Remove all falsy values from an array

Question

I would like to remove all falsy values from an array. Falsy values in JavaScript are false, null, 0, "", undefined, and NaN.

function bouncer(arr) {  arr = arr.filter(function (n) {      return (n !== undefined && n !== null && n !== false && n !== 0 && n !== "" && isNaN()!=NaN); });   return arr; }  bouncer([7, "ate", "", false, 9, NaN], ""); 

The above is getting satisfied for all except the NaN test case. Can someone help me check in the array whether it contains NaN or not?

Answer 1

You can use Boolean :

var myFilterArray = myArray.filter(Boolean); 

Answer 2

Since you want to get rid of "falsy" values, just let JavaScript do its thing:

function bouncer(arr) {   return arr.filter(function(v) { return !!v; }); } 

The double-application of the ! operator will make the filter callback return true when the value is "truthy" and false when it's "falsy".

(Your code is calling isNaN() but not passing it a value; that's why that test didn't work for you. The isNaN() function returns true if its parameter, when coerced to a number, is NaN, and false otherwise.)

edit — note that

function bouncer(arr) {   return arr.filter(Boolean); } 

would work too as LoremIpsum notes in another answer, because the built-in Boolean constructor does pretty much the exact same thing as !!.