Shortly, here is what I want :
it returns true when
arr=[a, b, c, d, e, f]
or
arr=[a, b, a, b, c, d, c, e, a, b]
they don't have to be in order.
and false when :
arr=[a, a, b, c, d, d, e, f]
Thank you.
2 Answers
You can use Array.prototype.every to test whether all elements in an array pass the test (that is a callback function returning a boolean value). In this case, make sure that the current element being checked is not the same as the previous one (index - 1), ignoring the very first element (!index).
var arr1 = ["a", "b", "c", "d", "e", "f"];
var arr2 = ["a", "a", "b", "c", "d", "d", "e", "f"];
function arrayHasConsecitiveRepetition(arrayIn) {
return arrayIn.every(function(element, index) {
return !index || element !== arrayIn[index - 1];
});
}
console.log(arrayHasConsecitiveRepetition(arr1)); // true
console.log(arrayHasConsecitiveRepetition(arr2)); // falseYou can of course also do it the other way around, using Array.prototype.some to check if at least one element matches the condition, which might actually be better in terms of performance, as it stops evaluating once the condition is true*:
var arr1 = ["a", "b", "c", "d", "e", "f"];
var arr2 = ["a", "a", "b", "c", "d", "d", "e", "f"];
function arrayHasConsecitiveRepetition(arrayIn) {
return !arrayIn.some(function(element, index) {
return index && element === arrayIn[index - 1];
});
}
console.log(arrayHasConsecitiveRepetition(arr1)); // true
console.log(arrayHasConsecitiveRepetition(arr2)); // false*Actually, it turns out that both methods return immediately once the condition is (not) met, so it's more a matter of taste or style than a matter of performance. I'd prefer to use the first approach in this case as it's clearer to me.
From the MDN pages:
If such an element is found, the
everymethod immediately returnsfalse.If such an element is found,
some()immediately returnstrue.
7 Comments
tsh
"which might actually be better in terms of performance, as it stops evaluating once the condition is true" Won't
Array#every stop evaluating once the condition is false?
Sarah Groß
That is indeed a good question, I hadn't thought of that. And it turns out you're right:
If such an element is found, the every method immediately returns false.! So I guess it's not a matter of performance at all, but a matter of taste.Sarah Groß
Updated the answer to address that
Wafae
so the .some and .every saves the time of making an if.. else ? right? I'm a newbie so I'm trying to understand what I'm reading and typing... Also if you don't mind, why wouldn't the answer of Monika work? that was the very first thing that came up to my mind... Thanks
Sarah Groß
It helps keep the code shorter. You could also loop over the array elements and check with if/else, yes.
|
One option:
const arr1 = ['a', 'b', 'c', 'd', 'e', 'f'];
const arr2 = ['a', 'a', 'b', 'c', 'd', 'd', 'e', 'f'];
function checkArr(arr) {
let previous;
const check = !arr.some(letter => {
if (letter === previous) {
return true;
}
previous = letter;
});
console.log(check);
}
checkArr(arr1); // true
checkArr(arr2); // false
1 Comment
Sarah Groß
Note that when the first element is
undefined, it will return false even if the second element is not undefined. You'd probably want to check for previous && letter === previous
a => a.every((n, i) => i === 0 || n !== a[i - 1])