I'm trying to create a symmetric n by n matrix where the symmetry line is linearly decreasing from n to 1.
For example a 5 by 5 would be:
5 4 3 2 1
4 4 3 2 1
3 3 3 2 1
2 2 2 2 1
1 1 1 1 1
Thanks
Using implicit expansion, the min function will generate a square matrix from the combination of a row and a column vector, so the result can be gained by doing:
N = 5;
A = min( (N:-1:1).', (N:-1:1) );
You may use:
numRows = 5;
mI = repmat((1:numRows)', 1, numRows);
mJ = repmat((1:numRows), numRows, 1);
mA = flip(flip(min(mI, mJ), 1), 2)
With the answer given by:
mA =
5 4 3 2 1
4 4 3 2 1
3 3 3 2 1
2 2 2 2 1
1 1 1 1 1
Related
I have been trying to write a code where the users enters two numbers in order to get 2 columns. It is very hard to explain by words what I am trying to achieve so here is an example:
If the user inputs a = 1 and b = 1, the following table should be created:
ans =
1 1
If the user inputs a = 2 and b = 2:
ans =
1 1
1 2
2 1
2 2
If the user inputs a = 2 and b = 5:
ans =
1 1
1 2
1 3
1 4
1 5
2 1
2 2
2 3
2 4
2 5
For other values of a and b, the matrix should be constructed according to the above shown sequence.
This can be achieved straight-forward by the use of repelem and repmat:
[repelem((1:a).',b),repmat((1:b).',a,1)]
A more elegant way is using meshgrid and reshape it after:
[A,B] = meshgrid(1:a,1:b);
[A(:),B(:)]
Let's create an anonymous function and test the first approach:
>> fun = #(a,b) [repelem((1:a).',b),repmat((1:b).',a,1)];
>> fun(1,1)
ans =
1 1
>> fun(2,2)
ans =
1 1
1 2
2 1
2 2
>> fun(2,5)
ans =
1 1
1 2
1 3
1 4
1 5
2 1
2 2
2 3
2 4
2 5
Here is an alternative way
for example a = 2 and b = 5
A(1:b*a,1) = reshape(mtimes((1:a).',ones(1,b)).',1,b*a)
A(1:b*a,2) = reshape(mtimes((1:b).',ones(1,a)),1,b*a)
A =
1 1
1 2
1 3
1 4
1 5
2 1
2 2
2 3
2 4
2 5
There is just one logic, in the code below you define a matrix of row size and a and column size b
>> mtimes((1:a).',ones(1,b))
ans =
1 1 1 1 1
2 2 2 2 2
and the next step simply reshapes the matrix column wise for a and row wise for b by taking a transpose
A(1:b*a,1) = reshape(mtimes((1:a).',ones(1,b)).',1,b*a)
A(1:b*a,2) = reshape(mtimes((1:b).',ones(1,a)),1,b*a)
As the title says, I want to find all rows in a Matlab matrix that in certain columns the values in the row are equal with the values in the previous row, or in general, equal in some row in the matrix. For example I have a matrix
1 2 3 4
1 2 8 10
4 5 7 9
2 3 6 4
1 2 4 7
and I want to find the following rows:
1 2 3 4
1 2 3 10
1 2 4 7
How do I do something like that and how do I do it generally for all the possible pairs in columns 1 and 2, and have equal values in previous rows, that exist in the matrix?
Here's a start to see if we're headed in the right direction:
>> M = [1 2 3 4;
1 2 8 10;
4 5 7 9;
2 3 6 4;
1 2 4 7];
>> N = M; %// copy M into a new matrix so we can modify it
>> idx = ismember(N(:,1:2), N(1,1:2), 'rows')
idx =
1
1
0
0
1
>> N(idx, :)
ans =
1 2 3 4
1 2 8 10
1 2 4 7
Then you can remove those rows from the original matrix and repeat.
>> N = N(~idx,:)
N =
4 5 7 9
2 3 6 4
this will give you the results
data1 =[1 2 3 4
1 2 8 10
4 5 7 9
2 3 6 4
1 2 4 7];
data2 = [1 2 3 4
1 2 3 10
1 2 4 7];
[exists,position] = ismember(data1,data2, 'rows')
where the exists vector tells you wheter the row is on the other matrix and position gives you the position...
a less elegant and simpler version would be
array_data1 = reshape (data1',[],1);
array_data2 = reshape (data2',[],1);
matchmatrix = zeros(size(data2,1),size(data1,1));
for irow1 = 1: size(data2,1)
for irow2 = 1: size(data1,1)
matchmatrix(irow1,irow2) = min(data2(irow1,:) == data1(irow2,:))~= 0;
end
end
the matchmatrix is to read as a connectivity matrix where value of 1 indicates which row of data1 matches with which row of data2
Supose there is a Matrix
A =
1 3 2 4
4 2 5 8
6 1 4 9
and I have a Vector containing the "class" of each column of this matrix for example
v = [1 , 1 , 2 , 3]
How can I sum the columns of the matrix to a new matrix as column vectors each to the column of their class? In this example columns 1 and 2 of A would added to the first column of the new matrix, column 2 to the 3 to the 2nd, column 4 the the 3rd.
Like
SUM =
4 2 4
6 5 8
7 4 9
Is this possible without loops?
One of the perfect scenarios to combine the powers of accumarray and bsxfun -
%// Since we are to accumulate columns, first step would be to transpose A
At = A.' %//'
%// Create a vector of linear IDs for use with ACCUMARRAY later on
idx = bsxfun(#plus,v(:),[0:size(A,1)-1]*max(v))
%// Use ACCUMARRAY to accumulate rows from At, i.e. columns from A based on the IDs
out = reshape(accumarray(idx(:),At(:)),[],size(A,1)).'
Sample run -
A =
1 3 2 4 6 0
4 2 5 8 9 2
6 1 4 9 8 9
v =
1 1 2 3 3 2
out =
4 2 10
6 7 17
7 13 17
An alternative with accumarray in 2D. Generate a grid with the vector v and then apply accumarray:
A = A.';
v = [1 1 2 3];
[X, Y] = ndgrid(v,1:size(A,2));
Here X and Y look like this:
X =
1 1 1
1 1 1
2 2 2
3 3 3
Y =
1 2 3
1 2 3
1 2 3
1 2 3
Then apply accumarray:
B=accumarray([X(:) Y(:)],A(:)),
SUM = B.'
SUM =
4 2 4
6 5 8
7 4 9
As you see, using [X(:) Y(:)] create the following array:
ans =
1 1
1 1
2 1
3 1
1 2
1 2
2 2
3 2
1 3
1 3
2 3
3 3
in which the vector v containing the "class" is replicated 3 times since there are 3 unique classes that are to be summed up together.
EDIT:
As pointed out by knedlsepp you can get rid of the transpose to A and B like so:
[X2, Y2] = ndgrid(1:size(A,1),v);
B = accumarray([X2(:) Y2(:)],A(:))
which ends up doing the same. I find it a bit more easier to visualize with the transposes but that gives the same result.
How about a one-liner?
result = full(sparse(repmat(v,size(A,1),1), repmat((1:size(A,1)).',1,size(A,2)), A));
Don't optimize prematurely!
The for loop performs fine for your problem:
out = zeros(size(A,1), max(v));
for i = 1:numel(v)
out(:,v(i)) = out(:,v(i)) + A(:,i);
end
BTW: With fine, I mean: fast, fast, fast!
I'd like to transform a vector
v = [1,2,3,4]
into a matrix (4x4) with diagonals like this:
1 2 3 4
2 1 2 3
3 2 1 2
4 3 2 1
You want to generate a Toeplitz matrix, so you should use the toeplitz function.
v = [ 1 2 3 4 ];
out = toeplitz(v)
out =
1 2 3 4
2 1 2 3
3 2 1 2
4 3 2 1
If A is your 4x4 array, just do:
A(1:5:end) = 1:4;
Best
v = [1 2 3 4];
A = gallery('circul',v);
B = triu(A,0)+triu(A,1)';
given matrix A of size 6 by 6 contain blocks of numbers,each block of size 2 by 2, and outher reference matrix R of size 2 by 12 also contain blocks of numbers, each block of size 2 by 2. the perpse of the whole process is to form a new matrix, called the Index matrix, contain index's that refer to the position of the blocks within the matrix A based on the order of the blocks within the reference matrix R. and here is an exemple
matrix A:
A =[1 1 2 2 3 3;
1 1 2 2 3 3;
1 1 3 3 4 4;
1 1 3 3 4 4;
4 4 5 5 6 6;
4 4 5 5 6 6 ]
matrix R:
R=[1 1 2 2 3 3 4 4 5 5 6 6;
1 1 2 2 3 3 4 4 5 5 6 6 ]
the new matrix is:
Index =[1 2 3;
1 3 4;
4 5 6]
any ideas ?
With my favourite three guys - bsxfun, permute, reshape for an efficient and generic solution -
blksz = 2; %// blocksize
num_rowblksA = size(A,1)/blksz; %// number of blocks along rows in A
%// Create blksz x blksz sized blocks for A and B
A1 = reshape(permute(reshape(A,blksz,num_rowblksA,[]),[1 3 2]),blksz^2,[])
R1 = reshape(R,blksz^2,1,[])
%// Find the matches with "bsxfun(#eq" and corresponding indices
[valid,idx] = max(all(bsxfun(#eq,A1,R1),1),[],3)
%// Or with PDIST2:
%// [valid,idx] = max(pdist2(A1.',reshape(R,blksz^2,[]).')==0,[],2)
idx(~valid) = 0
%// Reshape the indices to the shapes of blocked shapes in A
Index = reshape(idx,[],num_rowblksA).'
Sample run with more random inputs -
>> A
A =
2 1 1 2
1 2 2 1
1 1 1 1
2 2 2 2
1 2 2 1
1 2 1 1
>> R
R =
2 1 1 1 1 2 2 2 1 1 1 1
2 1 2 1 1 2 2 1 2 2 2 1
>> Index
Index =
0 0
5 5
3 0