If I have an array A = [1, 4, 3, 2] and B = [0, 2, 1, 2], and I want to return a new array (A - B) with values [1, 2, 2, 0], what is the most efficient approach to do this in JavaScript?
-
4Possible duplicate of What is the fastest or most elegant way to compute a set difference using Javascript arrays?jhpratt– jhpratt2017-07-27 05:37:19 +00:00Commented Jul 27, 2017 at 5:37
-
2A user with your rep should know the importance of sharing effort in question. SO is to get help for your problems and not solution for your requirementsRajesh Dixit– Rajesh Dixit2017-07-27 05:55:13 +00:00Commented Jul 27, 2017 at 5:55
-
Try this., stackoverflow.com/questions/1187518/javascript-array-differenceAnand– Anand2017-07-27 06:21:16 +00:00Commented Jul 27, 2017 at 6:21
-
Try this link: stackoverflow.com/questions/1187518/javascript-array-differenceAnand– Anand2017-07-27 06:23:06 +00:00Commented Jul 27, 2017 at 6:23
-
4The caluclated values in the example are wrong. It should be [1,2,2,0] and not [0,2,2,0].Adrian Dymorz– Adrian Dymorz2020-02-21 19:29:29 +00:00Commented Feb 21, 2020 at 19:29
Add a comment
|
7 Answers
If you want to find the set difference between two sets, that is, you want all the elements of A that are not in B:
const A = [1, 4, 3, 2]
const B = [0, 2, 1, 2]
console.log(A.filter(n => !B.includes(n)))If you want arithmetic differences between the elements of A and the corresponding elements of B, then please look to the other posted answers.
4 Comments
Melroy van den Berg
This is NOT what rrbest is asking for.
Andrii Gordiichuk
@danger89, agree with you, as a simple solution for this (in case both arrays have the same length):
ArrA.map((n, i) => n - ArrB[i]);
Sergiy Ostrovsky
By searching for "subtract array" I actually was looking for this answer, lol. Thanks
coppereyecat
It may not be what the original question was asking for, but it sure is what most people landing here from google are asking for, judging by the relative upvotes.
Use map method The map method takes three parameters in it's callback function like below
currentValue, index, array
var a = [1, 4, 3, 2],
b = [0, 2, 1, 2]
var x = a.map(function(item, index) {
// In this case item correspond to currentValue of array a,
// using index to get value from array b
return item - b[index];
})
console.log(x);
1 Comment
Winters
es6 version: var x = a.map((item, index) => item - b[index])
For Simple and efficient ever.
Check here : JsPref - For Vs Map Vs forEach
var a = [1, 4, 3, 2],
b = [0, 2, 1, 2],
x = [];
for(var i = 0;i<=b.length-1;i++)
x.push(a[i] - b[i]);
console.log(x);3 Comments
Avnesh Shakya
Both arrays length, suppose to be same. otherwise it can create a problem. Or handle it before.
Sankar
@AvneshShakya Yes. Both array should be same and that's what asked.
KCE
Strange, for me the JsPerf above doesn't show the for loop as fastest. So I made my own: jsperf.com/for-loop-vs-map-vs-for-each-kce and found that for large n the for loop is many times faster.
const A = [1, 4, 3, 2]
const B = [0, 2, 1, 2]
const C = A.map((valueA, indexInA) => valueA - B[indexInA])
console.log(C) // [1, 2, 2, 0]
Here the map is returning the substraction operation for each number of the first array.
Note: this will not work if the arrays have different lengths
1 Comment
Community
As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.
One-liner using ES6 for the array's of equal size in length:
let subResult = a.map((v, i) => v - b[i]); // [1, 2, 2, 0]
v = value, i = index
Comments
If you want to override values in the first table you can simply use forEach method for arrays forEach. ForEach method takes the same parameter as map method (element, index, array). It's similar with the previous answer with map keyword but here we are not returning the value but assign value by own.
var a = [1, 4, 3, 2],
b = [0, 2, 1, 2]
a.forEach(function(item, index, arr) {
// item - current value in the loop
// index - index for this value in the array
// arr - reference to analyzed array
arr[index] = item - b[index];
})
//in this case we override values in first array
console.log(a);Comments
function subtract(operand1 = [], operand2 = []) {
console.log('array1', operand1, 'array2', operand2);
const obj1 = {};
if (operand1.length === operand2.length) {
return operand1.map(($op, i) => {
return $op - operand2[i];
})
}
throw new Error('collections are of different lengths');
}
// Test by generating a random array
function getRandomArray(total){
const pool = []
for (let i = 0; i < total; i++) {
pool.push(Math.floor(Math.random() * total));
}
return pool;
}
console.log(subtract(getRandomArray(10), getRandomArray(10)))
Time Complexity is O(n)
You can also compare your answer with a big collection of arrays.
