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Creating Indicator Matrix
(6 answers)
Closed 5 years ago.
I have vector like V[5000,1] with values in range 1-10
How to get matrix W[5000,10]
V[1] W[1 0 0 0 0 0 0 0 0 0]
[2] W[0 1 0 0 0 0 0 0 0 0]
[10] W[0 0 0 0 0 0 0 0 0 1]
[7] W[0 0 0 0 0 0 1 0 0 0]
... ...
its like W have "1" in column indicated by vector V
Solution
You can use the following approach:
W = zeros(length(v),10);
W(sub2ind(size(W),1:length(v),v'))=1;
Results
v = [10,4,10,4,9]
W =
0 0 0 0 0 0 0 0 0 1
0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0
Related
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How can I index a MATLAB array returned by a function without first assigning it to a local variable?
(9 answers)
Matlab Error: ()-indexing must appear last in an index expression
(1 answer)
Closed 5 years ago.
I'm trying to realise the following Octave command in MATLAB:
M = eye(x)(y,:);
x is just a number (in my example 10) and y is a vector (here 8x1):
y = [1 3 4 5 7 10 9 10];
The Octave command would generate:
M =
1 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
The ones are kept very near to the diagonal.
The nearest I came with MATLAB is with the following commands:
n = size(y,1);
Y = eye(n, x);
but it would generate something still different. If the difference between rows and columns gets bigger, it would be very different.
M =
1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
How could I get the first matrix with MATLAB?
First you should find what this expression eye(x)(y,:) means. First create an identity matrix with the size of x by x, and then select rows with index in y. Therefore, the equivalent syntax would be:
E = eye(x);
M = E(y,:);
I have one binary image so it has only 2 value like 0 and 1. After, I convert this into a padded image of different values, like the image will have curve shape. I took a 3 X 3 matrix of value and if i get curve shape then I padded the image with 1, or any number. I use 15 different types shape values like junction point, end point etc.
After, I give the values 1 to 15 - or the appropriate number according its shape. As such, I am getting an image like:
Figure
0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0
I would like to count how many 1s there are in the image, followed by 2s, 3s, etc. up to 15. For example,
as shown in the figure, if the pad number was 5, the total number of pixels would be 3. If the pad number was 1, the total number of pixels would be 6.
Use histc:
>> im = [ 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 5 0 0 0 0 0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 ]; %// data
>> values = 1:15; %// possible values
>> count = histc(im(:), values)
count =
6 %// number of 1's
0 %// number of 2's, etc
0
0
3
0
0
0
0
0
0
0
0
0
0
Or compute it manually with bsxfun:
>> count = sum(bsxfun(#eq, im(:), values(:).'), 1)
>> count =
6 0 0 0 3 0 0 0 0 0 0 0 0 0 0
I can also suggest using accumarray:
im = [0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 5 0 0 0 0 0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 ]; %// data - borrowed from Luis Mendo
counts = accumarray(im(:) + 1, 1);
counts(1) = []
counts =
6
0
0
0
3
Note we have to offset by 1 as accumarray starts indexing the output array at 1. Because you want to disregard the 0s, I simply take the counts result and remove the first entry. This result agrees with what you are seeking. The first element is how many 1s we have encountered, which is 6. The last element is how many 5s you have encountered, which is 3. Because 5 is the largest number encountered in your image, we can say that all symbols after 5 (6, 7, 8, ..., 15) have a count of 0.
I need to construct the tech cycle constraint matrix Aa and the right side ba. The aim is building the technology cycle matrices in order to solve the scheduling linear problem constrained by Ax<=b. In this case -1 and +1 in A refers to the coefficients of the constraints of the problem such as starting times and precedences
TC = [1,2,3,4,6,7;1,2,5,4,6,7;2,5,6,7,0,0]; % Technology cycle
CT = [100,60,200,160,80,120;100,60,150,120,60,150;50,120,40,30,0,0]; % Cycle time
n_jb = size(TC,1); % number of jobs
n_op = sum(TC~=0,2); % number of operations for each job
N_op = sum(n_op); % total number of operations
c=1; % indice for constraints in Aa
Op=1; % counter for overall operation
n_tf = N_op - n_jb- sum(n_op==1); % number of job transfer between machines (also number of tech cycle constraint numbers)
Aa = zeros(n_tf,N_op); % Constraint matrx for tech cycle
ba = zeros(n_tf,1); % The right vector of the constraint function: Aa*x<=ba
for j=1:n_jb
if n_op(j)>1
for op=1:n_op(j)-1
Aa(c,Op)=-1;
Aa(c,Op+1)=1;
ba(c,1)=CT(j,op);
c=c+1;
Op=Op+1;
end
else
Op=Op+1;
end
Op=Op+1;
end
The output, like Aa is 3 """diagonal""" -1/+1 matrices:
-1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1
In order to be more precise in the following there is an image: showing the 3 different part of the matrix Aa. My question is: Is there a way to build the same this avoiding loops since A is not a 3x1 but will definitely become 30-50x1?
You can use diag to create the positive and negative ones. The second input to diag is to shift the diagonal to the side. In this case, 1 to the right.
Use cumsum to find the rows you want to remove. For n = [6, 6, 4], you want to remove the 6th, 12th and 16th row.
n = [6, 6, 4];
cols = sum(n);
A = -eye(cols) + diag(ones(cols-1,1), 1);
A(cumsum(n),:) = []
A =
-1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1
I'm using Matlab. I have a 2-D Binary image/array. like this
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 1 1 1 0 0 1 0 0
0 0 1 1 0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 1 1 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0
I want to find out center of very first white block/Circle with respect to y-axis
Answer of the above image will be.
0 1 0
1 1 1
0 1 0
Anyone who have have a simplest solution for this.
If you are looking for exact matches of the template, you can use a moving filter, one example is:
H=[0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 1 0 0 0 0 0 0;
0 0 0 0 0 1 1 1 0 0 1 0 0;
0 0 1 1 0 0 1 0 0 0 0 0 0;
0 0 0 1 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 1 0 0 0 0 0 0;
0 0 0 0 0 0 1 1 0 0 0 0 0;
0 0 0 0 0 0 1 0 0 0 0 0 0];
b=[0 1 0;
1 1 1;
0 1 0];
C=filter2(b,H, 'same');
[x,y]=find(C==max(max(C)));
x and y are the locations of your template in the order that it appears from the top left corner of your array.
Edit: if you have the Image Processing Toolbox and are looking for a less strict way of finding objects that have a roughly circular shape you can use regionprops with the 'Centroid' and 'Eccentricity' arguments with the bwconncomp function.
ObjectStats=regionprops(bwconncomp(H,4), 'Centroid', 'Eccentricity');
Objects with an 'Eccentricity' of 0 (or close to 0) will be the circles.
idx=find(cell2mat({ObjectStats.Eccentricity})==0); % Change ==0 to <0.2 or something to make it less strict.
ctrs={ObjectStats.Centroid};
>> ctrs{1,idx(1)}
ans =
7 3
Note that in your case, a lone pixel is an object with an eccentricity of 0, it is the smallest 'circle' that you can find. If you need to define a minimum size, use the 'Area' property of regionprops
You can do this with a simple 2 dimensional convolution. It will "overlay" the filter along a larger matrix and multiply the filter by the values it is overlaying. If the product is equal to the sum of the filter, then you know you found a match.
Here is some simple code.
mat = [0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 1 1 1 0 0 1 0 0
0 0 1 1 0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 1 1 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0];
filt = [0 1 0
1 1 1
0 1 0];
[row,col] = find(conv2(mat,filt,'same') == sum(filt(:)))
How to replace elements of a matrix by an another matrix in MATLAB?
Ex: let say if we have a matrix A, where
A=[1 0 0; 0 1 0; 1 0 1]
I want to replace all ones by
J=[1 0 0; 0 1 0; 0 0 1]
and zeros by
K=[0 0 0; 0 0 0; 0 0 0]
So that I can get 9x9 matrix. So how we will code it in MATLAB
Thanks
Sounds like you might want to take a look at the kronecker tensor product. This is not a general case but the idea should work for what you want
>> kron(A==1,J)+kron(A==0,K)
ans =
1 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0 0
1 0 0 0 0 0 1 0 0
0 1 0 0 0 0 0 1 0
0 0 1 0 0 0 0 0 1
which, for the example case, would simplify to a simpler command:
>> kron(A,J)
ans =
1 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0 0
1 0 0 0 0 0 1 0 0
0 1 0 0 0 0 0 1 0
0 0 1 0 0 0 0 0 1
You can do:
A2=imresize(A,size(A).*size(J),'nearest');
J2=repmat(J,size(A));
K2=repmat(K,size(A));
A2(A2==1)=J2(A2==1);
A2(A2==0)=K2(A2==0)