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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);
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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);
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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)
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How do I compare elements in an array using Javascript?
I want to know if cas[1], cas[2] and cas[3] have the same value.
cas = ["0","1", "2","3","4","5","6","7","8","9"];
if(cas[1]== cas[2] && cas[2]==cas[3]){
console.log("yea");
}
The first of all it's better to use ===, because 0 == false is true, but 0 === false is false.
=== works without type casting.
You don't need full cycle loop here, you have to compare only the first 3 elements so if statement is ok here.
If you want to do that with every 3 elements, you can do something like this.
const getThreeArrayElements = (array, startIndex) => array.slice(startIndex, startIndex + 3);
getThreeArrayElements([1, 2, 3, 3, 3, 3], 0); // [1, 2, 3]
getThreeArrayElements([1, 2, 3, 3, 3, 3], 3); // [3, 3, 3]
So you can easily get an array with 3 required elements.
The another task is how to compare them.
const areArrayElementsEqual = array => array.every(item => item === array[0]);
areArrayElementsEqual([1, 2, 3]); // false
areArrayElementsEqual([3, 3, 3]); // true
if you're looking for an algorithm to check the entire array :
let cas = ["0","1", "2","3","4","5","6","7","8","9"];
let checkCas = (cur = 1)=>{return cas[cur-1]===cas[cur]?true:cur<cas.length?checkCas(++cur):false;};
console.log(checkCas());
cas = ["0","1", "2","3","4","5","5","7","8","9"];
console.log(checkCas());
as for ticktacktoe :
let casa = ["0","1", "2","3","4","5","6","7","8","9"];
let allEqual = (cas,cur = 1)=>{return cas[cur-1]===cas[cur]?true:cur<cas.length?allEqual(cas,++cur):false;};
for(let i = 0;i<Math.floor(casa.length/3);i++)
{
let tempLengArr = [casa[i],casa[i+3],casa[i+6]];
if(allEqual(casa.slice(i*3,(i*3)+3))||allEqual(tempLengArr))
console.log("won");
}
if(allEqual([casa[0],casa[4],casa[9]])||
allEqual([casa[3],casa[4],casa[2]])) console.log("won");
might have messed up a bit on my allEqual call but it's an idea
This question already has answers here:
Zip arrays in JavaScript?
(5 answers)
Closed 3 years ago.
I wanted to merge two arrays like you can fold 2 packs of cards - in ugly code with for loop and assuming same length so no if's for safety it would look like this:
const arr = [1,2,3];
const rra = [4,5,6];
const result = [];
for(let i=0; i<arr.length; i++){
result.push(arr[i],rra[i]);
}
console.log(result); // Array(6) [ 1, 4, 2, 5, 3, 6 ]
I know there is something similar in String.raw() but it cuts off last element and returns a string, is there equivalent in array ?
String.raw({raw: [1,2,3]}, ...[4,5,6]); //"14253"
You can use .flatMap() for this - This is the same as .map(), followed by .flat()
const arr = [1,2,3];
const rra = [4,5,6];
let newArray = arr.flatMap((ele, i) => [ele, rra[i]]);
console.log(newArray);
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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);