Find all sets of length n combinations of m arrays - javascript

Objective: Find all sets of length n combinations of m arrays, such that index i of each item in a set is not the same as i in any other element of that set
I have the following arrays:
array1 = ['a', 'b', 'c', 'd'];
array2 = ['e', 'f', 'g', 'h'];
array3 = ['i', 'j', 'k', 'l'];
array4 = ['m', 'n', 'o', 'p'];
I would like to find every possible combination of these taking one element from each array, but then place those combinations into sets such that index i of any element in a given set is different to index i of another element in that same set. For instance, one set could be:
[
{ 1: "a", 2: "e", 3: "i", 4: "m" },
{ 1: "b", 2: "f", 3: "j", 4: "n" },
{ 1: "c", 2: "g", 3: "k", 4: "o" },
{ 1: "d", 2: "h", 3: "l", 4: "p" }
]
as every property '1' is different and taken from array1, every property '2' is different and taken from array2, etc.
Now I need to find every possible one of these.
I've tried looking at this post and implement it by creating combinations of combinations before filtering out everything invalid and cycling through to establish sets, but of course this missed many and took almost an hour to run on this example. Therefore, I need a more systematic approach to speed up the process and make it neater.

You basically want to find every permutation of every array and combine them. This can be done recursively:
function permutate(arr) {
// every array of length one is already permutated
if (arr.length == 1) return [ arr ];
let permutations = [];
for (let i = 0; i < arr.length; i++) {
// Remove the current element and permutate the rest
let sub = permutate(arr.splice(i, 1));
// Insert current element into every permutation
sub = sub.map(x => [arr[i], ...x]);
// Add permutations to list
permutations.push(...sub);
}
return permutations;
}
Next the combine function:
function combine(arrays, current = [], i = 0) {
if (i == arrays.length)
return [ current ];
let values = [];
for (let j = 0; j < arrays[i].length; j++) {
let temp = current.slice();
temp.push(arrays[i][j]);
values.push(...combine(arrays, temp, i + 1));
}
return values;
}
// If you get a call stack size exceeded (stackoverflow) error, you can replace
// this using nested for loops. For instance for 5 arrays with 5 elements each:
let sets = [];
for (let i = 0; i < permutations[0].length; i++) {
for (let j = 0; j < permutations[1].length; j++) {
for (let k = 0; k < permutations[2].length; k++) {
for (let l = 0; l < permutations[3].length; l++) {
for (let m = 0; m < permutations[4].length; m++) {
let set = [];
for (let n = 0; n < 5; n++) {
set.push([ permutations[0][i][n], permutations[1][j][n], permutations[2][k][n], permutations[3][l][n], permutations[4][m][n] ]);
}
sets.push(set);
}
}
}
}
}
By first permutating every array (which results in 24 different permutations for each one), then combining these (which is 24^4=331776 combinations), you'll get everything you need to construct the arrays. Just loop over every combination and put the elements at the same indices into the same set:
let permutations = [ array1, array2, array3, array4 ].map(arr => permutate(arr));
let sets = combine(permutations);
let out = [];
for (let i = 0; i < sets.length; i++) {
let set = [];
for (let j = 0; j < 4; j++) {
set.push([ sets[i][0][j], sets[i][1][j], sets[i][2][j], sets[i][3][j] ]);
}
out.push(set);
}
Working example:
array1 = ['a', 'b', 'c', 'd'];
array2 = ['e', 'f', 'g', 'h'];
array3 = ['i', 'j', 'k', 'l'];
array4 = ['m', 'n', 'o', 'p'];
function permutate(arr) {
if (arr.length == 1) return [ arr ];
let permutations = [];
for (let i = 0; i < arr.length; i++) {
let temp = arr.slice();
temp.splice(i, 1);
let sub = permutate(temp);
sub = sub.map(x => [arr[i], ...x]);
permutations.push(...sub);
}
return permutations;
}
function combine(arrays, current = [], i = 0) {
if (i == arrays.length)
return [ current ];
let values = [];
for (let j = 0; j < arrays[i].length; j++) {
let temp = current.slice();
temp.push(arrays[i][j]);
values.push(...combine(arrays, temp, i + 1));
}
return values;
}
let permutations = [ array1, array2, array3, array4 ].map(arr => permutate(arr));
console.log(permutations);
let sets = combine(permutations);
let out = [];
for (let i = 0; i < sets.length; i++) {
let set = [];
for (let j = 0; j < 4; j++) {
set.push([ sets[i][0][j], sets[i][1][j], sets[i][2][j], sets[i][3][j] ]);
}
out.push(set);
}
console.log(out);

Related

Combine each element of an array with the ones after it

I am trying to combine items in an array, with every item below it. It should make a set of the current character and each character below it, and iteratively walk down the array. For example, if I have an array like this:
var myArray = ['A','B','C','D']
I would like an output like this:
AB AC AD BC BD CD
The code I have is getting me close, but I am having a hard time figuring out the rest. Here is what I have so far:
var myArray = ['A', 'B', 'C', 'D']
var sql_parts = []
var string = "";
for (var i = 0; i < myArray.length; i++) {
recurse_function(string, i)
}
console.log(sql_parts)
function recurse_function(string_val, count) {
if ((myArray.length - count) == 0) {
return string_val;
} else {
string_val += myArray[count]
sql_parts.push(string_val)
recurse_function(string_val, count + 1)
}
}
But this produces:
["A", "AB", "ABC", "ABCD", "B", "BC", "BCD", "C", "CD", "D"]
Here is one solution:
Define the recursive function to take the array and an empty list initially to store the combinations
The base condition is when the array is empty or has one element
Otherwise, Remove the first element "start"
Iterate over the array to store its combinations with its following elements
Recur again with array and combinations updated
function recurse_function(array, combinations = []) {
if(array.length <= 1) return combinations;
const start = array.shift();
for(let i = 0; i < array.length; i++) combinations.push(`${start}${array[i]}`);
return recurse_function(array, combinations);
}
console.log( recurse_function(['A','B','C','D']) );
var myArray = ['A','B','C','D']
var sql_parts = []
for(var i =0; i< myArray.length; i++){
var a = myArray[i]
for(var j = i+1; j<myArray.length; j++){
var b=myArray[j]
var c= a+b
sql_parts.push(c)
}
}
I think this is something that you're looking for, it's a function that is very cheap on resources. Its not recursive (why you now would need something like that for this simple scenario)
var myArray = ['A','B','C','D']
let [a, b, iter] = [0, 1, 2]
let result = []
for (;a < myArray.length; a++) {
for (;b < myArray.length; b++) {
result.push(myArray[a]+myArray[b])
}
b = iter++
}
console.log(result)
I don't think you need to recurse in this particular case. You can simply combine the current element with the following ones with flatMap:
['A','B','C','D'].flatMap((x, i, xs) => xs.slice(i+1).map(y => x+y));
//=> ["AB", "AC", "AD", "BC", "BD", "CD"]
var myArray = ['A', 'B', 'C', 'D'];
var result = [];
myArray.forEach((item, index) => {
myArray.forEach((item2, index2) => (index < index2 ? result.push(item + item2) : ''));
});
console.log(result);

unexpected results which is empty array

when I run code, I see empty array, but I aim to have an array with some vowels.
let input = 'zai';
const vowels = ['a', 'u', 'o', 'e', 'i'];
let resultArray = [];
for (let inputindex = 0; inputindex < input.inputindex; inputindex++) {
for (let vowel = 0; vowel > vowels.length; vowel++) {
if (input[inputindex] === vowels[vowel]) {
resultArray.push(input[inputindex])
}
}
}
console.log(resultArray)
You have done a mistake by writing input.inputindex instead of input.length. Plus, you have wrote vowel > vowel.length instead of vowel < vowels.length. That's why your loop didn't run at least a single loop and ended up with an empty array. Correct the code as below and try again...
let input = 'zai';
const vowels = ['a', 'u', 'o', 'e', 'i'];
let resultArray = [];
for (let inputindex = 0; inputindex < input.length; inputindex++) {
for (let vowel = 0; vowel < vowels.length; vowel++) {
if (input[inputindex] === vowels[vowel]) {
resultArray.push(input[inputindex])
}
}
}
console.log(resultArray)
input string but you need array, i use split.
and some mistakes in loops
let input = 'zai';
let input2 = input.split('')
const vowels = ['a', 'u', 'o', 'e', 'i'];
let resultArray = [];
for(let inputindex = 0; inputindex < input2.length; inputindex++){
for(let vowel = 0; vowel < vowels.length; vowel++){
if(input2[inputindex] === vowels[vowel]){
resultArray.push(input2[inputindex])
}
}
}
display.log(resultArray)
Without array
let input = 'zai';
const vowels = ['a', 'u', 'o', 'e', 'i'];
let resultArray = [];
for(let inputindex = 0; inputindex < input2.length; inputindex++){
for(let vowel = 0; vowel < vowels.length; vowel++){
if(input.slice(inputindex, inputindex+1) === vowels[vowel]){
resultArray.push(input.slice(inputindex, inputindex+1))
}
}
}
display.log(resultArray)
looping through the inputs and comparing them with each value in vowels using the .some method as so:
let input = 'zai';
let vowels = ['a', 'u', 'o', 'e', 'i'];
let resultArray = [];
for (let inputindex = 0; inputindex < input.length; inputindex++) {
if (vowels.some(v => v === input[inputindex])) {
resultArray.push(input[inputindex])
}
}
console.log(resultArray)

Reverse array with for loops

Quick JavScript question. In the following pieces of code I'm reversing the array arr that's being passed to the function reverseArray.
In the first piece of code, it looks like that the local variable arr keeps changing even though inside the loop I am operating on the variable newArr which hold the initial value of arr. Hence the loop fails when arr.length reaches the values 3.
function reverseArray (arr) {
var newArr = [];
var inArr = arr;
console.log(inArr);
for (i = 0; i < arr.length; i++) {
newArr[i] = inArr.pop(i);
}
return newArr;
}
reverseArray(["A", "B", "C", "D", "E", "F"]);
// OUTPUT: ["F", "D", "E"]
On the other hand, if I store arr.length on local variable numArr, then it works perfectly and reverses the array.
function reverseArray (arr) {
var numArr = arr.length;
var newArr = [];
for (i = 0; i < numArr; i++) {
let inArr = arr;
newArr[i] = inArr.pop(i);
}
return newArr;
}
reverseArray(["A", "B", "C", "D", "E", "F"]);
// OUTPUT: ["F", "E", "D", "C", "B", "A"]
What am I missing?
pop (MDN, spec) is a mutator method: It changes the state of the array you call it on. So naturally, inArr.pop(1) modifies arr (in your first example), since inArr and arr both refer to the same array.
Probably worth noting as well that pop doesn't accept any parameters, so that 1 doesn't do anything.
In your first example, your best bet is to just assign another variable (say, j) the initial value arr.length - 1 and use arr[j] to get the value, then decrease j as you increase i. (Also, no point to inArr, and you need to declare i to avoid what I call The Horror of Implicit Globals:
function reverseArray (arr) {
var newArr = [];
for (var i = 0, j = arr.length - 1; i < arr.length; i++, j--) {
newArr[i] = arr[j];
}
return newArr;
}
console.log(reverseArray(["A", "B", "C", "D", "E", "F"]));
You can also just use arr[arr.length - i - 1] rather than a second variable:
function reverseArray (arr) {
var newArr = [];
for (var i = 0; i < arr.length; i++) {
newArr[i] = arr[arr.length - i - 1];
}
return newArr;
}
console.log(reverseArray(["A", "B", "C", "D", "E", "F"]));
You could take a copy of the array and use the length of the copy for checking the next pop/push command.
function reverseArray(array) {
var newArr = [],
inArr = array.slice(); // take copy of primitive values
while (inArr.length) { // check decrementing length
newArr.push(inArr.pop());
}
return newArr;
}
console.log(reverseArray(["A", "B", "C", "D", "E", "F"]));
To fullfill the condition, you could use a for statement as well.
function reverseArray(array) {
var newArr = [],
inArr = array.slice(); // take copy of primitive values
for(; inArr.length; ) { // check decrementing length
newArr.push(inArr.pop());
}
return newArr;
}
console.log(reverseArray(["A", "B", "C", "D", "E", "F"]));
Here It Goes using two For Loops:
let arr = [1, 12, 15, 16, 78, 89, 53];
for (let i = 0; i < arr.length; i++) {
for (let j = i + 1; j < arr.length; j++) {
let c, a, b;
b = arr[i];
arr[i] = arr[j];
arr[j] = b;
}
}
console.log(arr);
Example:
fruit=['banana', 'orange', 'mango', 'lemon'] //array
z=[] //empty array
for(i=(fruit.length-1);i>-1;i--){
z.push(fruit[i])
}
//i>-1 because the index of array ends at 0
console.log(z)
The problem with your first function is that when you pop an element from an array arr its actually decreasing the length of your array arr. That is why you see less results then the original number of elements.
so your code worked like this
1st pass: i=0 and arr.length = 6; // 0<6 true
2nd pass: i=1 and arr.length = 5; // 1<5 true
3rd pass: i=2 and arr.length = 4; // 2<4 true
4th pass: i=3 and arr.length = 3; // 3<3 false and it does not execute
so you see only three results.
let input = ["A", "B", "C", "D", "E", "F"];
let output = [];
for(let i = 1; i <= input.length; i++) {
// determine correct array index in respect to the current iteration
let index = input.length - i;
// push to new array
output.push(reverse[index]);
}
// verify result
console.log(output);
This is similar code to reverse an Array.
function reverseArray(arr) {
let newArr = [];
for (i = arr.length - 1; i >= 0; i--) {
newArr.push(arr[i]);
}
return newArr;
}
console.log(reverseArray(["A", "B", "C", "D", "E", "F"]));
//output: ["F","E","D","C","B","A"]
function reverseArray(a) {
let arrayLength = a.length
for (let i = 0; i< arrayLength/2; i ++){
let temp = a[i]
a[i] = a[arrayLength -(i+1)]
a[arrayLength - (i+1)] = temp
}
return a
}
reverseArray([1,2,3,4,5,6,7,8,9]) //expect: 9,8,7,6,5,4,3,2,1
You could try to use this code:
function reverseString(str) {
var newString = "";
for (var i = str.length - 1; i >= 0; i--) {
newString += str[i];
}
return newString;
}

Getting an array result not in order

function pairElement(str) {
var arr = [['G','C'],['C','G'],['A','T'],['T','A']],b=[]
for(var k=0;k<arr.length;++k){
var res = arr.filter(function(v){
return v;
})[k][0]
var j=0;
while(j<str.length){
if(str[j]===res){
b.push(arr[k])
}
j++;
}
}
return b;
}
console.log(pairElement("ATCGA"));
I want pairing result in order of the argument passed to the main function. This code's result should be
[['A','T'],['T','A'],['C','G'],['G','C'],['A','T']] but i'm getting as [['G','C'],['C','G'],['A','T'],['A','T'],['T','A']]
Your inner and outer loops are flipped. The outer loop should iterate over the string, and the inner loop should iterate over the array.
function pairElement(str) {
var arr = [['G', 'C'], ['C', 'G'], ['A', 'T'], ['T', 'A']],
b = [];
for (var k = 0; k < str.length; k++) {
for (var j = 0; j < arr.length; j++) {
if (str[k] === arr[j][0]) {
b.push(arr[j]);
}
}
}
return b;
}
console.log(pairElement("ATCGA"));
Your code can also be simplified using an object instead of a 2D array, and with Array#map:
function pairElement(str) {
var pairs = { 'G': 'C', 'C': 'G', 'A': 'T', 'T': 'A' };
return str.split('').map(ch => [ch, pairs[ch]]);
}
console.log(pairElement("ATCGA"));

How to access array in circular manner in JavaScript

I have an array like [A,B,C,D]. I want to access that array within a for loop like as
var arr = [A,B,C,D];
var len = arr.length;
for(var i = 0; i<len; i++){
0 - A,B,C
1 - B,C,D
2 - C,D,A
3 - D,A,B
}
I want to access that like in JavaScript, any ideas?
Answering to the main question, someone can access an array in a circular manner using modular arithmetic. That can be achieved in JavaScript with the modulus operator (%) and a workaround.
Given an array arr of a length n and a value val stored in it that will be obtained through an access index i, the circular manner, and safer way, to access the array, disregarding the value and sign of i, would be:
let val = arr[(i % n + n) % n];
This little trick is necessary -- someone can not use the modulus result straightforwardly -- because JavaScript always evaluates a modulus operation as the remainder of the division between dividend (the first operand) and divisor (the second operand) disconsidering their signs but assigning to the remainder the sign of the dividend. That behavior does not always result in the desired "wrap around" effect of the modular arithmetic and could result in a wrong access of a negative position of the array.
References for more information:
https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/what-is-modular-arithmetic
https://en.wikipedia.org/wiki/Modular_arithmetic
https://en.wikipedia.org/wiki/Modulo_operation
https://dev.to/maurobringolf/a-neat-trick-to-compute-modulo-of-negative-numbers-111e
Try this:
var arr = ["A","B","C","D"];
for (var i=0, len=arr.length; i<len; i++) {
alert(arr.slice(0, 3).join(","));
arr.push(arr.shift());
}
Without mutating the array, it would be
for (var i=0, len=arr.length; i<len; i++) {
var str = arr[i];
for (var j=1; j<3; j++)
str += ","+arr[(i+j)%len]; // you could push to an array as well
alert(str);
}
// or
for (var i=0, len=arr.length; i<len; i++)
alert(arr.slice(i, i+3).concat(arr.slice(0, Math.max(i+3-len, 0)).join(","));
Simply using modulus operator you can access array in circular manner.
var arr = ['A', 'B', 'C', 'D'];
for (var i = 0, len = arr.length; i < len; i++) {
for (var j = 0; j < 3; j++) {
console.log(arr[(i + j) % len])
}
console.log('****')
}
how about this one-liner I made ?
var nextItem = (list.indexOf(currentItem) < list.length - 1)
? list[list.indexOf(currentItem) + 1] : list[0];
for (var i = 0; i < arr.length; i++) {
var subarr = [];
for (var j = 0; j < 3; j++) {
subarr.push(arr[(i+j) % arr.length]);
}
console.log(i + " - " + subarr.join(','));
}
One line solution for "in place" circular shift:
const arr = ["A","B","C","D"];
arr.forEach((x,i,t) => {console.log(i,t); t.push(t.shift());});
console.log("end of cycle", arr); // control: cycled back to the original
logs:
0 Array ["A", "B", "C", "D"]
1 Array ["B", "C", "D", "A"]
2 Array ["C", "D", "A", "B"]
3 Array ["D", "A", "B", "C"]
"end of cycle" Array ["A", "B", "C", "D"]
If you want only the first 3 items, use:
arr.forEach((x,i,t) => {console.log(i,t.slice(0, 3)); t.push(t.shift());});
Another solutions:
var arr = ['A','B','C','D'];
var nextVal = function (arr) {
return arr[( ( ( nextVal.counter < ( arr.length - 1 ) ) ? ++nextVal.counter : nextVal.counter=0 ) )];
};
for(var i=0;i<arr.length;i++){
console.log(nextVal(arr)+','+nextVal(arr)+','+nextVal(arr));
}
And based on Modulo :
var arr = ['A','B','C','D'];
var len = arr.length;
var nextVal = function (arr, dir = 1) {
if ( dir < 0 ) { nextVal.counter--;}
let i = (nextVal.counter % len + len) % len;
if ( dir > 0 ) { nextVal.counter++; }
return arr[i];
};
nextVal.counter=0;
for(var i=0;i<arr.length;i++){
console.log(nextVal(arr)+','+nextVal(arr)+','+nextVal(arr));
}
// in reverse
console.log('-------------------');
nextVal.counter=0;
for(var i=0; i<10; i++) {
console.log(nextVal(arr, -1)+','+nextVal(arr, -1)+','+nextVal(arr, -1));
}
You could get the sliced part from index and the rest of slicing from start, if necessary.
This appraoch does not mutate the array.
const
getCircular = (array, size) => array.map((_, i, a) => [
...a.slice(i, i + size),
...a.slice(0, i + size < a.length ? 0 : i + size - array.length)
]);
console.log(getCircular(['A', 'B', 'C', 'D'], 3).map(a => a.join('')));
console.log(getCircular(['A', 'B', 'C', 'D'], 5).map(a => a.join('')));
.as-console-wrapper { max-height: 100% !important; top: 0; }
We can simply achieve this by using Array.splice() method along with the Destructuring assignment
Live Demo :
// Input array.
let arr = ['A', 'B', 'C', 'D'];
// Finding the length of an array.
const len = arr.length;
// Iterattion based on array length.
for(let i = 0; i < len; i++) {
const splittedArr = arr.splice(0, 3);
arr.push(splittedArr[0]);
arr = [splittedArr[1], splittedArr[2], ...arr]
console.log(splittedArr);
}

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