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true if the specified number of identical elements of the array is found

Let's say I have an array
var array = ["A", "A", "A", "B", "B", "A", "C", "B", "C"];
And I want to do a check, and get true if there are 4 identical elements "A"
If I use array.includes("A"), then it looks for only one such among all and in any case returns true, but I need when there are exactly 4 such elements
Or let's say in the same array I want to find 3 elements "B" and 2 elements "C", and return true only if there are as many of them as I'm looking for, if less, then return false
How can I do this?

Take a shot like this.
const myArr = ["A", "A", "A", "B", "B", "A", "C", "B", "C"];
const findExactly = (arr, val, q) => arr.filter(x => x == val).length == q;
// logs "true" as expected :)
console.log(findExactly(myArr, 'A', 4));
So the function findExactly receives an array, a value and a number X as the quantity. Returns a boolean if the array contains the value X times. So the example above works for the example you gave on the question "And I want to do a check, and get true if there are 4 identical elements "A"".

Pulling my answer from this on stackflow.
var array = ["A", "A", "A", "B", "B", "A", "C", "B", "C"];
const counts = {};
for (const el of arr) {
counts[el] = counts[el] ? counts[el] + 1 : 1;
}
console.log(counts["A"] > 4) // 4 or more "A"s have been found
console.log(counts["B"] === 3 && counts["C"] === 3) // exactly 3 "B"s and 2 "C"s have been found

Checking values of one array against another in JS

I'm trying to do a check that the first array contains the same values of the second array.
However I'm confused about my code.
First question is: why is my code running my else statement if all letters in the first array are contained in the second? it will run 2 lines of "this is not valid"
Second question is: if my first array contains a duplicate letter it will still pass the check e.g
["a", "b" , "a", "d", "e", "f"]; even though there is two a's in the first it will see the same "a" again. Anyone know a way around this.
Sorry for my long winded questions but I hope it makes sense. Thanks :)
var letters = ["a", "b" , "c", "d", "e", "f"];
var otherLetters = ["a","b", "c" , "d", "e", "f"];
var i = -1;
while(i<=letters.length){
i++;
if(otherLetters.includes(letters[i])){
console.log("This is valid");
}
else
console.log("This is not valid");
}

You didn't close the brackets. And your loop is very confusing, please use foreach. Here is a working example:
const letters = ["a", "b" , "c", "d", "e", "f"];
const otherLetters = ["a","b", "c" , "d", "e", "f"];
letters.forEach(el => {
if (otherLetters.includes(el)) {
console.log(el + 'is valid');
} else {
console.log(el + 'is not valid');
}
});

You are trying to access array elements which are out of bounds. The script runs 8 iterations over an array with 6 elements.

Nothing to worry, cpog90.
Try this solution.
var letters = ["a", "b" , "c", "d", "e", "f"];
var otherLetters = ["a","b", "c" , "d", "e", "f"];
var i = 0;
while(i<letters.length){
if(otherLetters.includes(letters[i])){
console.log("This is valid");
}
else {
console.log("This is not valid "+i);
}
i++;
}
What went wrong in your logic?
If you declare i = -1 and while(i<=letters.length), As 6 is length of letters, 8 iterations will be done as follows.
For first iteration (i = -1), 'while' condition returns true and checks for 'a'
output: This is valid
For second iteration (i = 0), 'while' condition returns true and checks for 'b'
output: This is valid
For third iteration (i = 1), 'while' condition returns true and checks for 'c'
output: This is valid
For fourth iteration (i = 2), 'while' condition returns true and checks for 'd'
output: This is valid
For fifth iteration (i = 3), 'while' condition returns true and checks for 'e'
output: This is valid
For sixth iteration (i = 4), 'while' condition returns true and checks for 'f'
output: This is valid
For seventh iteration (i = 5), 'while' condition returns true and checks for undefined value.
output: This is not valid
For eighth iteration (i = 6), 'while' condition returns true and checks for undefined value.
output: This is not valid

First of all you have set i = -1 which is confusing since array start position is 0.
The reason your loop is running two extra times is because loop started at -1 instead of 0 and next the condition i <= length.
Since [array length = last index + 1] your loop runs extra two times.
Just to make your code work assign var i = 0 and while condition i < letters.length

Simplest solution is using lodash. It has all optimizations out-of-the-box:
var letters = ["a", "b" , "c", "d", "e", "f"];
var otherLetters = ["f", "a","b", "c" , "d", "e"];
const finalLetters = _.sortBy(letters);
const finalOtherLetters = _.sortBy(otherLetters);
if (_.isEqual(finalLetters, finalOtherLetters)) {
console.log('Two arrays are equal.');
} else {
console.log('Two arrays are not equal.');
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js"></script>

Arrays are index based and starts from 0. So, the -1 and the less than letters.length check puts the code into out of bounds.
var letters = ["a", "b" , "c", "d", "e", "f"];
var otherLetters = ["a","b", "c" , "d", "e", "f"];
var i = 0;
while(i<letters.length)
{
if(otherLetters.includes(letters[i]))
{
console.log("This is valid");
}
else
{
console.log("This is not valid");
}
i++;
}

You can use a combination of Array.prototype.every with Array.prototype.includes, along with some extra guard clauses.
const areSequenceEqual = (arr1, arr2) => {
if (!arr1 || !arr2) {
return false;
}
if (arr1.length !== arr2.length) {
return false;
}
return arr1.every(x => arr2.includes(x));
};
const letters = ["a", "b", "c", "d", "e", "f"];
const otherLetters = ["a", "b", "c", "d", "e", "f"];
const someOtherLetters = ["a", "b", "c", "d", "e", "f", "g"];
console.log(areSequenceEqual(letters, otherLetters));
console.log(areSequenceEqual(letters, undefined));
console.log(areSequenceEqual(letters, someOtherLetters));

change order of array based on another array

Let say we have our array like this:
let myArray = ["A", "B", "C", "D"]
What if we want to modify the order of elements in myArray based on modifier array so that, if myArray includes any element of modifier then we send that element to the end of the myArray
Like this:
let modifier = ["B"]
myArray = ["A", "C", "D", "B"] // B is sent to the end of myArray
And if we have this:
let modifier = ["A", "C"]
myArray = ["B", "D", "A", "C"] // A and C are sent to the end of the array
I have tried looping and checking each array element against another but it went complicated...
Any help would be greatly appreciated.

Very simple.
Step-1: Remove elements of modifier array from original array
myArray = myArray.filter( (el) => !modifier.includes(el) );
Step-2: Push modifier array into original array
myArray = myArray.concat(modifier)
Update
As per demands in comments by seniors :) If use case is to move multiple data:
var myArray = ["A", "A", "A", "B", "B", "B", "C", "D", "E"];
var modifier = ["A", "B"];
// get static part
staticArray = myArray.filter( (el) => !modifier.includes(el) );
// get moving part
moveableArray = myArray.filter( (el) => modifier.includes(el) );
// merge both to get final array
myArray = staticArray.concat(moveableArray);
console.log(myArray);

You could sort the array and move the items of modifier to the end of the array.
function sort(array, lastValues) {
var last = Object.fromEntries(lastValues.map((v, i) => [v, i + 1]));
return array.sort((a, b) => (last[a] || - Infinity) - (last[b] || - Infinity));
}
var array = ["A", "B", "C", "D"];
console.log(...sort(array, ["B"]));
console.log(...sort(array, ["A", "C"]));

Simply use this to get desired result
let myArray = ["A", "B", "C", "D"];
let modifier = ["A", "C"];
for(let i=0;i<modifier.length;i++){
if(myArray.includes(modifier[i])){
myArray.splice(myArray.indexOf(modifier[i]), modifier[i]);
myArray.push(modifier[i]);
}
}
console.log(myArray);

How to print an array verbatim Javascript [duplicate]

This question already has answers here:
Convert array to JSON
(12 answers)
Closed 4 years ago.
Say I have an array "x", and x shows as [["A", "B", "C"], ["D", "E", "F"]] in the console. How would I convert this to a string verbatim, with all the brackets and quotes?
For example, this should output
"[["A", "B", "C"], ["D", "E", "F"]]"
(if possible, it would also be nice to add backslashes to the special characters, like "\[\[\"A\", \"B\", \"C\"\], \[\"D\", \"E\", \"F\"\]\]")

Try JSON.stringify:
var x = [["A", "B", "C"], ["D", "E", "F"]];
var string = JSON.stringify(x);

To print it like in your console you can use JSON.stringify().
const arr = [["A", "B", "C"], ["D", "E", "F"]];
const str = JSON.stringify(arr); // [["A","B","C"],["D","E","F"]]
If you also want to add a backslash to every bracket and quotation mark you could loop through every character of the string and add a backslash where it's needed.
// ES6 version
let escapedStr = "";
[...str].forEach(c => {
if (c === ']' || c === '[' || c === '"') c = `\\${c}`;
escapedStr += c;
});
console.log(escapedStr) // \[\[\"A\",\"B\",\"C\"\],\[\"D\",\"E\",\"F\"\]\]
If you do not want use ES6 you can do the same with a simple for loop
for (var i = 0, c=''; c = str.charAt(i); i++) {
if (c === ']' || c === '[' || c === '"') c = `\\${c}`;
escapedStr += c;
}
Edit: Using Regex you can solve the problem in one line
const arr = [["A", "B", "C"], ["D", "E", "F"]];
JSON.stringify(arr).replace(/(\[)|(\])|(\")/g, $1 => `\\${$1}`)

Removing Element From Array of Arrays with Regex

I'm using regex to test certain elements in an array of arrays. If an inner array doesn't follow the desired format, I'd like to remove it from the main/outer array. The regex I'm using is working correctly. I am not sure why it isn't removing - can anyone advise or offer any edits to resolve this problem?
for (var i = arr.length-1; i>0; i--) {
var a = /^\w+$/;
var b = /^\w+$/;
var c = /^\w+$/;
var first = a.test(arr[i][0]);
var second = b.test(arr[i][1]);
var third = c.test(arr[i][2]);
if ((!first) || (!second) || (!third)){
arr.splice(i,1);
}

When you cast splice method on an array, its length is updated immediately. Thus, in future iterations, you will probably jump over some of its members.
For example:
var arr = ['a','b','c','d','e','f','g']
for(var i = 0; i < arr.length; i++) {
console.log(i, arr)
if(i%2 === 0) {
arr.splice(i, 1) // remove elements with even index
}
}
console.log(arr)
It will output:
0 ["a", "b", "c", "d", "e", "f", "g"]
1 ["b", "c", "d", "e", "f", "g"]
2 ["b", "c", "d", "e", "f", "g"]
3 ["b", "c", "e", "f", "g"]
4 ["b", "c", "e", "f", "g"]
["b", "c", "e", "f"]
My suggestion is, do not modify the array itself if you still have to iterate through it. Use another variable to save it.
var arr = ['a','b','c','d','e','f','g']
var another = []
for(var i = 0; i < arr.length; i++) {
if(i%2) {
another.push(arr[i]) // store elements with odd index
}
}
console.log(another) // ["b", "d", "f"]
Or you could go with Array.prototype.filter, which is much simpler:
arr.filter(function(el, i) {
return i%2 // store elements with odd index
})
It also outputs:
["b", "d", "f"]

Your code seems to work to me. The code in your post was missing a } to close the for statement but that should have caused the script to fail to parse and not even run at all.
I do agree with Leo that it would probably be cleaner to rewrite it using Array.prototype.filter though.
The code in your question would look something like this as a filter:
arr = arr.filter(function (row) {
return /^\w+$/.test(row[0]) && /^\w+$/.test(row[1]) && /^\w+$/.test(row[2]);
});
jsFiddle
I'm assuming it is 3 different regular expressions in your actual code, if they are all identical in your code you can save a little overhead by defining the RegExp literal once:
arr = arr.filter(function (row) {
var rxIsWord = /^\w+$/;
return rxIsWord.test(row[0]) && rxIsWord.test(row[1]) && rxIsWord.test(row[2]);
});