I have a dataset the red line.
I'm trying to find the minimum points highlighted in yellow if I take the reflection/mirror image of a data set.
See example code / plot below I'm trying to find a way to find the minimum points highlighted in yellow of the reflection/mirror image of a dataset (the blue line) that is below the reflection line (the black line).
Please note this is just a simple dataset there will be much larger datasets around 100000+
PS: I'm using Octave 3.8.1 which is like matlab
clear all,clf, clc,tic
x1=[0.;2.04;4.08;6.12;8.16;10.2;12.24;14.28;16.32;18.36]
y1=[2;2.86;4;2;1;4;5;2;7;1]
x2=[0.;2.04;4.08;6.12;8.16;10.2;12.24;14.28;16.32;18.36]
y2=abs(y1-max(y1));
data1 = y2;
reflection_line=max(y1)/2
[pks3 idx3] = findpeaks(data1,"DoubleSided","MinPeakHeight",0.1);
line([min(x1) max(x1)], [reflection_line reflection_line]);
hold on;
plot(x1,reflection_line)
hold on;
plot(x1,y1,'-r',x2,y2,'-b')
I am not reflecting your original data, but rather find the local maxima of the original values, that are larger than your given line.
Using a DIY alternative to findpeaks:
A value is a local maximum, if it's larger than (or equal to) its predecessor and successor.
%% Setup
x1 = [0.;2.04;4.08;6.12;8.16;10.2;12.24;14.28;16.32;18.36];
y1 = [2;2.86;4;2;1;4;5;2;7;1];
reflection_line = max(y1)/2;
%% Sort by x value
[x1, I] = sort(x1);
y1 = y1(I);
%% Compute peaks
maxima = #(y) [true; y(2:end)>=y(1:end-1)] & ... % Value larger than predecessor
[y(1:end-1)>=y(2:end); true]; % Value larger than successor
maximaLargerThanLine = maxima(y1) & (y1>reflection_line);
%% Plotting
plot(x1,y1);
hold on;
plot(x1(maximaLargerThanLine),y1(maximaLargerThanLine),'rx');
line([min(x1) max(x1)], [reflection_line reflection_line]);
Related
I want to set the limit for X axis in this plot from 0 to 325. When i am using xlim to set the limits (commented in the code). It doesn't work properly. When i use xlim, the entire structure of plot changes. Any help will be appreciated.
figure
imagesc(transpose(all_area_for_visual));
colormap("jet")
colorbar('Ticks',0:3,'TickLabels',{'Home ','Field','Bad house','Good house'})
xlabel('Time (min)')
tickLocs = round(linspace(1,length(final_plot_mat_missing_part(2:end,1)),8));
timeVector = final_plot_mat_missing_part(2:end,1);
timeForTicks = (timeVector(tickLocs))./60;
xticks(tickLocs);
xticklabels(timeForTicks);
%xlim([0 325]);
ylabel('Car identity')
yticks(1:length(Ucolumnnames_fpm))
yticklabels([Ucolumnnames_fpm(1,:)])
If I get you right, you want to plot only part of the data in all_area_for_visual, given by a condition on tickLocs. So you should first condition the data, and then plot it:
% generate the vector of x values:
tickLocs = round(linspace(1,length(final_plot_mat_missing_part(2:end,1)),8));
% create an index vector (of logicals) that marks the columns to plot from the data matix:
validX = tickLocs(tickLocs<=325);
% plot only the relevant part of the data:
imagesc(transpose(all_area_for_visual(:,validX)));
% generate the correct ticks for the data that was plotted:
timeVector = final_plot_mat_missing_part(2:end,1);
timeForTicks = (timeVector(tickLocs(validX)))./60;
xticks(tickLocs(validX));
% here you continue with setting the labels, colormap and so on...
imagesc puts the data in little rectangles centered around integers 1:width and 1:height by default. You can specify what the x and y locations of each data point by adding two vectors to the call:
imagesc(x,y,transpose(all_area_for_visual));
where x and y are vectors with the locations along the x and y axes you want to place the data.
Note that xlim and xticks don’t change the location of the data, only the region of the axis shown, and the location of tick marks along the axis. With xticklabels you can change what is shown at each tick mark, so you can use that to “fake” the data locations, but the xlim setting still applies to the actual locations, not to the labels assigned to the ticks.
I think it is easier to plot the data in the right locations to start with. Here is an example:
% Fake your data, I'm making a small matrix here for illustration purposes
all_area_for_visual = min(floor(cumsum(rand(20,5)/2)),3);
times = linspace(0,500,20); % These are the locations along the time axis for each matrix element
car_id_names = [4,5,8,15,18]; % These are the labels to put along the y-axis
car_ids = 1:numel(car_id_names); % These are the locations to use along the y-axis
% Replicate your plot
figure
imagesc(times,car_ids,transpose(all_area_for_visual));
% ^^^ ^^^ NOTE! specifying locations
colormap("jet")
colorbar('Ticks',0:3,'TickLabels',{'Home ','Field','Bad house','Good house'})
xlabel('Time (min)')
ylabel('Car identity')
set(gca,'YTick',car_ids,'YTickLabel',car_id_names) % Combine YTICK and YTICKLABEL calls
% Now you can specify your limit, in actual time units (min)
xlim([0 325]);
I have a spectral data (1000 variables on xaxis, and peak intensities as y) and a list of peaks of interest at various specific x locations (a matrix called Peak) which I obtained from a function I made. Here, I would like to draw a line from the maximum value of each peaks to the xaxis - or, eventually, place a vertical arrow above each peaks but I read it is quite troublesome, so just a vertical line is welcome. However, using the following code, I get "Error using line Value must be a vector of numeric type". Any thoughts?
X = spectra;
[Peak,intensity]=PeakDetection(X);
nrow = length(Peak);
Peak2=Peak; % to put inside the real xaxis value
plot(xaxis,X);
hold on
for i = 1 : nbrow
Peak2(:,i) = round(xaxis(:,i)); % to get the real xaxis value and round it
xline = Peak2(:,i);
line('XData',xline,'YData',X,'Color','red','LineWidth',2);
end
hold off
Simple annotation:
Here is a simple way to annotate the peaks:
plot(x,y,x_peak,y_peak+0.1,'v','MarkerFaceColor','r');
where x and y is your data, and x_peak and y_peak is the coordinates of the peaks you want to annotate. The add of 0.1 is just for a better placing of the annotation and should be calibrated for your data.
For example (with some arbitrary data):
x = 1:1000;
y = sin(0.01*x).*cos(0.05*x);
[y_peak,x_peak] = PeakDetection(y); % this is just a sketch based on your code...
plot(x,y,x_peak,y_peak+0.1,'v','MarkerFaceColor','r');
the result:
Line annotation:
This is just a little bit more complicated because we need 4 values for each line. Again, assuming x_peak and y_peak as before:
plot(x,y);
hold on
ax = gca;
ymin = ax.YLim(1);
plot([x_peak;x_peak],[ymin*ones(1,numel(y_peak));y_peak],'r')
% you could write instead:
% line([x_peak;x_peak],[ymin*ones(1,numel(y_peak));y_peak],'Color','r')
% but I prefer the PLOT function.
hold off
and the result:
Arrow annotation:
If you really want those arrows, then you need to first convert the peak location to the normalized figure units. Here how to do that:
plot(x,y);
ylim([-1.5 1.5]) % only for a better look of the arrows
peaks = [x_peak.' y_peak.'];
ax = gca;
% This prat converts the axis unites to the figure normalized unites
% AX is a handle to the figure
% PEAKS is a n-by-2 matrix, where the first column is the x values and the
% second is the y values
pos = ax.Position;
% NORMPEAKS is a matrix in the same size of PEAKS, but with all the values
% converted to normalized units
normpx = pos(3)*((peaks(:,1)-ax.XLim(1))./range(ax.XLim))+ pos(1);
normpy = pos(4)*((peaks(:,2)-ax.YLim(1))./range(ax.YLim))+ pos(2);
normpeaks = [normpx normpy];
for k = 1:size(normpeaks,1)
annotation('arrow',[normpeaks(k,1) normpeaks(k,1)],...
[normpeaks(k,2)+0.1 normpeaks(k,2)],...
'Color','red','LineWidth',2)
end
and the result:
I am trying to simulate a network of mobile robots that uses artificial potential fields for movement planning to a shared destination xd. This is done by generating a series of m-files (one for each robot) from a symbolic expression, as this seems to be the best way in terms of computational time and accuracy. However, I can't figure out what is going wrong with my gradient computation: the analytical gradient that is being computed seems to be faulty, while the numerical gradient is calculated correctly (see the image posted below). I have written a MWE listed below, which also exhibits this problem. I have checked the file generating part of the code, and it does return a correct function file with a correct gradient. But I can't figure out why the analytic and numerical gradient are so different.
(A larger version of the image below can be found here)
% create symbolic variables
xd = sym('xd',[1 2]);
x = sym('x',[2 2]);
% create a potential function and a gradient function for both (x,y) pairs
% in x
for i=1:size(x,1)
phi = norm(x(i,:)-xd)/norm(x(1,:)-x(2,:)); % potential field function
xvector = reshape(x.',1,size(x,1)*size(x,2)); % reshape x to allow for gradient computation
grad = gradient(phi,xvector(2*i-1:2*i)); % compute the gradient
gradx = grad(1);grady=grad(2); % split the gradient in two components
% create function file names
gradfun = strcat('GradTester',int2str(i),'.m');
phifun = strcat('PotTester',int2str(i),'.m');
% generate two output files
matlabFunction(gradx, grady,'file',gradfun,'outputs',{'gradx','grady'},'vars',{xvector, xd});
matlabFunction(phi,'file',phifun,'vars',{xvector, xd});
end
clear all % make sure the workspace is empty: the functions are in the files
pause(0.1) % ensure the function file has been generated before it is called
% these are later overwritten by a specific case, but they can be used for
% debugging
x = 0.5*rand(2);
xd = 0.5*rand(1,2);
% values for the Stackoverflow case
x = [0.0533 0.0023;
0.4809 0.3875];
xd = [0.4087 0.4343];
xp = x; % dummy variable to keep x intact
% compute potential field and gradient for both (x,y) pairs
for i=1:size(x,1)
% create a grid centered on the selected (x,y) pair
xGrid = (x(i,1)-0.1):0.005:(x(i,1)+0.1);
yGrid = (x(i,2)-0.1):0.005:(x(i,2)+0.1);
% preallocate the gradient and potential matrices
gradx = zeros(length(xGrid),length(yGrid));
grady = zeros(length(xGrid),length(yGrid));
phi = zeros(length(xGrid),length(yGrid));
% generate appropriate function handles
fun = str2func(strcat('GradTester',int2str(i)));
fun2 = str2func(strcat('PotTester',int2str(i)));
% compute analytic gradient and potential for each position in the xGrid and
% yGrid vectors
for ii = 1:length(yGrid)
for jj = 1:length(xGrid)
xp(i,:) = [xGrid(ii) yGrid(jj)]; % select the position
Xvec = reshape(xp.',1,size(x,1)*size(x,2)); % turn the input into a vector
[gradx(ii,jj),grady(ii,jj)] = fun(Xvec,xd); % compute gradients
phi(jj,ii) = fun2(Xvec,xd); % compute potential value
end
end
[FX,FY] = gradient(phi); % compute the NUMERICAL gradient for comparison
%scale the numerical gradient
FX = FX/0.005;
FY = FY/0.005;
% plot analytic result
subplot(2,2,2*i-1)
hold all
xlim([xGrid(1) xGrid(end)]);
ylim([yGrid(1) yGrid(end)]);
quiver(xGrid,yGrid,-gradx,-grady)
contour(xGrid,yGrid,phi)
title(strcat('Analytic result for position ',int2str(i)));
xlabel('x');
ylabel('y');
subplot(2,2,2*i)
hold all
xlim([xGrid(1) xGrid(end)]);
ylim([yGrid(1) yGrid(end)]);
quiver(xGrid,yGrid,-FX,-FY)
contour(xGrid,yGrid,phi)
title(strcat('Numerical result for position ',int2str(i)));
xlabel('x');
ylabel('y');
end
The potential field I am trying to generate is defined by an (x,y) position, in my code called xd. x is the position matrix of dimension N x 2, where the first column represents x1, x2, and so on, and the second column represents y1, y2, and so on. Xvec is simply a reshaping of this vector to x1,y1,x2,y2,x3,y3 and so on, as the matlabfunction I am generating only accepts vector inputs.
The gradient for robot i is being calculated by taking the derivative w.r.t. x_i and y_i, these two components together yield a single derivative 'vector' shown in the quiver plots. The derivative should look like this, and I checked that the symbolic expression for [gradx,grady] indeed looks like that before an m-file is generated.
To fix the particular problem given in the question, you were actually calculating phi in such a way that meant you doing gradient(phi) was not giving the correct results compared to the symbolic gradient. I'll try and explain. Here is how you created xGrid and yGrid:
% create a grid centered on the selected (x,y) pair
xGrid = (x(i,1)-0.1):0.005:(x(i,1)+0.1);
yGrid = (x(i,2)-0.1):0.005:(x(i,2)+0.1);
But then in the for loop, ii and jj were used like phi(jj,ii) or gradx(ii,jj), but corresponding to the same physical position. This is why your results were different. Another problem you had was you used gradient incorrectly. Matlab assumes that [FX,FY]=gradient(phi) means that phi is calculated from phi=f(x,y) where x and y are matrices created using meshgrid. You effectively had the elements of phi arranged differently to that, an so gradient(phi) gave the wrong answer. Between reversing the jj and ii, and the incorrect gradient, the errors cancelled out (I suspect you tried doing phi(jj,ii) after trying phi(ii,jj) first and finding it didn't work).
Anyway, to sort it all out, on the line after you create xGrid and yGrid, put this in:
[X,Y]=meshgrid(xGrid,yGrid);
Then change the code after you load fun and fun2 to:
for ii = 1:length(xGrid) %// x loop
for jj = 1:length(yGrid) %// y loop
xp(i,:) = [X(ii,jj);Y(ii,jj)]; %// using X and Y not xGrid and yGrid
Xvec = reshape(xp.',1,size(x,1)*size(x,2));
[gradx(ii,jj),grady(ii,jj)] = fun(Xvec,xd);
phi(ii,jj) = fun2(Xvec,xd);
end
end
[FX,FY] = gradient(phi,0.005); %// use the second argument of gradient to set spacing
subplot(2,2,2*i-1)
hold all
axis([min(X(:)) max(X(:)) min(Y(:)) max(Y(:))]) %// use axis rather than xlim/ylim
quiver(X,Y,gradx,grady)
contour(X,Y,phi)
title(strcat('Analytic result for position ',int2str(i)));
xlabel('x');
ylabel('y');
subplot(2,2,2*i)
hold all
axis([min(X(:)) max(X(:)) min(Y(:)) max(Y(:))])
quiver(X,Y,FX,FY)
contour(X,Y,phi)
title(strcat('Numerical result for position ',int2str(i)));
xlabel('x');
ylabel('y');
I have some other comments about your code. I think your potential function is ill-defined, which is causing all sorts of problems. You say in the question that x is an Nx2 matrix, but you potential function is defined as
norm(x(i,:)-xd)/norm(x(1,:)-x(2,:));
which means if N was three, you'd have the following three potentials:
norm(x(1,:)-xd)/norm(x(1,:)-x(2,:));
norm(x(2,:)-xd)/norm(x(1,:)-x(2,:));
norm(x(3,:)-xd)/norm(x(1,:)-x(2,:));
and I don't think the third one makes sense. I think this could be causing some confusion with the gradients.
Also, I'm not sure if there is a reason to create the .m file functions in your real code, but they are not necessary for the code you posted.
I have an ECG signal and some special points of it calculated.
I want to have thicker LineWidth between those points (each pair). I did a sample with brush.
Here are my variables,
signal % the ECG signal
t % time
Q % location of red points
T % location of yellow points
Four of these pairs are visible in picture, but there are more.
Is it possible without loop _ hold on?
You can just use hold on and plot the data again on the region of interest:
% Some dummy data
x = 0:0.01:10;
y = sin(x);
plot(x,y)
% Data that we want emphasized
% You can also select a subset of your existing data
x_start = 2;
x_end = 4;
x_thick_line = x_start:0.01:x_end;
y_thick_line = sin(x_thick_line);
% Plot over the existing plot with thicker line
hold on
plot([x_start x_end],[y_thick_line(1) y_thick_line(end)],'ro',...
x_thick_line,y_thick_line,'Color','r','LineWidth',6')
This gives the following result in Octave, should be the same in MATLAB:
You should plot that function three times (assuming a..b shall be styled differently):
0..a - standard settings
a..b - use alternate color / line_width / ...
b..end
Hi i am trying to make a simple distribution histogram using some code from stack overflow
however i am unable to get it to work. i know that there are is a simple method for this using statistic toolbox but form a learning point of view i prefer a more explanatory code - can any one help me ?
%%
clear all
load('Mini Project 1.mat')
% Set data to var2
data = var2;
% Set the number of bins
nbins = 0:.8:8;
% Create a histogram plot of data sorted into (nbins) equally spaced bins
n = hist(data,nbins);
% Plot a bar chart with y values at each x value.
% Notice that the length(x) and length(y) have to be same.
bar(nbins,n);
MEAN = mean(data);
STD = sqrt(mean((data - MEAN).^2)); % can also use the simple std(data)
f = ( 1/(STD*sqrt(2*pi)) ) * exp(-0.5*((nbins-MEAN)/STD).^2 );
f = f*sum(nbins)/sum(f);
hold on;
% Plots a 2-D line plot(x,y) with the normal distribution,
% c = color cyan , Width of line = 2
plot (data,f, 'c', 'LineWidth', 2);
xlabel('');
ylabel('Firmness of apples after one month of storage')
title('Histogram compared to normal distribution');
hold of
You are confusing
hist
with
histc
Read up on both.
Also, you are not defining the number of bins, you are defining the bins themselves .
I don't have Matlab at hand now, but try the following:
If you want to compare a normal distribution to the bar plot bar(nbins,n), you should first normalize it:
bar(nbins,n/sum(n))
See if this solves your problem.
If not, try also removing the line f = f*sum(nbins)/sum(f);.