Given following array:
var arr = [undefined, undefined, 2, 5, undefined, undefined];
I'd like to get the count of elements which are defined (i.e.: those which are not undefined
). Other than looping through the array, is there a good way to do this?
This question is related to
javascript
undefined
Loop and count in all browsers:
var cnt = 0;
for (var i = 0; i < arr.length; i++) {
if (arr[i] !== undefined) {
++cnt;
}
}
In modern browsers:
var cnt = 0;
arr.foreach(function(val) {
if (val !== undefined) { ++cnt; }
})
No, the only way to know how many elements are not undefined is to loop through and count them. That doesn't mean you have to write the loop, though, just that something, somewhere has to do it. (See #3 below for why I added that caveat.)
How you loop through and count them is up to you. There are lots of ways:
for
loop from 0
to arr.length - 1
(inclusive).for..in
loop provided you take correct safeguards.some
, filter
, or reduce
, passing in an appropriate function. This is handy not only because you don't have to explicitly write the loop, but because using these features gives the JavaScript engine the opportunity to optimize the loop it does internally in various ways. (Whether it actually does will vary on the engine.)...but it all amounts to looping, either explicitly or (in the case of the new array features) implicitly.
Unfortunately, No. You will you have to go through a loop and count them.
EDIT :
var arrLength = arr.filter(Number);
alert(arrLength);
An array length
is not the number of elements in a array, it is the highest index + 1
. length
property will report correct element count only if there are valid elements in consecutive indices.
var a = [];
a[23] = 'foo';
a.length; // 24
Saying that, there is no way to exclude undefined elements from count without using any form of a loop.
Remove the values then check (remove null check here if you want)
const x = A.filter(item => item !== undefined || item !== null).length
With Lodash
const x = _.size(_.filter(A, item => !_.isNil(item)))
If the undefined's are implicit then you can do:
var len = 0;
for (var i in arr) { len++ };
undefined's are implicit if you don't set them explicitly
//both are a[0] and a[3] are explicit undefined
var arr = [undefined, 1, 2, undefined];
arr[6] = 3;
//now arr[4] and arr[5] are implicit undefined
delete arr[1]
//now arr[1] is implicit undefined
arr[2] = undefined
//now arr[2] is explicit undefined
Source: Stackoverflow.com