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
Related
let us suppose we have following code
function plot_test(x,y)
x_constucted=[ones(size(x)) x];
b = regress(y,x_constucted);
y_predicted=b(1)+b(2)*x;
scatter(x,y);
hold on
plot(x,y_predicted);
theString = sprintf('y = %.3f*x+%.3f ', b(2), b(1));
text(x(1), y_predicted(1), theString, 'FontSize', 8);
end
output of this equation is the following figure
my question is : how to align equation out of line? for instance on top left size? thanks in advance
If I understand you correctly, you want to move the printed equation out of the dots. Check out the text() function description. The first two values define the x and y position in your plot for the text.
x=1;
y=25;
To move it up, use the new variables in text(x,y,...). Hope that helps.
Some time ago I was looking for a solution for the same exact problem. As you may know, the legend command allows to specify a Location parameter and one of its many options is called best, described in the official Matlab documentation (here) as follows:
Inside axes where least conflict occurs with plot data
My workaround abuses this feature in order to find the best location to place a single text annotation inside the plot. The code below uses a build-in dataset since you didn't specify how your data looks like:
load carsmall;
x = [ones(size(Horsepower)) Horsepower];
y = MPG;
b = regress(y,x);
y_hat = b(1) + b(2) .* Horsepower;
scatter(Horsepower,y);
hold on;
plot(Horsepower,y_hat);
text_at_best(sprintf('y = %.3f*x+%.3f ',b(2),b(1)),'FontSize',12);
function h = text_at_best(txt,varargin)
l = legend(txt,[varargin{:}]);
t = annotation('textbox',varargin{:});
t.String = txt;
t.Position = l.Position;
t.LineStyle = 'None';
delete(l);
if nargout
h = t;
end
end
Here is the final result:
I don't know if this can fit your needs... but developing an algorithm for finding a non overlapping part of the plot in which to place a text looked like an overkill to me. Despite the text being quite far from the prediction line, it's still elegant, clear and comprehensible. The same goes with an even quicker workaround which consists in setting the regression equation as the plot title (blink blink).
folks. For my college project, I need to plot a half-wave rectifier with the sum of two sine waves. Thus, I have chosen MATLAB to use as a tool, but I am having this problem (after the code):
l=[0:10^-6:1/1500];
sig=8*sin(2*pi*100000*l)+6*sin(2*pi*10000*l);
subplot(211)
plot(sig);
for t=1:667
if (8.*sin(2.*pi.*100000.*l)+6.*sin(2.*pi.*10000.*l))<=0
sig(t)=0;
else
sig(t) = 2.*sin((2.*pi.*100000*l + 2.*pi.*10000*l)/2).*cos(2.*pi.*100000*l - 2.*pi.*10000*l);
end
end
The problem showed on the command screen is: "In an assignment A(:) = B, the number of elements in A and B must be the same". How do I solve this issue?
To obtain the rectified signal there are several forms, but the simplest and most compact way is to use the matrices, in this case it is the following:
l=[0:10^-6:1/1500];
sig = 8*sin(2*pi*100000*l)+6*sin(2*pi*10000*l);
sig_rect = sig.*(sig >= 0);
subplot(211)
plot(sig)
subplot(212)
plot(sig_rect)
If you want to use loops you must do the following:
sig_rect = zeros(length(sig));
for t=1:sig
if sig(t) <=0
sig_rect(t) = 0;
else
sig_rect(t) = sig(t);
end
end
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'm new here and this is my first post.. i want to know if there is a way to calculate all the possible convex hulls from 100 random points.. i've created a code that does this right but it gives me an error at convhull and everything beneath that part can't be calculated.. in other words i'd like to know if there is a way i can use function convhull inside a for loop without getting an error..
HERE IS THE CODE
hold on
N=100;
Points = zeros(N,2);
for i=1:N
Points(i,1)= rand(1,1)*100;
Points(i,2)= rand(1,1)*100;
end
SortedX = sortrows(Points,1);
NewVertices = SortedX;
X = reshape(SortedX(:,1),2,[]);
X = X(:);
Y = reshape(SortedX(:,2),2,[]);
Y = Y(:);
plot(X,Y,'*');
while length(NewVertices)>=3
NewVertices = NewVertices(NewVertices~=0);
rowsNV = length(NewVertices(:,1));
NewVertices = reshape(NewVertices,rowsNV/2,2);
XNV = reshape(NewVertices(:,1),rowsNV/2,[]);
XNV = XNV(:);
YNV = reshape(NewVertices(:,2),rowsNV/2,[]);
YNV = YNV(:);
K = convhull(XNV,YNV);
plot(XNV(K),YNV(K),'r-');
for i=1:length(K)-1
NewVertices(K(i),1)=0;
NewVertices(K(i),2)=0;
end
end
HERE IS THE ERROR
Error using convhull
Error computing the convex hull. Not
enough unique points specified.
Error in test2 (line 28)
K = convhull(XNV,YNV);
Thanks in advance
co2ark5
HERE IS THE FINAL CODE THAT WORKS CORRECTLY WITH DAN'S HELP
hold on
N=100;
Points = rand(N,2);
SortedX = sortrows(Points,1);
NewVertices = SortedX;
plot(SortedX(:,1),SortedX(:,2),'*');
while length(NewVertices)>=3
X=NewVertices(:,1);
Y=NewVertices(:,2);
K = convhull(X,Y);
plot(X(K),Y(K),'r-');
for i=1:length(K)-1
NewVertices(K(i),1)=0;
NewVertices(K(i),2)=0;
end
NewVertices = NewVertices(any(NewVertices,2),:);
end
The error is fairly expressive, you aren't specifying enough points (i.e. atleast 3 non colinear points). Please explain what you are trying to achieve with this code.
When do you get this error? I'm assuming not on the first run of the loop? After checking that there are more than 3 points you immediately potentially remove more points and then call convhull so it's quite possible that there are fewer than 3 points (and this is still ignoring the chance of colinearity). After the error, what is size(NewVertices)?
Some side comments: please explain that whole reshaping business before you plot (maybe add some comments). Also Points can be initialized much faster and simpler:
Points = rand(N, 2);
EDIT:
From reading your comments below it is clear to me that the logic of your loop is wrong. I cannot work out exactly how you are trying to remove points to create the new subset on which to find the hull, there are no comments and I don't get what you're doing with the reshape but I'm pretty sure all you need to do to fix this is to move the first line of your loop to the end like this:
while length(NewVertices)>=3
rowsNV = length(NewVertices(:,1));
NewVertices = reshape(NewVertices,rowsNV/2,2);
XNV = reshape(NewVertices(:,1),rowsNV/2,[]);
XNV = XNV(:);
YNV = reshape(NewVertices(:,2),rowsNV/2,[]);
YNV = YNV(:);
K = convhull(XNV,YNV);
plot(XNV(K),YNV(K),'r-');
for i=1:length(K)-1
NewVertices(K(i),1)=0;
NewVertices(K(i),2)=0;
end
NewVertices = NewVertices(NewVertices~=0);
end
This way you find your hull immediately after checking the number of points rather than checking, then removing points, which is why your loop runs on 1 - 2 points.
Thanks a lot DAN,
first i tried to change only the line you said but it didn't work again 'cause of the reshape functions i was using.. Then i decided to change them and i made it more simple and it did work!!
HERE IS THE FINAL CODE THAT WORKS CORRECTLY WITH DAN'S HELP
hold on
N=100;
Points = rand(N,2);
SortedX = sortrows(Points,1);
NewVertices = SortedX;
plot(SortedX(:,1),SortedX(:,2),'*');
while length(NewVertices)>=3
X=NewVertices(:,1);
Y=NewVertices(:,2);
K = convhull(X,Y);
plot(X(K),Y(K),'r-');
for i=1:length(K)-1
NewVertices(K(i),1)=0;
NewVertices(K(i),2)=0;
end
NewVertices = NewVertices(any(NewVertices,2),:);
end
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