I want to find how many edges are there in C1 and C2? I have stored their adjacency matrix like this :
1 2 3 4 5
1 0 1 1 0 0
2 1 0 0 0 1
3 0 0 0 1 0
4 0 0 0 0 1
5 0 0 0 1 0
If I give an input in an array as [1,3,4] O/P should be : 2
You can use your array to select the appropriate rows and columns from your adjacency matrix and then count the number of non-zeros in the result by using nnz
nnz(A([1 3 4], [1 3 4]))
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.
Let's have a M = [10 x 4 x 12] matrix. As example I take the M(:,:,4):
val(:,:,4) =
0 0 1 0
0 1 1 1
0 0 0 1
1 1 1 1
1 1 0 1
0 1 1 1
1 1 1 1
1 1 1 1
0 0 1 1
0 0 1 1
How can I obtain this:
val(:,:,4) =
0 0 3 0
0 2 2 2
0 0 0 4
1 1 1 1
1 1 0 1
0 2 2 2
1 1 1 1
1 1 1 1
0 0 3 3
0 0 3 3
If I have 1 in the first column then all the subsequent 1's should be 1.
If I have 0 in the first column but 1 in the second, all the subsequent 1's should be 2.
If I have 0 in the first and second column but 1 in the third then all the subsequent 1's should be 3.
If I have 0 in the first 3 columns but 1 in the forth then this one should be four.
Note: The logical matrix M is constructed:
Tab = [reshape(Avg_1step.',10,1,[]) reshape(Avg_2step.',10,1,[]) ...
reshape(Avg_4step.',10,1,[]) reshape(Avg_6step.',10,1,[])];
M = Tab>=repmat([20 40 60 80],10,1,size(Tab,3));
This is a very simple approach that works for both 2D and 3D matrices.
%// Find the column index of the first element in each "slice".
[~, idx] = max(val,[],2);
%// Multiply the column index with each row of the initial matrix
bsxfun(#times, val, idx);
This could be one approach -
%// Concatenate input array along dim3 to create a 2D array for easy work ahead
M2d = reshape(permute(M,[1 3 2]),size(M,1)*size(M,3),[]);
%// Find matches for each case, index into each matching row and
%// elementwise multiply all elements with the corresponding multiplying
%// factor of 2 or 3 or 4 and thus obtain the desired output but as 2D array
%// NOTE: Case 1 would not change any value, so it was skipped.
case2m = all(bsxfun(#eq,M2d(:,1:2),[0 1]),2);
M2d(case2m,:) = bsxfun(#times,M2d(case2m,:),2);
case3m = all(bsxfun(#eq,M2d(:,1:3),[0 0 1]),2);
M2d(case3m,:) = bsxfun(#times,M2d(case3m,:),3);
case4m = all(bsxfun(#eq,M2d(:,1:4),[0 0 0 1]),2);
M2d(case4m,:) = bsxfun(#times,M2d(case4m,:),4);
%// Cut the 2D array thus obtained at every size(a,1) to give us back a 3D
%// array version of the expected values
Mout = permute(reshape(M2d,size(M,1),size(M,3),[]),[1 3 2])
Code run with a random 6 x 4 x 2 sized input array -
M(:,:,1) =
1 1 0 1
1 0 1 1
1 0 0 1
0 0 1 1
1 0 0 0
1 0 1 1
M(:,:,2) =
0 1 0 1
1 1 0 0
1 1 0 0
0 0 1 1
0 0 0 1
0 0 1 0
Mout(:,:,1) =
1 1 0 1
1 0 1 1
1 0 0 1
0 0 3 3
1 0 0 0
1 0 1 1
Mout(:,:,2) =
0 2 0 2
1 1 0 0
1 1 0 0
0 0 3 3
0 0 0 4
0 0 3 0
say i have a matrix
A=zeros(10,3);
and a vector
ll=[1 1 1 2 2 2 3 1 3 2]';
and i want to assign the value in each row corresponding to the value in ll for that row to be 1
i.e output would be
A= 1 0 0
1 0 0
1 0 0
0 1 0
0 1 0
0 1 0
0 0 1
1 0 0
0 0 1
0 1 0
how i do it is using a for loop
for ii=1:length(ll)
A(ii,ll(ii)=1;
end
This should do the trick:
ll=[1 1 1 2 2 2 3 1 3 2]';
A=bsxfun(#eq,ll,1:max(ll))
I'm using bsxfun to check when the entry of ll is equal to an element of the row vector [1 2 3] (in this case). If the entry of ll is 1, it will be equal to the entry in the first column of the [1 2 3] vector and will give a 1 in the first column of A and zeros in the rest of the columns of that row.
Just convert to a linear index:
A((ll-1)*size(A,1) + (1:size(A,1)).') = 1;
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.