[Numpy-discussion] Counting the Colors of RGB-Image
a...@pdauf.de
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
Tony Yu
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
Nadav Horesh
From: numpy-discus...@scipy.org [numpy-discus...@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
Chris Barker
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
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a...@pdauf.de
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
Torgil Svensson
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
Chris Barker
> 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
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Christopher Barker, Ph.D.
Oceanographer
Emergency Response Division
NOAA/NOS/OR&R (206) 526-6959 voice
7600 Sand Point Way NE (206) 526-6329 fax
Seattle, WA 98115 (206) 526-6317 main reception
elodw
> 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 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