For given vector with values from y0 to yT I have to create following matrix:
1 y(P-1) y(P-2) ... y(0)
1 y(P) y(P-1) ... y(1)
1 y(P+1) y(P) ... y(2)
. . . ... .
. . . ... .
. . . ... .
1 y(T-1) y(T-2) ... y(T-p)
P means y(P) is a linear combination of P previous y's.
P is given (for example 100).
Is there any matlab function for this?
function to get the right y for given i,j is
getAIJ = #(i, j) y(p+(-(i+1))+(j-1));
So I just need to apply this function for every cell in matrix except the first column. Is there any matlab-way to do this? I am not sure how to use arrayfun on this.
Use buffer in signal processing toolbox. In this case
yy = buffer(y, P, P-1)';
Related
I am working on a Matlab project and I want to make the gradient of the following function in Matlab:
f(x) = c^T * x - sum (log(bi - (ai ^ T) * x)).
Where ai^T are the rows of a random A matrix nxm , where n=2, and m=20
c is random matrix nx1, and x is also random nx1.
b is random matrix mx1.
I've done the following but the results i get don't seem to be right..
function gc0 = gc(x, c, b, A)
for k = 1 : length(A(:,1))
f1(k) = sum(log(b - A(k,:)'*x(k)));
end
gradient(-f1)
gc0 = c - gradient(f1)';
Any ideas? I'd appreciate your help, I'm newbie in Matlab..
It seems that your loop contains a mistake. Looking at the formula above,
I think that the function evaluation should be
f1 = c'*x;
for k = 1 : length(A(1,:))
f1 = f1 - log(b(k) - A(:,k)'*x)
end
A shorter and faster notation for this in Matlab is
f = c'*x - sum(log(b - A' * x)) ;
The function 'gradient' does not calculate the gradient that I think you
want: it returns the differences of matrix entries, and your function f
is a scalar.
Instead, I suggest calculating the derivatives symbolically:
Gradf = c' + sum( A'./(b - A' * x) );
I have just started learning MATLAB .
Please find my codes below
m= ['A','B','C'];
cs=size(m,2);
for i=1:cs
for j=1:cs
if i~=j
s1=(m(i));s2=',';s3=(m(j));
s=strcat(s1,s2,s3);
disp(s);
end
end
end
It produces the following output on command window.
A,B
A,C
B,A
B,C
C,A
C,B
But , i want to wrap up all the outputs into a single matrix (or Cell Array ) , Lets say new_M .
So that the values of new_M shall contain the all above values like this .
new_M (6,1) =
[ A,B
A,C
B,A
B,C
C,A
C,B ]
Your help will be highly appreciatated . Thanks in advance.
The idiomatic way to do this would be to use nchoosek to get the indices you want and then use linear indexing:
m = ['A','B','C'] %// For a char array OR
m = {'A','B','C'} %// For a cell array
I = nchoosek(1:numel(m), 2)
new_M = m([I; I(:,end:-1:1)])
This will work. In 'c' you will find the values
m= ['A','B','C'];
cs=size(m,2);
c = cell(6,1)
t = 1;
for i=1:cs
for j=1:cs
if i~=j
s1=(m(i));s2=',';s3=(m(j));
s=strcat(s1,s2,s3);
disp(s)
c{t} = s;
t=t+1;
end
end
end
m= ['A','B','C'];
cs=size(m,2);
new_M = [];
for i=1:cs
for j=1:cs
if i~=j
s1=(m(i));s2=',';s3=(m(j));
s=strcat(s1,s2,s3);
new_M = [new_M;s];
end
end
end
The new_M matrix will contain all the values you need.
I'm trying to create a matrix such that if I define a random number between 0 and 1 and a random location in the matrix, I want all the values around that to "diffuse" out. Here's sort of an example:
0.214 0.432 0.531 0.631 0.593 0.642
0.389 0.467 0.587 0.723 0.654 0.689
0.421 0.523 0.743 0.812 0.765 0.754
0.543 0.612 0.732 0.843 0.889 0.743
0.322 0.543 0.661 0.732 0.643 0.694
0.221 0.321 0.492 0.643 0.521 0.598
if you notice, there's a peak at (4,5) = 0.889 and all the other numbers decrease as they move away from that peak.
I can't figure out a nice way to generate a code that does this. Any thoughts? I need to be able to generate this type of matrix with random peaks and a random rate of decrease...
Without knowing what other constraints you want to implement:
Come up with a function z = f(x,y) whose peak value is at (x0,y0) == (0,0) and whose values range between [0,1]. As an example, the PDF for the Normal distribution with mu = 0 and sigma = 1/sqrt(2*pi) has a peak at x == 0 of 1.0, and whose lower bound is zero. Similarly, a bivariate normal PDF with mu = {0,0} and determinate(sigma) == [1/(2*pi)]^2 will have similar characteristics.
Any mathematical function may have its domain shifted: f(x-x0, y-y0)
Your code will look something like this:
someFunction = #(x,y) theFunctionYouPicked(x,y);
[x0,y0,peak] = %{ you supply these values %};
myFunction = #(x,y) peak * someFunction(x - x0, y - y0);
[dimX,dimY] = %{ you supply these values %};
mymatrix = bsxfun( myFunction, 0:dimX, (0:dimY)' );
You can read more about bsxfun here; however, here's an example of how it works:
bsxfun( blah, [a b c], [d e f]' )
That should give the following matrix (or its transpose ... I don't have matlab in front of me):
[blah(a,d) blah(a,e) blah(a,f);
blah(b,d) blah(b,e) blah(b,f);
blah(c,d) blah(c,e) blah(c,f)]
Get a toy example working, then you can tinker with it to be more flexible. If the function dictating how it decreases is random (with the constraint that points closer to (x0,y0) are larger than more distant points), it won't be an issue to make a procedural function instead of using strictly mathematical ones.
In response to your answer:
Your equation could be thought of as a model for gravity where an object instantaneously induces a force on another mass, then stops exerting force. Following that logic, it could be modified to a naive vector formulation like this:
% v1 & v2 are vectors that point from the two peak points to the point [ii,jj]
theMatrix(ii,jj) = norm( (r1 / norm( v1 )) * v1 / norm( v1 ) ...
+ (r2 / norm( v2 )) * v2 / norm( v2 ) ...
);
The most extreme type of corner case you'll run into is one where v1 & v2 point in the same direction as in the following row:
[ . . A X1 X2 . . ]
... where you want a value for A w/respect to X1 & X2. Using the above expression it'll boil down to A = X1 / norm(v1) + X2 / norm(v2), which will definitely exceed the peak value at X1 because norm(v1) == 1. You could certainly do some dirty stuff to Band-Aid it, but personally I'd start looking for a different function.
Along those lines, if you used Newton's Law of Universal Gravitation with a few modifications:
You wouldn't need an analogue for G, so you could just assume G == 1
Treat each of the points in the matrix as having mass m2 == 1, so the equation reduces to: F_12 == -1 * (m1 / r^2) * RHAT_12
Sum the "force" vectors and calculate the norm to get each value
... you'll still run into the same problem. The corner case I laid out above would boil down to A = X1/norm(v1)^2 + X2/norm(v2)^2 == X1 + X2/4. Since it's inversely proportional to the square of the distances, it'd be easier to Band-Aid than the linear one, but I wouldn't recommend it.
Similarly, if you use polynomials it won't scale well; you can design one that won't ever exceed your chosen peaks, but there wouldn't be a lower bound.
You could use the logistic function to help with this:
1 / (1 + E^(-c*x))
Here's an example of using the logistic function on a degree 4 polynomial with peaks at points 2 & 4; you'll note I gave the polynomial a scaling factor to pull the polynomial down to relatively small values so calculated values aren't so close together.
I ended up creating a code that wraps the way I want based on a dimension, which I provide. Here's the code:
dims = 100;
A = zeros(dims);
b = floor(1+dims*rand(1));
c = floor(1+dims*rand(1));
d = rand(1);
x1 = c;
y1 = b;
A(x1,y1) = d;
for i = 1:dims
for j = i
k = 1-j;
while k <= j
if x1-j>0 && y1+k>0 && y1+k <= dims
if A(x1-j,y1+k) == 0
A(x1-j,y1+k) = eqn(d,x1-j,y1+k,x1,y1);
end
end
k = k+1;
end
end
for k = i
j = 1-k;
while j<=k
if x1+j>0 && y1+k>0 && y1+k <= dims && x1+j <= dims
if A(x1+j,y1+k)==0
A(x1+j, y1+k) = eqn(d,x1+j,y1+k,x1,y1);
end
end
j = j+1;
end
end
for j = i
k = 1-j;
while k<=j
if x1+j>0 && y1-k>0 && x1+j <= dims && y1-k<= dims
if A(x1+j,y1-k) == 0
A(x1+j,y1-k) = eqn(d,x1+j,y1-k,x1,y1);
end
end
k=k+1;
end
end
for k = i
j = 1-k;
while j<=k
if x1-j>0 && y1-k>0 && x1-j <= dims && y1-k<= dims
if A(x1-j,y1-k)==0
A(x1-j,y1-k) = eqn(d,x1-j,y1-k,x1,y1);
end
end
j = j+1;
end
end
end
colormap('hot');
imagesc(A);
colorbar;
If you notice, the code calls a function (I called it eqn), which provided the information for how to changes the values in each cell. The function that I settled on is d/distance (distance being computed using the standard distance formula).
It seems to work pretty well. I'm now just trying to develop a good way to have multiple peaks in the same square without one peak completely overwriting the other.
I need feval called with a bivariate function f and two vectors v1 and v2 (let us say that v1 are the x's and v2 the y's) to return a vector z=[f(v1(1),v2(1)) ... f(v1(n),v2(n)].
How can I do that ?
The following example does not work:
function z = f(x,y)
if x>0.5
z=x+y;
else
z=x+2*y;
end
end
indeed,
feval(f,[0.4 2 3],[4 5 6])
returns: [8.4 12 15]
instead of [8.4 7 9].
What is the correct syntax ?
The if condition doesn't work element-wise, as you seem to expect. Rather, the if branch is entered only if all the elements in the expression (x>0.5 in your case) evaluate to true.
To achieve what you want, change your function to
function z = f(x,y)
ind = x>.5;
z(ind) = x(ind) + y(ind);
z(~ind) = x(~ind) + 2*y(~ind);
end
Note that the logical index ind takes the place of your if.
For your particular function, the code could be simplified to
function z = f(x,y)
z = x + y + (x<=.5).*y;
end
I have a 30-vector, x where each element of x follows a standardised normal distribution.
So in Matlab,
I have:
for i=1:30;
x(i)=randn;
end;
Now I want to create 30*30=900 elements from vector, x to make a 900-vector, C defined as follows:
I am unable to do the loop for two variables (k and l) properly. I have:
for k=1:30,l=1:30;
C(k,l)=(1/30)*symsum((x(i))*(x(i-abs(k-l))),1,30+abs(k-l));
end
It says '??? Undefined function or method 'symsum' for input arguments of type
'double'.'
I hope to gain from this a 900-vector, C which I will then rewrite as a matrix. The reason I have using two indices k and l instead of one is because I eventually want these indices to denote the (k,l)-entry of such a matrix so it is important that that my 900-vector will be in the form of C = [ row 1 row 2 row 3 ... row 30 ] so I can use the reshape tool i.e.
C'=reshape(C,30,30)
Could anyone help me with the code for the summation and getting such a 900 vector.
Let's try to make this a bit efficient.
n = 30;
x = randn(n,1);
%# preassign C for speed
C = zeros(n);
%# fill only one half of C, since it's symmetric
for k = 2:n
for l = 1:k-1
%# shift the x-vector by |k-l| and sum it up
delta = k-l; %# k is always larger than l
C(k,l) = sum( x(1:end-delta).*x(1+delta:end) );
end
end
%# fill in the other half of C
C = C + C';
%# add the diagonal (where delta is 0, and thus each
%# element of x is multiplied with itself
C(1:n+1:end) = sum(x.^2);
It seems to me that you want a matrix C of 30x30 elements.
Given the formula that you provided I would do
x = randn(1,30)
C = zeros(30,30)
for k=1:30
for l=1:30
v = abs(k-l);
for i =1:30-v
C(k,l) = C(k,l) + x(i)*x(i+v);
end
end
end
if you actually need the vector you can obtain it from the matrix.