[1, 2, 3].indexOf(3) => 2
[1, 2, NaN].indexOf(NaN) => -1
[1, NaN, 3].indexOf(NaN) => -1
3 Answers
You can use Array.prototype.findIndex method to find out the index of NaN in an array
let index = [1,3,4,'hello',NaN,3].findIndex(Number.isNaN)
console.log(index)You can use Array.prototype.includes to check if NaN is present in an array or not. It won't give you the index though !! It will return a boolean value. If NaN is present true will be returned, otherwise false will be returned
let isNaNPresent = [1,2,NaN,'ball'].includes(NaN)
console.log(isNaNPresent)Don't use Array.prototype.indexOf
You can not use Array.Prototype.indexOf to find index of NaN inside an array.Because indexOf uses strict-equality-operator internally and NaN === NaN evaluates to false.So indexOf won't be able to detect NaN inside an array
[1,NaN,2].indexOf(NaN) // -1
Use Number.isNaN instead of isNaN :
Here i choose Number.isNaN over isNaN. Because isNaN treats string literal as NaN.On the other hand Number.isNaN treats only NaN literal as NaN
isNaN('hello world') // true
Number.isNaN('hello world') // false
Or, Write your own logic :
You can write your own logic to find NaN.As you already know that, NaN is the only value in javascript which is not equal to itself. That's the reason i suggested not to use Array.prototype.indexOf.
NaN === NaN // false
We can use this idea to write our own isNaN function.
[1,2,'str',NaN,5].findIndex(e=>e!=e) // 3
NaN is defined not to be equal to anything (not even itself). See here: http://www.w3schools.com/jsref/jsref_isNaN.asp
5 Comments
mcandre
Will upvote and accept when my timer runs out. That's just retarded.
geekosaur
Explain that to the IEEE; it's their floating point standard that defines
NaN to not be equal to anything. (That said, think of it as NULL in RDBMSes.)mcandre
-1 is also used as an error value. Would it make any sense for -1 != -1 ?
paxdiablo
@mcandre, it's not retarded. The whole point of NaN is to represent things that are considered not representable in the domain. That's not one type of thing, it's many. For example, you cannot possibly say that the square root of -1,
i, is the same as zero divided by itself (indeterminate). Yet they are both NaNs.mcandre
I get that. I do. But from a practical standpoint, if I want to, say, extract all the NaNs from an array, it would be useful to compare elements to NaN rather than having to call isNaN().
You have to look at each item to return an array of the indexes that are NaN values-
function findNaNs(arr){
return arr.map(function(itm, i){
if(isNaN(itm)) return i;
return false;
}).filter(function(itm){
return itm;
});
}
findNaNs([1, NaN, 3, 4, 'cat'/3])
//or to find the first one-
function firstNaN(arr){
var i= 0, L= arr.length;
while(i<L){
if(isNaN(arr[i])) return i;
++i;
}
return -1;
}
3 Comments
mcandre
Thank you, kennebec. It's not hard to code around this problem, just inconvenient.
Nelo Mitranim
Never use
isNaN, it's broken. It implicitly coerces value type. You get true for isNaN(undefined), isNaN({}) (but not isNaN([])), etc. Use comparison to itself instead: x !== x (NaN is the only value that doesn't equal self).
Ziyuan
The first routine cannot find NaN if it is the first element.
Array.prototype.includesif you just want to know if an array includes aNaN:[NaN].includes(NaN) /* true */.