[Numpy-discussion] efficient 3d histogram creation

Chris Colbert sccolbert@gmail....
Wed May 6 17:06:09 CDT 2009


I decided to hold myself over until being able to take a hard look at the
numpy histogramdd code:

Here is a quick thing a put together in cython. It's a 40x speedup over
histogramdd on Vista 32 using the minGW32 compiler. For a (480, 630, 3)
array, this executed in 0.005 seconds on my machine.

This only works for arrays with uint8 data types having dimensions (x, y, 3)
(common image format). The return array is a (16, 16, 16) equal width bin
histogram of the input.

If anyone wants the cython C-output, let me know and I will email it to you.


If there is interest, I will extend this for different size bins and aliases
for different data types.

Chris

import numpy as np

cimport numpy as np

DTYPE = np.uint8
DTYPE32 = np.int

ctypedef np.uint8_t DTYPE_t
ctypedef np.int_t DTYPE_t32

def hist3d(np.ndarray[DTYPE_t, ndim=3] img):
    cdef int x = img.shape[0]
    cdef int y = img.shape[1]
    cdef int z = img.shape[2]
    cdef int addx
    cdef int addy
    cdef int addz
    cdef np.ndarray[DTYPE_t32, ndim=3] out = np.zeros([16, 16, 16],
dtype=DTYPE32)
    cdef int i, j, v0, v1, v2


    for i in range(x):
        for j in range(y):
            v0 = img[i, j, 0]
            v1 = img[i, j, 1]
            v2 = img[i, j, 2]
            addx = (v0 - (v0 % 16)) / 16
            addy = (v1 - (v1 % 16)) / 16
            addz = (v2 - (v2 % 16)) / 16
            out[addx, addy, addz] += 1

    return out


On Tue, May 5, 2009 at 9:46 AM, David Huard <david.huard@gmail.com> wrote:

>
>
> On Mon, May 4, 2009 at 4:18 PM, <josef.pktd@gmail.com> wrote:
>
>> On Mon, May 4, 2009 at 4:00 PM, Chris Colbert <sccolbert@gmail.com>
>> wrote:
>> > i'll take a look at them over the next few days and see what i can hack
>> out.
>> >
>> > Chris
>> >
>> > On Mon, May 4, 2009 at 3:18 PM, David Huard <david.huard@gmail.com>
>> wrote:
>> >>
>> >>
>> >> On Mon, May 4, 2009 at 7:00 AM, <josef.pktd@gmail.com> wrote:
>> >>>
>> >>> On Mon, May 4, 2009 at 12:31 AM, Chris Colbert <sccolbert@gmail.com>
>> >>> wrote:
>> >>> > this actually sort of worked. Thanks for putting me on the right
>> track.
>> >>> >
>> >>> > Here is what I ended up with.
>> >>> >
>> >>> > this is what I ended up with:
>> >>> >
>> >>> > def hist3d(imgarray):
>> >>> >     histarray = N.zeros((16, 16, 16))
>> >>> >     temp = imgarray.copy()
>> >>> >     bins = N.arange(0, 257, 16)
>> >>> >     histarray = N.histogramdd((temp[:,:,0].ravel(),
>> >>> > temp[:,:,1].ravel(),
>> >>> > temp[:,:,2].ravel()), bins=(bins, bins, bins))[0]
>> >>> >     return histarray
>> >>> >
>> >>> > this creates a 3d histogram of rgb image values in the range 0,255
>> >>> > using 16
>> >>> > bins per component color.
>> >>> >
>> >>> > on a 640x480 image, it executes in 0.3 seconds vs 4.5 seconds for a
>> for
>> >>> > loop.
>> >>> >
>> >>> > not quite framerate, but good enough for prototyping.
>> >>> >
>> >>>
>> >>> I don't think your copy to temp is necessary, and use reshape(-1,3) as
>> >>> in the example of Stefan, which will avoid copying the array 3 times.
>> >>>
>> >>> If you need to gain some more speed, then rewriting histogramdd and
>> >>> removing some of the unnecessary checks and calculations looks
>> >>> possible.
>> >>
>> >> Indeed, the strategy used in the histogram function is faster than the
>> one
>> >> used in the histogramdd case, so porting one to the other should speed
>> >> things up.
>> >>
>> >> David
>>
>> is searchsorted faster than digitize and bincount ?
>>
>
> That depends on the number of bins and whether or not the bin width is
> uniform. A 1D benchmark I did a while ago showed that if the bin width is
> uniform, then the best strategy is to create a counter initialized to 0,
> loop through the data, compute i = (x-bin0) /binwidth and increment counter
> i by 1 (or by the weight of the data). If the bins are non uniform, then for
> nbin > 30 you'd better use searchsort, and digitize otherwise.
>
> For those interested in speeding up histogram code, I recommend reading a
> thread started by Cameron Walsh on the 12/12/06 named "Histograms of
> extremely large data sets" Code and benchmarks were posted.
>
> Chris, if your bins all have the same width, then you can certainly write
> an histogramdd routine that is way faster by using the indexing trick
> instead of digitize or searchsort.
>
> Cheers,
>
> David
>
>
>
>
>>
>> Using the idea of histogramdd, I get a bit below a tenth of a second,
>> my best for this problem is below.
>> I was trying for a while what the fastest way is to convert a two
>> dimensional array into a one dimensional index for bincount. I found
>> that using the return index of unique1d is very slow compared to
>> numeric index calculation.
>>
>> Josef
>>
>> example timed for:
>> nobs = 307200
>> nbins = 16
>> factors = np.random.randint(256,size=(nobs,3)).copy()
>> factors2 = factors.reshape(-1,480,3).copy()
>>
>> def hist3(factorsin, nbins):
>>    if factorsin.ndim != 2:
>>        factors = factorsin.reshape(-1,factorsin.shape[-1])
>>    else:
>>        factors = factorsin
>>    N, D = factors.shape
>>    darr = np.empty(factors.T.shape, dtype=int)
>>    nele = np.max(factors)+1
>>    bins = np.arange(0, nele, nele/nbins)
>>    bins[-1] += 1
>>    for i in range(D):
>>        darr[i] = np.digitize(factors[:,i],bins) - 1
>>
>>    #add weighted rows
>>    darrind = darr[D-1]
>>    for i in range(D-1):
>>        darrind += darr[i]*nbins**(D-i-1)
>>    return np.bincount(darrind)  # return flat not reshaped
>>
>
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>
>
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