I have real world 3D points which I want to project on a plane. The most of intensity [0-1] values fall in lower region (near zero).
Please see image 'before' his attched below.
I tried to normalize values
Col_=Intensity; % before
max(Col_)=0.46;min(Col_)=0.06;
Col=(Col_-min(Col_))/(max(Col_)-min(Col_));% after
max(Col)=1;min(Col)=0;
But still i have maximum values falling in lower region (near zero).
Please see second fig after normalization.
Result is still most of black region.Any suggestion. How can I strech my intensity information.
regards,!
It looks like you have already normalized as much as you can with linear scaling. If you want to get more contrast, you will have to give up preserving the original scaling and use a non-linear equalization.
For example: http://en.wikipedia.org/wiki/Histogram_equalization
If you have the image processing toolbox, matlab will do it for you:
http://www.mathworks.com/help/toolbox/images/ref/histeq.html
It looks like you have very few values outside the first bin, if you don't need to preserve the uniqueness of the intensities, you could just scale by a larger amount and clip the few that exceed 1.
When I normalize intensities I do something like this:
Col = Col - min(Col(:));
Col = Col/max(Col(:));
This will normalize your data points to the range [0,1].
Now, since you have many small values, you might be able to make out small changes better through log scaling.
Col_scaled = log(1+Col);
Linear scaling with such data rarely works for me. Using the log function is akin to tweaking gamma for visualization purposes.
I think the only thing you can do here is reduce the range.
After normalization do the following:
t = 0.1;
Col(Col > t) = t;
This will simply truncate the range of the data, which may be sufficient for what you are doing. Then you can re-normalize again if you wish.
Related
I have an image I which pixel intensities fall within the range of 0-1. I can calculate the image histogram by normalizing it but I found the curves is not exactly the same as the histogram of raw data. This will cause some issue for the later peaks finding process(See attached two images).
My question is in Matlab, is there any way I can plot the image histogram without normalization the data so that I can keep the curve shape unchanged? This will benefit for those raw images when their pixel intensities are not within 0-1 ranges. Currently, I cannot calculate their histogram if I don't normalize the data.
The Matlab code for normalization and histogram calculation is attached. Any suggestion will be appreciated!
h = imhist(mat2gray(I));
Documentation of imhist tells us that the function checks the data type of the input and scale the values accordingly. Therefore, without testing with your attached data, this may work:
h = imhist(uint8(I));
An alternatively you may scale the integer-representation to floating-representation, by either using argument of mat2gray
h = imhist(mat2gray(I, [0,255]));
or just divide it.
h = imhist(I/255);
The imhist answer in this thread describing normalizing or casting is completely correctly. Alternatively, you could use the histogram function in MATLAB which will work with unnormalized floating point data:
A = 255*rand(500,500);
histogram(A);
This might be considered a repost of this question however I am seeking a much deeper explanation on this matter and how to properly solve this problem.
I want to study the PSF/SRF of a voxel in a 44x44 matrix. For that I create a matrix 100x bigger (4400x4400) so 1 voxel in the smaller matrix corresponds to 100x100 voxels in the bigger one. I set the values to 1 of those 100^2 voxels.
Now I do a FFT of the big matrix and an IFFT of only the center portion (44x44) of the frequency space. This is the code:
A = zeros(4400,4400);
A(2201:2300,2201:2300) = 1;
B = fftshift(fft2(A));
C = ifft2(ifftshift(B(2179:2222,2179:2222)));
D = numel(C)/numel(B) * C;
figure, subplot(1,3,1), imshow(A), subplot(1,3,2), imshow(real(C)), subplot(1,3,3), imshow(real(D));
The problem is the following: I would expect the value in the voxel of the new 44x44 matrix to be 1. However, using this numel factor correction they decrease to 0.35. And if I don't apply the correction they go up to huge values.
For starters, let me try to clarify the scaling issue: For the DFT/IDFT there are various scaling conventions regarding the input size. You either need a factor of 1/N in the DFT or a factor of 1/N in the IDFT or a factor of 1/sqrt(N) in both. All have pros and cons and all are equally valid.
Matlab uses the 1/N in the IDFT convention, as you can see in the documentation.
In your example, the forward DFT has a size 4400, the backward IDFT a size of 44. Therefore the IDFT scaling is a factor 100 less than it should be to match the forward transformation and your values are a factor of 100 too large. Since you're doing a 2-D DFT/IDFT, the factor 100 is missing twice, so your rescaling should be 100^2. Your numel(C)/numel(B) does exactly that, I've just tried to give you the explanation for it.
A reason why you might not see the 1 is that you're plotting only the real part of the inverse DFT. Since you did some fftshifting you might have introduced a phase so that part of your signal is in the imaginary part.
edit: Another reason is that you truncate B to the central 44 by 44 window before transforming back. Since A is not bandlimited, B has energy also outside this window. By truncating you are losing a part of it. Therefore, it is not surprising that the resulting amplitude is lower.
Here is a zoom on the image of B to show this phenomenon:
The red square is what you keep, everything else is truncated. Due to Parsevals theorem, the total energy in image and Fourier domain is equal so by truncation you must also reduce the energy of your signal in the image domain.
I couldn't find an answer for RGB image.
How can someone get a value of SD,mean and Entropy of RGB image using MATLAB?
From http://airccse.org/journal/ijdms/papers/4612ijdms05.pdf TABLE3, it seems he got one answer so did he get the average of the RGB values?
Really in need of any help.
After reading the paper, because you are dealing with colour images, you have three channels of information to access. This means that you could alter one of the channels for a colour image and it could still affect the information it's trying to portray. The author wasn't very clear on how they were obtaining just a single value to represent the overall mean and standard deviation. Quite frankly, because this paper was published in a no-name journal, I'm not surprised how they managed to get away with it. If this was attempted to be published in more well known journals (IEEE, ACM, etc.), this would probably be rejected outright due to that very ambiguity.
On how I interpret this procedure, averaging all three channels doesn't make sense because you want to capture the differences over all channels. Doing this averaging will smear that information and those differences get lost. Practically speaking, if you averaged all three channels, should one channel change its intensity by 1, and when you averaged the channels together, the reported average would be so small that it probably would not register as a meaningful difference.
In my opinion, what you should perhaps do is treat the entire RGB image as a 1D signal, then perform the mean, standard deviation and entropy of that image. As such, given an RGB image stored in image_rgb, you can unroll the entire image into a 1D array like so:
image_1D = double(image_rgb(:));
The double casting is important because you want to maintain floating point precision when calculating the mean and standard deviation. The images will probably be of an unsigned integer type, and so this casting must be done to maintain floating point precision. If you don't do this, you may have calculations that get saturated or clamped beyond the limits of that data type and you won't get the right answer. As such, you can calculate the mean, standard deviation and entropy like so:
m = mean(image_1D);
s = std(image_1D);
e = entropy(image_1D);
entropy is a function in MATLAB that calculates the entropy of images so you should be fine here. As noted by #CitizenInsane in his answer, entropy unrolls a grayscale image into a 1D vector and applies the Shannon definition of entropy on this 1D vector. In a similar token, you can do the same thing with a RGB image, but we have already unrolled the signal into a 1D vector anyway, and so the input into entropy will certainly be well suited for the unrolled RGB image.
I have no idea how the author actually did it. But what you could do, is to treat the image as a 1D-array of size WxHx3 and then simply calculate the mean and standard deviation.
Don't know if table 3 is obtain in the same way but at least looking at entropy routine in image toolbox of matlab, RGB values are vectorized to single vector:
I = imread('rgb'); % Read RGB values
I = I(:); % Vectorization of RGB values
p = imhist(I); % Histogram
p(p == 0) = []; % remove zero entries in p
p = p ./ numel(I); % normalize p so that sum(p) is one.
E = -sum(p.*log2(p));
I'm currently trying to implement a method to generate TSP art, and for that I need a list of points (x,y), the local density of which is proportional to the gray scale pixel value of a given image.
My first thought was: well that works pretty much like Inverse Transform Sampling for statistics (you want to draw a sample that matches a given probability density function but you can only create a sample that is uniformly distributed).
I implemented this and it works fairly well, as evident by executing this code:
%% Load image, adjust it for our needs
im=imread('http://goo.gl/DDwV3t'); %load random headshot from google
im=imadjust(im,stretchlim(im,[.01,.65]),[]);
im=im2double(rgb2gray(im));
im=im(10:end-5,50:end-5);
figure;imshow(im);title('original');
im=1-im; %we want black dots on white background
im=flipud(im); %and we want it the right way up
%% process per row
imrow = cumsum(im,2);
imrow=imrow*size(imrow,1)./repmat(max(imrow,[],2),1,size(imrow,2));
y=1:size(imrow,2);
ximrow_i = zeros(size(imrow));
for i = 1:size(imrow,1)
mask =logical([diff(imrow(i,:))>=0.01,0]); %needed for interp
ximrow_i(i,:) = interp1(imrow(i,mask),y(mask),y);
end
y=1:size(ximrow_i,1);
y=repmat(y',1,size(ximrow_i,2));
y1=y(1:5:end,1:5:end); %downscale a bit
ximcol_i1=ximrow_i(1:5:end,1:5:end); %downscale a bit
figure('Color','w');plot(ximcol_i1(:),y1(:),'k.');title('Inverse Transform Sampling on rows');
axis equal;axis off;
%% process per column
imcol=cumsum(im,1);
imcol=imcol*size(imcol,2)./repmat(max(imcol,[],1),size(imcol,1),1);
y=1:size(imcol,1);
yimcol_i=zeros(size(imcol));
for i = 1:size(imcol,2)
mask =logical([diff(imcol(:,i))>=0.01;0]);
yimcol_i(:,i) = interp1(imcol(mask,i),y(mask),y);
end
y=1:size(imcol,2);
y=repmat(y,size(imcol,1),1);
y1=y(1:5:end,1:5:end);
yimcol_i1=yimcol_i(1:5:end,1:5:end);
figure('Color','w');plot(y1(:),yimcol_i1(:),'k.');title('Inverse Transform Sampling on cols');
axis equal;axis off;
It has the shortcoming that I can only use this per-row or per-column, but not both. The Inverse Transform Sampling method does not work for multivariate PDFs in general, and I'm fairly sure I wont be able to get it to work in this case.
Is there a simple method to achieve my goal that I haven't seen yet?
I am aware that an algorithm called Voronoi Stippler has been used to create the desired result and I will investigate that, but for the moment I liked the simplicity of Inverse Transform Sampling and would like to know if I can extend that method to match my needs.
It turns out this is fairly simple and can be done by Rejection Sampling.
For the special case where the instrumental distribution is U(0,1) it works like this (if I understood it correctly):
im=imread('http://goo.gl/DDwV3t'); %load random headshot from google
im=imadjust(im,stretchlim(im,[.01,.65]),[]);
im=im2double(rgb2gray(im));
im=im(10:end-5,50:end-5);
im=1-flipud(im);
d = im > .9*rand(size(im));
d=d&(rand(size(d))>.95); %randomly sieve out some more points
[i,j]=ind2sub(size(d),find(d));
figure('Color','w');plot(j,i,'k.');title('Rejection Sampling');
axis equal;axis off;
The sampling is done in one line:
d = im > .9*rand(size(im));
Since I ended up with too many points I randomly sampled the result thus reducing the number of points by approximately the factor 20.
This is pretty much the result I originally desired.
I'm trying to find some peaks in Matlab, but the function findpeaks.m doesn't have the width option. The peaks I want to be detected are in the balls. All the detected are in the red squares. As you can see they have a low width. Any help?
here's the code I use:
[pk,lo] = findpeaks(ecg);
lo2 = zeros(size(lo));
for m = 1:length(lo) - 1
if (ecg(m) - ecg(m+1)) > 0.025
lo2(m) = lo(m);
end
end
p = find(lo2 == 0);
lo2(p) = [];
figure, plot(ecg);
hold on
plot(lo, ecg(lo), 'rs');
By the looks of it you want to characterise each peak in terms of amplitude and width, so that you can apply thresholds (or simmilar) to these values to select only those meeting your criteria (tall and thin).
One way you could do this is to fit a normal distribution to each peak, pegging the mean and amplitude to the value you have found already, and using an optimisation function to find the standard deviation (width of normal distribution).
So, you would need a function which calculates a representation of your data based on the sum of all the gaussian distributions you have, and an error function (mean squared error perhaps) then you just need to throw this into one of matlabs inbuilt optimisation/minimisation functions.
The optimal set of standard deviation parameters would give you the widths of each peak, or at least a good approximation.
Another method, based on Adiel's comment and which is perhaps more appropriate since it looks like you are working on ecg data, would be to also find the local minima (troughs) as well as the peaks. From this you could construct an approximate measure of 'thinness' by taking the x-axis distance between the troughs on either side of a given peak.
You need to define a peak width first, determine how narrow you want your peaks to be and then select them accordingly.
For instance, you can define the width of a peak as the difference between the x-coordinates at which the y-coordinates equal to half of the peak's value (see here). Another approach, (which seems more appropriate here) is to measure the gradient at fixed distances from the peak itself, and selecting the peaks accordingly. In MATLAB, you'll probably use a gradient filter for that :
g = conv(ecg, [-1 0 1], 'same'); %// Gradient filter
idx = g(lo) > thr); %// Indices of narrow peaks
lo = lo(idx);
where thr is the threshold value that you need to determine for yourself. Lower threshold values mean more tolerance for wider peaks.
You need to define what it means to be a peak of interest, and what you mean by the width of that peak. Once you do those things, you are a step ahead.
Perhaps you might locate each peak using find peaks. Then locate the troughs, one of which should lie between each pair of peaks. A trough is simply a peak of -y. Make sure you worry about the first and last peaks/troughs.
Next, define the half height points as the location midway in height between each peak and trough. This can be done using a reverse linear interpolation on the curve.
Finally, the width at half height might be simply the distance (on the x axis) between those two half height points.
Thinking pragmatically, I suppose you could use something along the lines of this simple brute-force approach:
[peaks , peakLocations] = findpeaks(+X);
[troughs, troughLocations] = findpeaks(-X);
width = zeros(size(peaks));
for ii = 1:numel(peaks)
trough_before = troughLocations( ...
find(troughLocations < peakLocations(ii), 1,'last') );
trough_after = troughLocations( ...
find(troughLocations > peakLocations(ii), 1,'first') );
width(ii) = trough_after - trough_before;
end
This will find the distance between the two troughs surrounding a peak of interest.
Use the 'MinPeakHeight' option in findpeaks() to pre-prune your data. By the looks of it, there is no automatic way to extract the peaks you want (unless you somehow have explicit indices to them). Meaning, you'll have to select them manually.
Now of course, there will be many more details that will have to be dealt with, but given the shape of your data set, I think the underlying idea here can nicely solve your problem.