Hi all - If you have questions about isolating specific colors in ***brightfield*** images, check my video description above for a link to a new, more helpful video. Viewers who work with fluorescent microscopy will still probably find this video helpful.
Hi Juliana, No prob! If you want to cite something peer-reviewed, there are video publications via JoVE. I used the search terms "imagej gray value" and found something that might work for a peer-reviewed reference source: www.jove.com/v/58041/quantitative-cell-biology-neurodegeneration-drosophila-through , but there are many other items in that same search. Alternatively, you could cite my publication that uses this intensity measurement approach: doi.org/10.1371/journal.pone.0224301 . Let me know if you were looking for or need something else.
Thanks you very much on your tutorial. I have a question regarding the calculation for RGB brightfield image. I am using Oil red o staining, could you briefly walk me through how to get the mean overall intensity for the area of interest after you get Red, Green and blue mean values ? Is it to use the average of the sum of the 3 channels from the color histogram even if the stain is red in this case and white background - which mean I would have higher mean (R+G+B) for specimens with lower signal but lower mean value for those with red signal for the fat ? Thank you again !
Hi Olivia, I've been meaning to do another video tutorial on this. It seems that my approach for just obtaining the mean color values may be problematic for doing comparisons between brightfield images, due to how all three colors contribute to white. I think of how a good red stain could be compared to a poor/faded red stain, and it's not as straightforward as it seems. Indeed, a faded stain might have higher R,G, and/or B values due to being closer to white - whereas intuition would otherwise make us think that the R value should be higher (but that's not how it works!). I'm going to test a few things in ImageJ, put up a video this week, and then link it here in a reply once I'm done. You'll also note that another user or two asked about the Oil Red issue, so I think my video will focus on that as an example.
Hi again - I've now made and posted that new video tutorial with detailed explanations on how the color systems work. Please see here: ruclips.net/video/O0LMhS_JtGs/видео.html
Hi! Thanks for you complete and useful video! I have a question. My version of Image J does not show Color Histogram, only Histogram. I need the gray color distribution for greens, and the frequency, do you know how I can get those values? Thanks!
Hello! It may be the case that you are using the prior version of ImageJ that lacks the various plug-ins. The newer version is FIJI, or ImageJ2 with add-ons. imagej.net/software/fiji/ Give that a try and let me know if you still have issues.
Hey Kevin, thank you for this, this is amazing! Sorry for another question, but can I know if I can use this to compare the intensity of a colour vs blank. (I have a dye I use, I need to see how intense the colour is, can I use this to quantify the blue I get and how blue it is to another blue etc. ?
Hello! I did post a newer video on this channel: ruclips.net/video/O0LMhS_JtGs/видео.htmlsi=gZYk_dInTXcpL11K Take a look at that, and see if it might help you. I know I covered specific color staining intensity approaches there, but comment again if you're needing more pointers.
@@kruneuro Hey Kevin, thank you again for this video and your contribution towards science, this is greatly appreciated. Can I please know how to use ImageJ to get the different in colour intensity of two normal images. I have two images of teeth, both have being taken in accurate, camera settings, DSLR Macro lens and same lighting (the image to image difference is less than 2% and a lot of work has being done to ensure the accuracy of the two) My question is one image of the tooth I have is of just a normal tooth with active carries, the second image I take is stained with the dye we use. I need to find a way to quantify the colour blue vs the control which does not have the dye of the images of teeth. Can you please suggest me a way to help with this, thank you kindly!
@@javeentharka513Hi there, I think the portion of my other video (ruclips.net/video/O0LMhS_JtGs/видео.htmlsi=YbYk81-ctG6o8K5y) gives the particular method you could use. Specifically, the portion from 13:11 to 17:00 will cover what you need (
Did you mean total gray value? If so, I suspect you want the mean gray value. If you want to do it for the whole image, you make sure to either select the entire image or otherwise select nothing, and then use the Measure function, which will give the average gray value in the Mean column. Let me know if this is not what you were looking for.
Hi Kevin, thank you for sharing this lecture, it helps a lot. I have a question about my research analysis: I need to quantify the intensity of purple in a picture (which is the extracellular matrix), but I'm working with a metachromatic stain (toluidine blue), so the nuclei around my sample are very dark blue. Will I have to measure in a higher magnification to select only small pieces of matrix, so I don't get the interference of the blue nuclei? And in this case, I should look for the blue ou red values? If I convert to 8-bit and measure the mean of gray values, is it better? Sorry for so many questions, but I am really confused about the method I should use that better represents my data, and your experience would help me a lot 🙏
No worries about the questions. I wish I had been able to answer them more promptly, but this semester had a rough workload. I think that it can be difficult to filter out similar colors, so I usually advise for isolating the specific areas of interest (the EM in your case) via higher magnification. BUT, if going back and snapping more microscope pics is not feasible, then I think the color deconvolution can separate the blue and the purple out. Depending on the setting, it could isolate the image into blue and red or blue and purple components; I'm less certain that it can do the latter, though the latter is more of what you want in your case. I was testing out some things with an H&E stain, which is somewhat similar to what you are working with, even if not identical. Color deconvolution produced one image of only blue nuclei, one image of green haze (to be ignored) and one image of pink EM-type substance. Worth noting is that the pink image still did not have "holes" where the nuclei were; it was still pink in those areas. In this regard, I have concerns of trying to assess EM staining efficacy by selecting a broad area - including the cell nuclei - if the influence of the nuclear stain cannot be excised from your analysis. I guess a follow-up question you could answer is: When you use your EM stain alone, are there holes where nuclei exist? If so, I think that the higher magnification option will be a better fit, as color deconvolution doesn't seem to carefully excise the influence of the nuclear staining from the ME stain channel, only vice versa seems to work (no EM staining in the nuclear staining channel).
One additional thing I did find from further playing around is that the nuclei can be subtracted out of the image. After performing the above color deconvolution, I took the nuclear stain image, converted it to 8-bit grayscale via Image -> Type (might need to toggle to RGB and then 8-bit to get it to work), and then I inverted it via Edit -> Invert. I then took the EM stain image and converted it to an RGB image type. Finally, I went to Process -> Image Calculator, and then put the EM stain first, function Add, and the inverted nuclear stain next. The end result was that the EM stain now has white holes where the nuclei were, without otherwise affecting other parts of the image color or intensity-wise.
@@mellanieferreira3491 Hi again - I've created a new video that may be helpful for your continued image analysis. Please see here: ruclips.net/video/O0LMhS_JtGs/видео.html
Thanks a lot for the tutorial - really helpful so far! I'm using Oil red O staining and would like to compare the intensity of red between images/strains. Is it fair to divide mean(red) by mean(mean(red,blue,green)) and compare the resulting values to determine the intensity of red while not considering overall brightness? Again, thanks a lot!
Absolutely! I had been thinking about this and intent to make a brief video on it. I work mostly with fluorescence (black background), so I needed to re-think how to subtract white background in brightfield examples. Your example of dividing out the combined RBG means is what will do the trick, especially if applying it to a full image rather than a selection. I have a separate video on color deconvolution if that is helpful for you as well.
@@Mirabell97 I thought on that equation you devised, and I am unsure if it's the best way to isolate red values minus white in order to compare red values between images. I'll need some time to work on that idea, as there might be multiple ways to approach it. Some methods might involve either the Color Threshold tool or the Subtract option as in my other video, but I haven't gotten them to work in the right way yet. I'll try to keep you posted.
@@kruneuro that would be great! At least for the few images I've tested so far the values I get seem representative for the intensity of red I see - but since I don't really understand the theory behind it, I'd appreciate any feedback/advice!
@@Mirabell97 Hi again - I finally had time to investigate the weird color measurement issues we discussed. Please see my new video here, if you still have question on your analysis approach: ruclips.net/video/O0LMhS_JtGs/видео.html
i have an evan blue staining sample, in which BLUE is the positive signal. In this case measuring gray is not accurate and actually gives a obvious fault conclusion. how to only measure blue in this case?
Hi again: For your method, you'll want to use the "color histogram" function rather than the simpler "measure" function. The color histogram is best applied for color-picture light/brightfield microscopy situations like yours, and the measure function is best applied for either monochrome brightfield or single-channel fluorescence photos. So, go to 33:40, perform the steps, and specifically look at the "blue" row of the Results window. When you compare specimens where one is positive (much Evans blue staining) and one is negative (little to no Evans blue), you'll acquire the "blue" means for both specimens to quantify the difference between them.
@@kruneuro I tried, the issue I saw is that negative group has a higher mean value of blue color, while looking by eye, there is no blue staining......
@@gengpan Sorry about the delayed reply. That issue is odd: the blue value should be near or equal to zero if there is no blue in the image. However, other colors can contribute to the blue value, specifically whites/grays, yellows and purples. If such colors exist in the image, you can attempt color deconvolution to separate things out and re-attempt the intensity measurement or color histogram measurement. ruclips.net/video/xl6s0jho94M/видео.html You'd have to tweak the setting to your needs, but hopefully this resolves things.
@@kruneuro Hi. I had the same issue above. I think its because the positive is a dark blue and the negative is a light one, then if it is near to white it happens to be higher in the color histogram, right? how can we solve that, doing the calibration of optic density?
For optical density measurements, it seems that ImageJ will need to be calibrated first, then measurements within a selected will need to be taken. I think this video should be helpful: ruclips.net/video/ZCld9Qe2wZg/видео.html
@@hssz23 I believe the video I linked does work with IHC-positive cells. But, let me know if you are trying to do something specific - perhaps you mean isolating those cells by somehow dropping out the background, in order to get a purer "cell only" optical density?
The option should be under the Analyze menu. If it is not, you may have a different version of ImageJ. I recommend the "FIJI" version (a.k.a. "distribution") that has various plug-ins installed and calibrated. imagej.net/imaging/
So if I'm looking for quantifying the blue coloured region only, I've to go gor the Blue value? what about any colour other than RGB?? plz help me out with this!!
Yes - when you pull up the Color Histogram tool, the mean Blue value will be giving you the average for just shades of blue in the image *isolated* from red and green. The range of this number goes from 0-255 for each color (R, G, and B). Zero is black, 255 is fully that color (fully blue in this case) with no black tinge to it. If you're trying to measure particular colors other than R,G, or B, than it gets a bit tricky. ImageJ will treat any color as a combo of R+G+B. Thus, the color histogram breaks any color that you see into its RGB components when it does its measurement, so you wouldn't get a simple number to work with; you would instead get three color averages. Whether or not that would work for you depends on your goals.
@@kruneuro Incase of splitting the image in R G and B, I've seen that in R the red colour gets almost invisible, same with green in G.. so if I've to quantify the green colour I don't think i can select the G!? or I've to select G? moreover the imageJ I'm using ( don't know which version) doesn't have the option of histogram tool. if u could provide the link of your version!?
@@shibamdas915 To my knowledge, the up-to-date version with various useful plugins including the color histogram is via this link: imagej.net/Cookbook In case you want my full protocol, email me (find Kevin Urstadt at Occidental for my faculty profile and thus email address). I don't list direct links on here to my own stuff as I want to avoid being harassed by bots/spammers that comb through pages for links. What you are doing there is separating the image into three separate R, G, and B pieces. That can still work - you can select the same location in each piece (using Edit -> Selection -> Specify to get the exact selection box size and location). Then, if you just use the Analyze -> Measure function, it will give you the mean value within that box. So, if you are only using the blue piece, it will give only the mean blue value within your selection box. The Color Histogram tool shortcuts this process by not needing to split the image into three R,G,B pieces. And the reason why the red piece is nearly invisible is because there's probably very little red in the original image. Same goes for the green piece being nearly invisible. This tells me your image is mostly blue, with a lot of blue-similar colors like cyan, blue-green, and maybe something slightly purple-blue. I hope all of this is helpful!
@@shibamdas915 Hi again - I made a new video that discusses better ways to do the RGB split (or even alternate in-between colors) that may be helpful. I know it has been quite a while since our last correspondence on here, but I thought I'd let you know in vase you are continuing work on it: ruclips.net/video/O0LMhS_JtGs/видео.html
Hi there - I know it has been a while since we corresponded, but I posted a new video that has different approaches to color analysis. See here, in case it is helpful: ruclips.net/video/O0LMhS_JtGs/видео.html
I had to take a quick look through the video to see which part you were referring to; it is ~18:45, correct? I select the red coloring in this specific example because this particular stain (a diluted cresyl violet stain) only produces fluorescence in the red emission channel. And although I could capture it as red, I just convert it to grayscale for better visibility on screen. Whether I keep it red or I keep it gray does not affect how the measuring process or results go - both a red-only image (sometimes called "indexed") and a gray-only image still have shades from 0 to 255. To your question again: you can select any fluorescence color channel to perform these measurements or convert to grayscale - I just chose red in this example.
Hi Kevin, is this protocol (the google docs document) available anywhere? I would love to have it on hand, I'm analyzing RGB trichrome stained tissues so this would be awesome to have!
Hello Caroline, I can share it by email. I'm hesitant to post links on youtube (including my email) as bots may scavenge those links and harass me with spam or access requests. You can find my email through Occidental's Cognitive Science department profiles webpage. Sorry for the inconvenience!
Thanks for your interest. This is a slightly tricky question. I think for academic journals in biology, they may not accept references & protocols that are not from peer reviewed sources, such as youtube. A close video substitute may be found in JoVE; I used the search terms "imagej gray value" and found something that might work for a peer-reviewed reference source: www.jove.com/v/58041/quantitative-cell-biology-neurodegeneration-drosophila-through , but there are many other items in that same search. Alternatively, you could cite my publication that uses this intensity measurement approach: doi.org/10.1371/journal.pone.0224301 . Let me know if you have follow-up questions.
Hi all - If you have questions about isolating specific colors in ***brightfield*** images, check my video description above for a link to a new, more helpful video. Viewers who work with fluorescent microscopy will still probably find this video helpful.
Thanks a lot, very helpful!
Hi. Thanks for the video. Can you refer me to an article that I can cite in the methodology when using this evaluation technique?
Hi Juliana, No prob! If you want to cite something peer-reviewed, there are video publications via JoVE. I used the search terms "imagej gray value" and found something that might work for a peer-reviewed reference source: www.jove.com/v/58041/quantitative-cell-biology-neurodegeneration-drosophila-through , but there are many other items in that same search. Alternatively, you could cite my publication that uses this intensity measurement approach: doi.org/10.1371/journal.pone.0224301 . Let me know if you were looking for or need something else.
Thanks you very much on your tutorial.
I have a question regarding the calculation for RGB brightfield image. I am using Oil red o staining, could you briefly walk me through how to get the mean overall intensity for the area of interest after you get Red, Green and blue mean values ? Is it to use the average of the sum of the 3 channels from the color histogram even if the stain is red in this case and white background - which mean I would have higher mean (R+G+B) for specimens with lower signal but lower mean value for those with red signal for the fat ?
Thank you again !
Hi Olivia,
I've been meaning to do another video tutorial on this. It seems that my approach for just obtaining the mean color values may be problematic for doing comparisons between brightfield images, due to how all three colors contribute to white. I think of how a good red stain could be compared to a poor/faded red stain, and it's not as straightforward as it seems. Indeed, a faded stain might have higher R,G, and/or B values due to being closer to white - whereas intuition would otherwise make us think that the R value should be higher (but that's not how it works!). I'm going to test a few things in ImageJ, put up a video this week, and then link it here in a reply once I'm done.
You'll also note that another user or two asked about the Oil Red issue, so I think my video will focus on that as an example.
Hi again - I've now made and posted that new video tutorial with detailed explanations on how the color systems work. Please see here: ruclips.net/video/O0LMhS_JtGs/видео.html
Hi! Thanks for you complete and useful video! I have a question. My version of Image J does not show Color Histogram, only Histogram. I need the gray color distribution for greens, and the frequency, do you know how I can get those values? Thanks!
Hello! It may be the case that you are using the prior version of ImageJ that lacks the various plug-ins. The newer version is FIJI, or ImageJ2 with add-ons. imagej.net/software/fiji/
Give that a try and let me know if you still have issues.
@@kruneuro thank you!
Hey Kevin, thank you for this, this is amazing!
Sorry for another question, but can I know if I can use this to compare the intensity of a colour vs blank. (I have a dye I use, I need to see how intense the colour is, can I use this to quantify the blue I get and how blue it is to another blue etc. ?
Hello!
I did post a newer video on this channel: ruclips.net/video/O0LMhS_JtGs/видео.htmlsi=gZYk_dInTXcpL11K
Take a look at that, and see if it might help you. I know I covered specific color staining intensity approaches there, but comment again if you're needing more pointers.
@@kruneuro Hey Kevin, thank you again for this video and your contribution towards science, this is greatly appreciated.
Can I please know how to use ImageJ to get the different in colour intensity of two normal images.
I have two images of teeth, both have being taken in accurate, camera settings, DSLR Macro lens and same lighting (the image to image difference is less than 2% and a lot of work has being done to ensure the accuracy of the two)
My question is one image of the tooth I have is of just a normal tooth with active carries, the second image I take is stained with the dye we use.
I need to find a way to quantify the colour blue vs the control which does not have the dye of the images of teeth. Can you please suggest me a way to help with this, thank you kindly!
@@javeentharka513Hi there,
I think the portion of my other video (ruclips.net/video/O0LMhS_JtGs/видео.htmlsi=YbYk81-ctG6o8K5y) gives the particular method you could use. Specifically, the portion from 13:11 to 17:00 will cover what you need (
Thanks alot for this video, i would like to ask about , if i want to take the length for this part, how i can do it?
Could you clarify what you mean by "take the length" and which part you are referring to?
how to calculate total very value of Xray image in this software thanks
Did you mean total gray value?
If so, I suspect you want the mean gray value. If you want to do it for the whole image, you make sure to either select the entire image or otherwise select nothing, and then use the Measure function, which will give the average gray value in the Mean column. Let me know if this is not what you were looking for.
Hi Kevin, thank you for sharing this lecture, it helps a lot. I have a question about my research analysis: I need to quantify the intensity of purple in a picture (which is the extracellular matrix), but I'm working with a metachromatic stain (toluidine blue), so the nuclei around my sample are very dark blue. Will I have to measure in a higher magnification to select only small pieces of matrix, so I don't get the interference of the blue nuclei? And in this case, I should look for the blue ou red values?
If I convert to 8-bit and measure the mean of gray values, is it better?
Sorry for so many questions, but I am really confused about the method I should use that better represents my data, and your experience would help me a lot 🙏
No worries about the questions. I wish I had been able to answer them more promptly, but this semester had a rough workload.
I think that it can be difficult to filter out similar colors, so I usually advise for isolating the specific areas of interest (the EM in your case) via higher magnification. BUT, if going back and snapping more microscope pics is not feasible, then I think the color deconvolution can separate the blue and the purple out. Depending on the setting, it could isolate the image into blue and red or blue and purple components; I'm less certain that it can do the latter, though the latter is more of what you want in your case.
I was testing out some things with an H&E stain, which is somewhat similar to what you are working with, even if not identical. Color deconvolution produced one image of only blue nuclei, one image of green haze (to be ignored) and one image of pink EM-type substance. Worth noting is that the pink image still did not have "holes" where the nuclei were; it was still pink in those areas. In this regard, I have concerns of trying to assess EM staining efficacy by selecting a broad area - including the cell nuclei - if the influence of the nuclear stain cannot be excised from your analysis. I guess a follow-up question you could answer is: When you use your EM stain alone, are there holes where nuclei exist? If so, I think that the higher magnification option will be a better fit, as color deconvolution doesn't seem to carefully excise the influence of the nuclear staining from the ME stain channel, only vice versa seems to work (no EM staining in the nuclear staining channel).
One additional thing I did find from further playing around is that the nuclei can be subtracted out of the image. After performing the above color deconvolution, I took the nuclear stain image, converted it to 8-bit grayscale via Image -> Type (might need to toggle to RGB and then 8-bit to get it to work), and then I inverted it via Edit -> Invert. I then took the EM stain image and converted it to an RGB image type. Finally, I went to Process -> Image Calculator, and then put the EM stain first, function Add, and the inverted nuclear stain next. The end result was that the EM stain now has white holes where the nuclei were, without otherwise affecting other parts of the image color or intensity-wise.
@@kruneuro Thank you SO much. I will try all your suggestions and see what works best for my samples. You've helped me a lot, really. Happy holidays!
@@mellanieferreira3491 Hi again - I've created a new video that may be helpful for your continued image analysis. Please see here: ruclips.net/video/O0LMhS_JtGs/видео.html
Thanks a lot for the tutorial - really helpful so far! I'm using Oil red O staining and would like to compare the intensity of red between images/strains. Is it fair to divide mean(red) by mean(mean(red,blue,green)) and compare the resulting values to determine the intensity of red while not considering overall brightness? Again, thanks a lot!
Absolutely! I had been thinking about this and intent to make a brief video on it. I work mostly with fluorescence (black background), so I needed to re-think how to subtract white background in brightfield examples. Your example of dividing out the combined RBG means is what will do the trick, especially if applying it to a full image rather than a selection.
I have a separate video on color deconvolution if that is helpful for you as well.
@@kruneuro Thanks a lot for reassuring me that that approach might work!
@@Mirabell97 I thought on that equation you devised, and I am unsure if it's the best way to isolate red values minus white in order to compare red values between images. I'll need some time to work on that idea, as there might be multiple ways to approach it. Some methods might involve either the Color Threshold tool or the Subtract option as in my other video, but I haven't gotten them to work in the right way yet. I'll try to keep you posted.
@@kruneuro that would be great! At least for the few images I've tested so far the values I get seem representative for the intensity of red I see - but since I don't really understand the theory behind it, I'd appreciate any feedback/advice!
@@Mirabell97 Hi again - I finally had time to investigate the weird color measurement issues we discussed. Please see my new video here, if you still have question on your analysis approach: ruclips.net/video/O0LMhS_JtGs/видео.html
i have an evan blue staining sample, in which BLUE is the positive signal. In this case measuring gray is not accurate and actually gives a obvious fault conclusion. how to only measure blue in this case?
Hi again: For your method, you'll want to use the "color histogram" function rather than the simpler "measure" function. The color histogram is best applied for color-picture light/brightfield microscopy situations like yours, and the measure function is best applied for either monochrome brightfield or single-channel fluorescence photos. So, go to 33:40, perform the steps, and specifically look at the "blue" row of the Results window.
When you compare specimens where one is positive (much Evans blue staining) and one is negative (little to no Evans blue), you'll acquire the "blue" means for both specimens to quantify the difference between them.
@@kruneuro I tried, the issue I saw is that negative group has a higher mean value of blue color, while looking by eye, there is no blue staining......
@@gengpan Sorry about the delayed reply. That issue is odd: the blue value should be near or equal to zero if there is no blue in the image. However, other colors can contribute to the blue value, specifically whites/grays, yellows and purples. If such colors exist in the image, you can attempt color deconvolution to separate things out and re-attempt the intensity measurement or color histogram measurement.
ruclips.net/video/xl6s0jho94M/видео.html
You'd have to tweak the setting to your needs, but hopefully this resolves things.
@@kruneuro thank you
@@kruneuro Hi. I had the same issue above. I think its because the positive is a dark blue and the negative is a light one, then if it is near to white it happens to be higher in the color histogram, right? how can we solve that, doing the calibration of optic density?
so for chromogenic ihc how do i apply that ? for OD
optical density
For optical density measurements, it seems that ImageJ will need to be calibrated first, then measurements within a selected will need to be taken. I think this video should be helpful: ruclips.net/video/ZCld9Qe2wZg/видео.html
thanks alot for the link i just realized, however, for the current video, can we still apply it to ihc positive cells or so ?
@@hssz23 I believe the video I linked does work with IHC-positive cells. But, let me know if you are trying to do something specific - perhaps you mean isolating those cells by somehow dropping out the background, in order to get a purer "cell only" optical density?
hey howd you get the color histogram
The option should be under the Analyze menu. If it is not, you may have a different version of ImageJ. I recommend the "FIJI" version (a.k.a. "distribution") that has various plug-ins installed and calibrated. imagej.net/imaging/
It does not show color histogram option in analyze. Where can i find it?
I suspect you might have the regular version of ImageJ without the various add-ons. I recommend getting this version, called FIJI.
fiji.sc/
So if I'm looking for quantifying the blue coloured region only, I've to go gor the Blue value? what about any colour other than RGB??
plz help me out with this!!
Yes - when you pull up the Color Histogram tool, the mean Blue value will be giving you the average for just shades of blue in the image *isolated* from red and green. The range of this number goes from 0-255 for each color (R, G, and B). Zero is black, 255 is fully that color (fully blue in this case) with no black tinge to it.
If you're trying to measure particular colors other than R,G, or B, than it gets a bit tricky. ImageJ will treat any color as a combo of R+G+B. Thus, the color histogram breaks any color that you see into its RGB components when it does its measurement, so you wouldn't get a simple number to work with; you would instead get three color averages. Whether or not that would work for you depends on your goals.
@@kruneuro Incase of splitting the image in R G and B, I've seen that in R the red colour gets almost invisible, same with green in G.. so if I've to quantify the green colour I don't think i can select the G!? or I've to select G?
moreover the imageJ I'm using ( don't know which version) doesn't have the option of histogram tool. if u could provide the link of your version!?
@@shibamdas915 To my knowledge, the up-to-date version with various useful plugins including the color histogram is via this link: imagej.net/Cookbook
In case you want my full protocol, email me (find Kevin Urstadt at Occidental for my faculty profile and thus email address). I don't list direct links on here to my own stuff as I want to avoid being harassed by bots/spammers that comb through pages for links.
What you are doing there is separating the image into three separate R, G, and B pieces. That can still work - you can select the same location in each piece (using Edit -> Selection -> Specify to get the exact selection box size and location). Then, if you just use the Analyze -> Measure function, it will give you the mean value within that box. So, if you are only using the blue piece, it will give only the mean blue value within your selection box. The Color Histogram tool shortcuts this process by not needing to split the image into three R,G,B pieces.
And the reason why the red piece is nearly invisible is because there's probably very little red in the original image. Same goes for the green piece being nearly invisible. This tells me your image is mostly blue, with a lot of blue-similar colors like cyan, blue-green, and maybe something slightly purple-blue.
I hope all of this is helpful!
@@kruneuro Thank you so much
@@shibamdas915 Hi again - I made a new video that discusses better ways to do the RGB split (or even alternate in-between colors) that may be helpful. I know it has been quite a while since our last correspondence on here, but I thought I'd let you know in vase you are continuing work on it: ruclips.net/video/O0LMhS_JtGs/видео.html
Then How to measure the single color intensity, if dimmer blue actually gives a very high bMEAN value in the "color histogram"?
Hi there - I know it has been a while since we corresponded, but I posted a new video that has different approaches to color analysis. See here, in case it is helpful: ruclips.net/video/O0LMhS_JtGs/видео.html
@@kruneuro thank you.
why choose red channel for gray intensity analyse?
I had to take a quick look through the video to see which part you were referring to; it is ~18:45, correct? I select the red coloring in this specific example because this particular stain (a diluted cresyl violet stain) only produces fluorescence in the red emission channel. And although I could capture it as red, I just convert it to grayscale for better visibility on screen. Whether I keep it red or I keep it gray does not affect how the measuring process or results go - both a red-only image (sometimes called "indexed") and a gray-only image still have shades from 0 to 255. To your question again: you can select any fluorescence color channel to perform these measurements or convert to grayscale - I just chose red in this example.
Hi Kevin, is this protocol (the google docs document) available anywhere? I would love to have it on hand, I'm analyzing RGB trichrome stained tissues so this would be awesome to have!
Hello Caroline,
I can share it by email. I'm hesitant to post links on youtube (including my email) as bots may scavenge those links and harass me with spam or access requests. You can find my email through Occidental's Cognitive Science department profiles webpage. Sorry for the inconvenience!
Heads up - I sent replies to both of your emails. If you didn't receive either, check your spam filters.
How I can cite this protocol in my research? Also, I need previous studies cite the same protocol
Thanks for your interest. This is a slightly tricky question. I think for academic journals in biology, they may not accept references & protocols that are not from peer reviewed sources, such as youtube. A close video substitute may be found in JoVE; I used the search terms "imagej gray value" and found something that might work for a peer-reviewed reference source: www.jove.com/v/58041/quantitative-cell-biology-neurodegeneration-drosophila-through , but there are many other items in that same search. Alternatively, you could cite my publication that uses this intensity measurement approach: doi.org/10.1371/journal.pone.0224301 . Let me know if you have follow-up questions.