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);
Related
How to vectorize this code in MATLAB?
n = 3;
x = zeros(n);
y = x;
for i = 1:n
x(:,i) = i;
y(i,:) = i;
end
I am not able to vectorize it. Please help.
You can use meshgrid :
n = 3;
[x,y] = meshgrid(1:n,1:n)
x =
1 2 3
1 2 3
1 2 3
y =
1 1 1
2 2 2
3 3 3
n=3;
[x,y]=meshgrid(1:n);
This uses meshgrid which does this automatically.
Or you can use bsxfun as Divakar suggests:
bsxfun(#plus,1:n,zeros(n,1))
Just as a note on your initial looped code: it's bad practise to use i as a variable
If I can add something to the mix, create a row vector from 1 to n, then use repmat on this vector to create x. After, transpose x to get y:
n = 3;
x = repmat(1:n, n, 1);
y = x.';
Running this code, we get:
>> x
x =
1 2 3
1 2 3
1 2 3
>> y
y =
1 1 1
2 2 2
3 3 3
I have a vector that contains 5 numbers and I want to pad it with zeros. How can I do it?
A = [1 2 3 4 5].';
I want the zero padded vector to be like this:
A_new = [0 0 0 0 0 1 2 3 4 5].';
Also, for another case, I want to assign 1, 3, 4 to matrix W as follows, with all else being zeros. The length of W is 7. W = [0 1 0 0 3 0 4].
You can use following code
newA = [zeros(5,1); A]
About another case. You need something like
inds = [2 5 7];
elems = [1 3 4];
W = zeros(7,1);
W(inds) = elems
Is it possible to automatically add vectors that are not in the same length together for a matrix?
i.e:
a = [1 2 3 4]
b = [1 2]
How can I make C to be:
c = [1 2 3 4 ; 1 2 0 0]
or
c = [1 2 3 4 ; 1 2 NaN NaN]
or something like that
Thanks
This might help
a = [1 2 3 4];
b = [1 2];
c = a;
c(2,1:length(b)) = b;
c =
1 2 3 4
1 2 0 0
then, if you'd rather have NaN than 0, you could do what Dennis Jaheruddin suggests in a comment below.
Make a function like this
function out = cat2(a, b)
diff = length(a) - length(b)
if diff > 0
b = [b, nan(1, diff)];
else
a = [a, nan(1, -diff)];
end
out = [a;b];
end
(but also add a check to handle column vectors too)
cat2([1 2 3 4], [1 2])
ans =
1 2 3 4
1 2 NaN NaN
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've got a matrix A with the dimensions m X n. For every column i (i > 0and i <= n) I want to flip a coin and fill the whole column with 0 values with probability p. How can this be accomplished in MATLAB?
Example:
A = [1 2 3 4; 5 6 7 8] and p = 0.5 could result in
A' = [1 0 3 0; 5 0 7 0]
You can use the function rand() to generate an array of uniformly distributed random numbers, and use logical indexing to select colums where that array is less than p:
A = [1 2 3 4; 5 6 7 8];
p = 0.5;
A(:, rand(size(A,2), 1)<p) = 0
A =
0 2 0 0
0 6 0 0
You can do something like bsxfun(#times, A, rand(1, size(A, 2)) > p). Alex's answer is admittedly better, though.