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Comment by inklesspen | Hacker News Reader

uh, in python:

  import string, random
  ''.join(random.sample((string.letters+string.digits), 12))

I hope that's what you were asking for. If not, you might want to clarify your question.

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On morning's light, that won't do what you think. random.sample() gives you a unique sampling. No character will be repeated. try this instead:

    alphanum = string.letters+string.digits
    ''.join([alphanum[random.randint(0,61)] for i in xrange(12)])

You're quite right; I should have caught that.

You want to use an ID that can predictably be unique for something like this. You shouldn't use a random string.

What do you mean by "predictably unique"? Do you mean "guaranteed unique"?

As a way to generate unique ids this isn't horrible. 12^62 is something like 220 bits. The odds of a collision are even lower than with a UUID.

Guaranteed uniqueness is preferred, yes. But the level of effort needed to guarantee uniqueness across a large application / dataset / etc is much higher than "unique enough", just as it's a lot more expensive to prove a number is prime than to generate a number that is 99.9999% probably prime.

well, even 99.9999% possible, we should handle the 0.0001% exception ;) Can I understand that, if collision occurs, let the it generate the id again?

That would be guaranteeing uniqueness. Don't bother guaranteeing uniqueness. The 99.99999% was an understatement. 12^62 is 99.999999999999999999999999999999999999999999999999999999999992% unique. This is higher odds than being killed by a meteorite.

:) thanks

Hi, how to keep the string/id is unique? In another words, how to deal with the id conflict?

It's generating a big (BIG) random numbe. The odds of a conflict are many billion times more than the odds of getting hit by a meteorite.

http://en.wikipedia.org/wiki/UUID#Random_UUID_probability_of...

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