I'm having difficulties converting image pixels to coordinates and making them appear in my MATLAB workspace. For example, I have the image with pixel values as below (it's a binary image of size 4x4):
0 0 0 0
0 1 1 0
0 1 1 0
0 0 0 0
After getting the pixels, I want to read each value and if they're not equal to zero (which means 1), I want to read the coordinates of that value and save them in to my MATLAB workspace. For example, this is the idea that I thought of:
[x,y] = size(image)
for i=1:x
for j=1:y
if (image(i,j)~=0)
....
However, I am stuck. Can anyone give any suggestion on how to read the coordinates of the non-zero values and save them to my workspace?
Specifically, my expected result in the workspace:
2 2
2 3
3 2
3 3
Doing it with loops is probably not the most efficient way to do what you ask. Instead, use find. find determines the locations in a vector or matrix that are non-zero. In your case, all you have to do is:
[row,col] = find(image);
row and col would contain the row and column locations of the non-zero elements in your binary image. Therefore, with your example:
b = [0 0 0 0;
0 1 1 0;
0 1 1 0;
0 0 0 0];
We get:
>> disp([row, col]);
2 2
3 2
2 3
3 3
However, you'll see that the locations are not in the order you expect. This is because the locations are displayed in column-major order, meaning that the columns are traversed first. In your example, you are displaying them in row-major order. If you'd like to maintain this order, you would sort the results by the row coordinate:
>> sortrows([row, col])
ans =
2 2
2 3
3 2
3 3
However, if you really really really really... I mean really... want to use for loops, what you would do is keep two separate arrays that are initially empty, then loop through each pixel and determine whether it's non-zero. If it is, then you would add the x and y locations to these two separate arrays.
As such, you would do this:
row = []; col = [];
[x,y] = size(image);
for i=1:x
for j=1:y
if (image(i,j)~=0)
row = [row; i]; %// Concatenate row and column location if non-zero
col = [col; j];
end
end
end
This should give you the same results as find.
you can use meshgrid() to collect those coordinates. The function generates two outputs, first being x coordinates, second being y coordinates. you'd go like this:
[xcoord ycoord] = meshgrid( 1:x_size, 1:y_size);
zeros_coordsx = xcoord( image == 0);
zeros_coordsy = ycoord( image == 0);
this is way faster that nested looping and keeps you within matlab's natural vector operation space... these two outputs are in sync,meaning that
image( zeros_coordsy(1), zeros_coordsx(1))
is one of the zeros on the image
Related
Let's say I have 5 connected components (labelled objects) in an image called labelledImage from bwlabel. How can I manipulate labelledImage so that the objects that are labelled as 1 and 4 only display, while removing the objects that are labelled as 2, 3 and 5. Then, how can I manipulate the original RGB image so that the connected components that are labelled as 1 and 4 only display.
I know how to retain a single connected component by using this line of code below. However, I don't know how to do this for multiple labelled regions.
Works.
connectedComponent1 = (labelledImage == 1);
imshow(connectedComponent1)
Doesn't work.
connectedComponent1and4 = (labelledImage == [1 4]);
imshow(connectedComponent1and4)
You can't do logical indexing that way. The simplest way is to perhaps use Boolean statements to combine things.
connectedCompoonent1and4 = labelledImage == 1 | labelledImage == 4;
In general, supposing you had a vector of elements that denote which components you want to keep, you could use bsxfun, permute and any to help you with that. Something like this should work:
components = [1 4];
connected = any(bsxfun(#eq, labelledImage, permute(components, [1 3 2])), 3);
The above code uses matrix broadcasting to create a temporary 3D matrix where each slice i contains the ith value of the vector components which contain the desired labels you want to keep. labelledImage is also replicated in the third dimension so the result using bsxfun creates a 3D matrix where each slice i segments out the ith object you want to keep. We then combine all of the objects together using any and looking in the third dimension.
If you don't like one-liners, you could even use a simple for loop:
components = [1 4];
connected = false(size(labelledImage, 1), size(labelledImage, 2));
for ind = 1 : numel(components)
connected = connected | labelledImage == components(ind);
end
This creates an output image that is all false, then we loop through each value in the vector of components you want to keep and append those results on top of the result. The end will give you all of the components you want to keep.
Lastly, you could use also use ismember and determine those values in your matrix that can be found between the label matrix and the components vector and simply create your mask that way:
connected = ismember(labelledImage, components);
Now that you have a mask of objects you want to extract out, to use this on the original image, simply multiply each channel with the mask. Another use of bsxfun can do that for you. Assuming your image in RGB is called img, simply do the following:
outImg = bsxfun(#times, img, cast(connected, class(img)));
To perform element-wise multiplication, you must ensure that both matrices that are being multiplied have the same type. I convert the mask into the same class as whatever the input image is and perform the multiplication.
Use ismember.
Ex:
A = randi(5,5); % your connected component matrix
B = [1 4] % list of components you want to keep
A =
4 2 1 3 5
2 4 2 5 1
3 4 5 1 4
1 4 1 3 5
4 3 5 1 5
A(~ismember(A,B)) = 0
A =
4 0 1 0 0
0 4 0 0 1
0 4 0 1 4
1 4 1 0 0
4 0 0 1 0
For a project I am working on, I am loading in large image files, which Matlab inputs as LxWx3 arrays of uint8s. I have a function which concatenates these component-wise into a LxWx1 array of uint32s, but I can't find a way to do the reverse without using nested for loops, which is far too slow for the matrices I am working with.
Could anyone recommend a way to accomplish this effuciently? Basically, given a LxW Matrix of uint32s, I want to return a LxWx3 matrix of uint8s, where the (x, y,1:3) components are the three most significant bytes in the corresponding uint32.
You can do that with typecast:
A = uint32([2^16 2^30; 256 513]);
B = permute(reshape(typecast(A(:), 'uint8'), [], size(A,1), size(A,2)), [2 3 1]);
B = flipdim(B, 3); %// flip 3rd dim to bring MSB first, if needed (depends on computer)
B = B(:,:,2:4);
Example: for A = uint32([2^16 2^30; 256 513]);
A =
65536 1073741824
256 513
the result is
B(:,:,1) =
1 0
0 0
B(:,:,2) =
0 0
1 2
B(:,:,3) =
0 0
0 1
I have the following array:
AA = zeros(5,3);
AA(1,3)=1;
AA(3,3)=1;
AA(4,2)=1;
and I want to place the value one in the collumns defined by the following
vector a = [0; 2; 0; 0; 1]. Each value of this vector refers to the collumn
index that we want to change in each row. When zero apears no changes should be made.
Desired output:
0 0 1
0 1 0
0 0 1
0 1 0
1 0 0
Could you please suggest a way to do this without for loop? The goal is
a faster execution.
Thanks!!!
Approach 1
nrows = size(AA,1) %// Get the no. of rows, as we would use this parameter later on
%// Calculate the linear indices with `a` as the column indices and
%// [1:nrows] as the row indices
idx = (a-1)*nrows+[1:nrows]' %//'
%// Select the valid linear indices (ones that have the corresponding a as non-zeros
%// and use them to index into AA and set those as 1's
AA(idx(a~=0))=1
Code output with given AA -
>> AA
AA =
0 0 1
0 1 0
0 0 1
0 1 0
1 0 0
Approach 2
AA(sub2ind(size(AA),find(a~=0),a(a~=0)))=1
Breaking it down to few steps for explanation:
find(a~=0) and a(a~=0) gets us the VALID row and columns indices respectively as needed for sub2ind(size(),row,column) format.
sub2ind gets us the linear indices, which we can use to index into input matrix AA and set those in AA as 1's.
I have been trying to find all connected components using 8 neighbors in a binary image, without using the function "bwlabel".
For example, my input matrix is:
a =
1 1 0 0 0 0 0
1 1 0 0 1 1 0
1 1 0 0 0 1 0
1 1 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
I would to have something like this:
a =
1 1 0 0 0 0 0
1 1 0 0 2 2 0
1 1 0 0 0 2 0
1 1 0 0 0 0 0
0 0 0 0 0 3 0
0 0 0 0 0 0 0
There are 3 connected objects in this image.
This is a common problem in image processing. There are many variations, such as flood filling a region in an image, or finding what pixels belong to the same region. One common approach is to use depth first search. The idea is that you traverse your image from left to right and top to bottom and for any pixels encountered that are equal to 1, you add them to a stack. For each pixel in your stack, you pop off the stack, then look at the neighbouring pixels that are surrounding this pixel. Any pixels that are 1 you add to the stack. You need to keep an additional variable where any pixels you have already visited, you don't add these to the stack. When the stack is empty, we have found those pixels that are an entire region, so you mark these with a unique ID. You then repeat this procedure until you run out of regions in your image.
As such, given that your matrix is stored in A, this is the basic algorithm:
Initialize an array that's the same size as A that is logical. This will record which pixels we have examined or visited. Also initialize an output array B to all zeroes that gives you all of the connected components that you are seeking. Any locations that are zero in the end don't belong to any connected components. Also initialize an ID counter that keeps track of what connected component label each of these will have.
For each location that's in our matrix:
a. If the location is 0, mark this location as visited and continue.
b. If we have already visited this location, then continue.
c. If we have not visited this location... go to Step #3.
Add this unvisited location to a stack.
a. While this stack is not empty...
b. Pop this location off of the stack
c. If we have visited this location, then continue.
d. Else, mark this location as visited and mark this location with the connected components ID.
e. Given this location, look at the 8 neighbouring pixels.
f. Remove those pixels in this list that have been visited, not equal to 1 or out of bounds of the matrix
g. Whatever locations are remaining, add these to the stack.
Once the stack is empty, increment the counter, then go back to Step #2.
Keep going until we have visited all of the locations in our array.
Without further ado, here's the code.
%// Step #1
visited = false(size(A));
[rows,cols] = size(A);
B = zeros(rows,cols);
ID_counter = 1;
%// Step 2
%// For each location in your matrix...
for row = 1 : rows
for col = 1 : cols
%// Step 2a
%// If this location is not 1, mark as visited and continue
if A(row,col) == 0
visited(row,col) = true;
%// Step 2b
%// If we have visited, then continue
elseif visited(row,col)
continue;
%// Step 2c
%// Else...
else
%// Step 3
%// Initialize your stack with this location
stack = [row col];
%// Step 3a
%// While your stack isn't empty...
while ~isempty(stack)
%// Step 3b
%// Pop off the stack
loc = stack(1,:);
stack(1,:) = [];
%// Step 3c
%// If we have visited this location, continue
if visited(loc(1),loc(2))
continue;
end
%// Step 3d
%// Mark location as true and mark this location to be
%// its unique ID
visited(loc(1),loc(2)) = true;
B(loc(1),loc(2)) = ID_counter;
%// Step 3e
%// Look at the 8 neighbouring locations
[locs_y, locs_x] = meshgrid(loc(2)-1:loc(2)+1, loc(1)-1:loc(1)+1);
locs_y = locs_y(:);
locs_x = locs_x(:);
%%%% USE BELOW IF YOU WANT 4-CONNECTEDNESS
% See bottom of answer for explanation
%// Look at the 4 neighbouring locations
% locs_y = [loc(2)-1; loc(2)+1; loc(2); loc(2)];
% locs_x = [loc(1); loc(1); loc(1)-1; loc(1)+1];
%// Get rid of those locations out of bounds
out_of_bounds = locs_x < 1 | locs_x > rows | locs_y < 1 | locs_y > cols;
locs_y(out_of_bounds) = [];
locs_x(out_of_bounds) = [];
%// Step 3f
%// Get rid of those locations already visited
is_visited = visited(sub2ind([rows cols], locs_x, locs_y));
locs_y(is_visited) = [];
locs_x(is_visited) = [];
%// Get rid of those locations that are zero.
is_1 = A(sub2ind([rows cols], locs_x, locs_y));
locs_y(~is_1) = [];
locs_x(~is_1) = [];
%// Step 3g
%// Add remaining locations to the stack
stack = [stack; [locs_x locs_y]];
end
%// Step 4
%// Increment counter once complete region has been examined
ID_counter = ID_counter + 1;
end
end %// Step 5
end
With your example matrix, this is what I get for B:
B =
1 1 0 0 0 0 0
1 1 0 0 2 2 0
1 1 0 0 0 2 0
1 1 0 0 0 0 0
0 0 0 0 0 3 0
0 0 0 0 0 0 0
To search in a 4-connected neighbourhood
To modify the code to search in a 4-connected region, that is only North, East, West and South, the section where you see %// Look at the 8 neighbouring locations, that is:
%// Look at the 8 neighbouring locations
[locs_y, locs_x] = meshgrid(loc(2)-1:loc(2)+1, loc(1)-1:loc(1)+1);
locs_y = locs_y(:);
locs_x = locs_x(:);
To search in a 4-connected fashion, you simply have to modify this code to only give those cardinal directions:
%// Look at the 4 neighbouring locations
locs_y = [loc(2)-1; loc(2)+1; loc(2); loc(2)];
locs_x = [loc(1); loc(1); loc(1)-1; loc(1)+1];
The rest of the code remains untouched.
To match MATLAB's bwlabel function
If you want to match the output of MATLAB's bwlabel function, bwlabel searches for connected components in column major or FORTRAN order. The above code searches in row major or C order. Therefore you simply have to search along columns first rather than rows as what the above code is doing and you do this by swapping the order of the two for loops.
Specifically, instead of doing:
for row = 1 : rows
for col = 1 : cols
....
....
You would do:
for col = 1 : cols
for row = 1 : rows
....
....
This should now replicate the output of bwlabel.
I would like to calculate a couple of texture features (namely: small/ large number emphasis, number non-uniformity, second moment and entropy). Those can be computed from Neighboring gray-level dependence matrix. I'm struggling with understanding/implementation of this. There is very little info on this method (publicly available).
According to this paper:
This matrix takes the form of a two-dimensional array Q, where Q(i,j) can be considered as frequency counts of grayness variation of a processed image. It has a similar meaning as histogram of an image. This array is Ng×Nr where Ng is the number of possible gray levels and Nr is the number of possible neighbours to a pixel in an image.
If the image function f(i,j) is discrete, then it is easy to computer the Q matrix (for positive integer d, a) by counting the number of times the difference between each element in f(i,j) and its neighbours is equal or less than a at a certain distance d.
Here is the example from the same paper (d = 1, a = 0):
Input (image) matrix and output matrix Q:
I've been looking at this example for hours now and still can't figure out how they got that Q matrix. Anyone?
The method was originally created by C. Sun and W. Wee and was described in a paper called: "Neighboring gray level dependence matrix for texture classification" to which I got access, but can't download (after pressing download the page reloads and that's it).
In the example that you have provided, d=1 and a=0. When d=1, we consider pixels in an 8-pixel neighbourhood. When a=0, this means that we look for pixels that have the same value as the centre of the neighbourhood.
The basic algorithm is the following:
Initialize your NGLDM matrix to all zeroes. The total number of rows corresponds to the total number of possible intensities / values in your image. The total number of columns corresponds to how many pixels are in your neighbourhood plus 1. As such for d=1, we have an 8-pixel neighbourhood and so 8 + 1 = 9. Because there are 4 possible intensities (0,1,2,3), we thus have a 4 x 9 matrix. Let's call this matrix M.
For each pixel in your matrix, take note of this pixel. This goes in the Ng row.
Write out how many valid neighbours there are that surround this pixel.
Count how many times you see the neighbouring pixels matching that pixel in Step #1. This is your Nr column.
Once you figure out the numbers in Step #1 and Step #2, increment this location by 1.
Here's a slight gotcha: They ignore the border locations. As such, you don't do this procedure for the first row, last row, first column or last column. My guess is that they want to be sure that you have an 8-pixel neighbourhood all the time. This is also dictated by the distance d=1. You must be able to grab every valid pixel given a centre location at d=1. If d=2, then you would have to make sure that every pixel in the centre of the neighbourhood has a 25 pixel neighbourhood and so on.
Let's start from the second row, second column location of this matrix. Let's go through the steps:
Ng = 1 as the location is 1.
Valid neighbours - Starting from the top left pixel in this neighbourhood, and scanning left to right and omitting the centre, we have: 1, 1, 2, 0, 1, 0, 2, 2.
How many values are equal to 1? Three times. Therefore Nr = 3
M(Ng,Nr) += 1. Access row Ng = 1, and access row Nr = 3, and increment this spot by 1.
Want to know how I figured out they don't count the borders? Let's do the bottom left pixel. That location is 0, so Ng = 0. If you repeat the algorithm that I just said, you would expect Ng = 0, Nr = 1, and so you would expect at least one entry in that location in your matrix... but you don't! If you do similar checks around the border of the image, you'll see that entries that are supposed to be there... aren't. Take a look at the third row, fifth column. You would think that Ng = 1 and Nr = 1, but we don't see that in the matrix.
One more example. Why is M(Ng,Nr) = 4, Ng = 2, Nr = 4? Well, take a look at every pixel that has a 2 in it. The only valid locations where we can capture an 8 pixel neighbourhood successfully are the row=2, col=4, row=3, col=3, row=3, col=4, row=4, col=3, and row=4, col=4. By applying the same algorithm that we have seen, you'll see that for each of those locations, Nr = 4. As such, we see this combination of Ng = 2, Nr = 4 four times, and that's why the location is set to 4. However, in row=3, col=4, this actually is Nr = 5, as there are five 2s in that neighbourhood at that centre. That's why you see Ng = 2, Nr = 5, M(Ng,Nr) = 1.
As an example, let's do one of the locations. Let's do the 2 smack dab in the middle of the matrix (row=3, col=3):
Ng = 2
What are the valid neighbouring pixels? 1, 1, 2, 0, 2, 3, 2, 2 (omit the centre)
Count how many pixels equal to 2. There are four of them, so Nr = 4
M(Ng,Nr) += 1. Take Ng = 2, Nr = 4 and increment this spot by 1.
If you do this with the other valid locations that have 2, you'll see that Nr = 4 each time with the exception of the third row and fourth column, where Nr = 5.
So how would we implement this in MATLAB? What you can do is use im2col to transform each valid neighbourhood into columns. What I'm also going to do is extract the centre of each neighbourhood. This is actually the middle row of the matrix. We will then figure out how many pixels for each neighbourhood equal the centre, sum them up, and this will determine our Nr values. The Ng values will be the middle row values themselves. Once we do this, we can compute a histogram based on these values just like how the algorithm is doing to get our matrix. In other words, try doing this:
% // Your example
A = [1 1 2 3 1; 0 1 1 2 2; 0 0 2 2 1; 3 3 2 2 1; 0 0 2 0 1];
B = im2col(A, [3 3]); %//Convert neighbourhoods to columns - 3 x 3 means d = 1
C = bsxfun(#eq, B, B(5,:)); %//Figure out a logical matrix where each column tells
%//you how many elements equals the one in each centre
D = sum(C, 1) - 1; %// Must subtract by 1 to discount centre pixel
Ng = B(5,:).' + 1; % // We must make this into a column vector, and we also must
% // offset by 1 as MATLAB starts indexing by 1.
%// Column vector is for accumarray input
Nr = D.' + 1; %// Do the same for Nr. We could have simply left out the + 1 here and
%// took out the subtraction of -1 for D, but I want to explicitly show
%// the steps
Q = accumarray([Ng Nr], 1, [4 9]); %// 4 unique intensities, 9 possible locations (0-8)
... and here is our matrix:
Q =
0 0 1 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0
0 0 0 0 4 1 0 0 0
0 1 0 0 0 0 0 0 0
If you check this, you'll see this matches with Q.
Bonus
If you want to be able to accommodate for the algorithm in general, where you specify d and a, we can simply follow the guidelines of your text. For each neighbourhood, you find the difference between the centre pixel and all of the other pixels. You count how many pixels are <= a for any positive integer d. Note that this will create a 2*d + 1 x 2*d + 1 neighbourhood we need to examine. We can also make this into a function. Without further ado:
%// Set A up yourself, then use a and d as inputs
%// Precondition - a and d are both integers. a can be 0 and d is positive!
function [Q] = calculateGrayDepMatrix(A, a, d)
neigh = 2*d + 1; % //Calculate rows/columns of neighbourhood
numTotalNeigh = neigh*neigh; % //Calculate total number of pixels in neighbourhood
middleRow = ceil(numTotalNeigh / 2); %// Figure out which index the middle row is
B = im2col(A, [neigh neigh]); %// Make into columns
Cdiff = abs(bsxfun(#minus, B, B(middleRow,:))); %// For each neighbourhood, subtract with its centre
C = Cdiff <= a; %// For each neighbourhood, figure out which differences are <= a
D = sum(C, 1) - 1; % //For each neighbourhood, add them up
Ng = B(middleRow,:).' + 1; % // Determine Ng and Nr, and find Q
Nr = D.' + 1;
Q = accumarray([Ng Nr], 1, [max(Ng) numTotalNeigh]);
end
We can recreate the scenario we showed above with the example matrix by:
A = [1 1 2 3 1; 0 1 1 2 2; 0 0 2 2 1; 3 3 2 2 1; 0 0 2 0 1];
Q = calculateGrayDepMatrix(A, 0, 1);
Q is thus:
Q =
0 0 1 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0
0 0 0 0 4 1 0 0 0
0 1 0 0 0 0 0 0 0
Hope this helps!