2017-01-23 15:33 GMT+01:00 Robert Kern <robert.k...@gmail.com>: > On Mon, Jan 23, 2017 at 6:27 AM, Anne Archibald <peridot.face...@gmail.com> > wrote: > > > > On Wed, Jan 18, 2017 at 4:13 PM Nadav Har'El <n...@scylladb.com> wrote: > >> > >> On Wed, Jan 18, 2017 at 4:30 PM, <josef.p...@gmail.com> wrote: > >>> > >>>> Having more sampling schemes would be useful, but it's not possible > to implement sampling schemes with impossible properties. > >>> > >>> BTW: sampling 3 out of 3 without replacement is even worse > >>> > >>> No matter what sampling scheme and what selection probabilities we > use, we always have every element with probability 1 in the sample. > >> > >> I agree. The random-sample function of the type I envisioned will be > able to reproduce the desired probabilities in some cases (like the example > I gave) but not in others. Because doing this correctly involves a set of n > linear equations in comb(n,k) variables, it can have no solution, or many > solutions, depending on the n and k, and the desired probabilities. A > function of this sort could return an error if it can't achieve the desired > probabilities. > > > > It seems to me that the basic problem here is that the > numpy.random.choice docstring fails to explain what the function actually > does when called with weights and without replacement. Clearly there are > different expectations; I think numpy.random.choice chose one that is easy > to explain and implement but not necessarily what everyone expects. So the > docstring should be clarified. Perhaps a Notes section: > > > > When numpy.random.choice is called with replace=False and non-uniform > probabilities, the resulting distribution of samples is not obvious. > numpy.random.choice effectively follows the procedure: when choosing the > kth element in a set, the probability of element i occurring is p[i] > divided by the total probability of all not-yet-chosen (and therefore > eligible) elements. This approach is always possible as long as the sample > size is no larger than the population, but it means that the probability > that element i occurs in the sample is not exactly p[i]. > > I don't object to some Notes, but I would probably phrase it more like we > are providing the standard definition of the jargon term "sampling without > replacement" in the case of non-uniform probabilities. To my mind (or more > accurately, with my background), "replace=False" obviously picks out the > implemented procedure, and I would have been incredibly surprised if it did > anything else. If the option were named "unique=True", then I would have > needed some more documentation to let me know exactly how it was > implemented. > > FWIW, I totally agree with Robert

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