I want to plot just the first 50 points of the following function in MATLAB and I'm having difficulty finding out how to do this, I genuinely cannot find the command or much that is similar online. Can anyone help me out with this? Thank you in advance for all the help, anything is appreciated.
>> p=pi;
>> x=[0:0.1:4*p];
>> y = exp(-0.4*x).*sin(x);
>> plot(x,y), plot(x(1:50)) %%%this is one of my attempts
>> title ('MATLAB PRACTICE');
>> xlabel('x-Axis');
>> ylabel('y-Axis');
>> grid on
You are close, but it is instructive to break the problem into steps.
Think of it this way: you want to first select the first 50 values of x and y, and then plot those.
You can do that like so:
xSubset = x(1:50);
ySubset = y(1:50);
plot(xSubset, ySubset);
The syntax x(1:50) can be used to select a portion of x (or any vector).
We can accomplish the same thing without creating the temporary vectors, which may be desirable because it is more concise (and less typing):
plot(x(1:50), y(1:50));
Related
I am currently running improfile on MATLAB with a line going through the center of the picture below:
After doing so, I'm plotting the resultant using the code below:
xi = [1 size(d_Img,2) size(d_Img,2) 1];
yi = [ceil(size(d_Img,1)/2), ceil(size(d_Img,1)/2), ceil(size(d_Img,1)/2 ),ceil(size(d_Img,1)/2)];
c_d = improfile(d_Img,xi,yi);
c_c = improfile(c_Img,xi,yi);
c_d = c_d';
c_c = c_c';
size_c = size(c_d);
n = 1:size_c(2);
plot(n,c_d);
And here is the plot:
Why are the curves mirroring each other? I am asking to gain a better understanding of what exactly improfile seems to be achieving in MATLAB.
improfile computes something like a "path integral", it gives you the image intensity values around a user specified path. For example, if you use:
improfile(img,[1 1],[1 size(img,2)]);
It gives the same as img(:,1). This is because the path you are using in improfile is from (1,1) to (1,size(img,2)) , meaning the first row. However you could definitely add more complicated paths.
In your case you are going trow a path defined by 4 points. The points are, if I substitute your equation by the resultant numbers:
(1,79)->(134,79)->(134,79)->(1,79). Thus, looking at this, it is obvious that your result should be mirrored, because you are integrating over a line the way there, and back!
Sidenote:
you can do plot(c_d); and forget about that n
I want to draw a line on a graph to find the intersection point with another line. However, there's no response after I executed the script below. May I know what is the problem and how can I solve it?
x=1:2^20;
y2=2^24;
plot(x,y2);
Thanks!
What you want is to plot a line on 2^24. However, there are too many points for you computer probably, and you run out of memory
I am guessing that you'll need to plot your other inequality as well.
Something like
x=1:100:2^20;
% As Zoran and others suggested, You may not want all the points!
% It is too much memory
y2=2^24*ones(size(x)); % This ones is optional, but its good to know what you are doing (personal opinion)
plot(x,y2);
hold on
y1=(x+1).*log(x);
plot(x,y1);
However, you are still not there!
Another solution, which does not rely on plotting:
>> f = #(x) (x+1)*log(x)-2^24;
>> soln = fzero(f,1e6)
soln = 1.1987e+006
>> f(soln)
ans = 3.7253e-009
So your intersection point is at 1.1987e6.
Apparently,you have too many points for x, 2^20
Have to wait program to calculate, or plot,for example, every 100th point
This solution works for Matlab
x=1:100:2^20;
y2=2^2;
plot(x,y2,'o');
There is one more and maybe a bit smarter way: if you want to solve ((k+1)(ln k)<2^24) as you've commented above, use fsolve function to get just solution of an equation(!). Then use that solution to specify the area of your interest, so you won't have to plot the domain of 2^20.
(All functions are continuous so you don't have to worry about any wild singularities. Just examine the neigborhood of ks for which (k+1)(ln k)-2^24=0.)
I would like to interpolate a data set, but a given X can have multiple Y's, example given below:
A(:,1:2,1)=
95.2343 70.6159
96.4501 71.5573
97.4430 72.7315
98.9743 72.8699
100.0470 71.7690
100.3872 70.2699
100.7797 68.7837
102.1478 68.0814
103.6851 68.0521
105.0307 68.7966
105.8972 70.0666
106.7177 71.3665
107.7095 72.5416
108.9175 73.4924
110.3574 74.0309
111.8943 73.9859
113.3936 73.6446
114.6645 72.7794
115.5911 71.5522
116.2426 70.1591
116.3922 68.6286
116.3503 67.0914
116.7771 65.6147
117.9045 64.5692
119.4065 64.2425
120.9432 64.2923
122.2526 65.0975
122.9486 66.4682
122.8841 68.0043
122.5492 69.5051
122.2403 71.0109
122.0819 72.5402
These are points along the body of a snake, digitized from top-down video. Their posture frequently results in 2 or more Y points per X, and because they turn (often dramatically) while I need to maintain a consistent XY framework, I can't just rotate my X and Y.
I can make a spline function using cscvn, and when plotted using fnplt it looks great! But the moment I try to get values back using ppval, it all goes to crap and I get two curves that look nothing like the snake or what's shown in fnplt.
IMAGE HERE:
http://imgur.com/aYJ0Ftj
All I want is to fit a curve to those points, and convert that curve to a series of maybe 200 XY points.
I'm assuming that there is some obscenely simple answer, some command I've just never turned up in my searching, but I cannot find it for the life of me. This is quite far outside my usual skillset (I'm more used to crawling through swamps catching anything scaly or slimy), so I'm sure I'm overlooking something totally obvious to a skilled MATLAB user.
Thanks!
Perhaps this is what you already tried, but a comment was a bit small, so here is one idea that may help you
x = [1 2 3 4 5 6 7 6 5];
y = [1 4 4 3 2 2 2 3 4];
t = 1:length(x);
figure; plot(x,y)
xx = interp1(t,x,1:.01:length(x),'pchip');
yy = interp1(t,y,1:.01:length(x),'pchip');
hold on; plot(xx,yy,'g')
I avoided the unique values restriction on interp by interpolating x and y both as a function of t.
OK, thanks to Yvon and Trogdor, I managed to crudely cludge together some code:
ht=[1:32];
tx=linspace(1,32,500);
m=ht;
m(2:3,:)=squeeze(A(:,:,i))';
m1=m(1,:);
m2=m(2,:);
m3=m(3,:);
SFX=spline(m1,m2);
pSFX=ppval(SFX,tx);
SFY=spline(m1,m3);
pSFY=ppval(SFY,tx);
AS(:,1,i)=pSFX;
AS(:,2,i)=pSFY;
plot(AS(:,1,i),AS(:,2,i),'Marker','o');
It's riddled with pointless variables, steps, and redundancies, but I'm an anatomist, so that makes me comfortable (seriously, google 'recurrent laryngeal nerve').
How do I add reputation for y'all?
In Matlab, I've got several points in the 3D space. Those points represent a rope and I would like to draw a line linking all those points. Here is my problem: How the oraganization of those points should be done to have a "simple" and "more or less straigth line". In other words I would like to draw a line linking all the points from the first to the last one but not "going back". Maybe with a simple image i can explain better my problem:
This is what the code should do:
This is what the code shouldn't do:
Some idea of how can I achieve the intended result? How can I organize the points? I'm working with Matlab but if you know any paper where I can read how to do this it will be fine. Thank you.
If you just don't want to go back in the upper direction, the solution that #Dan suggested should be fine. Sort them in that direction before connecting.
However, if that is not a requirement but you are actually looking for a solution without ugly crossings that is as short as possible, you may want to look into solutions for the travelling salesman problem.
If you define the distance between 1 and 9 to be zero, the solution you find to the travelling salesman problem should give you a nice and short rope.
If you want to go for the TSP approach but get a little lost, just try a 'furthest insertion' start position or '2 opt' improvements as that is likely going to give a big improvement over random path selection already with minimal effort.
OK. I've found the right answer (by Gunther Struyf): Sort Coordinates Points in Matlab
In order to visualize the final result just add to Gunther's code the following lines:
row=data(:,1);
col=data(:,2);
figure
scatter(row,col,'r');
hold on
line(row(result),col(result));
The following code shows the same algorithm but for a 3D implementation:
close all;
data = [2,2, 2; 2,3,2 ; 1,2,3 ; 1,3,3 ; 2,1,3 ; 1,1,3 ; 3,2,4 ; 3,3,4 ; 3,1,5; 5,4,6; 7,3,8; 8,9,7; 9,4,7; 6,2,5; 5,8,6; 9,3,8;6,9,2];
row=data(:,1);
col=data(:,2);
frame=data(:,3);
dist = pdist2(data,data);
N = size(data,1);
result = NaN(1,N);
result(1) = 1; % first point is first row in data matrix
for ii=2:N
dist(:,result(ii-1)) = Inf;
[~, closest_idx] = min(dist(result(ii-1),:));
result(ii) = closest_idx;
end
figure
%Ploting the points directly (without sorting)
subplot(2,1,1);
scatter3(row,col,frame,'r');
hold on;
line(row,col,frame);
%Ploting after sorting
subplot(2,1,2);
scatter3(row,col,frame,'r');
hold on
line(row(result),col(result),frame(result));
Hey, I've got a problem plotting a function in Matlab.
I first run this:
format long
f = inline('-x.^2');
for i = 0:10
[I(i+1) h(i+1) tid(i+1)] = trapets(f,0,1,2^i);
end
trunk = I - log(2);
hold on
grid on
plot(log(h),log(trunk),'r+')
t = -7:0;
c = polyfit(log(h),log(trunk),1);
yy = polyval(c,t);
plot(t,yy)
grid off
hold off
koefficienter = real(c)
and after that I run this file:
hold on
plot(h,trunk,'r+:','linewidth',2)
axis([0 0.6 0 0.0014])
Thing is, I don't get any errors, and the plot windows pops up with axes and all, but there is no graph to be found. It's just an empty window with two axes.
Anyone got any ideas?
Edit:
Okay, so I'm new to this site and couldn't find the reply button, so I add a reply here instead.
#woodchips :
I just realized that I hadn't given you all the information for this problem.. Sorry about that, anyhow I would really appreciate it if someone had the time to help me with this, it would seriously make my week.
This is the part I accidentally left out:
function [ I,h,tid ] = trapets(
f,a,b,n )
h=(b-a)/n;
tic; I=(f(a)+f(b));
for k=2:2:n-2
I = I+2*f(a+k*h);
end
for k = 1:2:n-1
I = I + 4*f(a+k*h);
end
I = I * h/3;
tid = toc;
end
Edit 2: So, I think that the graph I'm seeking is actually getting plotted in the first code that I wrote, the problem is that the variabe 'I' is not changing, which I expect it to do, although the variabels 'n' and 'h' do change. If 'I' was working correctly, I would probably get the right graph (hopefully). Any ideas, anyone?
Unfortunately the home computer I had with Matlab on it died the other day so I can't test anything. First thing I can think of if to simply run step by step through the code and see if the results of the math are what you are expecting. For instance Matlab was primarily made and runs as a matrix calculator if I recall correctly. As such most of the simple math doesn't function as it would punching it in a calculator. An example would be that 2^i needs to be 2.^i to operate correctly in some cases. Same with .* and ./ to use the singular scalar verses the matrices math.
The best way to find out what is going wrong is to iterate through the math a few times to ensure that it is being performed as expected. Once that is verified then you can move on to looking at plotting formatting.