[Numpy-discussion] Optimized np.digitize for equidistant bins
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
Re: [Numpy-discussion] Optimized np.digitize for equidistant bins
Bin index is just value floor divided by the bin size. On Fri, Dec 18, 2020, 09:59 Martín Chalela 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
Re: [Numpy-discussion] Optimized np.digitize for equidistant bins
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 > 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 > ___ NumPy-Discussion mailing list NumPy-Discussion@python.org https://mail.python.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Optimized np.digitize for equidistant bins
There is: np.floor_divide. On Fri, Dec 18, 2020, 14:38 Martín Chalela wrote: > 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 >> 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 >> > ___ > 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
Re: [Numpy-discussion] Optimized np.digitize for equidistant bins
Thank you Joseph El vie, 18 dic 2020 a las 16:56, Joseph Fox-Rabinovitz (< jfoxrabinov...@gmail.com>) escribió: > There is: np.floor_divide. > > On Fri, Dec 18, 2020, 14:38 Martín Chalela > wrote: > >> 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 >>> 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 >>> >> ___ >> 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 > ___ NumPy-Discussion mailing list NumPy-Discussion@python.org https://mail.python.org/mailman/listinfo/numpy-discussion
