I'm new in MATLAB and using it for some medical analysis. I have a matrix which contains a circular shape in it. Here is a sample:
0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 1 0 1 1 0 0 0
0 0 1 0 0 0 0 1 0 0
0 1 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0
0 0 1 1 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 0
0 0 0 0 0 0 0 0 0 0
This matrix has been computed by this line:
egslbr= edge(slbr,'log');
Where slbr is my image. By merging the egslbr with my slbr I get the image bellow.
I want to cancel all the color pixel outside the greenish circle. Is there any way to do this?
You should be able to create a mask from your circle (assuming it's a complete circle -- the sample matrix you gave has a gap in it which makes this much more difficult, but I'm going to assume that was a mistake). Here's a simple way of doing it:
mask = ~cumsum(egslbr) | ~cumsum(egslbr,'reverse');
slbr(mask) = 0;
This should set every pixel outside the circle (not including the edge of the circle though) to zero.
Related
for i = 2:N
A(i,i-1:i+1) = [1, -2, 1];
end
Hello, matlab is telling me that this code can be faster by using spalloc for the matrix A (which I have) but also by vectorizing this for loop. I've tried to use the following:
i = 2:N
A(i, i-1:i+1)
but the result obviously turned out to be not what I want.
How can I solve this?
Thank you!
It looks like you're trying to get a second-order difference operator, except your loop winds up missing the first row and including an extra column. The normal (sparse) difference operator is generated like this:
N = 10;
v = ones(N, 1);
A = spdiags([v -2*v v], [-1 0 1], N, N);
full(A) % for display only
You'll see:
ans =
-2 1 0 0 0 0 0 0 0 0
1 -2 1 0 0 0 0 0 0 0
0 1 -2 1 0 0 0 0 0 0
0 0 1 -2 1 0 0 0 0 0
0 0 0 1 -2 1 0 0 0 0
0 0 0 0 1 -2 1 0 0 0
0 0 0 0 0 1 -2 1 0 0
0 0 0 0 0 0 1 -2 1 0
0 0 0 0 0 0 0 1 -2 1
0 0 0 0 0 0 0 0 1 -2
If that's not quite what you want (e.g., you really don't want the first row), then it's probably faster to generate it as above and then fix it up.
I am trying to find islands of numbers in a matrix.
By an island, I mean a rectangular area where ones are connected with each other either horizontally, vertically or diagonally including the boundary layer of zeros
Suppose I have this matrix:
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 1 0 0 0 0 0 0 0 0 1
0 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0
0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 1
0 0 0 1 0 1 0 1 1 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 0 0 0 0 0
0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0
0 0 1 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 0 0 0 0
By boundary layer, I mean row 2 and 7, and column 3 and 10 for island#1.
This is shown below:
I want the row and column indices of the islands. So for the above matrix, the desired output is:
isl{1}= {[2 3 4 5 6 7]; % row indices of island#1
[3 4 5 6 7 8 9 10]} % column indices of island#1
isl{2}= {[2 3 4 5 6 7]; % row indices of island#2
[12 13 14 15 16 17]}; % column indices of island#2
isl{3} ={[9 10 11 12]; % row indices of island#3
[2 3 4 5 6 7 8 9 10 11];} % column indices of island#3
It doesn't matter which island is detected first.
While I know that the [r,c] = find(matrix) function can give the row and column indices of ones but I have no clues on how to detect the connected ones since they can be connected in horizontal, vertical and diagonal order.
Any ideas on how to deal with this problem?
You should look at the BoundingBox and ConvexHull stats returned by regionprops:
a = imread('circlesBrightDark.png');
bw = a < 100;
s = regionprops('table',bw,'BoundingBox','ConvexHull')
https://www.mathworks.com/help/images/ref/regionprops.html
Finding the connected components and their bounding boxes is the easy part. The more difficult part is merging the bounding boxes into islands.
Bounding Boxes
First the easy part.
function bBoxes = getIslandBoxes(lMap)
% find bounding box of each candidate island
% lMap is a logical matrix containing zero or more connected components
bw = bwlabel(lMap); % label connected components in logical matrix
bBoxes = struct2cell(regionprops(bw, 'BoundingBox')); % get bounding boxes
bBoxes = cellfun(#round, bBoxes, 'UniformOutput', false); % round values
end
The values are rounded because the bounding boxes returned by regionprops lies outside its respective component on the grid lines rather than the cell center, and we need integer values to use as subscripts into the matrix. For example, a component that looks like this:
0 0 0
0 1 0
0 0 0
will have a bounding box of
[ 1.5000 1.5000 1.0000 1.0000 ]
which we round to
[ 2 2 1 1]
Merging
Now the hard part. First, the merge condition:
We merge bounding box b2 into bounding box b1 if b2 and the island of b1 (including the boundary layer) have a non-null intersection.
This condition ensures that bounding boxes are merged when one component is wholly or partially inside the bounding box of another, but it also catches the edge cases when a bounding box is within the zero boundary of another. Once all of the bounding boxes are merged, they are guaranteed to have a boundary of all zeros (or border the edge of the matrix), otherwise the nonzero value in its boundary would have been merged.
Since merging involves deleting the merged bounding box, the loops are done backwards so that we don't end up indexing non-existent array elements.
Unfortunately, making one pass through the array comparing each element to all the others is insufficient to catch all cases. To signal that all of the possible bounding boxes have been merged into islands, we use a flag called anyMerged and loop until we get through one complete iteration without merging anything.
function mBoxes = mergeBoxes(bBoxes)
% find bounding boxes that intersect, and merge them
mBoxes = bBoxes;
% merge bounding boxes that overlap
anyMerged = true; % flag to show when we've finished
while (anyMerged)
anyMerged = false; % no boxes merged on this iteration so far...
for box1 = numel(mBoxes):-1:2
for box2 = box1-1:-1:1
% if intersection between bounding boxes is > 0, merge
% the size of box1 is increased b y 1 on all sides...
% this is so that components that lie within the borders
% of another component, but not inside the bounding box,
% are merged
if (rectint(mBoxes{box1} + [-1 -1 2 2], mBoxes{box2}) > 0)
coords1 = rect2corners(mBoxes{box1});
coords2 = rect2corners(mBoxes{box2});
minX = min(coords1(1), coords2(1));
minY = min(coords1(2), coords2(2));
maxX = max(coords1(3), coords2(3));
maxY = max(coords1(4), coords2(4));
mBoxes{box2} = [minX, minY, maxX-minX+1, maxY-minY+1]; % merge
mBoxes(box1) = []; % delete redundant bounding box
anyMerged = true; % bounding boxes merged: loop again
break;
end
end
end
end
end
The merge function uses a small utility function that converts rectangles with the format [x y width height] to a vector of subscripts for the top-left, bottom-right corners [x1 y1 x2 y2]. (This was actually used in another function to check that an island had a zero border, but as discussed above, this check is unnecessary.)
function corners = rect2corners(rect)
% change from rect = x, y, width, height
% to corners = x1, y1, x2, y2
corners = [rect(1), ...
rect(2), ...
rect(1) + rect(3) - 1, ...
rect(2) + rect(4) - 1];
end
Output Formatting and Driver Function
The return value from mergeBoxes is a cell array of rectangle objects. If you find this format useful, you can stop here, but it's easy to get to the format requested with ranges of rows and columns for each island:
function rRanges = rect2range(bBoxes, mSize)
% convert rect = x, y, width, height to
% range = y:y+height-1; x:x+width-1
% and expand range by 1 in all 4 directions to include zero border,
% making sure to stay within borders of original matrix
rangeFun = #(rect) {max(rect(2)-1,1):min(rect(2)+rect(4),mSize(1));...
max(rect(1)-1,1):min(rect(1)+rect(3),mSize(2))};
rRanges = cellfun(rangeFun, bBoxes, 'UniformOutput', false);
end
All that's left is a main function to tie all of the others together and we're done.
function theIslands = getIslandRects(m)
% get rectangle around each component in map
lMap = logical(m);
% get the bounding boxes of candidate islands
bBoxes = getIslandBoxes(lMap);
% merge bounding boxes that overlap
bBoxes = mergeBoxes(bBoxes);
% convert bounding boxes to row/column ranges
theIslands = rect2range(bBoxes, size(lMap));
end
Here's a run using the sample matrix given in the question:
M =
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 1 0 0 0 0 0 0 0 0 1
0 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0
0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 1
0 0 0 1 0 1 0 1 1 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 0 0 0 0 0
0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0
0 0 1 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 0 0 0 0
>> getIslandRects(M)
ans =
{
[1,1] =
{
[1,1] =
9 10 11 12
[2,1] =
2 3 4 5 6 7 8 9 10 11
}
[1,2] =
{
[1,1] =
2 3 4 5 6 7
[2,1] =
3 4 5 6 7 8 9 10
}
[1,3] =
{
[1,1] =
2 3 4 5 6 7
[2,1] =
12 13 14 15 16 17
}
}
Quite easy!
Just use bwboundaries to get the boundaries of each of the blobs. you can then just get the min and max in each x and y direction of each boundary to build your box.
Use image dilation and regionprops
mat = [...
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 1 0 0 0 0 0 0 0 0 1;
0 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0;
0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 1;
0 0 0 1 0 1 0 1 1 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 0 0 0 0 0;
0 0 0 1 0 1 0 1 1 1 0 0 0 0 0 0 0;
0 0 1 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 0 0 0 0];
mat=logical(mat);
dil_mat=imdilate(mat,true(2,2)); %here we make bridges to 1 px away ones
l_mat=bwlabel(dil_mat,8);
bb = regionprops(l_mat,'BoundingBox');
bb = struct2cell(bb); bb = cellfun(#(x) fix(x), bb, 'un',0);
isl = cellfun(#(x) {max(1,x(2)):min(x(2)+x(4),size(mat,1)),...
max(1,x(1)):min(x(1)+x(3),size(mat,2))},bb,'un',0);
So say we have an empty grid of 0s:
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
And you can draw shapes on it. 1 represents a filled cell.
1 1 1 1 0 0 0 0
1 0 0 1 0 0 0 0
1 0 0 1 0 0 0 0
1 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0
0 1 1 0 0 1 1 1
1 0 0 1 0 1 0 0
0 1 1 0 0 1 0 0
A shape is considered closed if a four-directional flood-fill algorithm would not leak and fill any cells outside of the shape. A shape can not use the boundary of the grid as one of its sides. So if we filled in all of the closed shapes in this grid with 2s, we would have:
1 1 1 1 0 0 0 0
1 2 2 1 0 0 0 0
1 2 2 1 0 0 0 0
1 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0
0 1 1 0 0 1 1 1
1 2 2 1 0 1 0 0
0 1 1 0 0 1 0 0
Implementing the flood-fill algorithm is easy, but I can't figure out a way to (programatically) fill in all enclosed arbitrary shapes in a grid. Are there any type of algorithms or searches I could use for this?
What's wrong with flood-fill algorithm? It is simple end effective with complexity O(N).
At first scan edges for zero values and flood empty regions with mark value 3.
Then walk through inner place. If you find zero cell, flood-fill from this cell with value 2.
(Perhaps you are seeking something like connected-component labeling algorithm. It is intended to mark every connected region by unique mark value)
You can first find zero's which have a path to boundary:
Take arbitrary 0 cell in the boundary, mark it as -1, do this for all of its neighbouring cells recursively (neighbours of neighbours and so on all set to -1). Once none of the boundary cells is zero, turn all zero cells to 2. It means that they are surrounded by only 1's. After all turn all -1's to 0. This is O(n) which n is the number of cells in the grid.
Here is a (lazy) pseudo code, assuming we have n_1xn_2 grid:
function fill()
{
for int i=1..n_1
{
recursivecolor(i,1);
recursivecolor(i,n_2);
}
for int j=1..n_2
{
recursivecolor(1,j);
recursivecolor(n_1,j);
}
for i=1..n_1
for j=1 .. n_2
if (a[i][j] == 0)
a[i][j] = 2;
for i=1..n_1
for j=1 .. n_2
if (a[i][j] == -1)
a[i][j] = 0;
}
function recursivecolor(i,j)
{
if (a[i][j]!=0) return;
a[i][j] = -1;
if (a[i-1][j] == 0)
{
a[i-1][j] = -1;
recursivecolor(i-1,j);
}
// do this for all neighbours of i,j cell
// it also needs check for boundaries, e.g. i-1 should not be zero ...
}
I have a set with elements and the possible adjacent combinations for this are:
So the total possible combinations are c=11 which can be calculated with the formula:
I can model this using a as below whose elements can be represented as a(n,c) are:
I have tried to implement this in MATLAB, but since I have hard-coded the above math my code is not extensible for cases where n > 4:
n=4;
c=((n^2)/2)+(n/2)+1;
A=zeros(n,c);
for i=1:n
A(i,i+1)=1;
end
for i=1:n-1
A(i,n+i+1)=1;
A(i+1,n+i+1)=1;
end
for i=1:n-2
A(i,n+i+4)=1;
A(i+1,n+i+4)=1;
A(i+2,n+i+4)=1;
end
for i=1:n-3
A(i,n+i+6)=1;
A(i+1,n+i+6)=1;
A(i+2,n+i+6)=1;
A(i+3,n+i+6)=1;
end
Is there a relatively low complexity method to transform this problem in MATLAB with n number of elements of set N, following my above mathematical formulation?
The easy way to go about this is to take a bit pattern with the first k bits set and shift it down n - k times, saving each shifted column vector to the result. So, starting from
1
0
0
0
Shift 1, 2, and 3 times to get
|1 0 0 0|
|0 1 0 0|
|0 0 1 0|
|0 0 0 1|
We'll use circshift to achieve this.
function A = adjcombs(n)
c = (n^2 + n)/2 + 1; % number of combinations
A = zeros(n,c); % preallocate output array
col_idx = 1; % skip the first (all-zero) column
curr_col = zeros(n,1); % column vector containing current combination
for elem_count = 1:n
curr_col(elem_count) = 1; % add another element to our combination
for shift_count = 0:(n - elem_count)
col_idx = col_idx + 1; % increment column index
% shift the current column and insert it at the proper index
A(:,col_idx) = circshift(curr_col, shift_count);
end
end
end
Calling the function with n = 4 and 6 we get:
>> A = adjcombs(4)
A =
0 1 0 0 0 1 0 0 1 0 1
0 0 1 0 0 1 1 0 1 1 1
0 0 0 1 0 0 1 1 1 1 1
0 0 0 0 1 0 0 1 0 1 1
>> A = adjcombs(6)
A =
0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0 1
0 0 1 0 0 0 0 1 1 0 0 0 1 1 0 0 1 1 0 1 1 1
0 0 0 1 0 0 0 0 1 1 0 0 1 1 1 0 1 1 1 1 1 1
0 0 0 0 1 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1
0 0 0 0 0 1 0 0 0 0 1 1 0 0 1 1 0 1 1 1 1 1
0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0 1 1
Let Y be a vector of length N, containing numbers from 1 to 10. As example code you can use:
Y = vec(1:10);
I am writing the code which must create an N x 10 matrix, each row consisting of all zeros except for a 1 only in the position which corresponds to the number in vector Y. Thus, 1 in Y becomes 10000000000, 3 becomes 0010000000, and so on.
This approach works:
cell2mat(arrayfun(#(x)eye(10)(x,:), Y, 'UniformOutput', false))
My next idea was to "optimize", so eye(10) is not generated N times, and I wrote this:
theEye = eye(10);
cell2mat(arrayfun(#(x)theEye(x,:), Y, 'UniformOutput', false))
However, now Octave is giving me error:
error: can't perform indexing operations for diagonal matrix type
error: evaluating argument list element number 1
Why do I get this error? What is wrong?
Bonus questions — do you see a better way to do what I am doing? Is my attempt to optimize making things easier for Octave?
I ran this code in Octave and eye creates a matrix of a class (or whatever this is) known as a Diagonal Matrix:
octave:3> theEye = eye(10);
octave:4> theEye
theEye =
Diagonal Matrix
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
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
In fact, the documentation for Octave says that if the matrix is diagonal, a special object is created to handle the diagonal matrices instead of a standard matrix: https://www.gnu.org/software/octave/doc/interpreter/Creating-Diagonal-Matrices.html
What's interesting is that we can slice into this matrix outside of the arrayfun call, regardless of it being in a separate class.
octave:1> theEye = eye(10);
octave:2> theEye(1,:)
ans =
Diagonal Matrix
1 0 0 0 0 0 0 0 0 0
However, as soon as we put this into an arrayfun call, it decides to crap out:
octave:5> arrayfun(#(x)theEye(x,:), 1:3, 'uni', 0)
error: can't perform indexing operations for diagonal matrix type
This to me doesn't make any sense, especially since we can slice into it outside of arrayfun. One may suspect that it has something to do with arrayfun and since you are specifying UniformOutput to be false, a cell array of elements is returned per element in Y and perhaps something is going wrong when storing these slices into each cell array element.
However, this doesn't seem to be the culprit either. I took the first three rows of theEye, placed them into a cell array and merged them together using cell2mat:
octave:6> cell2mat({theEye(1,:); theEye(2,:); theEye(3,:)})
ans =
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
As such, I suspect that it may be some sort of internal bug (if you could call it that...). Thanks to user carandraug (see comment above), this is indeed a bug and it has been reported: https://savannah.gnu.org/bugs/?47510. What may also provide insight is that this code runs as expected in MATLAB.
In any case, one thing you can take away from this is that I would seriously refrain from using cell2mat. Just use straight up indexing:
Y = vec(1:10);
theEye = eye(10);
out = theEye(Y,:);
This would index into theEye and extract out the relevant rows stored in Y and create a matrix where each row is zero except for the corresponding value seen in each element Y.
Also, have a look at this post for a similar example: Replace specific columns in a matrix with a constant column vector
However, it is defined over the columns instead of the rows, but it's very similar to what you want to achieve.
Another approach; We start with the data:
>> len = 10; % max number
>> vec = randi(len, [1 7]) % vector of numbers
vec =
1 10 9 5 7 3 6
Now we build the indicator matrix:
>> I = full(sparse(1:numel(vec), vec, 1, numel(vec), len))
I =
1 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 1 0
0 0 0 0 1 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 1 0 0 0 0