I have an array composed by nine elements:
var ar = ['a','a','a','a','a','a','a','a','a'];
I need to control three elements at a time by a function that returns a true or false. If the first returns true, the others do not have to execute in order to obtain just one true result.
var one = fun(ar.slice(0, 3));
if (one != false) document.write(one);
var two = fun(ar.slice(3, 6));
if (two != false) document.write(two);
var three = fun(ar.slice(6, 9));
if (three != false) document.write(three);
How can I simplify this process? To avoid the creation of a multiarray.
Thanks !!
You could use an array with the slicing paramters and iterate with Array#some
function fun(array) {
return array.reduce(function (a, b) { return a + b; }) > 10;
}
var ar = [0, 1, 2, 3, 4, 5, 6, 7, 8];
[0, 3, 6].some(function (a) {
var result = fun(ar.slice(a, a + 3));
if (result) {
console.log(a, result);
return true;
}
});
Probably this is what you need.
Check out the demo below
var ar = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm'];
for(var i = 0; i < ar.length; i+=3){
console.log(ar.slice(i, i+3));
}
It sounds like some you want an iterator of sorts. An explicit example would be something like:
var input = [0, 1, 2, 3, 4, 5, 6, 7, 8];
// console.log(checkChunks(input, 2, checkChunkSum(10)));
console.log(checkChunks(input, 3, checkChunkSum(10)));
function checkChunks(arr, chunkLength, checker) {
var startIndex = 0;
var chunk;
while ((chunk = arr.slice(startIndex, startIndex += chunkLength)).length) {
var result = checker(chunk);
console.log('Checking chunk: ', chunk);
if (result !== false) {
return {
chunk: chunk,
result: result
};
}
}
return false;
}
function checkChunkSum(max) {
return function checkSum(arr) {
return arr.reduce(sum, 0) > max;
}
}
function sum(a, b) {
return a + b;
}
I'm trying to count duplicates in an array of dates and add them to a new array.
But i'm only getting the duplicates and the amount of times they exist in the array.
What I want is:
[a, a, b, c, c] => [a: 2, b: 1, c: 2]
Code:
$scope.result = { };
for(var i = 0; i < $scope.loginArray.length; ++i) {
if(! $scope.result[$scope.loginArray[i]]){
$scope.result[$scope.loginArray[i]] = 0;
++ $scope.result[$scope.loginArray[i]];}
}
Any suggestions?
You might need an object for this, not an array. So what you are doing is already great, but the if condition is messing up:
$scope.result = {};
for (var i = 0; i < $scope.loginArray.length; ++i) {
if (!$scope.result[$scope.loginArray[i]])
$scope.result[$scope.loginArray[i]] = 0;
++$scope.result[$scope.loginArray[i]];
}
Snippet
var a = ['a', 'a', 'b', 'c', 'c'];
var r = {};
for (var i = 0; i < a.length; ++i) {
if (!r[a[i]])
r[a[i]] = 0;
++r[a[i]];
}
console.log(r);
Or in better way, you can use .reduce like how others have given. A simple reduce function will be:
var a = ['a', 'a', 'b', 'c', 'c'];
var r = a.reduce(function(c, e) {
c[e] = (c[e] || 0) + 1;
return c;
}, {});
console.log(r);
For that, you can use .reduce:
var arr = ['a','a', 'b', 'c', 'c'];
var result = arr.reduce(function(p,c){
if(p[c] === undefined)
p[c] = 0;
p[c]++;
return p;
},{});
console.log(result);
You can use reduce and return object
var ar = ['a', 'a', 'b', 'c', 'c'];
var result = ar.reduce(function(r, e) {
r[e] = (r[e] || 0) + 1;
return r;
}, {});
console.log(result)
You can also first create Object and then use forEach add properties and increment values
var ar = ['a', 'a', 'b', 'c', 'c'], result = {}
ar.forEach(e => result[e] = (result[e] || 0)+1);
console.log(result)
lodash's countBy function will do the trick:
_.countBy(['a', 'a', 'b', 'c', 'c']) will evaluate to: {a: 2, b: 1, c: 2}
It does involve adding lodash as a dependency though.
I would do like this
var arr = ["a", "a", "b", "c", "c", "a"],
red = arr.reduce((p,c) => (p[c]++ || (p[c]=1),p),{});
console.log(red);
The question doesn't make much sense but not sure how to word it without an example. If someone can word it better, feel free to edit it.
Let's say I have an array of arrays such as this:
[ ['a', 'a', 'b', 'c'], [], ['d', 'a'], ['b', 'b', 'b', 'e'] ]
I would like the output to be:
['a', 'a', 'b', 'b', 'b', 'c', 'd', 'e']
Not sure if there is an easy way to do this in javascript/jquery/underscore. One way I could think of is to look through each of these arrays and count up the number of times each element shows up and keep track of the maximum amount of times it shows up. Then I can recreate it. But that seems pretty slow considering that my arrays can be very large.
You need to:
Loop over each inner array and count the values
Store each value and its count (if higher than current count) in a counter variable
In the end, convert the value and counts into an array
Following code shows a rough outline of the process. Remember to replace .forEach and for..in with appropriate code:
var input = [['a', 'a', 'b', 'c'], [], ['d', 'a'], ['b', 'b', 'b', 'e']],
inputCount = {};
input.forEach(function(inner) {
var innerCount = {};
inner.forEach(function(value) {
innerCount[value] = innerCount[value] ? innerCount[value] + 1 : 1;
});
var value;
for (value in innerCount) {
inputCount[value] = inputCount[value] ? Math.max(inputCount[value], innerCount[value]) : innerCount[value];
}
});
console.log(inputCount);
// Object {a: 2, b: 3, c: 1, d: 1, e: 1}
After messing around, I found a solution but not sure if I like it enough to use. I would probably use it if I can't think of another one.
I would use underscorejs countBy to get the count of all the elements.
var array = [ ['a', 'a', 'b', 'c'], [], ['d', 'a'], ['b', 'b', 'b', 'e'] ];
var count = _.map(array, function(inner) {
return _.countBy(inner, function(element) {
return element;
});
});
var total = {};
_.each(_.uniq(_.flatten(array)), function(element) {
var max = _.max(count, function(countedElement) {
return countedElement[element];
});
total[element] = max[element];
});
console.log(total); // {a: 2, b: 3, c: 1, d: 1, e: 1}
Then I would recreate the array with that total.
Here is example of simple nested loop approach:
var input = [ ['a', 'a', 'b', 'c'], [], ['d', 'a'], ['b', 'b', 'b', 'e'] ];
var countMap = {};
// iterate outer array
for (i=0; i < input.length; i++) {
// iterate inner array
for (j=0; j < input[i].length; j++) {
// increment map counter
var value = input[i][j];
if (countMap[input[i][j]] === undefined) {
countMap[value] = 1;
} else {
countMap[value]++;
}
}
}
console.log(countMap); // output such as {'a':2, 'b':4, 'c':1, 'd':1, 'e':1}
Not the most efficient solution but it should describe you the process:
var big = [ ['a', 'a', 'b', 'c'], [], ['d', 'a'], ['b', 'b', 'b', 'e'] ];
function map(arr){
var map = {}
for (var i=arr.length-1; i>-1; i--){
if(arr[i] in map) map[arr[i]]++;
else map[arr[i]] = 1;
}
return map;
}
function reduce(matrix){
var arrMap = {};
for (var i=matrix.length-1; i>-1; i--){
var arrRes = map(matrix[i]);
for (var key in arrRes){
if( !arrMap[key] || arrMap[key] < arrRes[key])
arrMap[key] = arrRes[key];
}
}
return arrMap;
}
function calc(matrix){
var res = [],
arrMap = reduce(matrix);
for (var key in arrMap){
while(arrMap[key] > 0 ){
res.push(key);
arrMap[key]--;
}
}
return res;
}
console.log(calc(big));
// Array [ "e", "b", "b", "b", "a", "a", "d", "c" ]
How can I produce all of the combinations of the values in N number of JavaScript arrays of variable lengths?
Let's say I have N number of JavaScript arrays, e.g.
var first = ['a', 'b', 'c', 'd'];
var second = ['e'];
var third = ['f', 'g', 'h', 'i', 'j'];
(Three arrays in this example, but its N number of arrays for the problem.)
And I want to output all the combinations of their values, to produce
aef
aeg
aeh
aei
aej
bef
beg
....
dej
EDIT: Here's the version I got working, using ffriend's accepted answer as the basis.
var allArrays = [['a', 'b'], ['c', 'z'], ['d', 'e', 'f']];
function allPossibleCases(arr) {
if (arr.length === 0) {
return [];
}
else if (arr.length ===1){
return arr[0];
}
else {
var result = [];
var allCasesOfRest = allPossibleCases(arr.slice(1)); // recur with the rest of array
for (var c in allCasesOfRest) {
for (var i = 0; i < arr[0].length; i++) {
result.push(arr[0][i] + allCasesOfRest[c]);
}
}
return result;
}
}
var results = allPossibleCases(allArrays);
//outputs ["acd", "bcd", "azd", "bzd", "ace", "bce", "aze", "bze", "acf", "bcf", "azf", "bzf"]
This is not permutations, see permutations definitions from Wikipedia.
But you can achieve this with recursion:
var allArrays = [
['a', 'b'],
['c'],
['d', 'e', 'f']
]
function allPossibleCases(arr) {
if (arr.length == 1) {
return arr[0];
} else {
var result = [];
var allCasesOfRest = allPossibleCases(arr.slice(1)); // recur with the rest of array
for (var i = 0; i < allCasesOfRest.length; i++) {
for (var j = 0; j < arr[0].length; j++) {
result.push(arr[0][j] + allCasesOfRest[i]);
}
}
return result;
}
}
console.log(allPossibleCases(allArrays))
You can also make it with loops, but it will be a bit tricky and will require implementing your own analogue of stack.
I suggest a simple recursive generator function as follows:
// Generate cartesian product of given iterables:
function* cartesian(head, ...tail) {
let remainder = tail.length ? cartesian(...tail) : [[]];
for (let r of remainder) for (let h of head) yield [h, ...r];
}
// Example:
const first = ['a', 'b', 'c', 'd'];
const second = ['e'];
const third = ['f', 'g', 'h', 'i', 'j'];
console.log(...cartesian(first, second, third));
You don't need recursion, or heavily nested loops, or even to generate/store the whole array of permutations in memory.
Since the number of permutations is the product of the lengths of each of the arrays (call this numPerms), you can create a function getPermutation(n) that returns a unique permutation between index 0 and numPerms - 1 by calculating the indices it needs to retrieve its characters from, based on n.
How is this done? If you think of creating permutations on arrays each containing: [0, 1, 2, ... 9] it's very simple... the 245th permutation (n=245) is "245", rather intuitively, or:
arrayHundreds[Math.floor(n / 100) % 10]
+ arrayTens[Math.floor(n / 10) % 10]
+ arrayOnes[Math.floor(n / 1) % 10]
The complication in your problem is that array sizes differ. We can work around this by replacing the n/100, n/10, etc... with other divisors. We can easily pre-calculate an array of divisors for this purpose. In the above example, the divisor of 100 was equal to arrayTens.length * arrayOnes.length. Therefore we can calculate the divisor for a given array to be the product of the lengths of the remaining arrays. The very last array always has a divisor of 1. Also, instead of modding by 10, we mod by the length of the current array.
Example code is below:
var allArrays = [first, second, third, ...];
// Pre-calculate divisors
var divisors = [];
for (var i = allArrays.length - 1; i >= 0; i--) {
divisors[i] = divisors[i + 1] ? divisors[i + 1] * allArrays[i + 1].length : 1;
}
function getPermutation(n) {
var result = "", curArray;
for (var i = 0; i < allArrays.length; i++) {
curArray = allArrays[i];
result += curArray[Math.floor(n / divisors[i]) % curArray.length];
}
return result;
}
Provided answers looks too difficult for me. So my solution is:
var allArrays = new Array(['a', 'b'], ['c', 'z'], ['d', 'e', 'f']);
function getPermutation(array, prefix) {
prefix = prefix || '';
if (!array.length) {
return prefix;
}
var result = array[0].reduce(function(result, value) {
return result.concat(getPermutation(array.slice(1), prefix + value));
}, []);
return result;
}
console.log(getPermutation(allArrays));
You could take a single line approach by generating a cartesian product.
result = items.reduce(
(a, b) => a.reduce(
(r, v) => r.concat(b.map(w => [].concat(v, w))),
[]
)
);
var items = [['a', 'b', 'c', 'd'], ['e'], ['f', 'g', 'h', 'i', 'j']],
result = items.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));
console.log(result.map(a => a.join(' ')));
.as-console-wrapper { max-height: 100% !important; top: 0; }
Copy of le_m's Answer to take Array of Arrays directly:
function *combinations(arrOfArr) {
let [head, ...tail] = arrOfArr
let remainder = tail.length ? combinations(tail) : [[]];
for (let r of remainder) for (let h of head) yield [h, ...r];
}
Hope it saves someone's time.
You can use a typical backtracking:
function cartesianProductConcatenate(arr) {
var data = new Array(arr.length);
return (function* recursive(pos) {
if(pos === arr.length) yield data.join('');
else for(var i=0; i<arr[pos].length; ++i) {
data[pos] = arr[pos][i];
yield* recursive(pos+1);
}
})(0);
}
I used generator functions to avoid allocating all the results simultaneously, but if you want you can
[...cartesianProductConcatenate([['a', 'b'], ['c', 'z'], ['d', 'e', 'f']])];
// ["acd","ace","acf","azd","aze","azf","bcd","bce","bcf","bzd","bze","bzf"]
Easiest way to find the Combinations
const arr1= [ 'a', 'b', 'c', 'd' ];
const arr2= [ '1', '2', '3' ];
const arr3= [ 'x', 'y', ];
const all = [arr1, arr2, arr3];
const output = all.reduce((acc, cu) => {
let ret = [];
acc.map(obj => {
cu.map(obj_1 => {
ret.push(obj + '-' + obj_1)
});
});
return ret;
})
console.log(output);
If you're looking for a flow-compatible function that can handle two dimensional arrays with any item type, you can use the function below.
const getUniqueCombinations = <T>(items : Array<Array<T>>, prepend : Array<T> = []) : Array<Array<T>> => {
if(!items || items.length === 0) return [prepend];
let out = [];
for(let i = 0; i < items[0].length; i++){
out = [...out, ...getUniqueCombinations(items.slice(1), [...prepend, items[0][i]])];
}
return out;
}
A visualisation of the operation:
in:
[
[Obj1, Obj2, Obj3],
[Obj4, Obj5],
[Obj6, Obj7]
]
out:
[
[Obj1, Obj4, Obj6 ],
[Obj1, Obj4, Obj7 ],
[Obj1, Obj5, Obj6 ],
[Obj1, Obj5, Obj7 ],
[Obj2, Obj4, Obj6 ],
[Obj2, Obj4, Obj7 ],
[Obj2, Obj5, Obj6 ],
[Obj2, Obj5, Obj7 ],
[Obj3, Obj4, Obj6 ],
[Obj3, Obj4, Obj7 ],
[Obj3, Obj5, Obj6 ],
[Obj3, Obj5, Obj7 ]
]
You could create a 2D array and reduce it. Then use flatMap to create combinations of strings in the accumulator array and the current array being iterated and concatenate them.
const data = [ ['a', 'b', 'c', 'd'], ['e'], ['f', 'g', 'h', 'i', 'j'] ]
const output = data.reduce((acc, cur) => acc.flatMap(c => cur.map(n => c + n)) )
console.log(output)
2021 version of David Tang's great answer
Also inspired with Neil Mountford's answer
const getAllCombinations = (arraysToCombine) => {
const divisors = [];
let permsCount = 1;
for (let i = arraysToCombine.length - 1; i >= 0; i--) {
divisors[i] = divisors[i + 1] ? divisors[i + 1] * arraysToCombine[i + 1].length : 1;
permsCount *= (arraysToCombine[i].length || 1);
}
const getCombination = (n, arrays, divisors) => arrays.reduce((acc, arr, i) => {
acc.push(arr[Math.floor(n / divisors[i]) % arr.length]);
return acc;
}, []);
const combinations = [];
for (let i = 0; i < permsCount; i++) {
combinations.push(getCombination(i, arraysToCombine, divisors));
}
return combinations;
};
console.log(getAllCombinations([['a', 'b'], ['c', 'z'], ['d', 'e', 'f']]));
Benchmarks: https://jsbench.me/gdkmxhm36d/1
Here's a version adapted from the above couple of answers, that produces the results in the order specified in the OP, and returns strings instead of arrays:
function *cartesianProduct(...arrays) {
if (!arrays.length) yield [];
else {
const [tail, ...head] = arrays.reverse();
const beginning = cartesianProduct(...head.reverse());
for (let b of beginning) for (let t of tail) yield b + t;
}
}
const first = ['a', 'b', 'c', 'd'];
const second = ['e'];
const third = ['f', 'g', 'h', 'i', 'j'];
console.log([...cartesianProduct(first, second, third)])
You could use this function too:
const result = (arrayOfArrays) => arrayOfArrays.reduce((t, i) => { let ac = []; for (const ti of t) { for (const ii of i) { ac.push(ti + '/' + ii) } } return ac })
result([['a', 'b', 'c', 'd'], ['e'], ['f', 'g', 'h', 'i', 'j']])
// which will output [ 'a/e/f', 'a/e/g', 'a/e/h','a/e/i','a/e/j','b/e/f','b/e/g','b/e/h','b/e/i','b/e/j','c/e/f','c/e/g','c/e/h','c/e/i','c/e/j','d/e/f','d/e/g','d/e/h','d/e/i','d/e/j']
Of course you can remove the + '/' in ac.push(ti + '/' + ii) to eliminate the slash from the final result. And you can replace those for (... of ...) with forEach functions (plus respective semicolon before return ac), whatever of those you are more comfortable with.
An array approach without recursion:
const combinations = [['1', '2', '3'], ['4', '5', '6'], ['7', '8']];
let outputCombinations = combinations[0]
combinations.slice(1).forEach(row => {
outputCombinations = outputCombinations.reduce((acc, existing) =>
acc.concat(row.map(item => existing + item))
, []);
});
console.log(outputCombinations);
let arr1 = [`a`, `b`, `c`];
let arr2 = [`p`, `q`, `r`];
let arr3 = [`x`, `y`, `z`];
let result = [];
arr1.forEach(e1 => {
arr2.forEach(e2 => {
arr3.forEach(e3 => {
result[result.length] = e1 + e2 + e3;
});
});
});
console.log(result);
/*
output:
[
'apx', 'apy', 'apz', 'aqx',
'aqy', 'aqz', 'arx', 'ary',
'arz', 'bpx', 'bpy', 'bpz',
'bqx', 'bqy', 'bqz', 'brx',
'bry', 'brz', 'cpx', 'cpy',
'cpz', 'cqx', 'cqy', 'cqz',
'crx', 'cry', 'crz'
]
*/
A solution without recursion, which also includes a function to retrieve a single combination by its id:
function getCombination(data, i) {
return data.map(group => {
let choice = group[i % group.length]
i = (i / group.length) | 0;
return choice;
});
}
function* combinations(data) {
let count = data.reduce((sum, {length}) => sum * length, 1);
for (let i = 0; i < count; i++) {
yield getCombination(data, i);
}
}
let data = [['a', 'b', 'c', 'd'], ['e'], ['f', 'g', 'h', 'i', 'j']];
for (let combination of combinations(data)) {
console.log(...combination);
}