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 <jsalvati@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-rep... @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.isaac@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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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 <jsalvati@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-rep... @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.isaac@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

On 12/20/2010 9:41 PM, josef.pktd@gmail.com wrote:

python has it in random

sample( population, k)

Yes, I mentioned this in my original post: http://www.mail-archive.com/numpy-discussion@scipy.org/msg29324.html But good simulation practice is perhaps to seed a simulation specific random number generator (not just rely on a global), and I don't want to pass around two different instances. So I want to get this functionality from numpy.random. Which reminds me of another question. numpy.random.RandomState accepts an int array as a seed: what is the *intended* use? Thanks, Alan

On 12/20/2010 10:49 PM, josef.pktd@gmail.com wrote:

What's the difference between a numpy Random and a python random.Random instance of separate states of the random number generators?

Sorry, I don't understand the question. The difference for my use is that a np.RandomState instance provides access to a different set of methods, which unfortunately does not include an equivalent to random.Random's sample method but which does include others I need. Would it be appropriate to request that an analog to random.sample be added to numpy.random? (It might sample only a range, since producing indexes would provide the base functionality.) Or is this functionality absent intentionally? Alan