How to compare two string arrays, case insensitive and independent about ordering - JS, ES6

Question

I want to compare two string arrays, but case insensitive and independent.

For the example:

['a', 'b', 'c'] === ['A', 'c', 'B'] -> TRUE

['a', 'b', 'c'] === ['a', 'b', 'd'] -> FALSE

TRUE when they are with the same length and same values (case insensitive ['A'] === ['a'] -> true) and independent, about ordering ['a', 'b'] === ['b', 'a'] -> true.

What I did for now is:

areEqual = (arr1, arr2) => {
    const equalLength = arr1.length === arr2.length;

    return arr2.every(arr2Item => {

        return arr1.includes(arr2Item.toLowerCase());

    }) && equalLength;
};

, but this is case sensitive.

I am using JS, ES6 with React.

Possible duplicate of [How to compare arrays in JavaScript?](https://stackoverflow.com/questions/7837456/how-to-compare-arrays-in-javascript) — Keno, Dec 20 '18 at 18:51
How can `['a', 'b', 'c'] === ['a', 'c'] -> TRUE` be true if `equalLength` is false? — Andy, Dec 20 '18 at 18:54
Edited, and yes, they can't be with different values and equal :). — gdfgdfg, Dec 20 '18 at 18:58
What is your expectation about `['a', 'a', 'b']` and `['b', 'b', 'a']`? Should they be equal? They have the same length and the same set of values, but each with different multiplicities. If you say they're equal, then shouldn't `['a', 'a', 'a', 'b']` also be equal to them? — Scott Sauyet, Dec 20 '18 at 18:59
Your third example still says TRUE for arrays of different lengths. — T.J. Crowder, Dec 20 '18 at 18:59
Yes, this is true `areEqual (['a', 'a', 'b'], ['b', 'b', 'a'])`, it must be `false`, but in all cases they will be with unique values. — gdfgdfg, Dec 20 '18 at 19:07

score 7 · Answer 1 · answered Dec 20 '18 at 18:55

7

You could normalise the strings to lower case and use a Set for checking the values.

function compare(a, b) {
    const lower = s => s.toLowerCase();
    return b
        .map(lower)
        .every(Set.prototype.has, new Set(a.map(lower)));
}

console.log(compare(['a', 'b', 'c'], ['A', 'c', 'B'])); //  true
console.log(compare(['a', 'b', 'c'], ['a', 'b', 'd'])); // false
console.log(compare(['a', 'b', 'c'], ['a', 'c']));      //  true

answered Dec 20 '18 at 18:55

Nina Scholz

376,160
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That is just pretty. – T.J. Crowder Dec 20 '18 at 18:57
1

This missing the check for same length. – Barmar Dec 20 '18 at 18:57
2

OP is contradicting his own conditions though – Keno Dec 20 '18 at 18:58
@KenoClayton - Very true, I see what you mean. OP's third example says they match (even though `a` is longer than `b`), but OP's code does a length **equality** check. – T.J. Crowder Dec 20 '18 at 18:58
His third example must be a typo. They're different length and the second array is missing `b`. – Barmar Dec 20 '18 at 19:00
That is a nice solution. – Bibberty Dec 20 '18 at 19:05
1

@MarkMeyer, op says, unique values, later not, more later unique. hm. i don't know. this solution assumes unique values. – Nina Scholz Dec 20 '18 at 20:52

score 1 · Answer 2 · answered Dec 20 '18 at 18:58

You need to replace arr1 with a lowercase copy of it. You should also return immediately if they're not the same length, rather than going through all the comparison work when it's not necessary.

areEqualCI = (arr1, arr2) => {
    if (arr1.length != arr2.length) {
        return false;
    }
    const arr1Lower = arr1.map(e => e.toLowerCase());
    return arr2.every(arr2Item => {
        return arr1Lower.includes(arr2Item.toLowerCase());
    });
};

It might also be better to sort the two arrays, then just compare them elementwise:

areEqualCI = (arr1, arr2) => {
    if (arr1.length != arr2.length) {
        return false;
    }
    const arr1Lower = arr1.map(e => e.toLowerCase()).sort();
    const arr2Lower = arr2.map(e => e.toLowerCase()).sort();
    for (let i = 0; i < arr1.length; i++) {
        if (arr1[i] != arr2[i]) {
            return false;
        }
    }
    return true;
}

How to compare two string arrays, case insensitive and independent about ordering - JS, ES6

2 Answers2