I have a couple of arrays:
x = [0:pi/16:2*pi]
y = [0:pi/16:2*pi]
And I want to make a matrix xy in this way:
xY = [(0,0) (0,pi/16) ... (0,2pi);
(pi/16,0) (pi/16,pi/16) ... (pi/16,2pi);
: : :
(2pi,0) (2pi,pi/16) ... (2pi,2pi)]
I've tried many things like this:
for i=1:length(x)
for j=1:length(y)
xy{i,j} = [{x(i),y(j)}];
end
end
, but I have encountered many errors.
I know it should be easy, and the biggest problem is that the title of my post (and because of that, the way I'm looking for help) is wrong, so I apologize for that.
I think I should mention that I'm trying to create a multi-layer perceptron that will get trained with that matrices and this formula:
fxy = cos(x)-3*sin(y);
Thanks in advance!
This is exactly what meshgrid is designed for.
The simplest way is to this is to create matrix of size length(x)-by-length(y)-by-2:
A = zeros(length(x), length(y), 2);
for i = 1 : length(x); for j = 1 : length(y); A(i, j, :) = [x(i), y(j)]; end; end;
In your case, matrix A will have size 33x33x2. To get pair using indexes i, j use the following code:
squeeze(A(i, j, :))
Or you can adjust your code to work with such 3-dimensional matrix.
Related
I have a random column matrix:
r = rand(1,300)';
I want to re-order it so that instead of having elements in the order of 1,2,3,...,300
I will have elements 1,11,21,31,...,291,2,12,22,32,...,292,3,13,33,...293,...,300.
In other words, I want to take every 10th value, beginning with 1 and put them in that order, then do the same for 2 with every 10th value. I know one way to do this is:
n = 10;
r = [r(1:n:numel(r)); r(2:n:numel(r)); r(3:n:numel(r));...;r(10:n:numel(r))]; % Skipped 4-9 in this example
But obviously, this is very cumbersome to do more than a couple of times. Is there something more efficient?
A loop should be easy, but I am not doing it correctly, it seems (I can see why this might not work, but I can't correct it).
(Here is what I tried:)
n = 10;
for i = 1:10
a = [r(i:n:numel(r))];
end
Any suggestions or help is greatly appreciated.
You can do it like this:
r = reshape(reshape(r, 10, 30)', 300, 1)
EDIT:
As pointed out by #LuisMendo on the comments, it's safer to use .' than ' to transpose the matrix, because if the matrix is complex, that could introduce a complex conjugation. Then, it would be safer to do it like this:
r = reshape(reshape(r, 10, 30).', 300, 1)
You could reshape it into 30x10 matrix, transpose, and take the flat index:
A = 1:300;
A = reshape(A,30,10);
A = A';
A = A(:);
Try this -
intv = 10; %%// Interval after which you intend to get the values consecutively
out = r(reshape(reshape(1:numel(r),intv,[])',1,[]))
Some of the other solutions posted are more efficient, but your idea was a good one. It requires a simple fix to work:
N = numel(r);
M = N/10;
a=[];
for ii = 1:M
a= [a r(ii:10:N)];
end
Hope this helps
I have a dimension problem by using mesh-plot.The following example works good but I want to plot mesh(zz,TT,u(:,:,2,1)) instead of mesh(u(:,:,2,1)). At this case, the dimensions do not agree and matlab gives error. How can I order this dimension problem?
clear;
z=linspace(0,10,5);
T=linspace(0,20,50);
for j=1:length(T)-1
for i=1:length(z)
u(i,j,2,1)=z(i)*T(j)+10;
end
end
figure(1)
[zz,TT]=meshgrid(z,T);
mesh(u(:,:,2,1))
The code can be simplified as:
z = linspace(0,10,5);
T = linspace(0,20,50);
[zz,TT] = ndgrid(z, T(1:end-1));
uu = zz.*TT + 10;
%u(:,:,2,1) = uu;
mesh(zz, TT, uu)
I take out one less element from T, because thats how you filled the matrix u. Also note the difference between MESHGRID and NDGRID
Your question is not clear at all. Is this what you're looking for?
z=linspace(0,10,5);
T=linspace(0,20,50);
for j=1:length(T)
for i=1:length(z)
u(i,j)=z(i)*T(j)+10;
end
end
[TT, zz]=meshgrid(T, z);
I am trying to write an algorithm in MatLab which takes as its input a lower triangular matrix. The output should be the inverse of this matrix (which also should be in lower triangular form). I have almost managed to solve this, but one part of my algorithm still leaves me scratching my head. So far I have:
function AI = inverse(A)
n = length(A);
I = eye(n);
AI = zeros(n);
for k = 1:n
AI(k,k) = (I(k,k) - A(k,1:(k-1))*AI(1:(k-1),k))/A(k,k);
for i = k+1:n
AI(i,k) = (I(i,k) - (??????????????))/A(i,i);
end
end
I have marked with question marks the part I am unsure of. I have tried to find a pattern for this part of the code by writing out the procedure on paper, but I just can't seem to find a proper way to solve this part.
If anyone can help me out, I would be very grateful!
Here is my code to get the inverse of a lower triangular matrix by using row transformation:
function AI = inverse(A)
len = length(A);
I = eye(len);
M = [A I];
for row = 1:len
M(row,:) = M(row,:)/M(row,row);
for idx = 1:row-1
M(row,:) = M(row,:) - M(idx,:)*M(row,idx);
end
end
AI = M(:,len+1:end);
end
You can see how it's done on Octave's source. This seems to be implemented in different places depending on the class of the matrix. For Float type Diagonal Matrix it's on liboctave/array/fDiagMatrix.cc, for Complex Diagonal matrix it's on liboctave/array/CDiagMatrix.cc, etc...
One of the advantages of free (as in freedom) software is that you are free to study how things are implemented ;)
Thanks for all the input! I was actually able to find a very nice and easy way to solve this problem today, given that the input is a lower triangular matrix:
function AI = inverse(A)
n = length(A);
I = eye(n);
AI = zeros(n);
for k = 1:n
for i = 1:n
AI(k,i) = (I(k,i) - A(k,1:(k-1))*AI(1:(k-1),i))/A(k,k);
end
end
I want to built a "travelling grid" MATLAB. Actually, I had to choose another MATLAB command instead of linspace to built my grid for any k. Is it possible with a MATLAB command?
for k=1:5
a=0;
b(k)=k.*3;
x=linspace(0,b(k),10);
y=linspace(0,30,10);
for z=1:length(x)
for t=1:length(y)
A(z,t,k)=x(z).*exp(-y(t));
end
end
end
Thanks for any help,
X = linspace(0,3,10);
XX(1,:,:) = bsxfun(#times,X,(1:5)')';
Y = exp(-linspace(0,30,10));
B = bsxfun(#times,Y',XX);
B = permute(B,[2,1,3]);
Your current code is working fine, so I'm not sure what the question is... Here is a slightly simpler implementation:
b = (1:5).*3;
A = zeros(10,10,5);
for k=1:5
[X,Y] = ndgrid(linspace(0,b(k),10), linspace(0,30,10));
A(:,:,k) = X.*exp(-Y);
end
If you also want the y-limits to change as well, the process is similar; you would have two loops and the result A being a 4D matrix
I have been unable to find a function in matlab or octave to do what I want.
I have a matrix m of two columns (x and y values). I know that I can extract the column by doing m(:,1) or m(:,2). I want to split it into smaller matricies of [potentially] equal size and and plot the mean of these matricies. In other words, I want to put the values into bins based on the x values, then find means of the bins. I feel like the hist function should help me, but it doesn't seem to.
Does anyone know of a built-in function to do something like this?
edit
I had intended to mention that I looked at hist and couldn't get it to do what I wanted, but it must have slipped my mind.
Example: Let's say I have the following (I'm trying this in octave, but afaik it works in matlab):
x=1:20;
y=[1:10,10:1];
m=[x, y];
If I want 10 bins, I would like m to be split into:
m1=[1:2, 1:2]
...
m5=[9:10, 9:10]
m6=[10:11, 10:-1:9]
...
m10=[19:20, 2:-1:1]
and then get the mean of each bin.
Update: I have posted a follow-up question here. I would greatly appreciate responses.
I have answered this in video form on my blog:
http://blogs.mathworks.com/videos/2009/01/07/binning-data-in-matlab/
Here is the code:
m = rand(10,2); %Generate data
x = m(:,1); %split into x and y
y = m(:,2);
topEdge = 1; % define limits
botEdge = 0; % define limits
numBins = 2; % define number of bins
binEdges = linspace(botEdge, topEdge, numBins+1);
[h,whichBin] = histc(x, binEdges);
for i = 1:numBins
flagBinMembers = (whichBin == i);
binMembers = y(flagBinMembers);
binMean(i) = mean(binMembers);
end