I am trying to apply an imrotate to an image however the angle of the image is unknown and changes with each image.
If i have converted the image to binary would it be possible to find the lowest y position '1' and the leftmost'1' and use the gradient/angle between them as the angle for my image rotate?
for example:
binary positions diagram
Using the angle between these two positions and aligning it with the x axis?
Current Progress - converted to binary and made the edges more distinguishable:
% convert to binary
greyImage = rgb2gray(C); % greyscale
cannyImage = edge(greyImage, 'canny'); % canny edge detection
% fill the gaps in the shape
se = strel('disk',2);
bw = imclose(cannyImage, se);
filled = imfill(bw, 'holes');
imshow(filled);
[~,lowerMostCol] = max(cumsum(sum(filled,2)));
[~,leftMostRow] = max(sum(filled,1)==1);
Approach #1
With a as the binary image, you could do something like this -
[~,lowermost] = max(cumsum(sum(a,2)));
lowermostpt = [lowermost,find(a(lowermost,:),1,'first')]
[~,rightmost] = max(cumsum(sum(a,1)));
rightmostpt = [find(a(:,rightmost),1,'first'),rightmost]
[~,topmost] = max(sum(a,2)==1);
topmostpt = [topmost,find(a(topmost,:),1,'first')]
[~,leftmost] = max(sum(a,1)==1);
leftmostpt = [find(a(:,leftmost),1,'first'),leftmost]
For performance efficiency, it might be a good idea to store the summations once and re-use later on.
Sample run -
a =
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
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 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 0 1 1 1 1 1 0 0 0
0 0 0 0 0 0 1 1 1 1 1 1 1 0 0
0 0 0 0 0 1 1 1 1 1 1 1 1 1 0
0 0 0 0 0 0 1 1 1 1 1 1 1 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 1 1 1 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 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 0 0 0 0 0 0 0 0
lowermostpt =
12 10
rightmostpt =
8 14
topmostpt =
4 10
leftmostpt =
8 6
Approach #2 Using bwboundaries from Image-processing toolbox -
idx = cell2mat(bwboundaries(a))
[~,p1] = min(idx(:,1))
topmostpt = idx(p1,:)
[~,p2] = max(idx(:,1))
lowermostpt = idx(p2,:)
[~,p3] = min(idx(:,2))
leftmostpt = idx(p3,:)
[~,p4] = max(idx(:,2))
rightmostpt = idx(p4,:)
Related
I have a matrix looks like:
0 0 0 0 0
1 0 0 0 0
0 2 0 0 0
0 0 2 0 0
0 0 0 1 0
1 0 0 0 1
0 4 0 0 0
0 0 3 0 0
6 0 0 4 0
0 3 0 0 2
0 0 5 0 0
It is 11x5 matrix.
I want to interpolate between the values vertically for each column.
Any help ?
Thanks.
M =[0 0 0 0 0
1 0 0 0 0
0 2 0 0 0
0 0 2 0 0
0 0 0 1 0
1 0 0 0 1
0 4 0 0 0
0 0 3 0 0
6 0 0 4 0
0 3 0 0 2
0 0 5 0 0];
xi = 1:size(M,1)
for colIdx = 1:size(M,2)
col = M(:,colIdx);
x = xi(~~col); %// Note that ~~col is a logical vector of elements that are not equal to zero. i.e. it's the same as col ~= 0
y = col(~~col);
M(:,colIdx) = interp1(x,y,xi);
end
then if you want the outer points to be 0 add this line after the loop:
M(isnan(M)) = 0;
I need to find all paths with a graph, and save these paths. My starting nodes are A, B or C, and the final node is G. My graphs have maximum 16 unweighted vertices.
I made the Matlab code below, but this has problems with bifurcations. Also, I don't know how to impose the starting and the final nodes. Can anyone help me with this?
path = cell(1,10) ; % initialize
% one_graph ={'AH','BO','CN','EG','EN','EO','HO','KN'} % (Graph example)
one_graph ={'AH','BN','DH','DN','GN'} % (Graph example)
for p = 1:length(one_graph)
edge = one_graph(p);
% In each graph there is only 1:1 conections
% detect node 1
existing_node1 = edge{1}(1) ;
Index_existing_node1 = strfind(allnodes, existing_node1) ;
[row1,col1] = find(ismember(allnodes, existing_node1));
% detect node 2
existing_node2 = edge{1}(2) ;
Index_existing_node2 = strfind(allnodes, existing_node2);
[row2,col2] = find(ismember(allnodes, existing_node2));
path_nonz = path(~cellfun('isempty',path)) ;
t = length(path_nonz) ;
if t>0 % save the first 2 nodes in the path
ttt = strcmp(allnodes(row1), path{t});
ttt2 = strcmp(allnodes(row2), path{t});
end;
if t==0
path{t+1} = allnodes{row1} ;
path{t+2} = allnodes{row2} ;
elseif ttt == 1
% disp('connect right')
path{t+1} = allnodes{row2} ;
elseif ttt2 == 1
% disp('connect right')
path{t+1} = allnodes{row1} ;
else
disp('Not next vertex')
end
end
For example, for
one_graph ={'AH','BN','DH','DN','GN'} % (Graph example)
I should save the following paths:
path1 = AHDNG
path2 = BNG
and for
one_graph ={'AH','BO','CN','EG','EN','EO','HO','KN'} % (Graph example)
I should save the following paths:
path1 = AHOEG
path2 = BOEG
path3 = CNEG
UPDATE 1:
From the adjacency matrix B(:,:,1)
B =
0 0 0 0 0 0 0 1 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 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 0 1 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 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
1 0 0 1 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 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
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 0 0
0 1 0 1 0 0 1 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 0 0 0 0 0 0 0 0 0 0 0
I derive the proper adjacency list:
Asparse = sparse(B(:,:,1));
Asparse =
(8,1) 1
(14,2) 1
(8,4) 1
(14,4) 1
(14,7) 1
(1,8) 1
(4,8) 1
(2,14) 1
(4,14) 1
(7,14) 1
Then, I tried to use the BFS algorithm found on Matlab Website
[distances,times,pred] = bfs(Asparse,1);
But, this doesn't save the paths. It just saves the previous node of each current node (in pred) and the distance from the initial node to each node (in distances). Any idea, how to save each path?
I've had to write a custom function to do this since 1) most BFS/DFS functions stop when the goal is reached and 2) they explicitly ignore cycles, which are required for multiple paths to the same target.
I believe this will get you what you need. I've made a slight modification to the adjacency matrix in your example to create an edge from {2,7} and {7,2} so that there would be two paths from 2 to 14. Note that this is a recursive function, so if you get around 500 nodes or so you're going to have problems and we'll have to come up with a version that uses an explicit stack.
function paths = findpaths(Adj, nodes, currentPath, start, target)
paths = {};
nodes(start) = 0;
currentPath = [currentPath start];
childAdj = Adj(start,:) & nodes;
childList = find(childAdj);
childCount = numel(childList);
if childCount == 0 || start == target
if start == target
paths = [paths; currentPath];
end
return;
end
for idx = 1:childCount
currentNode = childList(idx);
newNodes = nodes;
newNodes(currentNode) = 0;
newPaths = findpaths(Adj, newNodes, currentPath, currentNode, target);
paths = [paths; newPaths];
end
end
If you call this function like this:
A =[
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 1;
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 0 1;
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 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 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 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 0;
0 1 0 1 0 0 1 0 0 0 0 0 0 0];
unusedNodes=ones(1,size(A,1));
start=2;
target=14;
emptyPath=[];
allPaths = findpaths(A, unusedNodes, emptyPath, start, target)
the output should be:
allPaths =
{
[1,1] =
2 7 14
[2,1] =
2 14
}
Naturally, you need to call this for each starting node.
Actually, you don't have to call this multiple times. There was one more tip I forgot to tell you. If your graph has n nodes and you introduce a new node n+1 that has edges only to your candidate start nodes, you can call the function once with the new node as the start.
So if I add node 15 to the graph above with edges:
{15,1}, {15,2}
%// I wouldn't bother with {1,15} and {2,15}, they're totally unnecessary
and call the function with start = 15, here's what I get:
allPaths =
{
[1,1] =
15 1 8 4 14
[2,1] =
15 2 7 14
[3,1] =
15 2 14
}
You now have all of the paths with one call, although you need to remove the new node 15 from the head of each path.
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(:)))