Bin index is just value floor divided by the bin size. On Fri, Dec 18, 2020, 09:59 Martín Chalela <tinchochal...@gmail.com> wrote:
> Hi all! I was wondering if there is a way around to using np.digitize when > dealing with equidistant bins. For example: > bins = np.linspace(0, 1, 20) > > The main problem I encountered is that digitize calls np.searchsorted. > This is the correct way, I think, for generic bins, i.e. bins that have > different widths. However, in the special, but not uncommon, case of > equidistant bins, the searchsorted call can be very expensive and > unnecessary. One can perform a simple calculation like the following: > > def digitize_eqbins(x, bins): > """ > Return the indices of the bins to which each value in input array belongs. > Assumes equidistant bins. > """ > nbins = len(bins) - 1 > digit = (nbins * (x - bins[0]) / (bins[-1] - bins[0])).astype(np.int) > return digit + 1 > > Is there a better way of computing this for equidistant bins? > > Thank you! > Martin. > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@python.org > https://mail.python.org/mailman/listinfo/numpy-discussion >
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