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Compare 2 variables with comma separated values in JavaScript [closed]

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I have 2 variables say a=1,2,3,4 and b=1,2 , I want to compare these 2 variables and form a new variable with common values, please help me

Assuming you meant:
a="1,2,3,4"
b="1,2"
You could break those comma delimited values into arrays and then get the intersections of the arrays. Here's an example:
const ArrayA = a.split(",");
const ArrayB = b.split(",");
const intersection = ArrayA.filter(value => ArrayB.includes(value));
const commonCSV = intersection.join(",");

var a = [1, 2, 3, 4];
var b = [1, 2];
// Take unqiue values from a and concat with unique values from b
const occurrenceMap = [...new Set(a)].concat([...new Set(b)]).reduce((map, el) => {
map.set(el, map.get(el) ? map.get(el) + 1 : 1);
return map;
}, new Map());
const common = [];
// Iterate over the map entries and keep only the ones that occur more than once - these would be the intersection of the 2 arrays.
for (entry of occurrenceMap) {
const [key, val] = entry;
if (val > 1) {
common.push(key);
}
}
console.log(common);

arr1 = [1, 2, 3] const { length } = arr1 is going to output 3, how is this works? [duplicate]

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Javascript object bracket notation ({ Navigation } =) on left side of assign
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Closed 1 year ago.
so basically i found out that if we create array for example arr1 = [1, 2, 3] and new const { length } = arr1; will going to output 3 but how is this works exactly? is there any more operators instead of { length }? and what are the other use of using const { here some stuff } syntax? i do some research but i didnt find anything that points out this syntax. any assumptions?

Basically this is Object destructuring
Basic example:
const user = {
id: 42,
is_verified: true
};
const {id, is_verified} = user;
console.log(id); // 42
console.log(is_verified); // true
In your case:
const arr1 = [1, 2, 3]
arr1.length = 3
// Or,
const { length } = arr1
To know more: Object destructuring

Let examine each line of code :)
const arr1 = [1,2,3];
Here, you define an array "arr1", which has 3 elements, so arr1's length is 3.
In javascript, arr1 is an object, and 1 of its properties is "length", which reflects the length of the array. So, arr1.length = 3
Now the second line :
const {length} = arr1
Here, we use destructuring assignment to get the property "length" from object arr1. This line is equal to :
const length = arr1.length;
which give you length = 3
Make sense?
You can read more about the destructuring assignment here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment

JS .includes() with more than one values of array? [closed]

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I have an array of numbers like A = ['1', '34', '23', '55'] and want to use .includes() to find true or false. However, my tested values are in array like B = ['3', '1', '543', '33']
I'd tried to do A.includes(B) but it seems it is not working. A.includes('1', '123') returns true. How can I use my Array B to do the same?
I wanted to Check if the array A has at least one of the array B’s value then return true. Apology for missing this part!

If I understand you correctly, you're looking to do A.includes(B), but your inputs are stored in arrays. In this case, simply loop through the values and call includes() on the elements:
const A = ['1', '34', '23', '55'];
const B = ['3', '1', '543', '33'];
for (var i = 0; i < A.length; ++i)
console.log(A[i].includes(B[i]));

If you need to check if all values in B are in A, you can do it like below:
B.every(item => A.includes(item)) // true or false

includes() does not work this way. Per the Mozilla docs:
arr.includes(valueToFind[, fromIndex])
Where valueToFind is a single value and fromIndex is the index to start looking from. (In your case it looks like it ignored that because it was a string; otherwise it would have returned false.)
If what you want is to find if array A contains every element of array B, there's no standard library function that does exactly that. Here's what you can do:
function containsAll(container, contained) {
return contained.every(item => container.includes(item))
}
containsAll(A, B);

If I understand you correct you want to check if array B all of it's element includes
in array A or not it would be like this using every
B.every((el) => A.includes(el))
let A = [1, 2, 3], B = [2, 3, 4]
let result = B.every((el) => A.includes(el))
console.log(result)
and if you want at least one element from the second array to be in the first one you could do like this using some
B.some((el) => A.includes(el))
let A = [1, 2, 3], B = [2, 3, 4];
let result = B.some((el) => A.includes(el))
console.log(result)

How to push an array into another [closed]

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A very basic question but still learning.
I have a 1D array say = [a,b,c].
and another 2D array = [[1,2,3],[4,5,6],[7,8,9]].
How do I push the array into beginning of each row of 2D array so that my array result looks like this.
[[a,1,2,3],[b,4,5,6],[c,7,8,9]].

You can iterate each item from 2D array and add 1D array value into it.
For more help please check Unshift
var test = ['a', 'b', 'c'];
var tests = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];
var index = 0;
tests.map(item => { //Iterate each item from 2D araay
if (index < test.length) { // Check if we have any item in 1D array to avoid crash
item.unshift(test[index++]); // Unshift each item so you can add value at 0 index.
}
});
console.log(tests);

array1 = ["a","b","c"];
array2 = [[1,2,3],[4,5,6],[7,8,9]];
for(var i = 0; i< array2.length;i++){
array2[i].unshift(array1[i]);
}
console.log(array2);

You can loop over one of the array and use unshift to push value at 0 index.
var x = ['a','b','c'];
var y = [[1,2,3],[4,5,6],[7,8,9]];
y.forEach((item, index) => item.unshift(x[index]));
console.log(y);

Here a solution with .map() and the rest operator
let singleArr = ["a","b","c"];
let multiArr = [[1,2,3],[4,5,6],[7,8,9]];
let result = multiArr.map((el, i) => [singleArr[i],...el])
console.log(result);

Group by operation using one of arrays' values [closed]

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I have array of arrays in javascript that looks like: [[1, 23.34, -5.22], [1, 2.34, -52.22], [2, 0.34, -5.02], ...]. I need to group by by the first number in the arrays and produce 2 different ouputs:
Object (key: 2D array) that looks like this:
{1: [[23.34, -5.22], [2.34, -52.22]],
2: [[0.34, -5.02], ...],
...}
3D array that would just signify the grouping:
[[[23.34, -5.22], [2.34, -52.22]], [[0.34, -5.02], ...], ...]
I want to use ES6. Any suggestions would be greatly appreciated.

checkout this code
let array = [[1, 23.34, -5.22], [1, 2.34, -52.22], [2, 0.34, -5.02]];
let obj = {};
array.forEach(subArray =>
{
let key = subArray[0];
if(!obj[key])
obj[key] = [];
subArray.splice(0,1);
obj[key].push(subArray);
});
console.log(obj);
let keys = Object.keys(obj);
let array3D = [];
keys.forEach(key => {
array3D.push(obj[key])
});
console.log(array3D);

There's already a answer which uses map method but I find using forEach in this case more appropriate.
const a = [
[1, 23.34, -5.22], [1, 2.34, -52.22], [2, 0.34, -5.02]
];
let b = {};
let c = [];
a.forEach(x=>{
if(!b[x[0]]){
b[x[0]] = [];
}
b[x[0]].push(x.slice(1));
});
console.log(b);
Object.values(b).forEach(x => c.push(x));
console.log(c);