How do I sample randomly based on some probability(wightage)?

Scott David Daniels Scott.Daniels at Acm.Org
Wed May 27 17:08:53 CEST 2009


Sumitava Mukherjee wrote:
> I need to randomly sample from a list where all choices have weights
> attached to them. The probability of them being choosen is dependent
> on the weights.
I am not sure why everybody is normalizing by dividing the weights.
This isa possibility (I had fun writing it).

def sample_without_replacement(choices, weights):
     '''Yield elements sampled w/o replacement by weighting'''
     if len(weights) != len(choices):
         raise ValueError('%d choices, but %d weights?' % (
             len(choices), len(weights)))
     if min(weights) < 0:
         raise ValueError('Negative weights?: %s' % (
                     [(i, w) for i, w in enumerate(weights) if w < 0]))

     # look at only non-zero probabilities
     combined = [(w, v) for w, v in zip(weights, choices) if w > 0]

     # Go from highest probability down to reduce expected traversal
     combined.sort(key=operator.itemgetter(0), reverse=True)

     total = sum(w for w, v in combined) # sum(weights) also works
     while combined:
         spot = sample = random.random() * total
         for n, (weight, choice) in enumerate(combined):
             spot -= weight
             if spot <= 0:
                 break
         else:
             # n, (weight, choice) = 0, combined[0] # Highest probability
             raise ValueError('%f left after choosing %f/%f?: %s' % (
                                 spot, sample, total, combined))
         yield choice
         total -= weight
         if weight > total * 256: # arbitrary choice for recalculation
             # precision affected, rebuild
             total = sum(w for w, v in combined)
         del combined[n]
     raise ValueError('Samplng more than %d without replacement?' % (
                       sum(1 for w in weights if w > 0)))

for n in range(10):
     gen = sample_without_replacement('abcdef', [32,16,8,4,2,1])
     print gen.next(), gen.next(), gen.next()

--Scott David Daniels
Scott.Daniels at Acm.Org



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