I am trying to check if in a square matrix there is more than one true value in all possible diagonals and anti-diagonals, and return true, otherwise false.
So far I have tried as following but is not covering all possible diagonals:
n=8; %matrix dimension 8 x 8
diag= sum(A(1:n+1:end));
d1=diag>=2;
antiDiag=sum(A(n:n-1:end));
d2=antiDiag>=2;
if ~any(d1(:)) || ~any(d2(:))
res= true;
else
res=false;
end
this is a false:
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 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 1
this is a true:
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 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
Since these are my first steps in using Matlab, is there a specific function or a better way to achieve the result I am looking for?
To detect if there are more than one nonzero value in any diagonal or anti-diagonal (not just the main diagonal and antidiagonal): get the row and column indices of nonzero values, ii and jj; and then check if any value of ii-jj (diagonals) or ii+jj (anti-diagonals) is repeated:
[ii, jj] = find(A);
res = (numel(unique(ii-jj)) < numel(ii)) || (numel(unique(ii+jj)) < numel(ii));
One approach:
n=8; %// size of square matrix
A = logical(randi(2,n)-1); %// Create a logical matrix of 0s and 1s
d1 = sum(A(1:n+1:end)); %// sum all the values of Main diagonal
d2 = sum(A(n:n-1:end-1)); %// sum all the values of Main anti-diag
result = d1>=2 | d2>=2 %// result is true when any one of them is > than or = to 2
Sample run:
Inputs:
>> A
A =
0 1 1 1 1 0 1 0
0 1 1 1 1 1 0 0
0 1 0 1 1 0 0 1
0 1 1 0 1 1 0 0
0 1 0 1 1 0 0 1
1 0 0 0 1 1 0 1
1 1 1 1 1 1 0 0
1 1 1 1 0 0 0 1
Output:
result =
1
Note: This approach considers only the Main diag and Main Anti-Diag (considering the example you provided). If you want for all possible diags, the other answer from Luis Mendo is the way to go
Using #Santhan Salai's generating technique, we can use the diag function (to pull out the main diagonal of the matrix), the fliplr to flip over the center column and any to reduced to a single value.
n=8; %// size of square matrix
A = logical(randi(2,n)-1); %// Create a logical matrix of 0s and 1s
any([diag(A) ; diag(fliplr(A))])
Related
I created an array A by first using the command A = [1:10]'.
Then I created a 10x10 matrix, only containing 0's. I then overwrote this matrix with my A, resulting in this new matrix:
A =
1 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0 0 0
4 0 0 0 0 0 0 0 0 0
5 0 0 0 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0
7 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0 0
10 0 0 0 0 0 0 0 0 0
Now the problem is, when I run the command sum((A(1,1)):(A(1,end))), I keep getting 0 when I should be getting 1, as it is the sum of the first row. I tried running the same command on other matrices and they give me the correct answer, so why isn't it working here?
The term
(A(1,1)):(A(1,end))
creates and empty array, as A(1,1) = 1 and A(1,end) = 0 which makes it impossible for colon : to create a vector, so the sum over it is zero. But its not what you want anyway, I guess.
What you supposedly want is
sum(A(1,:))
or in respect to whole matrix, by specifying the dimension of the sum, e.g.
sum(A,2)
ans =
1
2
3
4
5
6
7
8
9
10
Edit
If you want to start from a different column index you can do the following:
sum( A(rowIndex,firstColumnIndex:lastColumnIndex) )
while end can be used as macro variable for the last index of the corresponding column or row.
Say we have a vector containing zeros interspersed blocks of ones of varying length, such as:
0 0 0 1 1 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0 0 1 1 1 0 0 0
I would like to transform this into a 2D array as follows. Each row contains zeros only and one of the blocks. I.e. the number of rows of the 2D array would be the number of blocks at the end. The above array would transform to:
0 0 0 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 0 0 0 0 0 0 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 0 0 0 0 0 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 0 0 0 0 0 0 1 1 1 0 0 0
I.e the first block ends up in the first row, the second block in the second row etc.
Question
This exercise is rather trivial using loops. My question is if there is a neat way using MATLAB matrix operations, MATLAB functions and array indexing to do this?
Off the top of my head you could use bwlabel (from the Image Processing Toolbox) to assign each cluster of 1's a unique value. You could then use bsxfun to check equality between the labeled version and the unique labels which will automatically expand it out into a matrix.
a = [0 0 0 1 1 0 0 0 0 1 1 1 0 0 0 1 1 0 0 0 0 1 1 1 0 0 0];
b = bwlabel(a);
out = bsxfun(#eq, (1:max(b))', b);
Without the image processing toolbox you could do effectively the same thing with:
C = cumsum(~a);
out = bsxfun(#eq, unique(C(logical(a))).', C);
I tried this one
N = 100; % set array size
A = randi(2,N,1)-1; % generate random array filled with 0 and 1
d = diff([0;A]); % if 1 : starting point of block
% if -1 : end point of block
ir = find(A); % Find A==1 which will be row index
ic = cumsum(d(ir)>0); % Set col index
% assemble array
% if you want output as full array
A_new = accumarray([ir,ic],ones(size(ir)),[length(A),ic(end)]);
% if you want output as sparse array
A_new = sparse(ir,ic,ones(size(ir)),length(A),ic(end),length(ir));
% display routine
figure;spy(A,'r');hold on;spy([zeros(size(A)),A_new]);
Turns out it is faster than #Suever 's solution (Compared tic toc time with size 10000, 1000 trial). Also, if you use sparse instead of accumarray, then it is much faster
Suever_time = 7~8 sec
Accumarray = 2~3 sec
Sparse = 0.2~0.3 sec
However, his one is much shorter and neat!
I have a matrix that I would like to split into two separate matrices based on a set of conditions.
The input matrix can be generated with the following code:
lbits = 8;
ntags = 10;
k = randi(lbits,1,ntags);
Tag = zeros(lbits,ntags);
Tag(lbits*(find(k)-1) + k)=1;
TagAnswer = Tag';
Which returns:
TagAnswer =
0 0 0 1 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 1 0 0 0 0 0 0
0 0 0 0 0 1 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 1
0 0 0 0 0 0 1 0
My conditions are:
If place of bit '1' is on position lbits/2 or higher, add the row to matrix A
If place of bit '1' is less then position lbits/2, add the row to matrix B
With the above TagAnswer I want the 2nd, 5th and 7th rows to be moved into B and the remaining rows moved into matrix A
Assuming my edit is correct, you can use the row and column outputs of find to index TagAnswer and pull the rows based on your conditions:
% Generate sample data
lbits = 8;
ntags = 10;
k = randi(lbits,1,ntags);
Tag = zeros(lbits,ntags);
Tag(lbits*(find(k)-1) + k)= 1;
TagAnswer = Tag';
% Find bit locations and distribute rows accordingly
[r, c] = find(TagAnswer);
A = TagAnswer(r(c>=(lbits/2)), :);
B = TagAnswer(r(c<(lbits/2)), :);
For my test case I have:
TagAnswer =
0 0 0 0 0 0 0 1
0 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 0 1 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 1 0 0
0 0 0 0 0 1 0 0
0 0 1 0 0 0 0 0
A =
0 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0
0 0 0 0 0 1 0 0
0 0 0 0 0 1 0 0
0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 1
B =
0 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 1 0 0 0 0 0
Edit: Because MATLAB stores data column-major, find also works column-major and will likely lose the row ordering. If it important to preserve the row ordering of TagAnswer in A and B you can use sort after the find call:
[r, sortidx] = sort(r);
c = c(sortidx);
I am working with iterative methods, and thus with large sparse matrices.
For instance, I want to set up a matrix like this:
1 1 0 0 1 0 0 0 0 0
1 1 1 0 0 1 0 0 0 0
0 1 1 1 0 0 1 0 0 0
0 0 1 1 1 0 0 1 0 0
1 0 0 1 1 1 0 0 1 0
0 1 0 0 1 1 1 0 0 1
So that only certain diagonals are non-zero. In my programming, I will be working with much larger matrix sizes, but Idea is the same: Only a few diagonals are non-zero, all other entries are zeros.
I know, how to do it in for loop, but it seems to be not effective, if the matrix size is large. Also I work with symmetric matrices.
I would appreciate, if you provide me a code for my sample matrix along with description.
You want spdiags:
m = 6; %// number of rows
n = 10; %// number of columns
diags = [-4 -1 0 1 4]; %// diagonals to be filled
A = spdiags(ones(min(m,n), numel(diags)), diags, m, n);
This gives:
>> full(A)
ans =
1 1 0 0 1 0 0 0 0 0
1 1 1 0 0 1 0 0 0 0
0 1 1 1 0 0 1 0 0 0
0 0 1 1 1 0 0 1 0 0
1 0 0 1 1 1 0 0 1 0
0 1 0 0 1 1 1 0 0 1
I have large array. Now I need a matrix with 8 elements in every row. My array looks like this:
A= Columns 1 through 18
0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 1 0 0
Columns 19 through 36
0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0
and so on. How I can get [nx8] matrix? For example:
B=[0 0 0 0 0 0 0 0
0 0 1 1 0 1 1 1
0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0]
I've tried reshape, but it didn't work correctly. I get one 1 where shouldn't be.
B=reshape(A,[],8)
You almost have it. The problem is that Matlab fills the matrix column-wise, whereas you seem to want it filled row-rise. So create an 8-row matrix and then transpose:
reshape(A,8,[]).'
What about vec2mat
vec2mat
vec2mat(A,8)