4

Suppose I have a list of strings. How do I generate a random one?

TIMEX
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    If you need to get random strings in a non-duplicative manner (not specified above), consider using a "shuffle" such as a Fisher-Yates. (After the shuffle you can just 'shift' off the front of the Array.) –  Dec 14 '10 at 01:23
  • possible duplicate of [Getting random value from an array](http://stackoverflow.com/questions/4550505/getting-random-value-from-an-array) – user Feb 04 '15 at 00:30

4 Answers4

21

You mean, get a random array member?

var strings = ['a', 'b', 'c'];

var randomIndex = Math.floor(Math.random() * strings.length);

var randomString = strings[randomIndex];

See it on jsFiddle.

If you mean a random string, it is a little different.

var randomStrLength = 16,
    pool = 'abcdefghijklmnopqrstuvwxyz0123456789',
    randomStr = '';


for (var i = 0; i < randomStrLength; i++) {
     var randomChar = pool.charAt(Math.floor(Math.random() * pool.length));
     randomStr += randomChar;   
}

See it on jsFiddle.

Of course, you can skip the pool variable and do String.fromCharCode() with a random number between 97 ('a'.charCodeAt(0)) and 122 ('z'.charCodeAt(0)) for lowercase letters, etc. But depending on the range you want (lowercase and uppercase, plus special characters), using a pool is less complex.

alex
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  • Wouldn't using Math.ceil() there cause one never to select the first element (0)? – Andrew Barber Dec 14 '10 at 01:09
  • @Andrew Barber: Not never, but extremely seldom. You would get a distribution over four indexes, where the first occurs only when the random number is exactly zero. – Guffa Dec 14 '10 at 01:13
  • @Guffa: extremely seldom -- you mean on an average 20% of the time ? – Mahesh Velaga Dec 14 '10 at 01:15
  • @Andrew Barber I changed that a while back when I realised my mistake :) – alex Dec 14 '10 at 01:18
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    @Mahesh Velaga: No, something closer to 0.00000000000001% of the time. – Guffa Dec 14 '10 at 01:20
  • @Guffa: Its a little tricky I think, what happens when the array size is low ? lets say 5 and what if the random number generated is less than 0.2 (which will be on an average 20% of the time)? – Mahesh Velaga Dec 14 '10 at 01:24
  • @Mahesh Velaga: The index will only be 0 for the random number 0.0, but for any value 0.0 < n <= 0.2, the index will be 1. For example `Math.ceil(0.0000000001*5) = 1`. – Guffa Dec 14 '10 at 01:31
  • @Guffa: Sorry, my bad .. I started off with taking ceil in my mind and in between I switched to floor, thanks for patiently answering my non-sense questions :) – Mahesh Velaga Dec 14 '10 at 01:34
8

Alex and Mahesh are right on, just wanted to demonstrate how I might implement their solutions if I felt like living dangerously. Which I do.

Array.prototype.chooseRandom = function() {
  return this[Math.floor(Math.random() * this.length)];
};
var a = [1, 2, 3, 4, 5];
a.chooseRandom(); // => 2
a.chooseRandom(); // => 1
a.chooseRandom(); // => 5
maerics
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2
var randomString = myStrings[Math.floor(Math.random() * myStrings.length)]
Guffa
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Mahesh Velaga
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0
var rand = 0; 
var newPic = []; 
var pic = [1,2,3] //length = 18 for ( var i=0; i<18; i++ ){ 
    rand = Math.floor(Math.random()*19); 
    newPic.push(pic[rand-1].slice()); 
} alert(newPic);
Alfred
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kara
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