Densitometry (or relative optical density)

Jason Wong wrote:

Hi, I’m trying to figure out how to use Photoshop for scientific image analysis purposes. […]

Might this program be helpfull? I don’t know.
http://www.reindeergraphics.com/foveapro/classify.shtml

Don wrote:

The number you are reading is the brightness of the spot from 0-255, not the density. The density in absolute terms is therefore porportional to the inverse of the number, or in density numbers to the log(10) of the inverse of the number/255.

Correct. Also, don’t forget gamma. This is an exponential funciton applied to the data before it is displayed. In practical terms this means an image value of 128 will be somewhat darker than 50% intensty.

You may set a custom working space with a gamma of 1.0 to eliminate this.

—

Mike Russell
http://www.curvemeister.com
http://www.zocalo.net/~mgr
http://geigy.2y.net

Well Fovea Pro sounds like a great program but at $800 I don’t think I can afford it! Is there no other way to do this manually? Like as someone said before can I take the log of the intensity to get density?

"Jason Wong" wrote in message

Well Fovea Pro sounds like a great program but at $800 I don’t think I can afford it! Is there no other way to do this manually? Like as someone said before can I take the log of the intensity to get density?

Photoshop unfortunately only reports in 8-bit/channel values, even for (15 or) 16-b/ch values. So you are limited to a relative (to the brightest pixel) Optical Density range of 0.0-2.4 . If that is enough for your purpose, there seems to be no other way than calculating by hand, or using a look-up table. As mentioned, make sure the image data is in Gamma 1 space.

Maybe, I’m not sure because my trial period expired, Picture Window (Pro for 16-b/ch) has such a functionality? Check http://www.dl-c.com for details.

Bart

(Jason Wong) writes

Well Fovea Pro sounds like a great program but at $800 I don’t think I can afford it!

If 8 bit data (effectively limiting you to densities below about 2.3) are sufficient for your needs, then the Image Processing Tool Kit (also from www.reindeergraphics.com) will give you the same functionality as Fovea Pro for about $250. To judge the price, you need to compare the package to "professional" image analysis software that does densitometry. These packages (e.g., Image Pro Plus) typically cost $3K and up.

Jason Wong wrote:

Hi, I’m trying to figure out how to use Photoshop
for scientific image analysis purposes.

Photoshop has some "enhancements" that makes it is rather difficult to utilize it for scientific purposes.

I need to analyze is the intensity in certain brain
regions created by a specific staining technique.

What is it that gives the "intensities"? X-ray film or something else?

I was thinking of doing is using the square marquee
of Fixed Size to anaylze a square from Region A, and
compare that to a same area in Region B. What confuses me the most is if I do Image –> Histogram, the value for Mean seems to be inversly proportional to the
staining intensity.

The relation depends on the working-space gamma, among other things.

i.e., if the region is darkly stained I
get a small Mean value and vice versa.

Yes, the relative whitepoint is RGB=255,255,255, from there on it gets darker and darker as you approach to level RGB=0,0,0.

Is there a specific formula I would use in order
to determin the relative optical density of a
certain region?

Yes there is but one needs to know what the source is, it has to have a known relation to density in the first place.

Timo Autiokari http://www.aim-dtp.net

In article <SPvXa.63741$>, Leonard Evens
wrote:

Photoshop starts with values in the range 0..255 (for each channel) obtained from some source. That source could be a digital camera or more likely in your case a scanner used to scan the Xray film. One would have to know just what that device is doing in creating the values
0..255. For example, Vuescan used in conjunction with one of the
standard film scanners can actually show what Ed Hamrick tells us is the density of the film at each point in the scan. I haven’t checked with a densitometer to see if his values are accurate, but they certainly seem to be in the right ball park of what I would expect. But his software, as does any scanning software, converts the density values to RGB values in the range 0..255. How it does it depends on some parameters, including the white and black points and a value of gamma. Since the aim in scanning is to get a usable image, whatever the density range in the source, it is going to be difficult to relate the latter values to density.

Most scanning software allows you to save scanning parameters. Might you prepare a slide with the maximum possible stain next to an unstained area? You could relate those densities to 0 and 255, respectively, and use that setting for your test slides. Tom Nelson
Tom Nelson Photography

Jason Wong wrote:

Hi, I’m trying to figure out how to use Photoshop for scientific image analysis purposes. I’m doing some neuroscience research, and what I need to analyze is the intensity in certain brain regions created by a specific staining technique. So what I was thinking of doing is using the square marquee of Fixed Size to anaylze a square from Region A, and compare that to a same area in Region B. What confuses me the most is if I do Image –> Histogram, the value for Mean seems to be inversly proportional to the staining intensity.
i.e., if the region is darkly stained I get a small Mean value and
vice versa.

Does anyone know what I’m talking about? Is there a specific formula I would use in order to determin the relative optical density of a certain region? Any help would be greatly appreciated!

I’ve been thinking some more about this.

Most, if not all, of the replies to your question have concentrated on the numeric conversion of RGB values to density. This response is centered on your need to calculate the average density of a selected area, thereby getting the value of the total amount of stain absorbed by a particular area of a cell.

Density is minus the log of the fraction of transmitted light, aka transparency. For example, clear film would have a transparencyof 1.0 and a density of -log(1.0) = 0.0. An area of film that transmits 1/100 of the light would have a density of -log(.01) = 2.0.

In black and white film density is proportional to the mass of silver per unit area, and this corresponds to your goal of measuring the mass of stain absorbed in a particular area of the cell. I cannot think of an exact way to do this in Photoshop because it deals in luminance directly, and not the log of luminance.

1) avoid gamma issues by converting the image to Lab mode.

2) You will need to calibrate the luminance values by matching a couple of density points, using using a specimen with known density values. This amounts to subtracting the background density from your measurements, and scaling to match a maximum density. Once this is done, you may read luminance values from your 50% means a 50% tranaparency value, or a density of -log(.5) = .30

3) selecting an area with the lasso will .cause Image>Histogram to display the average percent lightness of that area.

4) Subtracting that number from 100 and taking the log will give you an accurate density only if the stain for that area is uniform. If there is much variation in density, the result will be only an approximate average density.

Unfortunately, at the end of the day, the density number yielded by this method is based on an average transparency, and not an average density.

The problem with that is this. Consider a stained area consisting of a checkerboard of .1 and 1.0 transparency , corresponding to density values of
1.0 and 0.0, respectively. The average transparency will be .55, and the
density calculated from this would be -log(.55) = .25, and not the correct value of .5. If your subject does not vary much in density, the numbers just might be accurate enough using this method. My guess is not.

One solution to this is to save the image as a 16 bit grayscale (the raw file format is probably easiest to deal with), and write a C or Visual Basic program to take the logs. You may then open this image in Photoshop, and use the histogram tool to measure density values directly.

Another solution would be to have a custom plugin written to do this calculation. If this is a university setting, and you have student cycles available this could be very reasonable solution to your problem. Or hire a summer student.

There is also a Macintosh only program called NIH image that will do a variety of quantitative measurements of images:
see <http://rsb.info.nih.gov/nih-image/>

—

Mike Russell
http://www.curvemeister.com
http://www.zocalo.net/~mgr
http://geigy.2y.net

Thanks everyone for the helpful replies. However I’m still a little befuddled and confused as to what to do… I’ve switched to trying to learn ImageJ, a java based image analysis program which is the PC version of NIH Image (produced by the National Institute of Health). But I think I’m back to where I started, basically I don’t think I can convert my mean values to density without some kind of calibration step tablet or something like that. Here’s a copy of the message I just wrote to two other forums for help with ImageJ. Hopefully this will clear up what I wanted to do and maybe someone can tell me how to do it (easily preferrably!) in either photoshop or ImageJ. Thanks!

—-

Hello, I’m trying to analyze hippocampal brain slices which I have been immunostained for a neuronal marker (SNAP-25). All areas of the slice in which synaptic connections are occurring are stained brown, so the only unstained areas are where no tissue exists or where cell bodies are located. The hippocampus is composed of several regions of different neuron types (e.g. CA1, CA3, dentate gyrus), which are clearly demarked using this stain by differential staining intensities. However due to difficulties with the staining protocol, staining intensities were not uniform from day to day, brain to brain, and even parallel slice to slice. So to normalize this, I’m going to be dealing with ratios between one hippocampal region and another.

So far I’ve converted my RGB images to 8-bit grayscale by selecting Image -> Type -> 8-bit. I then openned up an image of a scale bar used to set the scale by selecting Analyze -> Set Scale. Next I drew a rectangle area selection tool to draw a square of fixed area (88 squared pixels). I then moved the square to the CA1 region, and recorded the mean gray value by selecting Analyze -> Measure. Finally I moved the selection square to record the mean gray value for a total of 3 regions per image.

What I’m mostly confused with is what the mean gray value actually MEANS. For the darker regions, I get a lower mean value but for the lighter regions, I get a greater one. That seems to go against what I would think of intuitively of what I actually want — which is pixel density. So if region A is darker than region C, I’d get a larger value for region A. Final analysis would involve taking the ration of A to C and comparing it with another region, such as B to C.

Someone mentioned something about calibrating my images. I’m not too sure what this means or entails, but I don’t think I have access to a calibrated optical density step tablet. I don’t know if that would help either because each image I took might have slight changes in brightness as I often fiddled around with the imaging system to produce the best "looking" picture. But I figure the ratios idea will help solve that dilema.

If anyone can understand what I’m trying to do and give me some help, please please please do! Thank you

"Jason Wong" wrote in message
SNIPped introduction

So far I’ve converted my RGB images to 8-bit grayscale by selecting Image -> Type -> 8-bit. I then openned up an image of a scale bar used to set the scale by selecting Analyze -> Set Scale. Next I drew a rectangle area selection tool to draw a square of fixed area (88 squared pixels). I then moved the square to the CA1 region, and recorded the mean gray value by selecting Analyze -> Measure. Finally I moved the selection square to record the mean gray value for a total of 3 regions per image.

Okay, you have now RGB Gray values. As far as I can see, all ImageJ (just like
Photoshop) does when you convert to 8-bit is take the unweighted average or mean (R+G+B)/3. These RGB values are a sort of brightness (higher values are brighter) on a scale from 0 to 255.

What I’m mostly confused with is what the mean gray value actually MEANS. For the darker regions, I get a lower mean value but for the lighter regions, I get a greater one. That seems to go against what I would think of intuitively of what I actually want — which is pixel density. So if region A is darker than region C, I’d get a larger value for region A. Final analysis would involve taking the ration of A to C and comparing it with another region, such as B to C.

You seem to be looking for Optical Density values, which would then be calculated as D=Log10(255/measured), assuming the brightest value in the image equals 255. This would give higher density values for darker tones, on a logarithmic scale, and the difference of two values in Log space is the same as a ratio in Linear space.

Someone mentioned something about calibrating my images. I’m not too sure what this means or entails, but I don’t think I have access to a calibrated optical density step tablet.

The best you can do with the above approach, is Relative density and not Absolute density. Relative density to a lightest measurement of 255 only gives a range of 2.4 density but that may be enough for your purpose. Because you only have 255 logarithmicly spaced values you will create gaps in the range at one end, and loss of precision at the other. Calibration only offers to get Absolute density instead of Relative density.

Hope that helps a bit.

Bart