Randomly picking numbers from a list according to a uniform distribution

Question

I have a list as such

a = [.5,.57,.67,.8,1,1.33,2,4]

Which looks like this when plotted:

I need to randomly pick a number in this list. In Python, I would normally go like this:

c = random.choice(a)

Except... Doing this biases the pick towards a lower value (the density is higher around 1 than it is around 4).

How would I go about picking a list entry according to a uniform distribution. As in c = random.random()*3.5+.5, but actually picking from the list?

Does the value being picked has to be element of the list? Otherwise depending on your data the eventual choice will still be biased... To some extent — Anzel, Oct 19 '15 at 16:29
Let's see, it's either discrete or continuous. Maybe you could pick a random float in that interval, and then retrieve the element on the list closest to that value. — mescarra, Oct 19 '15 at 16:30
You could separete between sections (e.g 0-4), randomize a section and only then randomize inside the section. This way, it won't matter how many numbers there are in each section. You might just need to check for empty sections. — FirstOne, Oct 19 '15 at 16:36
you can also do something like http://stackoverflow.com/questions/12141150/from-list-of-integers-get-number-closest-to-a-given-value then use `random.uniform` to get a random float number from 0.0-4.0 — R Nar, Oct 19 '15 at 16:38
"Doing this biases the pick towards a lower value". Do yo mean you want the expected value of the pick to be `(a[0]+a[-1]])/2`? — toine, Oct 19 '15 at 16:42

score 1 · Accepted Answer · answered Oct 19 '15 at 16:42

You could get floats from an uniform distribution and then choosing the one on your list that's closest to this generated value. Like so:

a = [.5,.57,.67,.8,1,1.33,2,4]
p = map(lambda x: abs(random.uniform(0,4) - x), a)
c = a[p.index(min(p))]

Of course, you could do it more efficently given that your list is sorted.

1 Answers1