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

[Joseph Fox-Rabinovitz]

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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