Jah_Alarm wrote:
> I've got a vector length n of integers (some of them are repeating),
> and I got a selection probability vector of the same length. How will
> I sample with replacement k (<=n) values with the probabilty vector.
> In Matlab this function is randsample. I couldn't find anything to
> this extent in Scipy or Numpy.
If all else fails you can do it yourself:
import random
import bisect
def iter_sample_with_replacement(values, weights):
_random = random.random
_bisect = bisect.bisect
acc_weights = []
sigma = 0
for w in weights:
sigma += w
acc_weights.append(sigma)
while 1:
yield values[_bisect(acc_weights, _random()*sigma)]
def sample_with_replacement(k, values, weights):
return list(islice(iter_sample_with_replacement(values, weights), k))
if __name__ == "__main__":
from itertools import islice
N = 10**6
values = range(4)
weights = [2, 3, 4, 1]
histo = [0] * len(values)
for v in islice(iter_sample_with_replacement(values, weights), N):
histo[v] += 1
print histo
print sample_with_replacement(30, values, weights)
Peter
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