So what I need to do is to apply an operation like
(x(i,j)-min(x)) / max(x(i,j)-min(x))
which basically converts each pixel value such that the values range between 0 and 1.
First of all, I realised that Matlab saves our image(rows * col * colour) in a 3D matrix on using imread,
Image = imread('image.jpg')
So, a simple max operation on image doesn't give me the max value of pixel and I'm not quite sure what it returns(another multidimensional array?). So I tried using something like
max_pixel = max(max(max(Image)))
I thought it worked fine. Similarly I used min thrice. My logic was that I was getting the min pixel value across all 3 colour planes.
After performing the above scaling operation I got an image which seemed to have only 0 or 1 values and no value in between which doesn't seem right. Has it got something to do with integer/float rounding off?
image = imread('cat.jpg')
maxI = max(max(max(image)))
minI = min(min(min(image)))
new_image = ((I-minI)./max(I-minI))
This gives output of only 1s and 0s which doesn't seem correct.
The other approach I'm trying is working on all colour planes separately as done here. But is that the correct way to do it?
I could also loop through all pixels but I'm assuming that will be time taking. Very new to this, any help will be great.
If you are not sure what a matlab functions returns or why, you should always do one of the following first:
Type help >functionName< or doc >functionName< in the command window, in your case: doc max. This will show you the essential must-know information of that specific function, such as what needs to be put in, and what will be output.
In the case of the max function, this yields the following results:
M = max(A) returns the maximum elements of an array.
If A is a vector, then max(A) returns the maximum of A.
If A is a matrix, then max(A) is a row vector containing the maximum
value of each column.
If A is a multidimensional array, then max(A) operates along the first
array dimension whose size does not equal 1, treating the elements as
vectors. The size of this dimension becomes 1 while the sizes of all
other dimensions remain the same. If A is an empty array whose first
dimension has zero length, then max(A) returns an empty array with the
same size as A
In other words, if you use max() on a matrix, it will output a vector that contains the maximum value of each column (the first non-singleton dimension). If you use max() on a matrix A of size m x n x 3, it will result in a matrix of maximum values of size 1 x n x 3. So this answers your question:
I'm not quite sure what it returns(another multidimensional array?)
Moving on:
I thought it worked fine. Similarly I used min thrice. My logic was that I was getting the min pixel value across all 3 colour planes.
This is correct. Alternatively, you can use max(A(:)) and min(A(:)), which is equivalent if you are just looking for the value.
And after performing the above operation I got an image which seemed to have only 0 or 1 values and no value in between which doesn't seem right. Has it got something to do with integer/float rounding off?
There is no way for us to know why this happens if you do not post a minimal, complete and verifiable example of your code. It could be that it is because your variables are of a certain type, or it could be because of an error in your calculations.
The other approach I'm trying is working on all colour planes separately as done here. But is that the correct way to do it?
This depends on what the intended end result is. Normalizing each colour (red, green, blue) seperately will result in a different result as compared to normalizing the values all at once (in 99% of cases, anyway).
You have a uint8 RGB image.
Just convert it to a double image by
I=imread('https://upload.wikimedia.org/wikipedia/commons/thumb/0/0b/Cat_poster_1.jpg/1920px-Cat_poster_1.jpg')
I=double(I)./255;
alternatively
I=im2double(I); %does the scaling if needed
Read about image data types
What are you doing wrong?
If what you want todo is convert a RGB image to [0-1] range, you are approaching the problem badly, regardless of the correctness of your MATLAB code. Let me give you an example of why:
Say you have an image with 2 colors.
A dark red (20,0,0):
A medium blue (0,0,128)
Now you want this changed to [0-1]. How do you scale it? Your suggested approach is to make the value 128->1 and either 20->20/128 or 20->1 (not relevant). However when you do this, you are changing the color! you are making the medium blue to be intense blue (maximum B channel, ) and making R way way more intense (instead of 20/255, 20/128, double brightness! ). This is bad, as this is with simple colors, but with combined RGB values you may even change the color itsef, not only the intensity. Therefore, the only correct way to convert to [0-1] range is to assume your min and max are [0, 255].
Related
I have two 3D arrays:
shape is a 240 x 121 x 10958 array
area is a 240 x 1 x 10958 array
The values of the arrays are of the type double. Both have NaN as fill values where there is no relevant data.
For each [240 x 121] page of the shape array, there are several elements filled by the same number. For example, there will be a block of 1s, a block of 2s, etc. For each corresponding page of the area array there is a single column of numeric values 240 rows long. What I need to do is progressively go through each page of the shape array (moving along the 3rd, 10958-long axis) and replace each numbered element in that page with the number that fills the row of the matching number in the area array.
For example, if I'm looking at shape(:,:,500), I want to replace all the 8s on that page with area(8,1,500). I need to do this for numbers 1 through 20, and I need to do it for all 10958 pages of the array.
If I extract a single page and only replace one number I can get it to work:
shapetest = shape(:,:,500);
shapetest(shapetest==8)=area(8,1,500);
This does exactly what I need for one page and for one number. Going through numbers 1-20 with a for loop doesn't seem like an issue, but I can't find a vectorized way to do this for all the pages of the original 3D array. In fact, I couldn't even get it work for a single page without extracting that page as its own matrix like I did above. I tried things like this to no avail:
shape(shape(:,:,500)==8)=area(8,1,500);
If I can't do it for one page, I'm even more lost as to how to do it for all at once. But I'm inexperienced in MATLAB, and I think I am just ignorant of the proper syntax.
Instead, I ended up using a cell array and the following very inefficient nested for loops:
MyCell=num2cell(shape,[2 1]);
shapetest3=reshape(MyCell,1,10958);
for w=1:numel(shapetest3)
test_result{1,w}=zeros(121,240)*NaN;
end
for k=1:10958
for i=1:29040 % 121 x 240
for n=1:20
if shapetest3{1,k}(i)==n
test_result{1,k}(i)=area(n,1,k);
end
end
end
end
This gets the job done, and I can easily turn it back to an array, but it is very slow, and I am confident there is a much better vectorized way. I'd appreciate any help or tips. Thanks in advance.
To vectorize the mapping operation, we can use shape as an index into area. But because the mapping is different for each plane, we need to loop over the planes to accomplish this. In short, it'll look like this:
test_result = zeros(size(shape)); % pre-allocate output
for k=1:size(area,3) % loop over planes
lut = area(:,1,k);
test_result(:,:,k) = lut(shape(:,:,k));
end
The above only works if shape only contains integer values in the range [1,N], where N = size(area,1). That is, for other values in shape we'll be doing wrong indexing. We will need to fix shape to avoid this. The question here is, what do we want to do with those out-of-range values?
As an example, preparing shape to deal with NaN values:
code = size(area,1) + 1; % this is an unused code word
shape(isnan(shape)) = code;
area(code,1,:) = NaN;
This replaces all NaN values in shape with the value code, which is one larger than any code value we were mapping. Then, we extend area to have one more value, a value for the input code. The value we fill in here is the value that the output test_result will have where shape is NaN. In this case, we write NaN, such that NaN in the input maps to NaN in the output.
Something similar can be done with values below 0 and above 240 (shape(shape<1 | shape>240) = code), or with non-integer values (shape(mod(shape,1)~=0) = code).
I have a column vector "distances", and I want to select a value randomly from this vector such that smaller values have a higher probability of being selected. So far I am using the following, where "possible_cells" is the randomly selected value:
w=(fliplr(1:numel(distances)))/100
possible_cells=randsample((sort(distances)),1,true,w)
Basically, I flipped the distance vector to create probabilities of selection "w" (if I am understanding randsample correctly), so that the smallest value has the probability of being selected equal to the highest value. To check how well this works, I randomly drew 50 values and by using a histogram, I see that the values are higher than I would expect. Does anyone have any idea on how else to do what I described above?
0 Comments
How about something like this?
let's start with 10 sample distances with lengths no greater than 20 just to demonstrate:
d = randi(20,10,1);
Next, since we want smaller values to be more likely, let's take the reciprocal of those distances:
d_rec = 1./d;
Now, let's normalize so we can create a distribution from which to select our distance:
d_rec_norm = d_rec ./ sum(d_rec);
This new variable reflects the probability with which to select each given distance. Now comes a little trick... we choose the distance like this:
d_i = find(rand < cumsum(d_rec_norm),1);
This will give us the index of our chosen distance. The logic behind this is that when cumulatively summing the normalized values associated with each distance (d_rec_norm) we create "bins" whose widths are proportional to the likelihood of selecting each distance. All that is left is to pick a random number between 0 and 1 (rand) and see which "bin" it falls in.
I'm a new poster here, so let me know if this is unclear and I can try to improve my explanation.
I'm looking through various online sources trying to learn some new stuff with matlab.
I can across a dilation function, shown below:
function rtn = dilation(in)
h =size(in,1);
l =size(in,2);
rtn = zeros(h,l,3);
rtn(:,:,1)=[in(2:h,:); in(h,:)];
rtn(:,:,2)=in;
rtn(:,:,3)=[in(1,:); in(1:h-1,:)];
rtn_two = max(rtn,[],3);
rtn(:,:,1)=[rtn_two(:,2:l), rtn_two(:,l)];
rtn(:,:,2)=rtn_two;
rtn(:,:,3)=[rtn_two(:,1), rtn_two(:,1:l-1)];
rtn = max(rtn,[],3);
The parameter it takes is: max(img,[],3) %where img is an image
I was wondering if anyone could shed some light on what this function appears to do and if there's a better (or less confusing way) to do it? Apart from a small wiki entry, I can't seem to find any documentation, hence asking for your help.
Could this be achieved with the imdilate function maybe?
What this is doing is creating two copies of the image shifted by one pixel up/down (with the last/first row duplicated to preserve size), then taking the max value of the 3 images at each point to create a vertically dilated image. Since the shifted copies and the original are layered in a 3-d matrix, max(img,[],3) 'flattens' the 3 layers along the 3rd dimension. It then repeats this column-wise for the horizontal part of the dilation.
For a trivial image:
00100
20000
00030
Step 1:
(:,:,1) (:,:,2) (:,:,3) max
20000 00100 00100 20100
00030 20000 00100 20130
00030 00030 20000 20030
Step 2:
(:,:,1) (:,:,2) (:,:,3) max
01000 20100 22010 22110
01300 20130 22013 22333
00300 20030 22003 22333
You're absolutely correct this would be simpler with the Image Processing Toolbox:
rtn = imdilate(in, ones(3));
With the original code, dilating by more than one pixel would require multiple iterations, and because it operates one dimension at a time it's limited to square (or possibly rectangular, with a bit of modification) structuring elements.
Your function replaces each element with the maximum value among the corresponding 3*3 kernel. By creating a 3D matrix, the function align each element with two of its shift, thus equivalently achieves the 3*3 kernel. Such alignment was done twice to find the maximum value along each column and row respectively.
You can generate a simple matrix to compare the result with imdilate:
a=magic(8)
rtn = dilation(a)
b=imdilate(a,ones(3))
Besides imdilate, you can also use
c=ordfilt2(a,9,ones(3))
to get the same result ( implements a 3-by-3 maximum filter. )
EDIT
You may have a try on 3D image with imdilate as well:
a(:,:,1)=magic(8);
a(:,:,2)=magic(8);
a(:,:,3)=magic(8);
mask = true(3,3,3);
mask(2,2,2) = false;
d = imdilate(a,mask);
I'm trying to find a way to find the sets of coordinates of the maximum value/s in a matrix of size [8,8], where the values in the matrix vary from 0 to 6 (generated through the rest of the script/function).
i.e. a matrix of zeros(8,8) where the value 1 is in [3,3], [3,5] and [5,3].
and I want to get returned something along the lines of ([3,3],[3,5],[5,3])
I have tried using things such as ind2sub, etc but with no luck (I keep getting things returned like [I,J] = [ [0,0,3,0,5,0,0,0] , [1,1,1,1,1,1,1,1] ])
Any ideas?
If more clarification is needed, just point out where you need it and I'd be glad to do so.
The problem you've been having with max so far is because it operates on one dimension. If you call it on a matrix, using its default parameters, it will return a single maximum element (and indices) for each column of the matrix. In your case, you want all maximums, and the global maximum at that.
Try this:
[I,J] = find(M == max(M(:)))
First, max(M(:)) finds the maximum element, then we construct a logical matrix M == max(M(:)) showing which elements are the maximum. Finally, you can use find to get the co-ordinates of those (if necessary).
Hi could someone please help I am using matlab to generate a disparity map. I have performed multi-wavelet transforms on two rectified stereo pairs and have used a stereo matching algorithm to combine the corresponding babsebands from each image to produce four intial disparity maps. However, I am now stuck and completely clueless on how to use a median operator to combine the values of these four disparity maps into one. Could someone please help me?
the four of my images are equal in size.
The previous code is irrelevant since it is in a different file(I have just saved the output from the previous file and now I am trying to code this in another file).
My thoughts were to:
1. Read the value of pixel p from each of the four basebands
2. Sort the values into ascending order
3. Calculate the median value of the pixel
4. Write the pixel value to a new image
5. Set p+1 and repeat steps 1-4 until last pixel is reached
Thank you
First, stack the images into a MxNx4 array:
bbstack = cat(3,bb1,bb2,bb3,bb4); % use bb{:} if they are in a single cell array
Then apply the median operator along the third dimension:
medbb = median(bbstack,3);