Say that we have the following random matrix:
1 2 3 4
5 6 7 8
9 8 7 6
5 4 3 2
I'd like to transform it into the following:
1 0 2 0 3 0 4 0
0 0 0 0 0 0 0 0
5 0 6 0 7 0 8 0
0 0 0 0 0 0 0 0
9 0 8 0 7 0 6 0
0 0 0 0 0 0 0 0
5 0 4 0 3 0 2 0
0 0 0 0 0 0 0 0
For some reason I cannot use mathjax format so it looks a bit awful, sorry for this. Point, is, that I want to add row and columns of zeros in between of my current rows and columns so that I increase its size 2x.
I came up with the following code, but it only works for very small matrixes if i use it on a a big image it cannot finish due to memory limitation problems.
clear all
I=imread('image.png');
I=rgb2gray(I);
B=zeros(2*size(I));
[x, y]=find(-inf<I<inf);
xy=[x,y];
nxy=xy;
%coord change
nxy=2*xy-1;
B(nxy(:,1),nxy(:,2))=I(xy(:,1),xy(:,2));
I expected to be fast because it is fully vectorised with maltlab functions but it fails miserably. Is there some other elegant way to do this?
If you take a look at your indexing vectors, this is something like I([1 1 2 2] ,[1 2 1 2] ); for a 2x2 matrix which means you index each row and column twice. The right solution is B(1:2:end,1:2:end)=I; which indexes every second row and every second column.
This can be also done via the one liner, say your original matrix is called A, then
kron(A,[1,0;0,0])
Related
I have written some code that compresses a matrix to remove zero columns and rows, but I can't work out how to reconstruct the original matrix.
Say I have a matrix:
A = [ 0 3 0 2 1 0 6
3 0 0 4 8 0 5
0 0 0 0 0 0 0
2 4 0 0 2 0 1
1 8 0 2 0 0 7
0 0 0 0 0 0 0
6 5 0 1 7 0 0 ]
Here rows/columns 3 and 6 are empty, so my compression function will give the output:
A_dash = [ 0 3 2 1 6
3 0 4 8 5
2 4 0 2 1
1 8 2 0 7
6 5 1 7 0 ]
A_map = [ 1 2 4 5 7]
Where A_map is a vector mapping the indicies of the rows/columns of A_dash to A. This means that if A_map(3) = 4, then row/column 4 of A is the same as row/column 3 of A_dash - ie. a row/column of zeroes must be inserted between columns/rows 2 and 3 in A_dash
What is the easiest way people can suggest for me to recreate matrix A from A_dash, using the information in A_map?
Here is what I have got so far:
% orig_size is original number of columns/rows
c_count = size(A_dash,1);
A = zeros(c_count, orig_size); % c_count rows to avoid dimension mismatch
for ii = 1:c_count
A(:,A_map(ii)) == A_dash(:,ii);
end
This gives me the right result column-wise:
A = [ 0 3 0 2 1 0 6
3 0 0 4 8 0 5
2 4 0 0 2 0 1
1 8 0 2 0 0 7
6 5 0 1 7 0 0 ]
However, I'm not sure how i should go about inserting the rows, i suppose i could copy the first 1:i rows into one matrix, i:end rows to a second matrix and concatenate those with a zero row in between, but that feels like a bit of a
clunky solution, and probably not very efficient for large sized matrices..
Otherwise, is there a better way that people can suggest I store the map information? I was thinking instead of storing the mapping between column/row indices, that I just store the indices of the zero columns/rows and then insert columns/rows of zeros where appropriate. Would this be a better way?
You've got the indices of the valid rows/columns. Now all you've got to do is put them in a new matrix of zeros the same size as A:
B=zeros(size(A));
B(A_map,A_map)=A_dash
B =
0 3 0 2 1 0 6
3 0 0 4 8 0 5
0 0 0 0 0 0 0
2 4 0 0 2 0 1
1 8 0 2 0 0 7
0 0 0 0 0 0 0
6 5 0 1 7 0 0
Just to check...
>> A==B
ans =
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
A and B are equal everywhere, so we've reconstructed A.
Using MATLAB, I have a matrix such as:
1 1 0
1 0 1
1 1 1
The aim is to represent the zero's as a mine in a minesweeper program and the values around the 0's should reflect how many mines are adjacent to it.
Therefore creating a vector like this:
1 2 0
1 0 2
1 1 1
I have thought to take elements around the zero as a sub matrix and then add 1, but then it will turn 0's into 1's.
How would I program such a task?
I think this can be achieved by simple convolution plus some post-processing on the resultant matrix as follows:
% Defining a 6x6 matrix of zeros and ones
mineMat=randi(2,6,6)-1;
numberOfMines=conv2(double(~mineMat),ones(3,3),'same').*mineMat;
% Result:
mineMat=
1 0 1 1 0 0
0 0 0 1 0 0
1 1 1 1 1 0
1 1 1 1 0 1
0 1 0 0 0 0
0 1 1 0 0 0
numberOfMines=
3 0 3 3 0 0
0 0 0 3 0 0
2 3 2 3 4 0
1 2 2 4 0 4
0 3 0 0 0 0
0 3 3 0 0 0
Parag's answer would be my first option. Another approach is to use blockproc (Image Processing Toolbox):
blockproc(~M, [1 1], #(x)sum(x.data(:)), 'Bordersize', [1 1], 'TrimBorder', 0).*M
Sounds like you are looking to apply a (two dimensional) filter:
M = [1 1 0; 1 0 1; 1 1 1]==0;
F = filter2(ones(3),M);
F(M)=0
The middle line basically does the work (applying the filter) to create the count. The last line ensures that the mines stay at value 0.
I'm trying to copy part of a matrix (matrix 1) in matlab to another empty matrix of zeros (matrix 2) so that the section I copy from matrix 1 has the same indices in matrix 2, e.g.
Matrix 1 (mat1):
0 3 0 0 2 4 1 2 6
1 3 4 2 0 0 0 2 0
0 2 6 1 3 6 6 1 1
0 0 0 2 1 3 3 1 0
1 4 5 2 3 3 0 0 1
Matrix 2 (mat2) desired output:
0 0 0 0 0 0 0 0 0
0 0 4 2 0 0 0 0 0
0 0 6 1 3 6 6 0 0
0 0 0 2 1 3 3 0 0
0 0 0 0 0 0 0 0 0
I've tried something like
mat2([2:4],[3:7]) = mat1([2:4],[3:7])
but of course it doesn't work... any ideas of an efficient way to do this? I couldn't find another thread to help with this problem.
Thanks!
It does work. You just need to create mat2 first:
mat2 = zeros(size(mat1));
mat2(2:4, 3:7) = mat1(2:4, 3:7);
Note that you don't need the square brackets on those ranges.
Do this:
mat2 = zeros(size(mat1));
Before copying over.
My problem is the following: I have a vector as
[3,4,5,6,7]
I want to create a matrix as
3 0 0 0 0
3 4 0 0 0
3 4 5 0 0
3 4 5 6 0
3 4 5 6 7
However, I don't want to use for loops because of the problem of size that I will eventually get.
I was thinking about using flipud, fliprl, hankel and toeplitz functions but cannot find a solution.
Try this:
x = [3,4,5,6,7]
tril(ones(length(x),1)*x)
ans =
3 0 0 0 0
3 4 0 0 0
3 4 5 0 0
3 4 5 6 0
3 4 5 6 7
If A is your vector, you can do
M=repmat(A, length(A), 1) .* tril(ones(length(A),length(A)),0)
I want to calculate the average slope or gradient at each iteration in such a matrix.
a=[ 10 9 8 7 6 5 5;
9 9 8 7 8 5 5;
8 8 7 7 5 5 5;
7 7 7 6 5 5 5;
6 6 6.6 5 5 5 5;
6 6 6.1 5 5 5 5;
6.3 5 5 5 5 5 5]
Where I am wanting to find the slope or gradient between the a(1,1) position during each step and at each point that boarders a value of 5. Each iteration the position of the 5's changes and so do the other values.
After doing so I will then average the slope. I haven't encountered a problem like this yet and I could not find a Matlab command to simplify.
First you must find out which the coast elements are. From your definition, an element is a coast element if it border (from the right) with a 5. If the sea level is 5, and is the lowest possible value i.e. no element goes beyond sea level, then you must first find all the land elements as,
land=a>5;
This returns,
ans =
1 1 1 1 1 0 0
1 1 1 1 1 0 0
1 1 1 1 0 0 0
1 1 1 1 0 0 0
1 1 1 0 0 0 0
1 1 1 0 0 0 0
1 0 0 0 0 0 0
Now, the coast elements are 1s that are followed by a 0. Take the column difference of the land matrix,
coastTmp=diff(land,1,2);
returning,
ans =
0 0 0 0 -1 0
0 0 0 0 -1 0
0 0 0 -1 0 0
0 0 0 -1 0 0
0 0 -1 0 0 0
0 0 -1 0 0 0
-1 0 0 0 0 0
and find the -1s,
coast=find(coastTmp==-1);
which are,
coast =
7
19
20
24
25
29
30
From here it is easy. The gradient is the difference of a(1,1) with all the coast elements, i.e.
slope=a(coast)-a(1,1); % negative slope here
giving,
slope =
-3.700000000000000
-3.400000000000000
-3.900000000000000
-3.000000000000000
-4.000000000000000
-4.000000000000000
-2.000000000000000
and of course the mean is,
mean(slope);