Develop Reference

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)

Find all sets of length n combinations of m arrays

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);

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);
}

Compare two Javascript Arrays and remove Duplicates

It is working well is there any other better way to remove duplicates from one array if it has elements of another array ?.
<script>
var array1 = new Array("a","b","c","d","e","f");
var array2 = new Array("c","e");
for (var i = 0; i<array2.length; i++) {
var arrlen = array1.length;
for (var j = 0; j<arrlen; j++) {
if (array2[i] == array1[j]) {
array1 = array1.slice(0, j).concat(array1.slice(j+1, arrlen));
}
}
}
alert(array1);
</script>

array1 = array1.filter(function(val) {
return array2.indexOf(val) == -1;
});
Or, with the availability of ES6:
array1 = array1.filter(val => !array2.includes(val));
filter() reference here
indexOf() reference here
includes() reference here

The trick, for reasons that are beyond me, is to loop the outer loop downwards (i--) and the inner loop upwards (j++).
See the example bellow:
function test() {
var array1 = new Array("a","b","c","d","e","f");
var array2 = new Array("c","e");
for (var i = array1.length - 1; i >= 0; i--) {
for (var j = 0; j < array2.length; j++) {
if (array1[i] === array2[j]) {
array1.splice(i, 1);
}
}
}
console.log(array1)
}
How do I know this? See the below:
for( var i =myArray.length - 1; i>=0; i--){
for( var j=0; j<toRemove.length; j++){
if(myArray[i] === toRemove[j]){
myArray.splice(i, 1);
}
}
}
or
var myArray = [
{name: 'deepak', place: 'bangalore'},
{name: 'chirag', place: 'bangalore'},
{name: 'alok', place: 'berhampur'},
{name: 'chandan', place: 'mumbai'}
];
var toRemove = [
{name: 'deepak', place: 'bangalore'},
{name: 'alok', place: 'berhampur'}
];
for( var i=myArray.length - 1; i>=0; i--){
for( var j=0; j<toRemove.length; j++){
if(myArray[i] && (myArray[i].name === toRemove[j].name)){
myArray.splice(i, 1);
}
}
}
alert(JSON.stringify(myArray));
On that note, would anyone be able to explain why the outer loop needs to be looped downwards (--)?
Good luck!

Using the Set.prototype Constructor: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set
let array1 = Array('a', 'b', 'c', 'd', 'e', 'f')
let array2 = Array('c', 'e', 'g')
let concat = array1.concat(array2) // join arrays => [ 'a', 'b', 'c', 'd', 'e', 'f', 'c', 'e', 'g' ]
// Set will filter out duplicates automatically
let set = new Set(concat) // => Set { 'a', 'b', 'c', 'd', 'e', 'f', 'g' }
// Use spread operator to extend Set to an Array
let result = [...set]
console.log(result) // => [ 'a', 'b', 'c', 'd', 'e', 'f', 'g' ]

You can try this
array1 = array1 .filter(val => {
return !array2.find((val2)=>{
// console.log({valueID:val.id+":"+val2.id});
return val.id===val2.id
})
});

use Array.splice()
var array1 = ['1', '2', '3', '4', '5'];
var array2 = ['4', '5'];
var index;
for (var i=0; i<array2.length; i++) {
index = array1.indexOf(array2[i]);
if (index > -1) {
array1.splice(index, 1);
}
}

This my solution
array1 = array1.filter(function(val) {
return array2.indexOf(val.toString()) == -1;
});

This is my solution to remove duplicate in ES6.
let foundDuplicate = false;
existingOptions.some(existingItem => {
result = result.filter(item => {
if (existingItem.value !== item.value) {
return item;
} else {
foundDuplicate = true;
}
});
return foundDuplicate;
});
I used this approach because in my case I was having array of objects and indexOf was having problem with it.

window.onload = function () {
var array1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm'];
var array2 = ['c', 'h', 'k'];
var array3 = [];
var SecondarrayIndexcount = 0;
for (var i = 0; i < array1.length; i++) {
for (var j = 0; j < array2.length; j++) {
if (array1[i] !== array2[j]) {
if (SecondarrayIndexcount === (array2.length - 1)) {
array3.push(array1[i]);
SecondarrayIndexcount = 0;
break;
}
SecondarrayIndexcount++;
}
}
}
for (var i in array3) {
alert(array3[i]);
}
}
</script>