Re: [Numpy-discussion] Counting the Colors of RGB-Image
On Wed, Jan 18, 2012 at 1:26 AM, a...@pdauf.de wrote: Your ideas are very helpfull and the code is very fast. I'm curios -- a number of ideas were floated here -- what did you end up using? -Chris -- Christopher Barker, Ph.D. Oceanographer Emergency Response Division NOAA/NOS/ORR (206) 526-6959 voice 7600 Sand Point Way NE (206) 526-6329 fax Seattle, WA 98115 (206) 526-6317 main reception chris.bar...@noaa.gov ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Counting the Colors of RGB-Image
Am 23.01.2012 18:17, schrieb Chris Barker: On Wed, Jan 18, 2012 at 1:26 AM,a...@pdauf.de wrote: Your ideas are very helpfull and the code is very fast. I'm curios -- a number of ideas were floated here -- what did you end up using? -Chris I'am sorry but when i see the code of Torgil Svenson, I think, the game is over. I use the follow. code: t0=clock() tt = n_im2.view() tt.shape = -1,3 ifl = tt[...,0].astype(np.int)*256*256 + tt[...,1].astype(np.int)*256 + tt[...,2].astype(np.int) colors, inv = np.unique(ifl,return_inverse=True) zus = np.array([colors[-1]+1]) colplus = np.hstack((colors,zus)) ccnt = np.histogram(ifl,colplus)[0] t1=clock() print (t1-t0) t0=t1 -- Christopher Barker, Ph.D. Oceanographer Emergency Response Division NOAA/NOS/ORR(206) 526-6959 voice 7600 Sand Point Way NE (206) 526-6329 fax Seattle, WA 98115 (206) 526-6317 main reception chris.bar...@noaa.gov ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Counting the Colors of RGB-Image
unique has an option to get indexes out which you can use in
combination with sort to get the actual counts out.
tab0 = zeros( 256*256*256 , dtype=int)
col=ravel(((im0[...,0].astype('u4')*256+im0[...,1])*256)+im0[...,2])
col,idx=unique(sort(col),True)
idx=hstack([idx,[2500*2500]])
tab0[col]=idx[1:]-idx[:-1]
tab0.shape=(256,256,256)
As Chris pointed out, if each pixel were 4 bytes you could probably
just use im0.view('u4') for histogram values.
//Torgil
On Wed, Jan 18, 2012 at 10:26 AM, a...@pdauf.de wrote:
Sorry,
that i use this way to send an answer to Tony Yu , Nadav Horesh , Chris
Barker.
When iam direct answering on Your e-mail i get an error 5.
I think i did a mistake.
Your ideas are very helpfull and the code is very fast.
Thank You
elodw
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Re: [Numpy-discussion] Counting the Colors of RGB-Image
Sorry, that i use this way to send an answer to Tony Yu , Nadav Horesh , Chris Barker. When iam direct answering on Your e-mail i get an error 5. I think i did a mistake. Your ideas are very helpfull and the code is very fast. Thank You elodw ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Counting the Colors of RGB-Image
Here's a thought: Too bad numpy doesn't have a 24 bit integer, but you could tack a 0 on, making your image 32 bit, then use histogram2d to count the colors. something like (untested): # create the 32 bit image 32bit_im = np.zeros((w, h), dtype = np.uint32) view = 32bit_im.view(dtype = np.uint8).reshape((w,h,4)) view[:,:,:3] = im # histogram it: bins = # this is the trick -- setting your bins right # remember that histrogram is designed for floats, so you're bin boundaries shold be between the inteer values you want. colors = np.histogram(32bit_im, bins=bins) NOTE: the image processing scikit may well have somethign already -- histogramming an image is a common process. -Chris On Sun, Jan 15, 2012 at 9:40 AM, Nadav Horesh nad...@visionsense.com wrote: im_flat = im0[...,0]*65536 + im[...,1]*256 +im[...,2] colours = np.unique(im_flat) Nadav From: numpy-discussion-boun...@scipy.org [numpy-discussion-boun...@scipy.org] On Behalf Of Tony Yu [tsy...@gmail.com] Sent: 15 January 2012 18:03 To: Discussion of Numerical Python Subject: Re: [Numpy-discussion] Counting the Colors of RGB-Image On Sun, Jan 15, 2012 at 10:45 AM, a...@pdauf.de wrote: Counting the Colors of RGB-Image, nameit im0 with im0.shape = 2500,3500,3 with this code: tab0 = zeros( (256,256,256) , dtype=int) tt = im0.view() tt.shape = -1,3 for r,g,b in tt: tab0[r,g,b] += 1 Question: Is there a faster way in numpy to get this result? MfG elodw Assuming that your image is made up of integer values (which I guess they'd have to be if you're indexing into `tab0`), then you could write: rgb_unique = set(tuple(rgb) for rgb in tt) I'm not sure if it's any faster than your loop, but I would assume it is. -Tony ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion -- -- Christopher Barker, Ph.D. Oceanographer Emergency Response Division NOAA/NOS/ORR (206) 526-6959 voice 7600 Sand Point Way NE (206) 526-6329 fax Seattle, WA 98115 (206) 526-6317 main reception chris.bar...@noaa.gov ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
[Numpy-discussion] Counting the Colors of RGB-Image
Counting the Colors of RGB-Image, nameit im0 with im0.shape = 2500,3500,3 with this code: tab0 = zeros( (256,256,256) , dtype=int) tt = im0.view() tt.shape = -1,3 for r,g,b in tt: tab0[r,g,b] += 1 Question: Is there a faster way in numpy to get this result? MfG elodw ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Counting the Colors of RGB-Image
On Sun, Jan 15, 2012 at 10:45 AM, a...@pdauf.de wrote: Counting the Colors of RGB-Image, nameit im0 with im0.shape = 2500,3500,3 with this code: tab0 = zeros( (256,256,256) , dtype=int) tt = im0.view() tt.shape = -1,3 for r,g,b in tt: tab0[r,g,b] += 1 Question: Is there a faster way in numpy to get this result? MfG elodw Assuming that your image is made up of integer values (which I guess they'd have to be if you're indexing into `tab0`), then you could write: rgb_unique = set(tuple(rgb) for rgb in tt) I'm not sure if it's any faster than your loop, but I would assume it is. -Tony ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Counting the Colors of RGB-Image
im_flat = im0[...,0]*65536 + im[...,1]*256 +im[...,2] colours = np.unique(im_flat) Nadav From: numpy-discussion-boun...@scipy.org [numpy-discussion-boun...@scipy.org] On Behalf Of Tony Yu [tsy...@gmail.com] Sent: 15 January 2012 18:03 To: Discussion of Numerical Python Subject: Re: [Numpy-discussion] Counting the Colors of RGB-Image On Sun, Jan 15, 2012 at 10:45 AM, a...@pdauf.demailto:a...@pdauf.de wrote: Counting the Colors of RGB-Image, nameit im0 with im0.shape = 2500,3500,3 with this code: tab0 = zeros( (256,256,256) , dtype=int) tt = im0.view() tt.shape = -1,3 for r,g,b in tt: tab0[r,g,b] += 1 Question: Is there a faster way in numpy to get this result? MfG elodw Assuming that your image is made up of integer values (which I guess they'd have to be if you're indexing into `tab0`), then you could write: rgb_unique = set(tuple(rgb) for rgb in tt) I'm not sure if it's any faster than your loop, but I would assume it is. -Tony ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
