if you're open to other libraries, this could be done extremely fast in numpy.
On my machine summing that whole array takes 75ms. On Thu, Jan 28, 2010 at 8:00 PM, Steven D'Aprano < st...@remove-this-cybersource.com.au> wrote: > 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 >
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