I need to transform thermographic images into gray-scale to detect temperature variation using matlab.
I tried to use the inbuilt function rgb2gray(), but it's inefficient for my image database, so I tried to compute the formula for this, using correlation between colours and temperature and then interpolating it, but then will be it possible to get the efficient mathematical formula?
My formula:
i=0.29 * rgb_img(:,:,1) + 0.59 * rgb_img(:,:,2) + 0.11 *
rgb_img(:,:,3);
and
i=0.6889*rgb_img(:,:,1)+.5211*rgb_img(:,:,2)+.4449*rgb_img(:,:,3)+72.8;
but both of these didn't help.
sample images are:
http://www.google.com/imgres?imgurl=http://www.breastthermography.com/images/AP-Breast-Case-2.jpg&imgrefurl=http://www.breastthermography.com/case_studies.htm&h=199&w=300&tbnid=48Yto8Y8RtRPjM:&zoom=1&docid=PX1nchTaFQpa3M&ei=ftVdVIatNZCQuATptoLgCA&tbm=isch
There is small vertical multicolor bar at the right edge of images. It represents palette which has been used for Gray->Color transformation. If you can no access to exact formula for this thansform, you can try approximation: extract 1-pixel wide vertical line to array from bottom to top.
If resulting array length is not equal to 256, adjust the range (interpolate if needed). Now you have map: Color -> Index of this Color = 8-bit Gray Value. Use it to construct grayscale image. If you need absolute temperature, you can adjust range to 280-350 (in 0.1 degrees) or another suitable range.
The formula used in JFIF is
Y = 0.299R +0.587G +0.114B
That's fairly close to what you have. So I do know what you mean when you say what you were using did not work.
Related
I have imported an image. I have parsed it to double precision and performed some filtering on it.
When I plot the result with imshow, the double image is too dark. But when I use imshowpair to plot the original and the final image, both images are correctly displayed.
I have tried to use uint8, im2uint8, multiply by 255 and then use those functions, but the only way to obtain the correct image is using imshowpair.
What can I do?
It sounds like a problem where the majority of your intensities / colour data are outside the dynamic range of what is accepted for imshow when showing double data.
I also see that you're using im2double, but im2double simply converts the image to double and if the image is already double, nothing happens. It's probably because of the way you are filtering the images. Are you doing some sort of edge detection? The reason why you're getting dark images is probably because the majority of your intensities are negative, or are hovering around 0. imshow whe displaying double type images assumes that the dynamic range of intensities is [0,1].
Therefore, one way to resolve your problem is to do:
imshow(im,[]);
This shifts the display so that range so the smallest value is mapped to 0, and the largest to 1.
If you'd like a more permanent solution, consider creating a new output variable that does this for you:
out = (im - min(im(:))) / (max(im(:)) - min(im(:)));
This will perform the same shifting that imshow does when displaying data for you. You can now just do:
imshow(out);
My x-axis is latitudes, y-axis is longitudes, and z-axis is the hist3 of the two. It is given by: z=hist3(location(:,1:2),[180,360]), where location(:,1) is the latitude column, and location(:,2) is the longitude column.
What I now want is, instead of plotting on a self-created XY plane, I want to plot the same on a worldmap. And instead of representing the frequency of each latitude-longitude pair with the height of the bars of hist3, I want to represent the frequency of each location by a heat map on top of the world map, corresponding to each latitude-longitude pair's frequency on the dataset. I have been searching a lot for this, but have not found much help. How to do this? I could only plot the skeleton of the worldmap like this:
worldmap world
load geoid
geoshow(geoid, geoidrefvec, 'DisplayType', 'texturemap');
load coast
geoshow(lat, long)
I don't know what the colour is being produced based on.
Additionally, if possible, I would also like to know how to plot the hist3 on a 3D map of the world (or globe), where each bar of the hist3 would correspond to the frequency of each location (i.e., each latitude-longitude pair). Thank you.
The hist3 documentation, which you can find here hist3, says:
Color the bars based on the frequency of the observations, i.e. according to the height of the bars. set(get(gca,'child'),'FaceColor','interp','CDataMode','auto');
If that's not what you need, you might wanna try it with colormap. More info about it here colormap. I haven't tried using colormap on histograms directly, so If colormap doesn't help, then you can try creating a new matrix manually which will have values in colors instead of the Z values the histogram originally had.
To do that, you need to first calculate the maximum Z value with:
maxZ=max(Z);
Then, you need to calculate how much of the colors should overlap. For example, if you use RGB system and you assign Blue for the lowest values of the histogram, then Green for the middle and Red for the High, and the green starts after the Blue with no overlap, than it will look artificial. So, if you decide that you will have, for example overlapping of 10 values, than, having in mind that every R, G and B component of the RGB color images have 255 values (8 bits) and 10 of each overlap with the former, that means that you will have 255 values (from the Blue) + 245 values (From the Green, which is 255 - 10 since 10 of the Green overlap with those of the Blue) + 245 (From the Red, with the same comment as for the Green), which is total amount of 745 values that you can assign to the new colored Histogram.
If 745 > maxZ there is no logic for you to map the new Z with more than maxZ values. Then you can calculate the number of overlaping values in this manner:
if 745 > maxZ
overlap=floor(255- (maxZ-255)/2)
end
At this point you have 10 overlapping values (or more if you still think that it doesn't looks good) if the maximum value of the Z is bigger than the total amount of values you are trying to assign to the new Z, or overlap overlapping values, if the maximum of Z is smaller.
When you have this two numbers (i.e. 745 and maxZ), you can write the following code so you can create the newZ.
First you need to specify that newZ is of the same size as Z. You can achieve that by creating a zero matrix with the same size as Z, but having in mind that in order to be in color, it has to have an additional dimension, which will specify the three color components (if you are working with RGB).
This can be achieved in the following manner:
newZ=zeros(size(Z),3)
The number 3 is here, as I said, so you would be able to give color to the new histogram.
Now you need to calculate the step (this is needed only if maxZ > The number of colors you wish to assign). The step can be calculated as:
stepZ=maxZ/Total_Number_of_Colors
If maxZ is, for example 2000 and Total_Number_of_Colors is (With 10 overlaping colours) 745, then stepZ=2.6845637583892617449664429530201. You will also need a counter so you would know what color you would assign to the new matrix. You can initialize it here:
count=0;
Now, finally the assignment is as follows:
For i=1:stepZ:maxZ
count=count+1;
If count>245
NewZ(Z==stepz,3)=count;
elseif count>245 && count<256
NewZ(Z==stepz,3)=count;
NewZ(Z==stepz,2)=count-245;
elseif count>255
NewZ(Z==stepz,2)=count-245;
elseif count>500 && count<511
NewZ(Z==stepz,2)=count-245;
NewZ(Z==stepz,1)=count-500;
else
NewZ(Z==stepz,1)=count-500;
end
end
At this point you have colored your histogram. Note that you can manually color it in different colors than red, green and blue (even if you are working in RGB), but it would be a bit harder, so if you don't like the colors you can experiment with the last bit of code (the one with the for loops), or check the internet of some other automatic way to color your newZ matrix.
Now, how do you think to superimpose this matrix (histogram) over your map? Do you want only the black lines to be shown over the colored histogram? If that's the case, than it can be achieved by resampling the NewZ matrix (the colored histogram) with the same precision as the map. For example, if the map is of size MxN, then the histogram needs to be adjusted to that size. If, on the other hand, their sizes are the same, then you can directly continue to the next part.
Your job is to find all pixels that have black in the map. Since the map is not binary (blacks and whites), it will be a bit more harder, but still achievable. You need to find a satisfactory threshold for the three components. All the lines under this threshold should be the black lines that are shown on the map. You can test these values with imshow(worldmap) and checking the values of the black lines you wish to preserve (borders and land edges, for example) by pointing the cross tool on the top of the figure, in the tools bar on every pixel which is of interest.
You don't need to test all black lines that you wish to preserve. You just need to have some basic info about what values the threshold should have. Then you continue with the rest of the code and if you don't like the result so much, you just adjust the threshold in some trial and error manner. When you have figured that this threshold is, for example, (40, 30, 60) for all of the RGB values of the map that you wish to preserve (have in mind that only values that are between (0,0,0) and (40,30,60) will be kept this way, all others will be erased), then you can add the black lines with the following few commands:
for i = 1:size(worldmap,1)
for j = 1:size(worldmap,2)
if worldmap(i,j,1)<40 && worldmap(i,j,2)<30 && worldmap(i,j,3)<60
newZ(i,j,:)=worldmap(i,j,:)
end
end
I want to note that I haven't tested this code, since I don't have Matlab near me atm, so It can have few errors, but those should be easily debugable.
Hopes this is what you need,
Cheers!
Just clarifying a point about the Otsu thresholding method that lacks definition in the documentation & wikipedia articles. If you apply the Otsu method (in matlab the function graythresh) it returns a threshold value between 0 and 1.
Given 2 hypothetical grayscale images:
dark (with pixel intensities in the range of 0 to 100) and
light (with pixel intensities in the range of 155 to 255)
If I got an Otsu threshold of 0.75 for both dark and light images respectively, what grayscale pixel intensity would it map to in each case?
dark -> 75 and light -> 231 E.g. relative to the range of values in each image
dark -> 75 and light -> 191 E.g. relative to the range 0 to max pixel value
dark -> 191 and light -> 191 E.g. relative to the full range of grayscale pixel values (0-255)?
The accepted answer by #Ratbert makes the incorrect claims that
The correct answer is the first one
and
graythresh uses the min and max values in the image as boundaries,
which is the most logical behavior.
and rayryeng appears to agree with it. David Parks appears to have empirically verified it.
The correct answer is given by Anand which strangely seems to have a negative vote. He explains very convincingly that
full range of grayscale pixel values' depends on the data type of the input image
As he explains,
this is the third option
except for the fact that the dark image could not possibly get a threshold of 0.75.
First let us clarify the difference between the claims for the simplest case, in clear MATLAB, so there is no confusion. For an image with values ranging from min to max, the question poses three possibilities which, when translated to an equation are:
threshold = min + (max-min) * graythresh
threshold = max * graythresh
threshold = 255 * graythresh
Suppose the image consists of just two points one with an intensity of 0, and the other with 100. This means dark = uint8([0 100]);. A second image light = dark+155;. When we compute 255*graythresh(dark) we get exactly 49.5. When we compute 255*graythresh(light) we get exactly 204.5. These answers make it patently obvious that the third option is the only possibility.
There is one further fine point. If you try 255*graythresh(uint8(1:2)) the answer is 1, and not 1.5. So it appears that if you are using greythresh to threshold an image, you should use image <= 255*graythesh(image) with a less-than-or-equal-to, rather than a plain less-than.
Your third answer seems most right to me, with the clarification that 'full range of grayscale pixel values' depends on the data type of the input image. For example, for a uint8 image, an Otsu threshold of 0.75 corresponds to around 191. For a uint16 image, this would correspond to 49151.
Well, for posterity sake I did a comparison of the approaches mentioned before. I took a typical image with a full range of grayscale pixel intensities, then took a dark and light version of the same image and got the graythresh value for each. I Applied the Otsu threshold using each of the above mappings.
The light image pretty clearly demonstrates that the algorithm is producing a threshold over the range of the images' pixel intensities. Were we to assume a full range of pixel intensities, the Otsu algorithm would be producing a threshold below any actually present in the image, which doesn't make much sense, at least presuming the existing black background is transparent/irrelevant.
I suppose someone could make an argument for assuming the full range of pixel intensities if you assume the existing black part of the image were relevant darkness. I would certainly welcome comments there.
Full size images seen below
Amending my words above: When I blacken all but the top half of the light image and take the Otsu threshold again I get the same threshold value of 0.3020. Were the dark parts of the image relevant to the Otsu threshold being produced the extra darkness would have affected the value, therefore, Ratbert's answer is empirically demonstrated as correct.
The correct answer is the first one :
dark = 75 and light = 230, relative to the range of values in each image
graythresh uses the min and max values in the image as boundaries, which is the most logical behavior.
My goal is to make a ridge(mountain)-like shape from the given line. For that purpose, I applied the gaussian filter to the given line. In this example below, one line is vertical and one has some slope. (here, background values are 0, line pixel values are 1.)
Given line:
Ridge shape:
When I applied gaussian filter, the peak heights are different. I guess this results from the rasterization problem. The image matrix itself is discrete integer space. The gaussian filter is actually not exactly circular (s by s matrix). Two lines also suffer from rasterization.
How can I get two same-peak-height nice-looking ridges(mountains)?
Is there more appropriate way to apply the filter?
Should I make a larger canvas(image matrix) and then reduce the canvas by interpolation? Is it a good way?
Moreover, I appreciate if you can suggest a way to make ridges with a certain peak height. When using gaussian filter, what we can do is deciding the size and sigma of the filter. Based on those parameters, the peak height varies.
For information, image matrix size is 250x250 here.
You can give a try to distance transform. Your image is a binary image (having only two type of values, 0 and 1). Therefore, you can generate similar effects with distance transform.
%Create an image similar to yours
img=false(250,250);
img(sub2ind(size(img),180:220,linspace(20,100,41)))=1;
img(1:200,150)=1;
%Distance transform
distImg=bwdist(img);
distImg(distImg>5)=0; %5 is set manually to achieve similar results to yours
distImg=5-distImg; %Get high values for the pixels inside the tube as shown
%in your figure
distImg(distImg==5)=0; %Making background pixels zero
%Plotting
surf(1:size(img,2),1:size(img,1),double(distImg));
To get images with certain peak height, you can change the threshold of 5 to a different value. If you set it to 10, you can get peaks with height equal to the next largest value present in the distance transform matrix. In case of 5 and 10, I found it to be around 3.5 and 8.
Again, if you want to be exact 5 and 10, then you may multiply the distance transform matrix with the normalization factor as follows.
normalizationFactor=(newValue-minValue)/(maxValue-minValue) %self-explanatory
Only disadvantage I see is, I don't get a smooth graph as you have. I tried with Gaussian filter too, but did not get a smooth graph.
My result:
I have seen a couple of images where they generally make a face out of numerous smaller images.
For example, say they tile 100 images in 10x10 grid, and somehow they vary hue/sat/col of the smaller images so that when you see the Big Picture, you see another image.
The question boils down to - say you have an image. What kind of algorithm would you apply to that image so that the average RGB value of that image is the one you have defined?
Calculate the hue/sat/value of every tile (use HSV because smaller differences here seem more "natural" to the human eye than in RGB space)
Now calculate the same values for each n*n tile of your big picture
Find the tiles with the closest HSV values (minimum of sqrt((h1-h2)^2 - (s1-s2)^2 - (v1-v2)^)) and stamp that tile scaled down to n*n into the result.
To find the HSV for a tile, it should be enough to sum up all RGB values and then divide them by the number of pixels and convert that final RGB triple into HSV. But to be save, I suggest that you try that both versions.
See which Wikipedia article for RGB <-> HSV conversions.
To refine the algorithm, you can split every tile into an mm and calculate the average HSV for each grid element. Then, when you look for a match, divide the big image into as usual but also calculate mm HSV values. Select which tile matches most of these m*m best. This allows the algorithm to select tiles which have the same structure as the big picture.
For that extra touch, try to create a gigapixel image.