Finding an item of an array of numbers without using a loop

Question

This is a stupid question, it feels like one. But mental block is bad right now. :(

My problem is I have an array consisting only of numbers. I want to use that arra

Answer 1

If you just want to check whether or not an item is contained by an Array, you can make usage of the nifty bitwise NOT operator along with the .indexOf() method:

if( ~a.indexOf(b) ) {
    // 2 was found, do domething here
}

It's always a good idea to use a shim or library to make sure methods like .indexOf() are available for your Javascript. @SLaks gave you the example for .indexOf() from MDC.

Short explanation why this works:

The bitwise not operator, negates all bits within a byte. That is also true for the positiv/negative bit. Basically it turns the result "-1" into "0". So, if nothing was found we have a falsy value which we want to have at this point. If we match something at the very beginning and get a result of "0" its translated into "-1" which is just fine since its not a falsy value. Any other possible return value is guaranteed a truthy value.

Answer 2

A simple loop seems called for here, but-

You can do it without a loop, and without extending IE, if you convert the array to a string.

var a = [2,4,6,8,10], n = 2;

if(RegExp('\\b'+n+'\\b').test(a.join(','))){
// n is in a
}

Answer 3

if (a.indexOf(2) >= 0)

Note that IE < 9 doesn't have indexOf, so you'll needto add it in case it doesn't exist:

if (!Array.prototype.indexOf)
{
  Array.prototype.indexOf = function(searchElement /*, fromIndex */)
  {
    "use strict";

    if (this === void 0 || this === null)
      throw new TypeError();

    var t = Object(this);
    var len = t.length >>> 0;
    if (len === 0)
      return -1;

    var n = 0;
    if (arguments.length > 0)
    {
      n = Number(arguments[1]);
      if (n !== n) // shortcut for verifying if it's NaN
        n = 0;
      else if (n !== 0 && n !== (1 / 0) && n !== -(1 / 0))
        n = (n > 0 || -1) * Math.floor(Math.abs(n));
    }

    if (n >= len)
      return -1;

    var k = n >= 0
          ? n
          : Math.max(len - Math.abs(n), 0);

    for (; k < len; k++)
    {
      if (k in t && t[k] === searchElement)
        return k;
    }
    return -1;
  };
}

Answer 4

You can simply use .includes() method of arrays:

let a = [2, 4, 6, 8, 10],
    b = 2;

if(a.includes(b)) {
    // your code goes here...
}

Answer 5

If you want a native look-up, use an object, not an array.

var a = {2:0,4:0,6:0,8:0,10:0},
    b = 2;

if (b in a) alert "yay!";

note that I use your array value as the key. Using 0 as the value is arbitrary, it does not matter what you put as the value when you use the in operator.

Use 1 or true if you want to be able to do

if (a[b]) alert "yay!";

but I'd recommend using in as this is both ideomatic and less error-prone.

---

EDIT: Regarding your notion that array lookup would be faster than object lookup. Try it.

console.log('begin building test array and object'); 
var x = [], y = {};
for (var i=0; i<1000; i++) {
  var n = Math.floor(Math.random() * 1000);
  x.push( n );
  y[n] = true;
}
console.log('finished building test array and object'); 

var foo = 0;

console.log('begin 1,000,000 array search rounds at ' + new Date());
for (var i=0; i<1000000; i++) {
  if (x.indexOf(i % 1000) > -1) foo++;
}
console.log('finished array search rounds at ' + new Date());


console.log('begin 1,000,000 object search rounds at ' + new Date());
for (var i=0; i<1000000; i++) {
  if ((i % 1000) in y) foo++;
}
console.log('finished object search rounds at ' + new Date());