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2D logical matrix from vector of coordinates (Basic matlab)
(2 answers)
Closed 9 years ago.
I would like to use values from a matrix to index an array. I will use a 3x2 matrix in the example but it could be a matrix of any height in the actual code. The array will be 5x5 in the example but could be a square array of any size. The size of the array and height of the matrix have no relationship.
Here is my code
X =
2 1
4 3
1 4
Grid=zeros(5,5)
Grid =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
So i would like to access points 2,1 4,3 and 1,4 and add one to the value in that location.
I have tried the following code
Grid(X(:,1),X(:,2))=Grid(X(:,1),X(:,2))+1
Which gives this result
Grid =
1 0 1 1 0
1 0 1 1 0
0 0 0 0 0
1 0 1 1 0
0 0 0 0 0
Which is not what I require.
I have tried other ways with no luck, I think i could use a loop or create a FLAT array but don't really want to, I think there must be a more efficient way.
Anyone have any ideas? I'm using Matlab 2012b.
As always thanks for your time and any help you may be able to give.
Edit-1 Required Result
This is the result I would like
Grid =
0 0 0 1 0
1 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 0 0
Edit-2
The coordinate matrix may hold duplicate values, so I would like the value in the relative location in the Array (Grid in the example) to show how many times this coordinate occurs. So my solution is
Grid(sub2ind(size(Grid),X(:,1),X(:,2)))=Grid(sub2ind(size(Grid),X(:,1),X(:,2)))+1
Using the answer to 2D logical matrix from vector of coordinates (Basic matlab) that Oleg pointed me to. I managed to solve my question by converting subscripts to linear indexes:
pos = sub2ind(size(Grid), X(:,1), X(:,2));
Grid(pos) = 1;
Related
I have used the following function for finding confusion matrix c=confusionmat(x,y)
it gives me output like that ,
C =
0 4 0 0
0 0 0 0
0 0 0 0
0 0 4 0
now i want to find accuracy and f_measure from this confusion matrix
I have got a 2D matrix. There is some region in the matrix where the elements are non-zero, in particular everywhere around the edge they are zero.
I plot the matrix using image as a colorplot and would like to add the curve that shows the boundary between non-zero values to zero values in the matrix. Is there any neat way to do this without loops?
This looks like a job for convhull :
To illustrate this code i'll take a dummy example :
A=zeros(10);
B=binornd(1,0.5,8,8);
A(2:end-1,2:end-1)=B
A =
0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 1 1 1 1 1 1 0 0
0 0 1 1 0 0 0 0 1 0
0 0 0 1 0 0 0 1 0 0
0 1 0 0 0 0 0 1 0 0
0 0 0 1 1 1 1 1 1 0
0 0 1 0 1 1 1 1 0 0
0 1 0 1 1 1 1 0 1 0
0 0 0 0 0 0 0 0 0 0
1/ Find the locations of all non zero entries :
[row,col]=find(A);
2/ Take the convex hull of these locations
k=convhull(row,col);
3/ Plot the convex hull (I plot the non zero points aswell but in your problem it will be your image points)
plot(row(k),col(k),'r-',row,col,'b*')
Result :
Another option is using the image processing toolbox and the bwperim function. This will work if you know that your area is completely closed (i.e. has no holes in the boundary)
This is an example using a black and white image, and you have 2 options: fill the inner gaps before, or not. You can see in the result the differences.
A = imread('circles.png');
Afill=imfill(A,'holes'); % optional
Abound1=bwperim(Afill);
Abound2=bwperim(A);
imshow([A,Abound, Abound2])
You can plot one on top of the other with:
[x,y]= find(Abound2);
hold on
image(A*255) %// If A is logical, else use just A (not *255)
colormap('gray')
plot(y,x,'r.')
hold off
axis tight
If you have a gray-scale image (or a matrix with a single value in each position (2D matrix), then you can binarize it first by either:
If you know everything outside your object is EXACTLY zero
A=yourA>0;
If you want to separate your object from the background, and the background is not exactly zero by A=im2bw(yourA,level), by choosing your own level, or letting Otsu do it for you with level=graythresh(yourA)
I'm writing a MATLAB code, I encountered a problem: I have a (2N+1)*(2N+1) matrix for example 7*7. I want to assign coordinate system to it such that the matrix center is the origin of coordinate system. I mean I want to assign (0,0) to row 4 and column 4 of matrix, (1,0) to row 4 and column 5 of matrix and so on. please help me
Thank you in advance
I want to generate a line of ones in all possible directions in a square matrix like this:
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 1 0
0 0 0 1 0 0 0
0 1 0 0 0 0 0
1 0 0 0 0 0 0
0 0 0 0 0 0 0
center of matrix is the origin. this line has 30 degree from horizontal axis.
What you want is a simple mapping from the original matrix counting system to a customized one. Here I have built two cell matrices, representing the coordinates of the elements in the matrix.
Here I have done a simple mapping as follows:
for ii = 1:7
for jj=1:7
D{ii,jj} = C{ii,jj} - [4,4];
end
end
Generally, for matrix of size 2*N+1, you will do the following:
for ii = 1:2*N+1
for jj = 1:2*N+1
D{ii,jj} = C{ii,jj} - [N+1,N+1];
end
end
where C is the original matrix and D is the mapped matrix. After you well-understood what I have done here, you can then replace the for-loops with more efficient functions such as bsxfun.
So I'm trying to create an adjacency matrix, and I'm confused on the difference between accumarray(matrix+1,1) and accumarray(matrix,1).
I did:
matrix = [ 1 3
4 2
1 3
3 1]
adMatrix1 = accumarray(matrix,1);
adMatrix1=adMatrix1~=0;
adMatrix1 = [0 0 1
0 0 0
1 0 0
0 1 0]
and then:
adMatrix2 = accumarray(matrix+1,1);
adMatrix2=adMatrix2~=0;
adMatrix2 = [0 0 0 0
0 0 0 1
0 0 0 0
0 1 0 0
0 0 1 0]
I know that with the "matrix+1", there's an extra row and column of zero's, but I don't understand why you would do it that way. When I looked it up, according to this post I should use "matrix+1", and the best explanation that I got for that was that "because indexing in matlab starts at 1".
I don't understand that at all... if I was trying to create an adjacency matrix, which way is correct? Any help would be greatly appreciated, thanks!
If your node IDs are 0 indexed you need the +1, otherwise you don't. So the question you need to ask is, are your node IDs 0 indexed or 1 indexed?
Does your matrix accept multiple links? If yes, then both results with accumarray above are not correct, since node 1 and 3 are connected 2 times.
Btw, you can consider sparse and full:
full(sparse(matrix(:,1), matrix(:,2), ones(1, size(matrix, 1))))
ans =
0 0 2
0 0 0
1 0 0
0 1 0
ones(1, size(matrix, 1)) is actually a vector of weight.
I am very interested in fingerprint verification and studying minutia extraction at present. I have found the following code online and wonder if someone would be kind enough to explain it? I have looked up centroid, regionprops etc, I understand these a little but the code below has me puzzled!
fun=#minutie;
L = nlfilter(K,[3 3],fun);
%% Termination
LTerm=(L==1);
imshow(LTerm)
LTermLab=bwlabel(LTerm);
propTerm=regionprops(LTermLab,'Centroid');
CentroidTerm=round(cat(1,propTerm(:).Centroid));
imshow(~K)
set(gcf,'position',[1 1 600 600]);
hold on
plot(CentroidTerm(:,1),CentroidTerm(:,2),'ro')
%% Bifurcation
LBif=(L==3);
LBifLab=bwlabel(LBif);
propBif=regionprops(LBifLab,'Centroid','Image');
CentroidBif=round(cat(1,propBif(:).Centroid));
plot(CentroidBif(:,1),CentroidBif(:,2),'go')
The code first filters the binary image with a neighborhood of 3x3 pixels. nfilter is a moving filter function. It will go through all the pixels in the image given as argument and apply an operation based on the values of the neighboring pixels.
I don't know the exact content of the minutie filter, but judging by the rest of the code, it probably counts the pixels with a value of 1 in the neighborhood of all 1s. In other words it will be equal to one at the end of a segment, and equal to 3 when there are 3 branches (a bifurcation).
Example:
Let a filter sum up the ones in the neighborhood, like this:
sum(block(1,1:3), block(3,1:3), block(2,1), block(2,3))*block(2, 2);
where block denotes a neighborhood around each pixel of the binary image.
In the left matrix below (if you ignore the boundary exceptions) there is one position with a one that has exactly one 1 in its 3x3 neighborhood, in the right matrix, there is one position with a one that has exactly three 1s in its 3x3 neighborhood.
[0 0 0 0 0 [0 0 1 0 0
0 0 0 0 0 0 0 1 0 0
0 0 1 0 0 1 1 1 0 0
0 0 1 0 0 0 0 1 0 0
0 0 1 0 0] 0 0 1 0 0]
The filtered output would be:
[0 0 0 0 0 [0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 3 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0] 0 0 0 0 0]
It found a termination in the left matrix, and a bifurcation in the right matrix.
The filtered image are then thresholded at the value 1 and 3, then the use of bwlabel and regionprops is somewhat mysterious to me† since bifurcations and terminations are single points, their position is simply their index. I think you could simply achieve the detection of the coordinates of the terminations and bifurcation using something like:
[It Jt]= find(L==1);
[Ib Jb]= find(L==3);
† one reason I can think of is that coordinates in images and arrays are different in matlab, and these two function output coordinates in the image format, which is easier to plot on top of the original image.