We often need to generate more than one such sample from an array, e.g. for permutation tests. If we shuffle an array x of size N and use x[:M] as a random sample "without replacement", we just need to put them back randomly to get the next sample (cf. Fisher-Yates shuffle). That way we get O(M) amortized complexity for each sample of size M. Only the first sample will have complexity O(N).

Sturla > I know this question came up on the mailing list some time ago > (19/09/2008), and the conclusion was that yes, you can do it more or > less efficiently in pure python; the trick is to use two different > methods. If your sample is more than, say, a quarter the size of the > set you're drawing from, you permute the set and take the first few. > If your sample is smaller, you draw with replacement, then redraw the > duplicates, and repeat until there aren't any more duplicates. Since > you only do this when your sample is much smaller than the population > you don't need to repeat many times. > > Here's the code I posted to the previous discussion (not tested this > time around) with comments: > > ''' > def choose_without_replacement(m,n,repeats=None): > """Choose n nonnegative integers less than m without replacement > > Returns an array of shape n, or (n,repeats). > """ > if repeats is None: > r = 1 > else: > r = repeats > if n>m: > raise ValueError, "Cannot find %d nonnegative integers less > than %d" % (n,m) > if n>m/2: > res = np.sort(np.random.rand(m,r).argsort(axis=0)[:n,:],axis=0) > else: > res = np.random.random_integers(m,size=(n,r)) > while True: > res = np.sort(res,axis=0) > w = np.nonzero(np.diff(res,axis=0)==0) > nr = len(w[0]) > if nr==0: > break > res[w] = np.random.random_integers(m,size=nr) > > if repeats is None: > return res[:,0] > else: > return res > > For really large values of repeats it does too much sorting; I didn't > have the energy to make it pull all the ones with repeats to the > beginning so that only they need to be re-sorted the next time > through. Still, the expected number of trips through the while loop > grows only logarithmically with repeats, so it shouldn't be too bad. > ''' > > Anne > > On 20 December 2010 12:13, John Salvatier <jsalv...@u.washington.edu> > wrote: >> 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 >> >> >> _______________________________________________ >> NumPy-Discussion mailing list >> NumPy-Discussion@scipy.org >> http://mail.scipy.org/mailman/listinfo/numpy-discussion >> >> > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion > _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion