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
Related
if i have a matrix, say:
A = [ 0 2 4 0
2 0 5 0
4 5 0 3
0 0 3 0 ]
and i want to find the maximum value in the matrix i can type:
max(max(A))
or
max(A(:))
if i only want to find the maximum of rows 1 and 2 and columns 3 and 4 i can do this:
a = [1 2]
b = [3 4]
max(max(A(a,b))
but what if i want to find the indices of the rows and columns that correspond to that value?
according to the matlab documentation, if i am using the whole matrix i can use the ind2sub function:
[val,idx] = max(A(:))
[row,col] = ind2sub(size(A),idx)
but how can i get that working for my example where i am using vectors a and b to determine the rows and columns it finds the values over?
here is the only way i have been able to work it out so far:
max_val = 0;
max_idx = [1 1];
for ii = a
[val,idx] = max(A(ii,b))
if val > max_val
max_val = val
max_idx = [ii idx]
but that seems rather clunky to me.. any ideas?
Assuming that the submatrix A(a,b) is contiguous (like in your example):
A = [ 0 2 4 0
2 0 5 0
4 5 0 3
0 0 3 0 ]
a = [1 2]; b = [3 4];
B = A(a,b)
[val,idx] = max(B(:));
[row,col] = ind2sub(size(B),idx);
maxrow = row + a(1) - 1;
maxcol = col + b(1) - 1;
You are finding the relative index in the submatrix B. Which is equivalent to the additional rows and columns from the upper left corner of the submatrix.
Now assuming that a and b result in a set of rows and columns that are NOT a contiguous submatrix, e.g. a = [1 3], b = [3 4], the result is very similar. "row" and "col" are the index in the a and b vectors:
A = [ 0 2 4 0
2 0 5 0
4 5 0 3
0 0 3 0 ]
a = [1 3]; b = [3 4];
B = A(a,b)
[val,idx] = max(B(:));
[row,col] = ind2sub(size(B),idx);
maxrow = a(row);
maxcol = b(col);
Now you're working in the index of indexes.
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])
Suppose I have a matrix A, and I'd like to get the matrix [A 0; 0 1]. Is there a built function to do this?
So if my matrix is [2 3; 1 4], I'd get back [2 3 0; 1 4 0; 0 0 1]
The easiest way is:
newA = A;
newA(end+1,end+1) = 1;
This works because you can index outside an array for assignments, because end indicates the last element (here in row and column), and because Matlab pads with zeros when you grow an array. If you just want to grow A, you can even skip the creation of newA, of course.
I always use matrix concatenation for problems like this
So for your example:
A = [2 3; 1 4]
A = [A A(:,1)*0; A(1,:)*0 1]
produces
A =
2 3 0
1 4 0
0 0 1
The nice thing about this trick is that its very flexible and you can do all sorts of tranformations
very easily. For example
A = [2 3; 1 4]
A = [1 A(1,:)*0; A(:,1)*0 A]
produces
A =
1 0 0
0 2 3
0 1 4
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);
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.