[Numpy-discussion] Optimized np.digitize for equidistant bins

[Martín Chalela]

Thank you Joseph

El vie, 18 dic 2020 a las 16:56, Joseph Fox-Rabinovitz (<
jfoxrabinovitz at gmail.com>) escribió:

> There is: np.floor_divide.
>
> On Fri, Dec 18, 2020, 14:38 Martín Chalela <tinchochalela at gmail.com>
> 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 (<
>> jfoxrabinovitz at gmail.com>) escribió:
>>
>>> Bin index is just value floor divided by the bin size.
>>>
>>> On Fri, Dec 18, 2020, 09:59 Martín Chalela <tinchochalela at 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.
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