nonuniform sampling with replacement

[Aram Ter-Sarkissov]

On 22 мар, 01:27, Peter Otten <__pete... at web.de> wrote:
> 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

thanks a lot,

Alex