I want to see if two arrays of strings are equal.
Eg:
compare(["abc", "def"], ["def", "abc"])
should return true and similarly,
compare(["abc", "def"], ["def", "ghi"])
should return false.
What is the best way to do this?
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Please provide the code where your are attempting thisunicorn2– unicorn22016-01-22 07:27:56 +00:00Commented Jan 22, 2016 at 7:27
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1Either sort the arrays and compare a[n] to b[n], or skip sorting and look for a[n] in every position of b.Culme– Culme2016-01-22 07:28:52 +00:00Commented Jan 22, 2016 at 7:28
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2Possible duplicate of How to compare arrays in JavaScript?Ashish Rajput– Ashish Rajput2016-01-22 07:31:51 +00:00Commented Jan 22, 2016 at 7:31
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7 Answers
JSON.stringify(array1.sort()) === JSON.stringify(array2.sort())
2 Comments
Papasmile
very cool for primitives, but watch out for many pitfalls masteringjs.io/tutorials/fundamentals/compare-arrays
rafark
This is potentially terrible for performance, for example when you're executing your function many times like on hover or on scroll.
JavaScript doesn't have a Set or Multiset data structure (at least, not one with wide browser support), which is what you'd normally use for testing that two sets of items are the same regardless of order. So I recommend sorting the arrays and checking that their contents are equal. If you know the arrays contain only strings, you can check the items with simple equality:
function compare(array1, array2) {
if (array1.length != array2.length) {
return false;
}
array1 = array1.slice();
array1.sort();
array2 = array2.slice();
array2.sort();
for (var i = 0; i < array1.length; i++) {
if (array1[i] != array2[i]) {
return false;
}
}
return true;
}
console.log(compare(["abc", "def"], ["def", "abc"])); // true
console.log(compare(["abc", "def"], ["def", "ghi"])); // false
For more general cases, you'll need a more complex definition of equality, and I recommend browsing the answers to this question.
5 Comments
Royi Namir
I think you have a problem with the logic in
if (array1[i] != array2[i]) { return false; }. You don;t want to break if match not found, you should keep looking a match
Davide Rossi
it looks they have to be considered equals if they contain the same items, even if in a different order. This won't pass your check. Gonna be a n^2 iteration... or will need sorting as you implemented.
Abhijith S
Thank You for the answer. Is this the most efficient way? Doesn't JS have any built in function for this operation?
alltom
I think I addressed your comments. I originally misread the question and expected the arrays to be exactly equal. Now I use the same approach as Culme and sort the arrays.
alltom
Abhijith: I usually use this approach. In extreme cases I will put all the items from
array1 as keys in an Object (the closest thing JavaScript has to a Set data structure) to get constant-time lookup when iterating through array2, but I've hardly ever needed to.Naive algorithm: O(N^2)
function compare(array1, array2){
for (var i = 0; i < array1.length; i++){
if (array2.indexOf(array1[i]) < 0) return false;
}
return true;
}
Better one: (uses sorting, O(NLOGN))
function compare(array1, array2){
array1.sort();
array2.sort();
for (var i = 0; i < array1.length; i++){
if (array1[i] !== array2[i]) return false;
}
return true;
}
1 Comment
Ethan
Both of these need
if(array1.length !== array2.length) { return false } at the start for them to work.I would suggest following ES6-oneliner
const compare = (a1, a2) =>
(a1 = new Set(a1)) &&
(a2 = new Set(a2)) &&
a1.size === a2.size &&
[...a1].every(v => a2.has(v));
- Remove duplicates by converting arrays to Sets (for
compare(['a', 'a'], ['a', 'b'])should returnfalse). - Length comparison (for
compare(['a', 'b'], ['a', 'b', 'c'])should returnfalse). - Check if the first set items are present in the second set.
2 Comments
Stephen Harris
This would return true for
a1 = ['foo', 'foo'] and a2 = ['foo', 'bar'];
dhilt
@StephenHarris Thank you! I applied
Set converting to resolve duplicates situation.Unified solution:
function compare(a, b){
var isEqual = false;
if (Array.isArray(a) && Array.isArray(b) && a.length == b.length){
a.sort();
b.sort();
var i;
for (i = 0; i < arr1.length; i++){
if (a[i] === b[i]){
isEqual = true;
} else{
isEqual = false;
break;
}
}
}
return isEqual;
}
var arr1 = ["def", "abc"], arr2 = ["abc", "def"];
console.log(compare(arr2,arr1)); // gives 'true'
console.log(compare(["abc", "def"], ["def", "ghi"])); // gives 'false'
Comments
function compare(arr1, arr2){
var match = true
if(arr1.length != arr2.length){
match = false
}
arr1 = arr1.slice();
arr1.sort();
arr2.slice();
arr2.sort();
for(var i = 0; i < arr1.length; i++){
if(arr1[i] != arr2[i]){
match = false;
}
}
return match;
}
console.log(compare(["abc", "def"], ["def", "abc"])); // it will return true
console.log(compare(["abc", "def"], ["def", "ghi"])); // it will return false
Comments
Verbose typescript example for ordered or unordered
(obviously inspired by @dhilts answer above)
const VERBOSE = console.log // ()=>void
const areDeduplicatedArraysEqualUnordered = (a1: any[] | Set<any>, a2: any[] | Set<any>) => {
(a1 = new Set(a1)); (a2 = (new Set(a2))); VERBOSE('checking arrays as Sets', { a1, a2 })
return !!(a1.size === a2.size && [...a1].every(v => (a2 as Set<any>).has(v)))
}
const areDeduplicatedArraysEqualOrdered = (a1: any[], a2: any[]) => {
(a1 = [...(new Set(a1))] as any[]) && (a2 = [...(new Set(a2))] as any[]) && VERBOSE('checking deduplicated arrays', { a1, a2 })
return !!(a1.length === a2.length && a1.every((v: any, i) => a2[i] === v))
}
const a2 = [3,5,8]
const a1 = [8,5,3]
const notEqual = areDeduplicatedArraysEqualOrdered(a1,a2)
const yesEqual = areDeduplicatedArraysEqualUnordered(a1,a2)
VERBOSE('results', { notEqual, yesEqual})
