I have a matrix of data. I used the polarplot command in MATLAB to plot this matrix with respect to theta.
The data oscillates between 3999.20 and 4001.52 As you can see in the following plot, the order of magnitude of oscillation of data is too small to see.
How can I modify my polar plot to see small oscillations?
My code is as below:
yf=[%750 data point]
theta = 0:4*pi/749:4*pi;
rho = yf
pax = polaraxes;
polarplot(theta,rho)
pax.ThetaDir = 'counterclockwise';
pax.ThetaZeroLocation='right'
pax.ThetaColor='r'
pax.ThetaTick=[0 30 60 90 120 150 180 210 240 270 300 330 ];
pax.RDir='normal';
pax.RTick=[3999.34 3999.67 4000 4000.33 4000.66 4000.99 4001.33 ]
pax.FontSize = 12;
Desired output:
Actual output
2-axis plot
To give an example of setting the r limits as suggested by #beaker
The following code uses the same data with using rlim to set manual limits in the second example. This scales the polar axis so that it only plots values between [3999,4000], exaggerating the scale of the oscillation.
theta = 0:0.01:2*pi;
rho = sin(2*theta).*cos(2*theta) + 3999 %To approximate your data
figure;
subplot(1,2,1)
polarplot(theta,rho)
title('Automatic r-limits')
subplot(1,2,2)
polarplot(theta,rho)
rlim([3999, 4000])
title('rlim([3999, 4000])')
Something like that maybe, where you subtract the mean of the data and scale the amplitude by a factor of 10?
yf=[%750 data point]
amp = yf - mean(yf);
amp = amp*10; % choose whatever scaling factor works for you
theta = 0:4*pi/749:4*pi;
rho = mean(yf) + amp;
Without the actual data, it's difficult to say what this will look like, but the general principle should work.
Related
I am trying to fit a distribution curve to the histogram of some data. (I have used some model data here instead because it is difficult to upload the actual data. I have included the complete code after my question.)
Because the histogram looks normally distributed when I plotted the x-axis in logscale, I transform the data first before fitting a normal distribution to it and I got the following results:
>>pdn=fitdist(log(data),'Normal')
pdn =
Normal distribution
mu = -0.334458 [-0.34704, -0.321876]
sigma = 0.351478 [0.342804, 0.360605]
When I plotted out the pdf with the histogram, I got this:
The result seems reasonable to me. Then I discovered that in the Matlab fitdist(), it already has a 'Lognormal' option and I don't really need the transform my data first and this is what I got:
>>pdln = fitdist(data,'Lognormal')
pdln =
Lognormal distribution
mu = -0.334458 [-0.34704, -0.321876]
sigma = 0.351478 [0.342804, 0.360605]
Exactly the same mean and standard deviation as I have got before. However, when I plotted it out with the histogram, I got a different curve:
This curve fits better to the data but the positions of the mean and the mean+/-std points are not as I have expected (i.e. mean at the peak and the mean+/-std at the same levels).
Which come to my question, why would fitdist(data,'Lognormal') give the same result as fitdist(log(data),'Normal') but a different plot? I have looked through the Matlab help pages and I still could not understand why, or where are my mistakes, please help.
My aim for all this is to get some numerical parameters about the distributions of my data under different conditions and compare them to see if there is any difference. At the moment, I am not certain which way would give me reliable estimates of the means and standard deviations.
The code for the graphs is below:
%random data in lognormal distribution
mu=-0.335742;
sigma=0.35228;
data=lognrnd(mu,sigma,[3000 1]);
%make histogram
interval=0.1;
svalue=sort(data);
bx(1)=interval/2;
i=2;
while bx(i-1)<=max(svalue)
bx(i)=bx(i-1)+interval;
i=i+1;
end
by=hist(svalue,bx);
subplot(211)
h = bar(bx,by,'hist');
set(h,'FaceColor',[.9 .9 .9]);
set(gca,'xlim',[0.05 10]);
xticks=[0.05 0.1 0.2 0.5 1 2 5 10];
set(gca,'xscale','log','xminortick','on')
set(gca,'xtick',xticks)
ylabel('counts')
subplot(212)
h = bar(bx,by,'hist');
set(h,'FaceColor',[.9 .9 .9]);
set(gca,'xlim',[0.05 10]);
xticks=[0.05 0.1 0.2 0.5 1 2 5 10];
set(gca,'xscale','log','xminortick','on')
set(gca,'xtick',xticks)
ylabel('counts')
% fit distribution curves
pdf_x = 0:0.01:max(data);
max_by=max(by); % for scaling the pdf to the histogram
% case 1 - PDF fitted using fitdist(log(data),'Normal')
subplot(211)
hold on
pdn = fitdist(log(data),'Normal')
pdf_y = pdf(pdn,log(pdf_x));
h1=plot(pdf_x,pdf_y./max(pdf_y).*max_by,'-k');
range=[exp(pdn.mu-pdn.sigma) exp(pdn.mu+pdn.sigma)];
h2=plot(exp(pdn.mu),pdf(pdn,(pdn.mu))./max(pdf_y).*max_by,'sk') ;
h3=plot(range,pdf(pdn,log(range))./max(pdf_y).*max_by,'ok') ;
title('PDF fitted using fitdist(log(data),''Normal'')');
legend([h1 h2 h3],'pdf','mean','meam+/-std');
% case 2 - PDF fitted using fitdist(data,'Lognormal')
subplot(212)
hold on
pdln = fitdist(data,'Lognormal')
pdf_y = pdf(pdln,pdf_x);
h1=plot(pdf_x,pdf_y./max(pdf_y).*max_by,'-b');
range=[exp(pdln.mu-pdln.sigma) exp(pdln.mu+pdln.sigma)];
h2=plot(exp(pdln.mu),pdf(pdln,exp(pdln.mu))./max(pdf_y).*max_by,'sb');
h3=plot(range,pdf(pdln,range)./max(pdf_y).*max_by,'ob') ;
title('PDF fitted using fitdist(data,''Lognormal'')');
legend([h1 h2 h3],'pdf','mean','meam+/-std');
How can I find the exact y-coordinate of the red gaussian at x = 0.5 without using Data Cursor?
I want the blue line to end when it touches the gaussian. But I need to find the point of intersection between the gaussian of the histfit shown in red, and the blue line at 0.5. I can access the data points of the histfit plot as follows:
C = get(get(gca, 'Children'), 'YData');
C{1,1}
line([0.5 0.5],[0 max(C{1,1})],'Color','b');
However, the data points of the gaussian don't relate to this axes. Meaning, x-axis of C{1,1} is from 1 - 100 and not from 0.1 to 0.9.
Whats the easiest way to find the y-coordinate of the gaussian at 0.5 so that I can replace max(C{1,1}) by that?
Getting XData as well should give you the right x-values:
C = get(get(gca, 'Children'), 'XData');
Alternatively, the values of YData should be at regular intervals, even if not on the correct scale (since it originated from hist), so you could probably find the y-value corresponding to x=0.5 in the plot.
The point x=0.5 between 0.1 and 0.85 (approximately, from the plot) scales to the point x=53.33 between 1 and 100. If the y-value at x=53 isn't accurate enough for plotting, you can just interpolate the value between 53 and 54 and that should be enough.
Here is some code to that should do the job:
XPlotRange = [0.1 0.85];
XDataRange = [1 100];
XPlotToInterp = 0.5;
XDataToInterp = XDataRange(1) + (XPlotToInterp - XPlotRange(1))*diff(XDataRange)/diff(XDataRange);
XData1 = floor(XDataToInterp);
XData2 = ceil(XDataToInterp);
YInterp = interp1([XData1 XData2], [YData(XData1) YData(XData2)], XDataToInterp);
Here YInterp is the interpolated y-value for the corresponding x-value.
Assume y is a vector with random numbers following the distribution f(x)=sqrt(4-x^2)/(2*pi). At the moment I use the command hist(y,30). How can I plot the distribution function f(x)=sqrt(4-x^2)/(2*pi) into the same histogram?
Instead of normalizing numerically, you could also do it by finding a theoretical scaling factor as follows.
nbins = 30;
nsamples = max(size(y));
binsize = (max(y)-min(y)) / nsamples
hist(y,nbins)
hold on
x1=linspace(min(y),max(y),100);
scalefactor = nsamples * binsize
y1=scalefactor * sqrt(4-x^2)/(2*pi)
plot(x1,y1)
Update: How it works.
For any dataset that is large enough to give a good approximation to the pdf (call it f(x)), the integral of f(x) over this domain will be approximately unity. However we know that the area under any histogram is precisely equal to the total number of samples times the bin-width.
So a very simple scale factor to bring the pdf into line with the histogram is Ns*Wb, the total number of sample point times the width of the bins.
Let's take an example of another distribution function, the standard normal. To do exactly what you say you want, you do this:
nRand = 10000;
y = randn(1,nRand);
[myHist, bins] = hist(y,30);
pdf = normpdf(bins);
figure, bar(bins, myHist,1); hold on; plot(bins,pdf,'rx-'); hold off;
This is probably NOT what you actually want though. Why? You'll notice that your density function looks like a thin line at the bottom of your histogram plot. This is because a histogram is counts of numbers in bins, while a density function is normalized to integrate to one. If you have hundreds of items in a bin, there is no way that the density function will match that in scale, so you have a scaling or normalization problem. Either you have to normalize the histogram, or plot a scaled distribution function. I prefer to scale the distribution function so that my counts are sensical when I look at the histogram:
normalizedpdf = pdf/sum(pdf)*sum(myHist);
figure, bar(bins, myHist,1); hold on; plot(bins,normalizedpdf,'rx-'); hold off;
Your case is the same, except you'll use the function f(x) you specified instead of the normpdf command.
Let me add another example to the mix:
%# some normally distributed random data
data = randn(1e3,1);
%# histogram
numbins = 30;
hist(data, numbins);
h(1) = get(gca,'Children');
set(h(1), 'FaceColor',[.8 .8 1])
%# figure out how to scale the pdf (with area = 1), to the area of the histogram
[bincounts,binpos] = hist(data, numbins);
binwidth = binpos(2) - binpos(1);
histarea = binwidth*sum(bincounts);
%# fit a gaussian
[muhat,sigmahat] = normfit(data);
x = linspace(binpos(1),binpos(end),100);
y = normpdf(x, muhat, sigmahat);
h(2) = line(x, y*histarea, 'Color','b', 'LineWidth',2);
%# kernel estimator
[f,x,u] = ksdensity( data );
h(3) = line(x, f*histarea, 'Color','r', 'LineWidth',2);
legend(h, {'freq hist','fitted Gaussian','kernel estimator'})
The function called DicePlot simulates rolling 10 dice 5000 times.
The function calculates the sum of values of the 10 dice of each roll, which will be a 1 ⇥ 5000 vector, and plot relative frequency histogram with edges of bins being selected in where each bin in the histogram represents a possible value of for the sum of the dice.
The mean and standard deviation of the 1 ⇥ 5000 sums of dice values will be computed, and the probability density function of normal distribution (with the mean and standard deviation computed) on top of the relative frequency histogram will be plotted.
Below is my code so far - What am I doing wrong? The graph shows up but not the extra red line on top? I looked at answers like this, and I don't think I'll be plotting anything like the Gaussian function.
% function[]= DicePlot()
for roll=1:5000
diceValues = randi(6,[1, 10]);
SumDice(roll) = sum(diceValues);
end
distr=zeros(1,6*10);
for i = 10:60
distr(i)=histc(SumDice,i);
end
bar(distr,1)
Y = normpdf(X)
xlabel('sum of dice values')
ylabel('relative frequency')
title(['NumDice = ',num2str(NumDice),' , NumRolls = ',num2str(NumRolls)]);
end
It is supposed to look like
But it looks like
The red line is not there because you aren't plotting it. Look at the documentation for normpdf. It computes the pdf, it doesn't plot it. So you problem is how do you add this line to the plot. The answer to that problem is to google "matlab hold on".
Here's some code to get you going in the right direction:
% Normalize your distribution
normalizedDist = distr/sum(distr);
bar(normalizedDist ,1);
hold on
% Setup your density function using the mean and std of your sample data
mu = mean(SumDice);
stdv = std(SumDice);
yy = normpdf(xx,mu,stdv);
xx = linspace(0,60);
% Plot pdf
h = plot(xx,yy,'r'); set(h,'linewidth',1.5);
I have a simple loglog curve as above. Is there some function in Matlab which can fit this curve by segmented lines and show the starting and end points of these line segments ? I have checked the curve fitting toolbox in matlab. They seems to do curve fitting by either one line or some functions. I do not want to curve fitting by one line only.
If there is no direct function, any alternative to achieve the same goal is fine with me. My goal is to fit the curve by segmented lines and get locations of the end points of these segments .
First of all, your problem is not called curve fitting. Curve fitting is when you have data, and you find the best function that describes it, in some sense. You, on the other hand, want to create a piecewise linear approximation of your function.
I suggest the following strategy:
Split manually into sections. The section size should depend on the derivative, large derivative -> small section
Sample the function at the nodes between the sections
Find a linear interpolation that passes through the points mentioned above.
Here is an example of a code that does that. You can see that the red line (interpolation) is very close to the original function, despite the small amount of sections. This happens due to the adaptive section size.
function fitLogLog()
x = 2:1000;
y = log(log(x));
%# Find section sizes, by using an inverse of the approximation of the derivative
numOfSections = 20;
indexes = round(linspace(1,numel(y),numOfSections));
derivativeApprox = diff(y(indexes));
inverseDerivative = 1./derivativeApprox;
weightOfSection = inverseDerivative/sum(inverseDerivative);
totalRange = max(x(:))-min(x(:));
sectionSize = weightOfSection.* totalRange;
%# The relevant nodes
xNodes = x(1) + [ 0 cumsum(sectionSize)];
yNodes = log(log(xNodes));
figure;plot(x,y);
hold on;
plot (xNodes,yNodes,'r');
scatter (xNodes,yNodes,'r');
legend('log(log(x))','adaptive linear interpolation');
end
Andrey's adaptive solution provides a more accurate overall fit. If what you want is segments of a fixed length, however, then here is something that should work, using a method that also returns a complete set of all the fitted values. Could be vectorized if speed is needed.
Nsamp = 1000; %number of data samples on x-axis
x = [1:Nsamp]; %this is your x-axis
Nlines = 5; %number of lines to fit
fx = exp(-10*x/Nsamp); %generate something like your current data, f(x)
gx = NaN(size(fx)); %this will hold your fitted lines, g(x)
joins = round(linspace(1, Nsamp, Nlines+1)); %define equally spaced breaks along the x-axis
dx = diff(x(joins)); %x-change
df = diff(fx(joins)); %f(x)-change
m = df./dx; %gradient for each section
for i = 1:Nlines
x1 = joins(i); %start point
x2 = joins(i+1); %end point
gx(x1:x2) = fx(x1) + m(i)*(0:dx(i)); %compute line segment
end
subplot(2,1,1)
h(1,:) = plot(x, fx, 'b', x, gx, 'k', joins, gx(joins), 'ro');
title('Normal Plot')
subplot(2,1,2)
h(2,:) = loglog(x, fx, 'b', x, gx, 'k', joins, gx(joins), 'ro');
title('Log Log Plot')
for ip = 1:2
subplot(2,1,ip)
set(h(ip,:), 'LineWidth', 2)
legend('Data', 'Piecewise Linear', 'Location', 'NorthEastOutside')
legend boxoff
end
This is not an exact answer to this question, but since I arrived here based on a search, I'd like to answer the related question of how to create (not fit) a piecewise linear function that is intended to represent the mean (or median, or some other other function) of interval data in a scatter plot.
First, a related but more sophisticated alternative using regression, which apparently has some MATLAB code listed on the wikipedia page, is Multivariate adaptive regression splines.
The solution here is to just calculate the mean on overlapping intervals to get points
function [x, y] = intervalAggregate(Xdata, Ydata, aggFun, intStep, intOverlap)
% intOverlap in [0, 1); 0 for no overlap of intervals, etc.
% intStep this is the size of the interval being aggregated.
minX = min(Xdata);
maxX = max(Xdata);
minY = min(Ydata);
maxY = max(Ydata);
intInc = intOverlap*intStep; %How far we advance each iteraction.
if intOverlap <= 0
intInc = intStep;
end
nInt = ceil((maxX-minX)/intInc); %Number of aggregations
parfor i = 1:nInt
xStart = minX + (i-1)*intInc;
xEnd = xStart + intStep;
intervalIndices = find((Xdata >= xStart) & (Xdata <= xEnd));
x(i) = aggFun(Xdata(intervalIndices));
y(i) = aggFun(Ydata(intervalIndices));
end
For instance, to calculate the mean over some paired X and Y data I had handy with intervals of length 0.1 having roughly 1/3 overlap with each other (see scatter image):
[x,y] = intervalAggregate(Xdat, Ydat, #mean, 0.1, 0.333)
x =
Columns 1 through 8
0.0552 0.0868 0.1170 0.1475 0.1844 0.2173 0.2498 0.2834
Columns 9 through 15
0.3182 0.3561 0.3875 0.4178 0.4494 0.4671 0.4822
y =
Columns 1 through 8
0.9992 0.9983 0.9971 0.9955 0.9927 0.9905 0.9876 0.9846
Columns 9 through 15
0.9803 0.9750 0.9707 0.9653 0.9598 0.9560 0.9537
We see that as x increases, y tends to decrease slightly. From there, it is easy enough to draw line segments and/or perform some other kind of smoothing.
(Note that I did not attempt to vectorize this solution; a much faster version could be assumed if Xdata is sorted.)