Matlab im2col function - matlab

This is a bit more of a general question, but, no matter how many times I read the description of MATLAB's im2col function, I cannot fully understand it. I need it for the computational efficiency because MATLAB is awful with nested for loops. Here's what I'm attempting to do, but using nested for loops:
[TRIMMED]=TM_FILTER(IMAGE, FILTER_SIZE, PERCENT)
Takes a 2-D array and returns the array, filtered with a
square trimed mean filter with length/width equal to FILTER_SIZE and percent equal to PERCENT.
%}
function [trimmed]=tm_filter(image, filter_size, percent)
if rem(filter_size, 2)==0 %make sure filter has a center pixel
error('filter size must be odd numbered'); %error and return if number is odd
return
end
if percent > 100 || percent < 0
error('Percentage must be ? [0, 100]');
return
end
[rows, columns]=size(image); %figure out pixels needed
n=(filter_size-1)/2; %n is pixel distance from center pixel to boundaries
padded=(padarray(image, [n,n],128)); %padding on boundaries so center pixel always has neighborhood
for i=1+n:rows %rows from first non-padded entry to last nonpadded entry
for j=1+n:columns %colums from first non-padded entry to last nonpadded entry
subimage=padded(i-n:i+n,j-n:j+n); %neighborhood same size as filter
average=trimmean(trimmean(subimage, percent), percent); %computes trimmed mean of neighborhood as trimmed mean of vector of trimmed means
trimmed(i-n, j-n)=average; %stores averaged pixel in new array
end
end
trimmed=uint8(trimmed); %converts image to gray levels from 0-255

Here is the code you want: note the entire nested loop was replaced with a single statement.
[TRIMMED]=TM_FILTER(IMAGE, FILTER_SIZE, PERCENT)
Takes a 2-D array and returns the array, filtered with a
square trimed mean filter with length/width equal to FILTER_SIZE and percent equal to PERCENT.
%}
function [trimmed]=tm_filter(image, filter_size, percent)
if rem(filter_size, 2)==0 %make sure filter has a center pixel
error('filter size must be odd numbered'); %error and return if number is odd
return
end
if percent > 100 || percent < 0
error('Percentage must be ? [0, 100]');
return
end
trimmed = (uint8)trimmean(im2col(image, filter_size), percent);
Explanation:
the im2col function turns each region of filter_size into a column. Your trimmean function can then operate on each of the regions (columns) in a single operation - much more efficient than extracting each shape in turn. Also note this requires only a single application of trimmean - in your original you first do it on the columns, then again on the rows, which will actually cause a more severe trim than I think you intended (exclude 50% first time, then 50% again - feels like excluding 75%. Not exactly true but you get my point). Also you will find that changing the order of operations (row, then column vs column, then row) would change the result because the filter is nonlinear.
For example
im = reshape(1:9, [3 3]);
disp(im2col(im,[2 2])
results in
1 2 4 5
2 3 5 6
4 5 7 8
5 6 8 9
since you took each of the 4 possible blocks of 2x2 from this matrix:
1 4 7
2 5 8
3 6 9
and turned them into columns
Note - with this technique (applied to the unpadded image) you do lose some pixels on the edge; your method added some padding so that every pixel (even ones on the edge) has a complete neighborhood, and as such the filter returns an image that is the same size as the original (but it's not clear what the effect of padding/filtering will be near the margin, and especially the corner: you have almost 75% percent of pixels fixed at 128 and that is likely to dominate the behavior in the corner).

why im2col? why not nlfilter?
>> trimmed = nlfilter( image, [filter_size filter_size],...
#(x) treimmean( trimmean(x, percent), percent ) );
Are you sure you process the entire image?
i and j only goes up to rows and columns respectively. However, when you update trimmed you access i-n and j-n. What about the last n rows and columns?
Why do you apply trimmean twice for each block? Isn't it more appropriate to process the block at once, as in trimmean( x(:), percent)?
I believe the results of trimmean( trimmean(x, percent), percent) will be different than those of trimmean( x(:), percent). Have you give it a thought?
A small remark, it is best not to use i and j as variable names in matlab.

Related

Convolution using 'valid' in Matlab's conv() function

Here is an example of convolution given:
I have two questions here:
Why is the vector 𝑥 padded with two 0s on each side? As, the length of kernel ℎ is 3. If 𝑥 is padded with one 0 on each side, the middle element of convolution output would be within the range of the length of 𝑥, why not one 0 on each side?
Explain the following output to me:
>> x = [1, 2, 1, 3];
>> h = [2, 0, 1];
>> y = conv(x, h, 'valid')
y =
3 8
>>
What is valid doing here in the context of the previously shown mathematics on vectors 𝑥 and ℎ?
I can't speak as to the amount of zero padding that is proper .... That being said, any zero padding is making up data that is not there. This isn't necessarily wrong, but you should be aware that the values computing this information may be biased. Sometimes you care about this, sometimes you don't. Introducing 1 zero (in this case) would leave the middle kernel value always in the data, but why should that be a stopping criteria? Importantly, adding on 2 zeros still leaves one multiplication of values that are actually present in the data and the kernel (the x[0]*h[0] and x[3]*h[2] - using 0 based indexing). Adding on a 3rd zero (or more) would just yield zeros in the output since 3 is the length of the kernel. In other words zero padding will always yield an output that is partially based on the actual data (but not completely) for any zero padding from n=1 to n = length(h)-1 (in this case either 1 or 2).
Even though zero padding with length 2 or 1 still has multiplications based on real data, some values are summed over "fake" data (those multiplied with a padded zero). In this case Matlab gives you 3 options for how you want the data returned. First, you can get the full convolution, which includes values that are biased because they include adding in 0 values that aren't really in the data. Alternatively you can get same, which means the length of the output is the length of the data y = [4 3 8 1]. This corresponds to 1 zero but note that for longer kernels you could technically get other lengths between full and same, Matlab just doesn't return those for you.
Finally, and probably most important to understand out of all this, you have the valid option. In your example only 2 samples of the output are computed from summations that occur only from multiplications over real data (i.e. from multiplying samples of the kernel with samples from x and not from zeros). More specifically:
y[2] = h[2]*x[0] + h[1]*x[1] + h[2]*x[2] = 3 //0 based indexing like example
y[3] = h[2]*x[1] + h[1]*x[2] + h[2]*x[3] = 8
Note none of the other y values are computed with only h and x, they all involve a padded zero which is not necessarily indicative of the real data. For example:
y[4] = h[2]*x[2] + h[1]*x[3] + h[2]*0 <= padded zero

Converting subsripts and indices

I'm having a hard time figuring out transformation between spaces using ind2sub and sub2ind. Could someone help? The problem is as follows:
I have a mask Y (or region-of-interest) in which voxel values are either 1 or zero: Y=72x72x33 double. I then find all the voxels with value of 1 (there are 15 of them) then use ind2sub to get the x y z coordinates for these voxels:
indx = find(Y>0);
[x,y,z] = ind2sub(size(Y),indx);
XYZ = [x y z]';
Since there are 15 voxels with the value of 1, I end up with XYZ=3x15 double, containing coordinates of these 15 voxels, something like this:
25 26 24 25 26 ...26
28 28 29 29 29 ...30
8 8 8 8 8 ...9
Based on some arbitrary criteria, I remove 6 voxels so XYZ become 3x9 double. Let's call this new_XYZ. Now I would like to transform this new_XYZ back into a mask (let's call it new_Y). I tried this:
new_Y=sub2ind(size(Y),new_XYZ);
Here, I probably did something wrong with the sub2ind since new_Y didn't give me what I expected. The dimensions are also not 72x72x33. The old mask is a sphere so I expect the new mask to be close to that. Instead, I get a straight line. Can someone help me with the transformation?
Thanks.
A.
There are a couple of things wrong in your approach. To detail I'll use a slightly smaller sample data set, but the explanation can scale up to any size.
Y = randi(10,5,5,3)-5 ;
This create a 5x5x3 containing random integers numbers from -5 to 5 (so a good chance to have about half of them positive (>0).
Before I go further, you can get a direct mask of your condition in the same shape of your matrix:
mask_positive_Y = Y>0 ;
>> whos mask_positive_Y
Name Size Bytes Class Attributes
mask_positive_Y 5x5x3 75 logical
This gives you an array of logical (boolean), the same size of your matrix, containing 0 (false) everywhere and 1 (true) where your condition is validated (Y>0).
If you can work directly with this mask, you do not need to use ind2sub and sub2ind back and forth. For example, doing Y(mask_positive_Y) = NaN; would replace all the values indicated by the mask with NaN.
Since you want to modify the mask (remove some points), you may still need to get their indices, in which case you can simply call:
indx = find( mask_positive_Y ) ; %// will give the same result as: indx = find(Y>0);
Now let's assume you got your indices the way you specified, removed the 6 points and got your new_XYZ matrice. The way to rebuild the mask is as follow.
ind2sub gave you 3 vectors as output, and right enough, sub2ind will expect 3 vectors as input (not a 3 column matrix as you did). So you will have to decompose your new_XYZ into 3 vectors before you send it to sub2ind.
new_indx = sub2ind(size(Y), new_XYZ(:,1) , new_XYZ(:,2) , new_XYZ(:,3) );
But don't forget that you did transpose your result matrix when you did XYZ = [x y z]';, so make sure your new_XYZ is also transposed back with new_XYZ = new_XYZ.' ; before you decompose it in the code line above (or simply send the lines of new_XYZ instead of the columns as showed).
By the way, the proper transpose shorthand notation is .' (and not simply ' which is the Complex conjugate transpose).
Now this new_indx is only a new vector of linear indices, homogeneous to the indx you had earlier. You could already use this to assign values under the mask, but if you want a new mask the same shape than your matrix Y, you have to go a bit further:
new_Ymask = false( size(Y) ) ; %// build an empty mask (false everywhere)
new_Ymask(new_indx) = true ; %// assign true to the masked values
This will be the same size as your initial matrix Y, but also the same size as the first boolean mask I showed you mask_positive_Y.

Calculating the Local Ternary Pattern of an image?

I am calculating the Local Ternary Pattern of an image. My code is given below. Am I going in the right direction or not?
function [ I3 ] = LTP(I2)
m=size(I2,1);
n=size(I2,2);
for i=2:m-1
for j=2:n-1
J0=I2(i,j);
I3(i-1,j-1)=I2(i-1,j-1)>J0;
end
end
I2 is the image LTP is applied to.
This isn't quite correct. Here's an example of LTP given a 3 x 3 image patch and a threshold t:
(source: hindawi.com)
The range that you assign a pixel in a window to 0 is when the threshold is between c - t and c + t, where c is the centre intensity of the pixel. Therefore, because the intensity is 34 in the centre of this window, the range is between [29,39]. Any values that are beyond 39 get assigned 1 and any values that are below 29 get assigned -1. Once you determine the ternary codes, you split up the codes into upper and lower patterns. Basically, any values that get assigned a -1 get assigned 0 for upper patterns and any values that get assigned a -1 get assigned 1 for lower patterns. Also, for the lower pattern, any values that are 1 from the original window get mapped to 0. The final pattern is reading the bit pattern starting from the east location with respect to the centre (row 2, column 3), then going around counter-clockwise. Therefore, you should probably modify your function so that you're outputting both lower patterns and upper patterns in your image.
Let's write the corrected version of your code. Bear in mind that I will not give an optimized version. Let's get a basic algorithm working, and it'll be up to you on how you want to optimize this. As such, change your code to something like this, bearing in mind all of the stuff I talked about above. BTW, your function is not defined properly. You can't use spaces to define your function, as well as your variables. It will interpret each word in between spaces as variables or functions, and that's not what you want. Assuming your neighbourhood size is 3 x 3 and your image is grayscale, try something like this:
function [ ltp_upper, ltp_lower ] = LTP(im, t)
%// Get the dimensions
rows=size(im,1);
cols=size(im,2);
%// Reordering vector - Essentially for getting binary strings
reorder_vector = [8 7 4 1 2 3 6 9];
%// For the upper and lower LTP patterns
ltp_upper = zeros(size(im));
ltp_lower = zeros(size(im));
%// For each pixel in our image, ignoring the borders...
for row = 2 : rows - 1
for col = 2 : cols - 1
cen = im(row,col); %// Get centre
%// Get neighbourhood - cast to double for better precision
pixels = double(im(row-1:row+1,col-1:col+1));
%// Get ranges and determine LTP
out_LTP = zeros(3, 3);
low = cen - t;
high = cen + t;
out_LTP(pixels < low) = -1;
out_LTP(pixels > high) = 1;
out_LTP(pixels >= low & pixels <= high) = 0;
%// Get upper and lower patterns
upper = out_LTP;
upper(upper == -1) = 0;
upper = upper(reorder_vector);
lower = out_LTP;
lower(lower == 1) = 0;
lower(lower == -1) = 1;
lower = lower(reorder_vector);
%// Convert to a binary character string, then use bin2dec
%// to get the decimal representation
upper_bitstring = char(48 + upper);
ltp_upper(row,col) = bin2dec(upper_bitstring);
lower_bitstring = char(48 + lower);
ltp_lower(row,col) = bin2dec(lower_bitstring);
end
end
Let's go through this code slowly. First, I get the dimensions of the image so I can iterate over each pixel. Also, bear in mind that I'm assuming that the image is grayscale. Once I do this, I allocate space to store the upper and lower LTP patterns per pixel in our image as we will need to output this to the user. I have decided to ignore the border pixels where when we consider a pixel neighbourhood, if the window goes out of bounds, we ignore these locations.
Now, for each valid pixel that is within the valid borders of the image, we extract our pixel neighbourhood. I convert these to double precision to allow for negative differences, as well as for better precision. I then calculate the low and high ranges, then create a LTP pattern following the guidelines we talked about above.
Once I calculate the LTP pattern, I create two versions of the LTP pattern, upper and lower where any values of -1 for the upper pattern get mapped to 0 and 1 for the lower pattern. Also, for the lower pattern, any values that were 1 from the original window get mapped to 0. After, this, I extract out the bits in the order that I laid out - starting from the east, go counter-clockwise. That's the purpose of the reorder_vector as this will allow us to extract those exact locations. These locations will now become a 1D vector.
This 1D vector is important, as we now need to convert this vector into character string so that we can use bin2dec to convert the value into a decimal number. These numbers for the upper and lower LTPs are what are finally used for the output, and we place those in the corresponding positions of both output variables.
This code is untested, so it'll be up to you to debug this if it doesn't work to your specifications.
Good luck!

Find the number of zero elements in a matrix in MATLAB [duplicate]

This question already has answers here:
Find specific value's count in a vector
(4 answers)
Closed 8 years ago.
I have a NxM matrix for example named A. After some processes I want to count the zero elements.
How can I do this in one line code? I tried A==0 which returns a 2D matrix.
There is a function to find the number of nonzero matrix elements nnz. You can use this function on a logical matrix, which will return the number of true.
In this case, we apply nnz on the matrix A==0, hence the elements of the logical matrix are true, if the original element was 0, false for any other element than 0.
A = [1, 3, 1;
0, 0, 2;
0, 2, 1];
nnz(A==0) %// returns 3, i.e. the number of zeros of A (the amount of true in A==0)
The credits for the benchmarking belong to Divarkar.
Benchmarking
Using the following paramters and inputs, one can benchmark the solutions presented here with timeit.
Input sizes
Small sized datasize - 1:10:100
Medium sized datasize - 50:50:1000
Large sized datasize - 500:500:4000
Varying % of zeros
~10% of zeros case - A = round(rand(N)*5);
~50% of zeros case - A = rand(N);A(A<=0.5)=0;
~90% of zeros case - A = rand(N);A(A<=0.9)=0;
The results are shown next -
1) Small Datasizes
2. Medium Datasizes
3. Large Datasizes
Observations
If you look closely into the NNZ and SUM performance plots for medium and large datasizes, you would notice that their performances get closer to each other for 10% and 90% zeros cases. For 50% zeros case, the performance gap between SUM and NNZ methods is comparatively wider.
As a general observation across all datasizes and all three fraction cases of zeros,
SUM method seems to be the undisputed winner. Again, an interesting thing was observed here that the general case solution sum(A(:)==0) seems to be better in performance than sum(~A(:)).
some basic matlab to know: the (:) operator will flatten any matrix into a column vector , ~ is the NOT operator flipping zeros to ones and non zero values to zero, then we just use sum:
sum(~A(:))
This should be also about 10 times faster than the length(find... scheme, in case efficiency is important.
Edit: in the case of NaN values you can resort to the solution:
sum(A(:)==0)
I'll add something to the mix as well. You can use histc and compute the histogram of the entire matrix. You specify the second parameter to be which bins the numbers should be collected at. If we just want to count the number of zeroes, we can simply specify 0 as the second parameter. However, if you specify a matrix into histc, it will operate along the columns but we want to operate on the entire matrix. As such, simply transform the matrix into a column vector A(:) and use histc. In other words, do this:
histc(A(:), 0)
This should be equivalent to counting the number of zeroes in the entire matrix A.
Well I don't know if I'm answering well the question but you could code it as follows :
% Random Matrix
M = [1 0 4 8 0 6;
0 0 7 4 8 0;
8 7 4 0 6 0];
n = size(M,1); % Number of lines of M
p = size(M,2); % Number of columns of M
nbrOfZeros = 0; % counter
for i=1:n
for j=1:p
if M(i,j) == 0
nbrOfZeros = nbrOfZeros + 1;
end
end
end
nbrOfZeros

Histogram Equalization method without use of histeq

I am new to Matlab and am trying to implement code to perform the same function as histeq without actual use of the function. In my code the image colour I get changes drastically when it should not change that much. The average intensity in the image (ranging between 0 and 255) is 105.3196. The image is of an open source pollen particle.
Any help would be much appreciated. The sooner the better! Please could any help be simplified as my Matlab understanding is limited. Thanks.
clc;
clear all;
close all;
pollenJpg = imread ('pollen.jpg', 'jpg');
greyscalePollen = rgb2gray (pollenJpg);
histEqPollen = histeq(greyscalePollen);
averagePollen = mean2 (greyscalePollen)
sizeGreyScalePollen = size(greyscalePollen);
rowsGreyScalePollen = sizeGreyScalePollen(1,1);
columnsGreyScalePollen = sizeGreyScalePollen(1,2);
for i = (1:rowsGreyScalePollen)
for j = (1:columnsGreyScalePollen)
if (greyscalePollen(i,j) > averagePollen)
greyscalePollen(i,j) = greyscalePollen(i,j) + (0.1 * averagePollen);
if (greyscalePollen(i,j) > 255)
greyscalePollen(i,j) = 255;
end
elseif (greyscalePollen(i,j) < averagePollen)
greyscalePollen(i,j) = greyscalePollen(i,j) - (0.1 * averagePollen);
if (greyscalePollen(i,j) > 0)
greyscalePollen(i,j) = 0;
end
end
end
end
figure;
imshow (pollenJpg);
title ('Original Image');
figure;
imshow (greyscalePollen);
title ('Attempted Histogram Equalization of Image');
figure;
imshow (histEqPollen);
title ('True Histogram Equalization of Image');
To implement the equalisation algorithm described on the Wikipedia page, follow these these steps:
Decide on a binSize to group greyscale values. (This is a tweakable, the larger the bin, the less accurate the result from the ideal case, but I think it can cause problems if chosen too small on real images).
Then, calculate the probability of a pixel being a shade of grey:
pixelCount = imageWidth * imageHeight
histogram = all zero
for each pixel in image at coordinates i, j
histogram[floor(pixel / 255 / 10) + 1] += 1 / pixelCount // 1-based arrays, not 0-based
// Note a technicality here: you may need to
// write special code to handle pixels of 255,
// because they will fall in their own bin. Or instead use rounding with an offset.
The histogram in this calculation is scaled (divided by the pixel count) so that the values make sense as probabilities. You can of course factor the division out of the for loop.
Now you need to calculate the accumulative sum of this:
histogramSum = all zero // length of histogramSum must be one bigger than histogram
for i == 1 .. length(histogram)
histogramSum[i + 1] = histogramSum[i] + histogram[i]
Now you have to invert this function and this is the tricky part. The best is to not calculate an explicit inverse, but calculate it on the spot, and apply it on the image. The basic idea is to search for the pixel value in the histogramSum (find the closest index below), and then do a linear interpolation between the index and the next index.
foreach pixel in image at coordinates i, j
hIndex = findIndex(pixel, histogramSum) // You have to write findIndex, it should be simple
equilisationFactor = (pixel - histogramSum[hIndex])/(histogramSum[hIndex + 1] - histogramSum[hIndex]) * binSize
// This above is the linear interpolation step.
// Notice the technicality that you need to handle:
// histogramSum[hIndex + 1] may be out of bounds
equalisedImage[i, j] = pixel * equilisationFactor
Edit: without drilling into the maths, I can't be 100% sure, but I think that division by 0 errors are possible. These can occur if one bin is empty, so consecutive sums are equal. So you need special code to handle this case too. The best you can do is take the value for the factor as halfway between hIndex, hIndex + n, where n is the highest value for which histogramSum[hIndex + n] == histogramSum[hIndex].
And that should be it, once you have dealt with all the technicalities.
The above algorithm is slow (especially in the findIndex step). You may be able to optimize this with a special lookup datastructure. But only do that when it's working, and only if necessary.
One more thing about your Matlab code: the rows and columns are inverted. Because of the symmetry in the algorithm, the result is the same, but it can cause puzzling bugs in other algorithms, and be very confusing if you examine pixel values during debugging. In the pseudocode above I used them the same as you, though.
Relatively few (5) lines of code can do this. I used a low contrast file called 'pollen.jpg' that I found at http://commons.wikimedia.org/wiki/File%3ALepismium_lorentzianum_pollen.jpg
I read it in using your code, run all the above, then do the following:
% find out the index of pixels sorted by intensity:
[gv gi] = sort(greyscalePollen(:));
% create a table of "approximately equal" intensity values:
N = numel(gv);
newVals = repmat(0:255, [ceil(N/256) 1]);
% perform lookup:
% the pixels in sorted order need new values from "equal bins" table:
newImg = zeros(size(greyscalePollen));
newImg(gi) = newVals(1:N);
% if the size of the image doesn't divide into 256, the last bin will have
% slightly fewer pixels in it than the others
When I run this algorithm, and then create a composite of the four images (original, your attempt, my attempt, and histeq), you get the following:
I think it's convincing. The images are not exactly identical - I believe that is because the matlab histeq routine ignores all pixels with value 0. Since it is fully vectorized it is also pretty fast (although not nearly as fast as histeq by about a factor 15 on my image.
EDIT: a bit of explanation might be in order. The repmat command I use to create the newVals matrix creates a matrix that looks like this:
0 1 2 3 4 ... 255
0 1 2 3 4 ... 255
0 1 2 3 4 ... 255
...
0 1 2 3 4 ... 255
Since matlab stores matrices in "first index first" order, if you read this matrix with a single index (as I do in the line newVals(1:N)), you access first all the zeros, then all the ones, etc:
0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 ...
So - when I know the indices of the pixels in the order of their intensity (as returned by the second argument of the sort command, which I called gi), then I can easily assign the value 0 to the first N/256 pixels, the value 1 to the next N/256 etc, with the command I used:
newImg(gi) = newVals(1:N);
I hope this makes the code a little easier to understand.