Right! I just thought there would/should be a "digitize" function that did this.
El vie, 18 dic 2020 a las 14:16, Joseph Fox-Rabinovitz (< jfoxrabinov...@gmail.com>) escribió: > 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 >> > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@python.org > https://mail.python.org/mailman/listinfo/numpy-discussion >
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