I think this is not possible to do efficiently with just numpy. If you want to do this efficiently, I wrote a no-replacement sampler in Cython some time ago (below). I hearby release it to the public domain.

''' Created on Oct 24, 2009 http://stackoverflow.com/questions/311703/algorithm-for-sampling-without-replacement @author: johnsalvatier ''' from __future__ import division import numpy def random_no_replace(sampleSize, populationSize, numSamples): samples = numpy.zeros((numSamples, sampleSize),dtype=int) # Use Knuth's variable names cdef int n = sampleSize cdef int N = populationSize cdef i = 0 cdef int t = 0 # total input records dealt with cdef int m = 0 # number of items selected so far cdef double u while i < numSamples: t = 0 m = 0 while m < n : u = numpy.random.uniform() # call a uniform(0,1) random number generator if (N - t)*u >= n - m : t += 1 else: samples[i,m] = t t += 1 m += 1 i += 1 return samples On Mon, Dec 20, 2010 at 8:28 AM, Alan G Isaac <alan.is...@gmail.com> wrote: > I want to sample *without* replacement from a vector > (as with Python's random.sample). I don't see a direct > replacement for this, and I don't want to carry two > PRNG's around. Is the best way something like this? > > permutation(myvector)[:samplesize] > > Thanks, > Alan Isaac > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >

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