I am very new to mat lab and I am trying to get to grips with integrating under a curve.
I wanted to see this difference between using the 'trapz(y)' and 'trapz(x,y)' to find the area under a curve of a Gaussian function what I can not seem to understand is why I am getting two different area values and I am trying to figure which one is more accurate.
dataset = xlsread('Lab 3 Results 11.10.18 (1).xlsx','Sheet3','C6:D515');
x=dataset(:,1);
a1=38.38;
b1=1179;
c1=36.85;
d1=6.3
y=a1*exp(-((x-b1)/c1).^2)-d1;
int1=trapz(x,y)
int2=trapz(y)
So when I run this code I get int1=1738.3 and int2=5.78.4 but when I integrated this function by hand using the trapezium rule my ans came out to be nearer int1 rather that int2 is there anyone that could shed some light on this if possible? I just cant imagen visulay how matlab is using the trapz rule two different ways,
Neither implementation is more accurate, but trapz(y) assumes unit spacing of each data point (e.g., spacing between data points is uniformly x = 1). See trapz documentation.
Since you know the x-coordinates, use trapz(x,y).
Related
I have three vectors (22257x1). I want to plot these in a 3D mesh/wireframe with a colour corresponding to the 'Z'-value.
My efforts so far:
plot3(): allows for only one colour
scatter3(): does not connect the points
mesh/surf: require matrixes instead of vectors (for Z at least)
magic()/reshape() for fitting the vectors to the mesh/surf functions did not work for me, also for gridding, memory was not enough...
Many snippets I could find on the www (inc. StackOverflow)
Could anyone please help?
Here is the given system I want to plot and obtain the vector field and the angles they make with the x axis. I want to find the index of a closed curve.
I know how to do this theoretically by choosing convenient points and see how the vector looks like at that point. Also I can always use
to compute the angles. However I am having trouble trying to code it. Please don't mark me down if the question is unclear. I am asking it the way I understand it. I am new to matlab. Can someone point me in the right direction please?
This is a pretty hard challenge for someone new to matlab, I would recommend taking on some smaller challenges first to get you used to matlab's conventions.
That said, Matlab is all about numerical solutions so, unless you want to go down the symbolic maths route (and in that case I would probably opt for Mathematica instead), your first task is to decide on the limits and granularity of your simulated space, then define them so you can apply your system of equations to it.
There are lots of ways of doing this - some more efficient - but for ease of understanding I propose this:
Define the axes individually first
xpts = -10:0.1:10;
ypts = -10:0.1:10;
tpts = 0:0.01:10;
The a:b:c syntax gives you the lower limit (a), the upper limit (c) and the spacing (b), so you'll get 201 points for the x. You could use the linspace notation if that suits you better, look it up by typing doc linspace into the matlab console.
Now you can create a grid of your coordinate points. You actually end up with three 3d matrices, one holding the x-coords of your space and the others holding the y and t. They look redundant, but it's worth it because you can use matrix operations on them.
[XX, YY, TT] = meshgrid(xpts, ypts, tpts);
From here on you can perform whatever operations you like on those matrices. So to compute x^2.y you could do
x2y = XX.^2 .* YY;
remembering that you'll get a 3d matrix out of it and all the slices in the third dimension (corresponding to t) will be the same.
Some notes
Matlab has a good builtin help system. You can type 'help functionname' to get a quick reminder in the console or 'doc functionname' to open the help browser for details and examples. They really are very good, they'll help enormously.
I used XX and YY because that's just my preference, but I avoid single-letter variable names as a general rule. You don't have to.
Matrix multiplication is the default so if you try to do XX*YY you won't get the answer you expect! To do element-wise multiplication use the .* operator instead. This will do a11 = b11*c11, a12 = b12*c12, ...
To raise each element of the matrix to a given power use .^rather than ^ for similar reasons. Likewise division.
You have to make sure your matrices are the correct size for your operations. To do elementwise operations on matrices they have to be the same size. To do matrix operations they have to follow the matrix rules on sizing, as will the output. You will find the size() function handy for debugging.
Plotting vector fields can be done with quiver. To plot the components separately you have more options: surf, contour and others. Look up the help docs and they will link to similar types. The plot family are mainly about lines so they aren't much help for fields without creative use of the markers, colours and alpha.
To plot the curve, or any other contour, you don't have to test the values of a matrix - it won't work well anyway because of the granularity - you can use the contour plot with specific contour values.
Solving systems of dynamic equations is completely possible, but you will be doing a numeric simulation and your results will again be subject to the granularity of your grid. If you have closed form solutions, like your phi expression, they may be easier to work with conceptually but harder to get working in matlab.
This kind of problem is tractable in matlab but it involves some non-basic uses which are pretty hard to follow until you've got your head round Matlab's syntax. I would advise to start with a 2d grid instead
[XX, YY] = meshgrid(xpts, ypts);
and compute some functions of that like x^2.y or x^2 - y^2. Get used to plotting them using quiver or plotting the coordinates separately in intensity maps or surfaces.
I am trying to build a receiver operating characteristic (ROC) curves to evaluate the discriminating ability of my classifier to correctly classify diseased and non-diseased subjects.
I understand that the closer the curve follows the left-hand border and then the top border of the ROC space, the more accurate the test. My experiments gave me quite desirable value of area under curve (auc), i.e. 0.86458. However, the ROC curve (in which I included the cut-off points for tracing purposes) seems quite strange as it gave me straight lines as below:
... and not a curve I expected and as I normally see from any references like this:
Does it hav something to do with the number of observations used? (in this case I only have 50 samples). Or is this just fine as long as the the auc value is high and that the 'curve' comes above the 45-degree diagonal of the ROC space? I would be glad if someone can share their thoughts about it. Thank you!
By the way, I used the perfcurve() function in matlab:
% ROC comparison between the proposed approach and the baseline
[X1,Y1,T1,auc1,OPTROCPT1,SUBY5,SUBYNAMES1] = perfcurve(testLabel,predlabel_prop,1);
[X2,Y2,T2,auc2,OPTROCPT2,SUBY2,SUBYNAMES2] = perfcurve(testLabel,predLabel_base,1);
figure;
plot(X1,Y1,'-r*',X2,Y2,'--ko');
legend('proposed approach','baseline','Location','east');
xlabel('False positive rate'); ylabel('True positive rate')
title('ROC comparison of the proposed approach and the baseline')
text(0.6,0.3,{'* - proposed method',strcat('Area Under Curve = ',...
num2str(auc1))},'EdgeColor','r');
text(0.6,0.15,{'o - baseline',strcat('Area Under Curve = ',num2str(auc2))},'EdgeColor','k');
You probably have too litte data.
You curve indicates your data set has 13 negative and 5 positive examples (in your test set?)
Furthermore, all but 4 have exactly the same score (maybe 0)? Or is that your cutoff?
Given this small sample size, I would not accept the hypothesis that your proposed method is better than the baseline, but accept the alternative - the methods perform as good as the other: the difference of 0.04 is much too small for this tiny sample size, the results are virtually identical. Any variation within the cut-off area (the diagonal part) can be much larger than this 0.04... On a different run, a different test set, the results may be the other way around.
Shape of your curve is just a result of high explanatory power of your model and limited number of observations (e.g. take a look at the example here http://nl.mathworks.com/help/stats/perfcurve.html).
I have some data which I wish to model in order to be able to get relatively accurate values in the same range as the data.
To do this I used polyfit to fit a 6th order polynomial and due to my x-axis values it suggested I centred and scaled it to get a more accurate fit which I did.
However, now I want to find the derivative of this function in order to model the velocity of my model.
But I am not sure how the polyder function interacts with the scaled and fitted polyfit which I have produced. (I don't want to use the unscaled model as this is not very accurate).
Here is some code which reproduces my problem. I attempted to rescale the x values before putting them into the fit for the derivative but this still did no fix the problem.
x = 0:100;
y = 2*x.^2 + x + 1;
Fit = polyfit(x,y,2);
[ScaledFit,s,mu] = polyfit(x,y,2);
Deriv = polyder(Fit);
ScaledDeriv = polyder(ScaledFit);
plot(x,polyval(Deriv,x),'b.');
hold on
plot(x,polyval(ScaledDeriv,(x-mu(1))/mu(2)),'r.');
Here I have chosen a simple polynomial so that I could fit it accurate and produce the actual derivative.
Any help would be greatly appreciated thanks.
I am using Matlab R2014a BTW.
Edit.
Just been playing about with it and by dividing the resulting points for the differential by the standard deviation mu(2) it gave a very close result within the range -3e-13 to about 5e-13.
polyval(ScaledDeriv,(x-mu(1))/mu(2))/mu(2);
Not sure quite why this is the case, is there another more elegant way to solve this?
Edit2. Sorry for another edit but again was mucking around and found that for a large sample x = 1:1000; the deviation became much bigger up to 10. I am not sure if this is due to a bad polyfit even though it is centred and scaled or due to the funny way the derivative is plotted.
Thanks for your time
A simple application of the chain rule gives
Since by definition
it follows that
Which is exactly what you have verified numerically.
The lack of accuracy for large samples is due to the global, rather then local, Lagrange polynomial interpolation which you have done. I would suggest that you try to fit your data with splines, and obtain the derivative with fnder(). Another option is to apply the polyfit() function locally, i.e. to a moving small set of points, and then apply polyder() to all the fitted polynomials.
I want to plot some equations and inequalities like x>=50, y>=0,4x-5y>=8,x=40,x=60,y=25, y=45 in matlab and want to get the area produced by intersecting these equations and inequalities. Is it possible using matlab? If yes can someone provide me some manual? If not, is there some other software that can do this?
Integrals would work for your purposes, provided you know the points at which the curves intersect (something Matlab is also able to compute). Take a look at the documentation on the integral function.
q = integral(fun,xmin,xmax) approximates the integral of function fun
from xmin to xmax using global adaptive quadrature and default error
tolerances.
EDIT: As an additional resource, take a look at the code provided by user Grzegorz Konz on the Mathworks blog.
EDIT #2: I'm not familiar with any Matlab functions that'll take a vector of functions and return the points of intersection (if any) between all the curves. Users have produced functions that return the set of intersection points between two curves. You could run this function for each pair of equations in your list and use a function like polyarea to compute the area of the enclosed region if the curves are all straight lines.