| Re: Does IDL has histogram matching function? [message #32786] |
Fri, 08 November 2002 15:41  |
Pavel A. Romashkin
Messages: 531
Registered: November 2000
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Senior Member |
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David Fanning wrote:
>
> OK, the first feedback is from a disgruntled user
> who takes my typing literally (if you can imagine!)
Uh, no wonder the autoreply on your vacation e-mail message is "And if
you don't like it you're damn welcome to write it all by yourself!" :-)
Cheers,
Pavel
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| Re: Does IDL has histogram matching function? [message #32794 is a reply to message #32786] |
Fri, 08 November 2002 09:48  |
David Fanning
Messages: 11724
Registered: August 2001
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Senior Member |
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David Fanning (david@dfanning.com) writes:
> Try this:
>
> a = LonGen(255)
> b = a#b
> b = BytScl(b)
> Window, 1
> Plot, Histogram(b, Min=0, Max=255)
> Window, 2, XSize=256, YSize=256)
> TV, HistoMatch(image, Histogram(b, Min=0, Max=255))
>
> Still looks good, I think.
>
> OK, I'm waiting for feedback. :-)
OK, the first feedback is from a disgruntled user
who takes my typing literally (if you can imagine!)
Should be:
b = a#a
Cheers,
David
--
David W. Fanning, Ph.D.
Fanning Software Consulting, Inc.
Phone: 970-221-0438, E-mail: david@dfanning.com
Coyote's Guide to IDL Programming: http://www.dfanning.com/
Toll-Free IDL Book Orders: 1-888-461-0155
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| Re: Does IDL has histogram matching function? [message #32795 is a reply to message #32794] |
Fri, 08 November 2002 09:35 |
David Fanning
Messages: 11724
Registered: August 2001
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Senior Member |
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David Fanning (david@dfanning.com) writes:
> I expect it might take a day or so to write the code.
> Do you have any money? :-)
Ah, forget the money. This turned out to be too easy. :-)
Here is a routine, named HISTOMATCH, that takes an image
and a histogram that you would like to perform histogram
matching to.
;*********************************************************
FUNCTION HistoMatch, image, histogram_to_match
; Perform histogram matching according to the method of
; Gonzales and Woods in Digital Image Processing, pp 94-102
; image - The input image.
; histogram_to_match - The histogram used for histogram matching.
; Calculate the histogram of the input image.
h = Histogram(Byte(image), Binsize=1, Min=0, Max=255)
totalPixels = Float(N_Elements(image))
; Find a mapping from the input pixels to s.
s = FltArr(256)
FOR k=0,255 DO BEGIN
s[k] = Total(h(0:k) / totalPixels)
ENDFOR
; Find a mapping from input histogram to v.
v = FltArr(256)
FOR q=0,255 DO BEGIN
v[q] = Total(histogram_to_match(0:q) / totalPixels)
ENDFOR
; Find z from v and s.
z = BytArr(256)
FOR j=0,255 DO BEGIN
I = Where(v LT s[j], count)
IF count GT 0 THEN z[j] = (Reverse(I))[0] ELSE z[j]=0
ENDFOR
; Create the matched image.
matchedImage = z[Byte(image)]
RETURN, matchedImage
END
;*********************************************************
I'm certain JD or someone will point out to me how to
use another Histogram to eliminate the Where function,
but, hey, this is for free. I'm trying to make a living
here. :-(
Does it work!? I think so. I'm not sure.
Try this. Let's see if we can match am image to the
histogram formed by calculating the histogram of
the histogram equalized image. (The result should
be the same as the histogram equalized image, more
or less.)
;*********************************************************
PRO TestIt
filename = Filepath('ctscan.dat', Subdir=['examples', 'data'])
OpenR, lun, filename, /Get_Lun
image = BytArr(256, 256)
ReadU, lun, image
Free_Lun, lun
Window, XSize=3*256, YSize=256
TV, image, 0
TV, Hist_Equal(image), 1
TV, HistoMatch(image, Histogram(Hist_Equal(image), Min=0, Max=255)), 2
END
;*********************************************************
IDL> TestIt
Wow! And this was on the *first* try. *That* doesn't happen too
often. :-)
Try this:
a = LonGen(255)
b = a#b
b = BytScl(b)
Window, 1
Plot, Histogram(b, Min=0, Max=255)
Window, 2, XSize=256, YSize=256)
TV, HistoMatch(image, Histogram(b, Min=0, Max=255))
Still looks good, I think.
OK, I'm waiting for feedback. :-)
Cheers,
David
--
David W. Fanning, Ph.D.
Fanning Software Consulting, Inc.
Phone: 970-221-0438, E-mail: david@dfanning.com
Coyote's Guide to IDL Programming: http://www.dfanning.com/
Toll-Free IDL Book Orders: 1-888-461-0155
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| Re: Does IDL has histogram matching function? [message #32956 is a reply to message #32795] |
Mon, 18 November 2002 15:09 |
aardvark62
Messages: 6
Registered: November 2002
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Junior Member |
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There is an undocumented keyword to HIST_EQUAL that looks like it
might do the same thing as Davids HistoMatch. Here is an example:
filename = filepath('ctscan.dat', subdir=['examples', 'data'])
image = read_binary(filename, data_dims=[256, 256])
desired_hist = histogram(hist_equal(image), min=0, max=255)
window, xsize=3*256, ysize=256
tv, image, 0
tv, hist_equal(image), 1
tv, hist_equal(image, fcn=total(desired_hist, /cumulative)), 2
end
David Fanning <david@dfanning.com> wrote in message news:<MPG.1835a3e2693e7288989a0b@news.frii.com>...
> David Fanning (david@dfanning.com) writes:
>
>> I expect it might take a day or so to write the code.
>> Do you have any money? :-)
>
> Ah, forget the money. This turned out to be too easy. :-)
>
> Here is a routine, named HISTOMATCH, that takes an image
> and a histogram that you would like to perform histogram
> matching to.
>
> ;*********************************************************
> FUNCTION HistoMatch, image, histogram_to_match
>
> ; Perform histogram matching according to the method of
> ; Gonzales and Woods in Digital Image Processing, pp 94-102
>
> ; image - The input image.
> ; histogram_to_match - The histogram used for histogram matching.
>
> ; Calculate the histogram of the input image.
>
> h = Histogram(Byte(image), Binsize=1, Min=0, Max=255)
> totalPixels = Float(N_Elements(image))
>
> ; Find a mapping from the input pixels to s.
>
> s = FltArr(256)
> FOR k=0,255 DO BEGIN
> s[k] = Total(h(0:k) / totalPixels)
> ENDFOR
>
> ; Find a mapping from input histogram to v.
>
> v = FltArr(256)
> FOR q=0,255 DO BEGIN
> v[q] = Total(histogram_to_match(0:q) / totalPixels)
> ENDFOR
>
> ; Find z from v and s.
>
> z = BytArr(256)
> FOR j=0,255 DO BEGIN
> I = Where(v LT s[j], count)
> IF count GT 0 THEN z[j] = (Reverse(I))[0] ELSE z[j]=0
> ENDFOR
>
> ; Create the matched image.
>
> matchedImage = z[Byte(image)]
> RETURN, matchedImage
> END
> ;*********************************************************
>
> I'm certain JD or someone will point out to me how to
> use another Histogram to eliminate the Where function,
> but, hey, this is for free. I'm trying to make a living
> here. :-(
>
> Does it work!? I think so. I'm not sure.
>
> Try this. Let's see if we can match am image to the
> histogram formed by calculating the histogram of
> the histogram equalized image. (The result should
> be the same as the histogram equalized image, more
> or less.)
>
> ;*********************************************************
> PRO TestIt
> filename = Filepath('ctscan.dat', Subdir=['examples', 'data'])
> OpenR, lun, filename, /Get_Lun
> image = BytArr(256, 256)
> ReadU, lun, image
> Free_Lun, lun
>
> Window, XSize=3*256, YSize=256
> TV, image, 0
> TV, Hist_Equal(image), 1
> TV, HistoMatch(image, Histogram(Hist_Equal(image), Min=0, Max=255)), 2
> END
> ;*********************************************************
>
> IDL> TestIt
>
> Wow! And this was on the *first* try. *That* doesn't happen too
> often. :-)
>
> Try this:
>
> a = LonGen(255)
> b = a#b
> b = BytScl(b)
> Window, 1
> Plot, Histogram(b, Min=0, Max=255)
> Window, 2, XSize=256, YSize=256)
> TV, HistoMatch(image, Histogram(b, Min=0, Max=255))
>
> Still looks good, I think.
>
> OK, I'm waiting for feedback. :-)
>
> Cheers,
>
> David
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| Re: Does IDL has histogram matching function? [message #32977 is a reply to message #32795] |
Sat, 23 November 2002 00:34 |
tianyf_cn
Messages: 19
Registered: November 2002
|
Junior Member |
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Maybe I want to process float type images. Or the output data values
are in a narrow range. Does anyone have some ideas?
Thanks.
Yours,
Tian.
David Fanning <david@dfanning.com> wrote in message news:<MPG.1835a3e2693e7288989a0b@news.frii.com>...
> David Fanning (david@dfanning.com) writes:
>
>> I expect it might take a day or so to write the code.
>> Do you have any money? :-)
>
> Ah, forget the money. This turned out to be too easy. :-)
>
> Here is a routine, named HISTOMATCH, that takes an image
> and a histogram that you would like to perform histogram
> matching to.
>
> ;*********************************************************
> FUNCTION HistoMatch, image, histogram_to_match
>
> ; Perform histogram matching according to the method of
> ; Gonzales and Woods in Digital Image Processing, pp 94-102
>
> ; image - The input image.
> ; histogram_to_match - The histogram used for histogram matching.
>
> ; Calculate the histogram of the input image.
>
> h = Histogram(Byte(image), Binsize=1, Min=0, Max=255)
> totalPixels = Float(N_Elements(image))
>
> ; Find a mapping from the input pixels to s.
>
> s = FltArr(256)
> FOR k=0,255 DO BEGIN
> s[k] = Total(h(0:k) / totalPixels)
> ENDFOR
>
> ; Find a mapping from input histogram to v.
>
> v = FltArr(256)
> FOR q=0,255 DO BEGIN
> v[q] = Total(histogram_to_match(0:q) / totalPixels)
> ENDFOR
>
> ; Find z from v and s.
>
> z = BytArr(256)
> FOR j=0,255 DO BEGIN
> I = Where(v LT s[j], count)
> IF count GT 0 THEN z[j] = (Reverse(I))[0] ELSE z[j]=0
> ENDFOR
>
> ; Create the matched image.
>
> matchedImage = z[Byte(image)]
> RETURN, matchedImage
> END
> ;*********************************************************
>
> I'm certain JD or someone will point out to me how to
> use another Histogram to eliminate the Where function,
> but, hey, this is for free. I'm trying to make a living
> here. :-(
>
> Does it work!? I think so. I'm not sure.
>
> Try this. Let's see if we can match am image to the
> histogram formed by calculating the histogram of
> the histogram equalized image. (The result should
> be the same as the histogram equalized image, more
> or less.)
>
> ;*********************************************************
> PRO TestIt
> filename = Filepath('ctscan.dat', Subdir=['examples', 'data'])
> OpenR, lun, filename, /Get_Lun
> image = BytArr(256, 256)
> ReadU, lun, image
> Free_Lun, lun
>
> Window, XSize=3*256, YSize=256
> TV, image, 0
> TV, Hist_Equal(image), 1
> TV, HistoMatch(image, Histogram(Hist_Equal(image), Min=0, Max=255)), 2
> END
> ;*********************************************************
>
> IDL> TestIt
>
> Wow! And this was on the *first* try. *That* doesn't happen too
> often. :-)
>
> Try this:
>
> a = LonGen(255)
> b = a#b
> b = BytScl(b)
> Window, 1
> Plot, Histogram(b, Min=0, Max=255)
> Window, 2, XSize=256, YSize=256)
> TV, HistoMatch(image, Histogram(b, Min=0, Max=255))
>
> Still looks good, I think.
>
> OK, I'm waiting for feedback. :-)
>
> Cheers,
>
> David
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| Re: Does IDL has histogram matching function? [message #33133 is a reply to message #32977] |
Mon, 09 December 2002 14:12 |
aardvark62
Messages: 6
Registered: November 2002
|
Junior Member |
|
|
tianyf_cn@yahoo.com.cn (TIAN Yunfeng) wrote in message news:<42e9d2cb.0211230034.560a064e@posting.google.com>...
> Maybe I want to process float type images. Or the output data values
> are in a narrow range. Does anyone have some ideas?
>
> Thanks.
>
> Yours,
> Tian.
>
Tain,
Are you wanting to specify your desired curve algebraically? As is
done with QSIMP for example? If not, I think you will have to group
your data into bins as is done with HIST_EQUAL, FCN. The algorithm
that I posted Friday for the FCN keyword, or David's algorithm, might
be a start. As they stand, these algorithms limit you to 256 bins and
BYTSCLed results. But they probably could be translated to higher
resolution by substituting your own algebra where they call BYTSCL, or
simply scaling your result to fit OMIN and OMAX.
-Paul Sorenson
> David Fanning <david@dfanning.com> wrote in message news:<MPG.1835a3e2693e7288989a0b@news.frii.com>...
>> David Fanning (david@dfanning.com) writes:
>>
>>> I expect it might take a day or so to write the code.
>>> Do you have any money? :-)
>>
>> Ah, forget the money. This turned out to be too easy. :-)
>>
>> Here is a routine, named HISTOMATCH, that takes an image
>> and a histogram that you would like to perform histogram
>> matching to.
>>
>> ;*********************************************************
>> FUNCTION HistoMatch, image, histogram_to_match
>>
>> ; Perform histogram matching according to the method of
>> ; Gonzales and Woods in Digital Image Processing, pp 94-102
>>
>> ; image - The input image.
>> ; histogram_to_match - The histogram used for histogram matching.
>>
>> ; Calculate the histogram of the input image.
>>
>> h = Histogram(Byte(image), Binsize=1, Min=0, Max=255)
>> totalPixels = Float(N_Elements(image))
>>
>> ; Find a mapping from the input pixels to s.
>>
>> s = FltArr(256)
>> FOR k=0,255 DO BEGIN
>> s[k] = Total(h(0:k) / totalPixels)
>> ENDFOR
>>
>> ; Find a mapping from input histogram to v.
>>
>> v = FltArr(256)
>> FOR q=0,255 DO BEGIN
>> v[q] = Total(histogram_to_match(0:q) / totalPixels)
>> ENDFOR
>>
>> ; Find z from v and s.
>>
>> z = BytArr(256)
>> FOR j=0,255 DO BEGIN
>> I = Where(v LT s[j], count)
>> IF count GT 0 THEN z[j] = (Reverse(I))[0] ELSE z[j]=0
>> ENDFOR
>>
>> ; Create the matched image.
>>
>> matchedImage = z[Byte(image)]
>> RETURN, matchedImage
>> END
>> ;*********************************************************
>>
>> I'm certain JD or someone will point out to me how to
>> use another Histogram to eliminate the Where function,
>> but, hey, this is for free. I'm trying to make a living
>> here. :-(
>>
>> Does it work!? I think so. I'm not sure.
>>
>> Try this. Let's see if we can match am image to the
>> histogram formed by calculating the histogram of
>> the histogram equalized image. (The result should
>> be the same as the histogram equalized image, more
>> or less.)
>>
>> ;*********************************************************
>> PRO TestIt
>> filename = Filepath('ctscan.dat', Subdir=['examples', 'data'])
>> OpenR, lun, filename, /Get_Lun
>> image = BytArr(256, 256)
>> ReadU, lun, image
>> Free_Lun, lun
>>
>> Window, XSize=3*256, YSize=256
>> TV, image, 0
>> TV, Hist_Equal(image), 1
>> TV, HistoMatch(image, Histogram(Hist_Equal(image), Min=0, Max=255)), 2
>> END
>> ;*********************************************************
>>
>> IDL> TestIt
>>
>> Wow! And this was on the *first* try. *That* doesn't happen too
>> often. :-)
>>
>> Try this:
>>
>> a = LonGen(255)
>> b = a#b
>> b = BytScl(b)
>> Window, 1
>> Plot, Histogram(b, Min=0, Max=255)
>> Window, 2, XSize=256, YSize=256)
>> TV, HistoMatch(image, Histogram(b, Min=0, Max=255))
>>
>> Still looks good, I think.
>>
>> OK, I'm waiting for feedback. :-)
>>
>> Cheers,
>>
>> David
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| Re: Does IDL has histogram matching function? [message #33190 is a reply to message #33133] |
Thu, 12 December 2002 14:11 |
Paul Sorenson
Messages: 48
Registered: May 2002
|
Member |
|
|
David Fanning writes:
> Oddly enough, I was just thinking about histogram matching
> because I was re-reading that section of the book in
> Digital Image Processing by Gonzales and Woods. (Have I
> mentioned what a great book this is!?) I think I finally
> understand how to do this.
There are some things in that Gonzalez and Woods book that really hurt my
brain. On page 96, they say that G(z) = T(r). Get out of town! :-) How
can this be? They don't *look* the same in the plots shown on page 98
(Figure 3.19). G(z) is the desired cumulative distribution and T(r) is the
cumulative distribution of the input image. All of this appears in their
discussion of Histogram Matching.
-Paul Sorenson
"Paul Sorenson" <aardvark62@msn.com> wrote in message
news:8270ac8d.0212091412.51094acd@posting.google.com...
> tianyf_cn@yahoo.com.cn (TIAN Yunfeng) wrote in message
news:<42e9d2cb.0211230034.560a064e@posting.google.com>...
>> Maybe I want to process float type images. Or the output data values
>> are in a narrow range. Does anyone have some ideas?
>>
>> Thanks.
>>
>> Yours,
>> Tian.
>>
> Tain,
>
> Are you wanting to specify your desired curve algebraically? As is
> done with QSIMP for example? If not, I think you will have to group
> your data into bins as is done with HIST_EQUAL, FCN. The algorithm
> that I posted Friday for the FCN keyword, or David's algorithm, might
> be a start. As they stand, these algorithms limit you to 256 bins and
> BYTSCLed results. But they probably could be translated to higher
> resolution by substituting your own algebra where they call BYTSCL, or
> simply scaling your result to fit OMIN and OMAX.
>
> -Paul Sorenson
>> David Fanning <david@dfanning.com> wrote in message
news:<MPG.1835a3e2693e7288989a0b@news.frii.com>...
>>> David Fanning (david@dfanning.com) writes:
>>>
>>>> I expect it might take a day or so to write the code.
>>>> Do you have any money? :-)
>>>
>>> Ah, forget the money. This turned out to be too easy. :-)
>>>
>>> Here is a routine, named HISTOMATCH, that takes an image
>>> and a histogram that you would like to perform histogram
>>> matching to.
>>>
>>> ;*********************************************************
>>> FUNCTION HistoMatch, image, histogram_to_match
>>>
>>> ; Perform histogram matching according to the method of
>>> ; Gonzales and Woods in Digital Image Processing, pp 94-102
>>>
>>> ; image - The input image.
>>> ; histogram_to_match - The histogram used for histogram matching.
>>>
>>> ; Calculate the histogram of the input image.
>>>
>>> h = Histogram(Byte(image), Binsize=1, Min=0, Max=255)
>>> totalPixels = Float(N_Elements(image))
>>>
>>> ; Find a mapping from the input pixels to s.
>>>
>>> s = FltArr(256)
>>> FOR k=0,255 DO BEGIN
>>> s[k] = Total(h(0:k) / totalPixels)
>>> ENDFOR
>>>
>>> ; Find a mapping from input histogram to v.
>>>
>>> v = FltArr(256)
>>> FOR q=0,255 DO BEGIN
>>> v[q] = Total(histogram_to_match(0:q) / totalPixels)
>>> ENDFOR
>>>
>>> ; Find z from v and s.
>>>
>>> z = BytArr(256)
>>> FOR j=0,255 DO BEGIN
>>> I = Where(v LT s[j], count)
>>> IF count GT 0 THEN z[j] = (Reverse(I))[0] ELSE z[j]=0
>>> ENDFOR
>>>
>>> ; Create the matched image.
>>>
>>> matchedImage = z[Byte(image)]
>>> RETURN, matchedImage
>>> END
>>> ;*********************************************************
>>>
>>> I'm certain JD or someone will point out to me how to
>>> use another Histogram to eliminate the Where function,
>>> but, hey, this is for free. I'm trying to make a living
>>> here. :-(
>>>
>>> Does it work!? I think so. I'm not sure.
>>>
>>> Try this. Let's see if we can match am image to the
>>> histogram formed by calculating the histogram of
>>> the histogram equalized image. (The result should
>>> be the same as the histogram equalized image, more
>>> or less.)
>>>
>>> ;*********************************************************
>>> PRO TestIt
>>> filename = Filepath('ctscan.dat', Subdir=['examples', 'data'])
>>> OpenR, lun, filename, /Get_Lun
>>> image = BytArr(256, 256)
>>> ReadU, lun, image
>>> Free_Lun, lun
>>>
>>> Window, XSize=3*256, YSize=256
>>> TV, image, 0
>>> TV, Hist_Equal(image), 1
>>> TV, HistoMatch(image, Histogram(Hist_Equal(image), Min=0, Max=255)), 2
>>> END
>>> ;*********************************************************
>>>
>>> IDL> TestIt
>>>
>>> Wow! And this was on the *first* try. *That* doesn't happen too
>>> often. :-)
>>>
>>> Try this:
>>>
>>> a = LonGen(255)
>>> b = a#b
>>> b = BytScl(b)
>>> Window, 1
>>> Plot, Histogram(b, Min=0, Max=255)
>>> Window, 2, XSize=256, YSize=256)
>>> TV, HistoMatch(image, Histogram(b, Min=0, Max=255))
>>>
>>> Still looks good, I think.
>>>
>>> OK, I'm waiting for feedback. :-)
>>>
>>> Cheers,
>>>
>>> David
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