Here is an example matrix (but the result shouldn't be constrained to only working on this):
a=zeros(7,7);
a(5,3:6)=1;
a(2,2)=1;
a(2,4)=1;
a(7,1:2)=1
a=
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
1 1 0 0 0 0 0
I want to get rid of all the 1's that are alone (the noise), such that I only have the line of 1's on the fifth row.
rules:
-the 1's are in 'connected lines' if there are adjacent 1's (including diagonally) e.g.:
0 0 0 1 0 0 1 0 1
1 1 1 0 1 0 0 1 0
0 0 0 0 0 1 0 0 0
(The connected lines are what I want to keep. I want to get rid of all the 1's that are not in connected lines, the connected lines can intersect each other)
the 'connected lines need to be at least 3 elements long. So in the 7x7 example, there would only be one line that matches this criteria. If a(7,3) was set to 1, then there would be a connected line at the bottom left also
I am currently looking at this through a column by column approach, and here is the first draft of my code so far:
for nnn=2:6
rowPoss=find(a(:,nnn)==1);
rowPoss2=find(a(:,nnn+1)==1);
for nn=1:length(rowPoss)
if myResult(rowPoss(nn)-1:rowPoss(nn)+1,n-1)==0 %
%then?
end
end
end
My difficulty is, during this column by column process, I'd have to enable a way to recognise the beginning of the connected line, the middle of the connected line, and when a connected line ends. The same rules for this, when applied to noise (the lone 1's), would just ignore the lone 1's.
The output I want is basically:
b=
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 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
If you have image processing toolbox, try bwareaopen
b = bwareaopen(a, 3);
Sample Run #1:
>> a
a =
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
1 1 0 0 0 0 0
>> b
b =
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 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
Sample Run #2:
>> a
a =
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
1 1 0 0 0 0 0
>> b
b =
0 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 1 1 1 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
Related
I have an adjacency matrix describing two trees that merges in the middle. An exemple for 10 nodes:
The corresponding adjacency matrix is a 10x10 matrix where the first row correspond to the first node (start of the first tree, node #1) and the last row to the root of the second tree (end of the second tree, node #10).
Here is the adjacency matrix corresponding to a larger example with 22 nodes:
0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0
0 0 1 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 0 0 0 0 1 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 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 0 1 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 1 0 0 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 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 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0
I'm trying to plot this adjacency matrix as the picture showed above using Matlab. Matlab as some tools for plotting trees, for example the code given in:
https://blogs.mathworks.com/cleve/2017/03/20/morse-code-binary-trees-and-graphs/
However, using the previous matrix (let's label it 'A') and the following code:
G = digraph(A);
Gp = plot(G);
does not produce a tree, but a graph (not ordered as a tree).
Thus, how to produce a picture of a tree (as showed above) using 'A' in Matlab?
Please note that I also have matrices describing trees where the degree between the children nodes are 3 (or more) rather than 2.
Zero-out half of your adjacency matrix to make the connection one-way.
By default, MATLAB tries to decide the layout of graphs automatically, based on the structure of your graph. There are a few different graph layouts you can choose from.
The option you wanted is the 'Layered' layout; but I tested it with your example, and it definitely still doesn't look like a tree. The reason being that your adjacency matrix is symmetrical, and the connection is two-way. This confuses MATLAB when placing the nodes, and it doesn't think it's a tree.
The easy fix, is you can zero-out the lower triangular half of your adjacency matrix. I used the function tril for this.
% Create a lower triangular matrix with the dimension of A,
idx = tril(ones(size(A)));
% Make it a logical array to select matrix elements with
idx = logical(idx);
% Select the defined lower triangular part, and set that to zero
A(idx) = 0;
% Generate-Plot graph as you did
G = digraph(A);
plot(G)
Result
I want to create a MxN matrix as shown below:
[1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1]
I have window size, let's say, 5 and it moves 3 in every row. Is it possible to create such a matrix without using for loops? Or is there any optimum way to do it?
This is a one line solution:
reshape([reshape([ones(5,6);zeros(21,6)], 1,[]), ones(1,5)],[],7).'
note:
The desired matrix can be seen as concatenation of a [6, 5+21] matrix:
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
and a [1 ,5] matrix:
1 1 1 1 1
that reshaped to a [7 , 23] matrix.
Other solution using repelem + bsxfun + accumarray:
r = repelem (1:7,5);
c= bsxfun(#plus, ((1:5)-3).',3*(1:7));
out = accumarray([r(:) c(:)] ,1)
Indices of rows and columns of 1 s can be generated and accumarray can be used to create the desired matrix.
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)