On Thu, 28 Jan 2010 15:52:14 -0800, elsa wrote:
> Now, what I need to do is randomly choose one myList[i], however the
> distribution of my random choice needs to be proportional to the values
> of myList[i][0]. So, for this list above, I'd have a much higher chance
> of choosing myList[0] than myList[1].
>
> Here is how I'm doing it at the moment:
>
> def chooseI(myList):
> mySum=0
> choice = random.choice(range(1,sum([i[0] for i in myList])+1))
> for i in range(len(myList)):
> mySum+=myList[i][0]
> if mySum>=choice:
> return i
> break
This isn't related to your problem, but you don't need the break after
the return -- the return will leave the loop and the function, and the
break will never be reached.
You could probably speed the code up a little by changing all the calls
to range into xrange. range actually generates a list of integers, which
is time consuming, while xrange is a lazy generator which just produces
each integer one at a time. (You can ignore this if you are using Python
3.0 or 3.1.)
Another small optimization you could use is to use a generator expression
instead of a list comprehension inside the call to sum. That should
generate the sum lazily, instead of calculating a giant list first and
then adding it up.
But I'd try something like this:
# Untested.
def chooseI(myList):
total = sum(i[0] for i in myList)
mySum = 0
choice = random.randrange(total)
for i in myList:
mySum += i[0]
if mySum >= choice:
return i
--
Steven
--
http://mail.python.org/mailman/listinfo/python-list
