Is there a matlab function which allows me to do the following operation?
x = [1 2 2 3];
and then based on x I want to build the matrix m = [1 2 2 3; 1 2 2 3; 1 2 2 3; 1 2 2 3]
You are looking for the REPMAT function:
x = [1 2 2 3];
m = repmat(x,4,1);
You can also use indexing to repeat the rows:
m = x(ones(4,1),:);
or even outer-product:
m = ones(4,1)*x;
and also using BSXFUN:
m = bsxfun(#times, x, ones(4,1))
You could try using vertcat, like this:
x = [1 2 2 3];
m = vertcat(x,x,x,x);
Or even simply:
x = [1 2 2 3];
m = [x;x;x;x];
EDIT:
for multiples of x, you can do:
x = [1 2 2 3];
m = [x;2*x;3*x]; % [1 2 2 3; 2 4 4 6; 3 6 6 9]
EDIT2:
For an arbitrary number of x's in m...
n = 3; % number of repetitions...
x = [1 2 2 3];
m = [];
for i=1:n
m = [m;x];
end
Related
Is there a matlab function which allows me to do the following operation?
x = [1 2 2 3];
and then based on x I want to build the matrix m = [1 2 2 3; 1 2 2 3; 1 2 2 3; 1 2 2 3]
You are looking for the REPMAT function:
x = [1 2 2 3];
m = repmat(x,4,1);
You can also use indexing to repeat the rows:
m = x(ones(4,1),:);
or even outer-product:
m = ones(4,1)*x;
and also using BSXFUN:
m = bsxfun(#times, x, ones(4,1))
You could try using vertcat, like this:
x = [1 2 2 3];
m = vertcat(x,x,x,x);
Or even simply:
x = [1 2 2 3];
m = [x;x;x;x];
EDIT:
for multiples of x, you can do:
x = [1 2 2 3];
m = [x;2*x;3*x]; % [1 2 2 3; 2 4 4 6; 3 6 6 9]
EDIT2:
For an arbitrary number of x's in m...
n = 3; % number of repetitions...
x = [1 2 2 3];
m = [];
for i=1:n
m = [m;x];
end
Given this vector
a = [1 2 3 4]
I want to create a matrix like this
b = [1 0 0 0;
2 1 0 0;
3 2 1 0;
4 3 2 1;
0 4 3 2;
0 0 4 3;
0 0 0 4]
in a vectorized way not using loops.
Hint: use conv2 (hover mouse to see code):
a = [1 2 3 4];
b = conv2(a(:), eye(numel(a)));
Or, in a similar mood, you can use convmtx (from the Signal Processing Toolbox):
a = [1 2 3 4];
b = convmtx(a(:), numel(a));
One way to do it:
a = [1 2 3 4]
n = numel(a);
%// create circulant matrix from input vector
b = gallery('circul',[a zeros(1,n-1)]).' %'
%// crop the result
c = b(:,1:n)
Another way:
b = union( tril(toeplitz(a)), triu(toeplitz(fliplr(a))),'rows','stable')
or its slightly variation
b = union( toeplitz(a,a.*0),toeplitz(fliplr(a),a.*0).','rows','stable')
and probably even faster:
b = [ toeplitz(a,a.*0) ; toeplitz(fliplr(a),a.*0).' ]
b(numel(a),:) = []
With bsxfun -
na = numel(a)
b = zeros(2*na-1,na)
b(bsxfun(#plus,[1:na]',[0:na-1]*2*na)) = repmat(a(:),1,na)
If you are looking for a faster pre-allocation, you can do -
b(2*na-1,na) = 0;.
Another bsxfun -
a=[1 2 3 4];
m=numel(a);
b=[a,zeros(1,m-1)].';
Q=bsxfun(#circshift, b, [0:m-1])
A = [1 3 5 8]
B = [1 2 3 4 5 6 7 8]
I would like to create a vector C which returns the rownumber of the element in vector A with the smallest non-negative difference to each element in vector B.
So, given the example above, it should return:
C = [1 2 2 3 3 4 4 4]
I'm sure there are many ways to do this. Here's one:
A = [1 3 5 8]
B = [1 2 3 4 5 6 7 8]
%create matrices of the values to subtract
[a,b] = meshgrid(A,B);
%subtract
aLessB = a-b;
%make sure we don't use the negative values
aLessB(aLessB < 0) = Inf;
%sort the subtracted matrix
[dum, idx] = sort(aLessB, 2, 'ascend');
idx(:,1) is the solution you are looking for.
An alternative solution:
D = bsxfun(#minus, A', B);
D(D < 0) = Inf;
[~, C] = min(D, [], 1);
I have a matrix A of size nRows x nCols.
I have a nx2 matrix B which contains indices of the matrix A.
I want to get the values of A at the indices given in B.
lets say,
B = [1, 2;
2, 3;
3, 4]
A(1,2) = 1
A(2,3) = 2
A(3,4) = 1
I want to know any Matlab command which gives the following, given A and B (I don't want to use loops):
[1 2 1]
I guess this is what you are looking for:
A(sub2ind(size(A),B(:,1),B(:,2)))
This is what you want:
A = [1,2; 3, 4; 5, 6; 7,8; 9,0]; % this is your N by 2 matrix
B = [1,1; 1,2; 2,1; 3, 1; 4,2]; % these are your indexes
A(sub2ind(size(A), B(:,1), B(:,2)))
A =
1 2
3 4
5 6
7 8
9 0
B =
1 1
1 2
2 1
3 1
4 2
ans =
1
2
3
5
8
I have a matrix 'x' and a row vector 'v'; the number of elements in the row vector is the same as the number of columns in the matrix. Is there any predefined function for doing the following operation?
for c = 1 : columns(x)
for r = 1 : rows(x)
x(r, c) -= v(c);
end
end
bsxfun(#minus,x,v)
Here's an octave demonstration:
octave> x = [1 2 3;2 3 4]
x =
1 2 3
2 3 4
octave> v = [2 0 1]
v =
2 0 1
octave>
octave> z=bsxfun(#minus,x,v)
z =
-1 2 2
0 3 3
If you are using Octave 3.6.0 or later, you don't have to use bsxfun since Octave performs automatic broadcasting (note that this is the same as actually using bsxfun, just easier on the eye). For example:
octave> x = [1 2 3; 2 3 4]
x =
1 2 3
2 3 4
octave> v = [2 0 1]
v =
2 0 1
octave> z = x - v
z =
-1 2 2
0 3 3
Alternatively, you can replicate your vector and directly subtract it from the matrix
z = x-repmat(v, size(x, 1), 1);