array)[last] = value;
}
var array = [[1, 2, [6, 7, 8]], 4, 5];
setValue(array, [0, 2, 2 ], 'hello'); // targetting 8
console.log(array);
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Changing rows to columns javascript

I wanted to change the rows into columns of an array.
[
[1],
[1,2],
[1,2,3],
[4,2,3],
[4,5,3],
[4,5,6]
]
to
[
[1,1,1,4,4,4],
[2,2,2,5,5],
[3,3,3,6]
]
I tried
var res = [];
for(i in this.fields) {
for(j in this.fields[i].value) {
if(i === 0) res[j] = [];
res[j][i] = this.fields[i].value[j];
}
}
this gives me empty set.

Create this function:
function transpose(arr) {
return Object.keys(arr[0]).map(function (c) {
return arr.map(function (r) {
return r[c];
});
});
}
and then:
var transposedArray = transpose(originalArray);

What you're asking looks a little weird because you have different lengths and you're ignoring undefined values, but it is still achievable.
Don't use for..in loops for Array, use a normal for. Also, you'll need to know how many items you'll have in your new parent Array, which is the max of the lengths of the original child Arrays.
var arrR = [ // will refer to "down" and "across" as in this literal
[1],
[1, 2],
[1, 2, 3],
[4, 2, 3],
[4, 5, 3],
[4, 5, 6]
];
function r2c(arr) {
var arrC = [], // next get the longest sub-array length
x = Math.max.apply(Math, arr.map(function (e) {return e.length;})),
y = arr.length,
i, j;
for (i = 0; i < x; ++i) { // this is the loop "down"
arrC[i] = [];
for (j = 0; j < y; ++j) // and this is the loop "across"
if (i in arr[j])
arrC[i].push(arr[j][i]);
}
return arrC;
}
var arrC = r2c(arrR);
/* [
[1, 1, 1, 4, 4, 4],
[2, 2, 2, 5, 5],
[3, 3, 3, 6]
] */
You should still consider if you're happy with [[1], [1, 2], [1]] becoming [[1, 1, 1], [2]], which I would consider unexpected (the position of 2 is completely lost), but seems to be what you intend.

Similar to Pauls but doesn't need to get the max length first:
function transpose(arr) {
// Loop over arrays as long as one has values
// Arrays should be contiguous, may fail if sparse
for (var result = [], i=0, more; more; i++) {
more = false;
// Get the ith element of each array (if there is one)
for (var j=0, jLen=arr.length; j<jLen; j++) {
// Don't add missing members
if (arr[j].hasOwnProperty(i)) {
// Add array for result if not already there
result[i] = result[i] || [];
// Do transpose
result[i][j] = arr[j][i];
// Only keep going while there is data
more = true;
}
}
}
return result;
}
BTW, a fixed version of your original function is:
function transpose2(fields) {
// Make sure the result array is initialised
var res = [];
// Don't forget to keep counters local - declare them
// I've removed *this* as it's a plain function, use it if
// it's an instance method
for(var i in fields) {
// Values are read directly, there is no "value" accessor
for(var j in fields[i]) {
// Don't rely on order of enumeration - may not start at 0
if(!res[j]) res[j] = [];
// Do the transpose
res[j][i] = fields[i][j];
}
}
return res;
}
But as noted above, for..in is not liked for arrays, particularly as there are many libraries that extend built-ins like Array.prototype so you will traverse those properties too. But if you're cool with that, this is a good way to deal with sparse arrays. You can add a hasOwnProperty test to avoid inherited enumerables.
Note also that the order of enumeration isn't necessarily from '0' or in any particular order, hence changed way of initialising res[j].

Counting the occurrences / frequency of array elements

In Javascript, I'm trying to take an initial array of number values and count the elements inside it. Ideally, the result would be two new arrays, the first specifying each unique element, and the second containing the number of times each element occurs. However, I'm open to suggestions on the format of the output.
For example, if the initial array was:
5, 5, 5, 2, 2, 2, 2, 2, 9, 4
Then two new arrays would be created. The first would contain the name of each unique element:
5, 2, 9, 4
The second would contain the number of times that element occurred in the initial array:
3, 5, 1, 1
Because the number 5 occurs three times in the initial array, the number 2 occurs five times and 9 and 4 both appear once.
I've searched a lot for a solution, but nothing seems to work, and everything I've tried myself has wound up being ridiculously complex. Any help would be appreciated!
Thanks :)

You can use an object to hold the results:
const arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const counts = {};
for (const num of arr) {
counts[num] = counts[num] ? counts[num] + 1 : 1;
}
console.log(counts);
console.log(counts[5], counts[2], counts[9], counts[4]);
So, now your counts object can tell you what the count is for a particular number:
console.log(counts[5]); // logs '3'
If you want to get an array of members, just use the keys() functions
keys(counts); // returns ["5", "2", "9", "4"]

const occurrences = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4].reduce(function (acc, curr) {
return acc[curr] ? ++acc[curr] : acc[curr] = 1, acc
}, {});
console.log(occurrences) // => {2: 5, 4: 1, 5: 3, 9: 1}

const arr = [2, 2, 5, 2, 2, 2, 4, 5, 5, 9];
function foo (array) {
let a = [],
b = [],
arr = [...array], // clone array so we don't change the original when using .sort()
prev;
arr.sort();
for (let element of arr) {
if (element !== prev) {
a.push(element);
b.push(1);
}
else ++b[b.length - 1];
prev = element;
}
return [a, b];
}
const result = foo(arr);
console.log('[' + result[0] + ']','[' + result[1] + ']')
console.log(arr)

One line ES6 solution. So many answers using object as a map but I can't see anyone using an actual Map
const map = arr.reduce((acc, e) => acc.set(e, (acc.get(e) || 0) + 1), new Map());
Use map.keys() to get unique elements
Use map.values() to get the occurrences
Use map.entries() to get the pairs [element, frequency]
var arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
const map = arr.reduce((acc, e) => acc.set(e, (acc.get(e) || 0) + 1), new Map());
console.info([...map.keys()])
console.info([...map.values()])
console.info([...map.entries()])

If using underscore or lodash, this is the simplest thing to do:
_.countBy(array);
Such that:
_.countBy([5, 5, 5, 2, 2, 2, 2, 2, 9, 4])
=> Object {2: 5, 4: 1, 5: 3, 9: 1}
As pointed out by others, you can then execute the _.keys() and _.values() functions on the result to get just the unique numbers, and their occurrences, respectively. But in my experience, the original object is much easier to deal with.

Don't use two arrays for the result, use an object:
a = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
result = { };
for(var i = 0; i < a.length; ++i) {
if(!result[a[i]])
result[a[i]] = 0;
++result[a[i]];
}
Then result will look like:
{
2: 5,
4: 1,
5: 3,
9: 1
}

How about an ECMAScript2015 option.
const a = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const aCount = new Map([...new Set(a)].map(
x => [x, a.filter(y => y === x).length]
));
aCount.get(5) // 3
aCount.get(2) // 5
aCount.get(9) // 1
aCount.get(4) // 1
This example passes the input array to the Set constructor creating a collection of unique values. The spread syntax then expands these values into a new array so we can call map and translate this into a two-dimensional array of [value, count] pairs - i.e. the following structure:
Array [
[5, 3],
[2, 5],
[9, 1],
[4, 1]
]
The new array is then passed to the Map constructor resulting in an iterable object:
Map {
5 => 3,
2 => 5,
9 => 1,
4 => 1
}
The great thing about a Map object is that it preserves data-types - that is to say aCount.get(5) will return 3 but aCount.get("5") will return undefined. It also allows for any value / type to act as a key meaning this solution will also work with an array of objects.
function frequencies(/* {Array} */ a){
return new Map([...new Set(a)].map(
x => [x, a.filter(y => y === x).length]
));
}
let foo = { value: 'foo' },
bar = { value: 'bar' },
baz = { value: 'baz' };
let aNumbers = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4],
aObjects = [foo, bar, foo, foo, baz, bar];
frequencies(aNumbers).forEach((val, key) => console.log(key + ': ' + val));
frequencies(aObjects).forEach((val, key) => console.log(key.value + ': ' + val));

I think this is the simplest way how to count occurrences with same value in array.
var a = [true, false, false, false];
a.filter(function(value){
return value === false;
}).length

const data = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
function count(arr) {
return arr.reduce((prev, curr) => (prev[curr] = ++prev[curr] || 1, prev), {})
}
console.log(count(data))

2021's version
The more elegant way is using Logical nullish assignment (x ??= y) combined with Array#reduce() with O(n) time complexity.
The main idea is still using Array#reduce() to aggregate with output as object to get the highest performance (both time and space complexity) in terms of searching & construct bunches of intermediate arrays like other answers.
const arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
const result = arr.reduce((acc, curr) => {
acc[curr] ??= {[curr]: 0};
acc[curr][curr]++;
return acc;
}, {});
console.log(Object.values(result));
Clean & Refactor code
Using Comma operator (,) syntax.
The comma operator (,) evaluates each of its operands (from left to
right) and returns the value of the last operand.
const arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
const result = arr.reduce((acc, curr) => (acc[curr] = (acc[curr] || 0) + 1, acc), {});
console.log(result);
Output
{
"2": 5,
"4": 1,
"5": 3,
"9": 1
}

If you favour a single liner.
arr.reduce(function(countMap, word) {countMap[word] = ++countMap[word] || 1;return countMap}, {});
Edit (6/12/2015):
The Explanation from the inside out.
countMap is a map that maps a word with its frequency, which we can see the anonymous function. What reduce does is apply the function with arguments as all the array elements and countMap being passed as the return value of the last function call. The last parameter ({}) is the default value of countMap for the first function call.

ES6 version should be much simplifier (another one line solution)
let arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
let acc = arr.reduce((acc, val) => acc.set(val, 1 + (acc.get(val) || 0)), new Map());
console.log(acc);
// output: Map { 5 => 3, 2 => 5, 9 => 1, 4 => 1 }
A Map instead of plain Object helping us to distinguish different type of elements, or else all counting are base on strings

Edit 2020: this is a pretty old answer (nine years). Extending the native prototype will always generate discussion. Although I think the programmer is free to choose her own programming style, here's a (more modern) approach to the problem without extending Array.prototype:
{
// create array with some pseudo random values (1 - 5)
const arr = Array.from({length: 100})
.map( () => Math.floor(1 + Math.random() * 5) );
// frequencies using a reducer
const arrFrequencies = arr.reduce((acc, value) =>
({ ...acc, [value]: acc[value] + 1 || 1}), {} )
console.log(arrFrequencies);
console.log(`Value 4 occurs ${arrFrequencies[4]} times in arrFrequencies`);
// bonus: restore Array from frequencies
const arrRestored = Object.entries(arrFrequencies)
.reduce( (acc, [key, value]) => acc.concat(Array(value).fill(+key)), [] );
console.log(arrRestored.join());
}
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The old (2011) answer: you could extend Array.prototype, like this:
{
Array.prototype.frequencies = function() {
var l = this.length,
result = {
all: []
};
while (l--) {
result[this[l]] = result[this[l]] ? ++result[this[l]] : 1;
}
// all pairs (label, frequencies) to an array of arrays(2)
for (var l in result) {
if (result.hasOwnProperty(l) && l !== 'all') {
result.all.push([l, result[l]]);
}
}
return result;
};
var freqs = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4].frequencies();
console.log(`freqs[2]: ${freqs[2]}`); //=> 5
// or
var freqs = '1,1,2,one,one,2,2,22,three,four,five,three,three,five'
.split(',')
.frequencies();
console.log(`freqs.three: ${freqs.three}`); //=> 3
// Alternatively you can utilize Array.map:
Array.prototype.frequencies = function() {
var freqs = {
sum: 0
};
this.map(function(a) {
if (!(a in this)) {
this[a] = 1;
} else {
this[a] += 1;
}
this.sum += 1;
return a;
}, freqs);
return freqs;
}
}
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Based on answer of #adamse and #pmandell (which I upvote), in ES6 you can do it in one line:
2017 edit: I use || to reduce code size and make it more readable.
var a=[7,1,7,2,2,7,3,3,3,7,,7,7,7];
alert(JSON.stringify(
a.reduce((r,k)=>{r[k]=1+r[k]||1;return r},{})
));
It can be used to count characters:
var s="ABRACADABRA";
alert(JSON.stringify(
s.split('').reduce((a, c)=>{a[c]++?0:a[c]=1;return a},{})
));

A shorter version using reduce and tilde (~) operator.
const data = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
function freq(nums) {
return nums.reduce((acc, curr) => {
acc[curr] = -~acc[curr];
return acc;
}, {});
}
console.log(freq(data));

If you are using underscore you can go the functional route
a = ['foo', 'foo', 'bar'];
var results = _.reduce(a,function(counts,key){ counts[key]++; return counts },
_.object( _.map( _.uniq(a), function(key) { return [key, 0] })))
so your first array is
_.keys(results)
and the second array is
_.values(results)
most of this will default to native javascript functions if they are available
demo : http://jsfiddle.net/dAaUU/

So here's how I'd do it with some of the newest javascript features:
First, reduce the array to a Map of the counts:
let countMap = array.reduce(
(map, value) => {map.set(value, (map.get(value) || 0) + 1); return map},
new Map()
)
By using a Map, your starting array can contain any type of object, and the counts will be correct. Without a Map, some types of objects will give you strange counts.
See the Map docs for more info on the differences.
This could also be done with an object if all your values are symbols, numbers, or strings:
let countObject = array.reduce(
(map, value) => { map[value] = (map[value] || 0) + 1; return map },
{}
)
Or slightly fancier in a functional way without mutation, using destructuring and object spread syntax:
let countObject = array.reduce(
(value, {[value]: count = 0, ...rest}) => ({ [value]: count + 1, ...rest }),
{}
)
At this point, you can use the Map or object for your counts (and the map is directly iterable, unlike an object), or convert it to two arrays.
For the Map:
countMap.forEach((count, value) => console.log(`value: ${value}, count: ${count}`)
let values = countMap.keys()
let counts = countMap.values()
Or for the object:
Object
.entries(countObject) // convert to array of [key, valueAtKey] pairs
.forEach(([value, count]) => console.log(`value: ${value}, count: ${count}`)
let values = Object.keys(countObject)
let counts = Object.values(countObject)

var array = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
function countDuplicates(obj, num){
obj[num] = (++obj[num] || 1);
return obj;
}
var answer = array.reduce(countDuplicates, {});
// answer => {2:5, 4:1, 5:3, 9:1};
If you still want two arrays, then you could use answer like this...
var uniqueNums = Object.keys(answer);
// uniqueNums => ["2", "4", "5", "9"];
var countOfNums = Object.keys(answer).map(key => answer[key]);
// countOfNums => [5, 1, 3, 1];
Or if you want uniqueNums to be numbers
var uniqueNums = Object.keys(answer).map(key => +key);
// uniqueNums => [2, 4, 5, 9];

Here's just something light and easy for the eyes...
function count(a,i){
var result = 0;
for(var o in a)
if(a[o] == i)
result++;
return result;
}
Edit: And since you want all the occurences...
function count(a){
var result = {};
for(var i in a){
if(result[a[i]] == undefined) result[a[i]] = 0;
result[a[i]]++;
}
return result;
}

Solution using a map with O(n) time complexity.
var arr = [2, 2, 2, 2, 2, 4, 5, 5, 5, 9];
const countOccurrences = (arr) => {
const map = {};
for ( var i = 0; i < arr.length; i++ ) {
map[arr[i]] = ~~map[arr[i]] + 1;
}
return map;
}
Demo: http://jsfiddle.net/simevidas/bnACW/

There is a much better and easy way that we can do this using ramda.js.
Code sample here
const ary = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
R.countBy(r=> r)(ary)
countBy documentation is at documentation

I know this question is old but I realized there are too few solutions where you get the count array as asked with a minimal code so here is mine
// The initial array we want to count occurences
var initial = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
// The count array asked for
var count = Array.from(new Set(initial)).map(val => initial.filter(v => v === val).length);
// Outputs [ 3, 5, 1, 1 ]
Beside you can get the set from that initial array with
var set = Array.from(new Set(initial));
//set = [5, 2, 9, 4]

My solution with ramda:
const testArray = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
const counfFrequency = R.compose(
R.map(R.length),
R.groupBy(R.identity),
)
counfFrequency(testArray)
Link to REPL.

Using MAP you can have 2 arrays in the output: One containing the occurrences & the other one is containing the number of occurrences.
const dataset = [2,2,4,2,6,4,7,8,5,6,7,10,10,10,15];
let values = [];
let keys = [];
var mapWithOccurences = dataset.reduce((a,c) => {
if(a.has(c)) a.set(c,a.get(c)+1);
else a.set(c,1);
return a;
}, new Map())
.forEach((value, key, map) => {
keys.push(key);
values.push(value);
});
console.log(keys)
console.log(values)

This question is more than 8 years old and many, many answers do not really take ES6 and its numerous advantages into account.
Perhaps is even more important to think about the consequences of our code for garbage collection/memory management whenever we create additional arrays, make double or triple copies of arrays or even convert arrays into objects. These are trivial observations for small applications but if scale is a long term objective then think about these, thoroughly.
If you just need a "counter" for specific data types and the starting point is an array (I assume you want therefore an ordered list and take advantage of the many properties and methods arrays offer), you can just simply iterate through array1 and populate array2 with the values and number of occurrences for these values found in array1.
As simple as that.
Example of simple class SimpleCounter (ES6) for Object Oriented Programming and Object Oriented Design
class SimpleCounter {
constructor(rawList){ // input array type
this.rawList = rawList;
this.finalList = [];
}
mapValues(){ // returns a new array
this.rawList.forEach(value => {
this.finalList[value] ? this.finalList[value]++ : this.finalList[value] = 1;
});
this.rawList = null; // remove array1 for garbage collection
return this.finalList;
}
}
module.exports = SimpleCounter;

Using Lodash
const values = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
const frequency = _.map(_.groupBy(values), val => ({ value: val[0], frequency: val.length }));
console.log(frequency);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js"></script>

To return an array which is then sortable:
let array = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]
let reducedArray = array.reduce( (acc, curr, _, arr) => {
if (acc.length == 0) acc.push({item: curr, count: 1})
else if (acc.findIndex(f => f.item === curr ) === -1) acc.push({item: curr, count: 1})
else ++acc[acc.findIndex(f => f.item === curr)].count
return acc
}, []);
console.log(reducedArray.sort((a,b) => b.count - a.count ))
/*
Output:
[
{
"item": 2,
"count": 5
},
{
"item": 5,
"count": 3
},
{
"item": 9,
"count": 1
},
{
"item": 4,
"count": 1
}
]
*/

Check out the code below.
<html>
<head>
<script>
// array with values
var ar = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4];
var Unique = []; // we'll store a list of unique values in here
var Counts = []; // we'll store the number of occurances in here
for(var i in ar)
{
var Index = ar[i];
Unique[Index] = ar[i];
if(typeof(Counts[Index])=='undefined')
Counts[Index]=1;
else
Counts[Index]++;
}
// remove empty items
Unique = Unique.filter(function(){ return true});
Counts = Counts.filter(function(){ return true});
alert(ar.join(','));
alert(Unique.join(','));
alert(Counts.join(','));
var a=[];
for(var i=0; i<Unique.length; i++)
{
a.push(Unique[i] + ':' + Counts[i] + 'x');
}
alert(a.join(', '));
</script>
</head>
<body>
</body>
</html>

function countOcurrences(arr){
return arr.reduce((aggregator, value, index, array) => {
if(!aggregator[value]){
return aggregator = {...aggregator, [value]: 1};
}else{
return aggregator = {...aggregator, [value]:++aggregator[value]};
}
}, {})
}

You can simplify this a bit by extending your arrays with a count function. It works similar to Ruby’s Array#count, if you’re familiar with it.
Array.prototype.count = function(obj){
var count = this.length;
if(typeof(obj) !== "undefined"){
var array = this.slice(0), count = 0; // clone array and reset count
for(i = 0; i < array.length; i++){
if(array[i] == obj){ count++ }
}
}
return count;
}
Usage:
let array = ['a', 'b', 'd', 'a', 'c'];
array.count('a'); // => 2
array.count('b'); // => 1
array.count('e'); // => 0
array.count(); // => 5
Gist
Edit
You can then get your first array, with each occurred item, using Array#filter:
let occurred = [];
array.filter(function(item) {
if (!occurred.includes(item)) {
occurred.push(item);
return true;
}
}); // => ["a", "b", "d", "c"]
And your second array, with the number of occurrences, using Array#count into Array#map:
occurred.map(array.count.bind(array)); // => [2, 1, 1, 1]
Alternatively, if order is irrelevant, you can just return it as a key-value pair:
let occurrences = {}
occurred.forEach(function(item) { occurrences[item] = array.count(item) });
occurences; // => {2: 5, 4: 1, 5: 3, 9: 1}

sort array keys by value

I'm using javascript, and I have an array containing multiple values, which may be non-unique. I'd like to take this array and generate a new array, or ordered list, of its keys in ascending order of value. For example, if I have [ 2, 2, 4, 5, 1, 6 ], I'd like to generate [ 5, 4, 0, 1, 2, 3 ].
I was thinking of iterating over the original list and inserting each value into the new list while checking for proper placement by comparing to the existing values of the new list every time an insertion is performed. This seems wasteful, though, as I'd have to (potentially) check every value of the new list for every insertion.
Anyone have a simpler method for this?

I think you meant [ 4, 0, 1, 2, 3, 5 ].
function GetSortedKeys(values) {
var array_with_keys = [];
for (var i = 0; i < values.length; i++) {
array_with_keys.push({ key: i, value: values[i] });
}
array_with_keys.sort(function(a, b) {
if (a.value < b.value) { return -1; }
if (a.value > b.value) { return 1; }
return 0;
});
var keys = [];
for (var i = 0; i < array_with_keys.length; i++) {
keys.push(array_with_keys[i].key);
}
return keys;
}
var array = [2, 2, 4, 5, 1, 6];
alert(GetSortedKeys(array));
This is the simplest method I can come up with on Javascript, unfortunately.

Using the nice Underscore.JS:
var get_sorted_keys = function(values) {
var keys_idx = [], i;
for (i = 0; i < values.length; i++) {
keys_idx.push(i);
}
var keys = _.sortBy(keys_idx, function(idx){ return values[idx]; });
return keys;
};
var array = [2, 2, 4, 5, 1, 6];
console.log("Sorted keys:", get_sorted_keys(array));
Output:
Sorted keys: [4, 0, 1, 2, 3, 5]