I'm using random.random() to get a random float (obviously!). But what I really want to do is something like:
there's a 30% chance my app does this:
pass
else:
pass
Can you guys help me structure this?
2 Answers
if random.random() > 0.5:
# your app does this
pass
else:
# your app does that
pass
15 Comments
irrelephant
It's probably easier to use
random() < x since x would directly correspond to the probability of the condition's being true.
temporary_user_name
But does this answer the question? The asker wants to make it so that if he flips a coin repeatedly, it gives randomly ordered results, skewed 40% (or whatever) towards heads. Your answer could feasibly give 0% heads.
temporary_user_name
@irrelephant Yes, but that could just keep returning something greater than 0.5, 100% of the time. It is not guaranteed to give an X% chance, like Tyler wants. Unless..and here my understanding of the random module falls short...the distribution of random.random() is completely uniform, so exactly half the results will be greater than 0.5 and half of them will be less? In that case, this answer is correct, and I'm mistaken.
abarnert
@Aerovistae: Even a truly random sequence could just keep returning something greater than 0.5, 100% of the time. That's a perfectly reasonable result for something that has a 50% chance of happening. It's not the same thing as happening 50% of the times when run N times (except as N approaches infinity), but the OP asked for the former, not the latter. In other words, it's perfectly reasonable for something with a 50% chance of happening to happen, say, 9 times in 20 runs, or even 0 (it's just not that likely—the odds are 1/2^N).
abarnert
@TylerSeymour: Yes, he's used .5 to mean a 50% chance; change it to .3 to mean a 30% chance. Most people would write this as
random.random() < .3, but of course .3 > random.random() means the exact same thing, as would the answer's random.random() > .7. |
Try this:
if random.randint(1, 10) in (1, 2, 3):
print '30% chance'
else:
print '70% chance'
Here randint will generate a number between 1-10, there's a 30% chance that it's between 1-3 and a 70% chance that it's between 4-10
2 Comments
Danica
Less flexible than @hd1's method, since this can only do equal probabilities, but convenient. Note that you could also skip the
== 1 if you wanted.abarnert
This is a better solution when your probabilities are best described as "1 in 2" or "1 in 3". Of course for a rational number, you have a choice of describing it either way, so
if random.randrange(10) < 3 is just as reasonable as random.random() < .3; it's just a matter of which one seems clearer to you.
hd1's answer; if you're thinking of it as "3 out of 10", accept Óscar López's answer. They're both correct, and equivalent, it's just a matter of which is more readable or seems more "right" in your case.