I have a 2 dimensional matrix and I want to get the data along a particular line. Similar to what 'Slice' does to a 3D matrix. Is there a a way to do a similar thing on a 2D matrix.
Thanks in advance.
Extracting all values of a column or a line:
>> M = magic(4)
M =
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
>> particular_row = 3;
>> M(particular_row,:)
ans =
9 7 6 12
>> particular_column = 2;
>> M(:,particular_column)
ans =
2
11
7
14
Extracting values along a diagonal:
What if I want to get the data along any direction say along a line joining matrix index (1,1) to (4,4) of a 5x5 matrix?
I'd use linear indexing and the sub2ind function for this task. Demo:
(1,1) to (4,4):
>> M = magic(5)
M =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
>> M(sub2ind(size(M), 1:4, 1:4))
ans =
17 5 13 21
Another example: (1,2) to (3,4):
M(sub2ind(size(M), 1:3, 2:4))
ans =
24 7 20
Related
I have a matrix A in Matlab of dimension Nx(N-1), e.g.
N=5;
A=[1 2 3 4;
5 6 7 8;
9 10 11 12;
13 14 15 16;
17 18 19 20];
I want to rearrange the elements of A in a certain way. Specifically I want to create a matrix B of dimension (N-1)xN such that:
for i=1,...,N,
B(:,i) collects
1) the first i-1 elements of the i-1th column of A and
2) the last N-i elements of the ith column of A.
Notice that for i=1 the i-1th column of A does not exist and therefore 1) is skipped; similarly, for i=N theith column of A does not exist and therefore 2) is skipped.
In the example above
B=[5 1 2 3 4
9 10 6 7 8
13 14 15 11 12
17 18 19 20 16];
This code does what I want. I am asking your help to vectorise it in an efficient way.
B=zeros(N-1,N);
for i=1:N
if i>1 && i<N
step1=A(1:i-1,i-1);
step2=A(i+1:N,i);
B(:,i)=[step1;step2];
elseif i==1
B(:,i)=A(i+1:N,i);
elseif i==N
B(:,i)=A(1:i-1,i-1);
end
end
Extract the lower and upper triangular matrices of A. Then reassemble them with a "diagonal shift":
u = triu(A);
l = tril(A,-1);
B = padarray(u(1:end-1,:),[0 1],'pre') + padarray(l(2:end,:),[0 1],'post');
Another valid approach using logical indexing combined with tril and triu:
B = zeros(size(A'));
B(tril(true(size(B)))) = A(tril(true(size(A)), -1));
B(triu(true(size(B)), 1)) = A(triu(true(size(A))));
Result:
>> B
B =
5 1 2 3 4
9 10 6 7 8
13 14 15 11 12
17 18 19 20 16
The input is an N-by-1 matrix. I need to reshape it to L-by-M matrix. The following is an example.
Input:
b =
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Set length = 18, Output:
X =
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9
8 9 10
9 10 11
10 11 12
11 12 13
12 13 14
13 14 15
14 15 16
15 16 17
16 17 18
17 18 19
18 19 20
Because I have a very big matrix, using a loop to reshape is very inefficient. How can I improve the reshape speed?
Your example output matrix X is the perfect matrix to index a vector of length N to get what you want. It's also very easy to create using bsxfun:
N = 20;
b = rand(N,1);
M = 3; %// number of columns
L = N-M; %// Note that N-M is an upper limit for L!
idx = bsxfun(#plus, (0:L)', 1:M)
X = b(idx)
That's exactly what im2col (from the Image Processing Toolbox) does:
b = (1:20).'; %'// example data
L = 18; % // desired length of sliding blocks
x = im2col(b, [L 1]); % // result
I'd use horzcat. For example:
function X = reshaper(b,len)
diff = length(b) - len + 1;
X = b(1:len);
for i=2:diff
X = horzcat(X,b(i:len+(i-1)));
end
You could probably remove the for loop with some further thought.
I have two 3D matrices A(kl,1,r) and B(1,rs,r). kl=rs.
I need to get a new matrix C(kl,rs,r) which should have the product of column vector of A(kl,1) by the row vector of B(1,rs) for every page r without for loop
C=zeros(size(A,1),size(B,2),r);
for rr=1:size(A,3)
dummy=squeeze(A(:,:,rr))*squeeze(B(:,:,rr))';
C(:,:,rr)=dummy;
end
can anyone help with that? :)
Using bsxfun, you could do that directly in one line
out = bsxfun(#times, A, B);
Sample Inputs:
>> A
A(:,:,1) =
6
10
3
A(:,:,2) =
2
2
1
>> B
B(:,:,1) =
5 5 4
B(:,:,2) =
8 7 8
Results:
out(:,:,1) =
30 30 24
50 50 40
15 15 12
out(:,:,2) =
16 14 16
16 14 16
8 7 8
I want to write a Matlab script.
In my example I have a vector A=[1 3 4 5 7 8 9 10 11 13 14 15 16 17 19 20 21]
Now I want to cut the vector automatically at the points where a number is missing(here the numbers 2, 6, 12, 18 are missing).
As a result I want to have the vectors [1] and [3 4 5] and [7 8 9 10 11] and [13 14 15 16 17] and [19 20 21]. So as you can see the new vectors have different lenghts.
I thought about using a for loop, but I am not sure how to write these new vectors.
Thank you for your help :)
One liner with diff, cumsum & accumarray -
out = accumarray(cumsum([0 ; diff(A(:))~=1])+1,A(:),[],#(x) {x})
Sample run -
>> A
A =
1 3 4 5 7 8 9 10 ...
11 13 14 15 16 17 19 20 21
>> celldisp(out)
out{1} =
1
out{2} =
3
4
5
out{3} =
7
8
9
10
11
out{4} =
13
14
15
16
17
out{5} =
19
20
21
This is one approach:
s = [find(diff(A(:).')>1) numel(A)]; %'// detect where consecutive difference exceeds 1
s = [s(1) diff(s)]; %// sizes of groups
result = mat2cell(A(:).', 1, s); %'// split into cells according to those sizes
In your example, this gives
>> celldisp(result)
result{1} =
1
result{2} =
3 4 5
result{3} =
7 8 9 10 11
result{4} =
13 14 15 16 17
result{5} =
19 20 21
Another approach (computes group sizes differently):
s = diff([0 sum(bsxfun(#lt, A(:), setdiff(1:max(A(:).'), A(:).')), 1) numel(A)]);
result = mat2cell(A(:).', 1, s);
I made a matrix in Matlab, say,
A = magic(5);
A =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Now I found the indices I want using the find function as:
ind = find(A(:,5)>3 & A(:,4)>= 8);
ind =
1
2
3
Now if I want to get a subset of matrix A for those indices using B = A(ind) function, I only get the first column of the matrix:
B = A(ind)
B =
17
23
4
How can I get all the columns as subset??
Oops ... I got it
B = A(ind,:);