I am trying to reduce dimensionality of a training set using PCA.
I have come across two approaches.
[V,U,eigen]=pca(train_x);
eigen_sum=0;
for lamda=1:length(eigen)
eigen_sum=eigen_sum+eigen(lamda,1);
if(eigen_sum/sum(eigen)>=0.90)
break;
end
end
train_x=train_x*V(:, 1:lamda);
Here, I simply use the eigenvalue matrix to reconstruct the training set with lower amount of features determined by principal components describing 90% of original set.
The alternate method that I found is almost exactly the same, save the last line, which changes to:
train_x=U(:,1:lamda);
In other words, we take the training set as the principal component representation of the original training set up to some feature lamda.
Both of these methods seem to yield similar results (out of sample test error), but there is difference, however minuscule it may be.
My question is, which one is the right method?
The answer depends on your data, and what you want to do.
Using your variable names. Generally speaking is easy to expect that the outputs of pca maintain
U = train_x * V
But this is only true if your data is normalized, specifically if you already removed the mean from each component. If not, then what one can expect is
U = train_x * V - mean(train_x * V)
And in that regard, weather you want to remove or maintain the mean of your data before processing it, depends on your application.
It's also worth noting that even if you remove the mean before processing, there might be some small difference, but it will be around floating point precision error
((train_x * V) - U) ./ U ~~ 1.0e-15
And this error can be safely ignored
Related
There is a [Q,R] = qr(A,0) function in Matlab, which, according to documentation, returns an "economy" version of qr-decomposition of A. norm(A-Q*R) returns ~1e-12 for my data set. Also Q'*Q should theoretically return I. In practice there are small nonzero elements above and below the diagonal (of the order of 1e-6 or so), as well as diagonal elements that are slightly greater than 1 (again, by 1e-6 or so). Is anyone aware of a way to control precision of qr(.,0), or quality(orthogonality) of resulting Q, either by specifying epsilon, or via the number of iterations ? The size of the data set makes qr(A) run out of memory so I have to use qr(A,0).
When I try the non- economy setting, I actually get comparable results for A-Q*R. Even for a tiny matrix containing small numbers as shown here:
A = magic(20);
[Q, R] = qr(A); %Result does not change when using qr(A,0)
norm(A-Q*R)
As such I don't believe the 'economy' is the problem as confirmed by #horchler in the comments, but that you have just ran into the limits of how accurate calculations can be done with data of type 'double'.
Even if you change the accuracy somehow, you will always be dealing with an approximation, so perhaps the first thing to consider here is whether you really need greater accuracy than you already have. If you need more accuracy there may always be a way, but I doubt whether it will be a straightforward one.
Why is the fitfunction from Matlab so slow? I'm trying to fit a gauss4 so I can get the means of the gaussians.
here's my plot,
I want to get the means from the blue data and red data.
I'm fitting a gaussian there but this function is really slow.
Is there an alternative?
fa = fit(fn', facm', 'gauss4');
acm = [fa.b1 fa.b2 fa.b3 fa.b4];
a_cm = sort(acm, 'ascend');
I would apply some of the options available with fit. These include smoothing by setting SmoothingParam (your data is quite noisy, the alternative of applying a time domain filter may also help*), and setting the values of your initial parameter estimates, with StartPoint. Your fits may also not be converging because you set your tolerances (TolFun, TolX) too low, although from inspection of your fits that does not appear to be the case, in fact the opposite is likely, you probably want to increase the MaxIter and/or MaxFunEvals.
To figure out how to get going you can also try the Spectr-O-Matic toolbox. It requires Matlab 7.12. It includes a script called GaussFit.m to fit gauss4 to data, but it also uses the fit routine and provides examples on how to set and get parameters.
Note that smoothing will of course broaden your peaks, but you can subtract the contribution after the fact. The effect on the mean should not be deleterious, on the contrary, since you are presumably removing noise this should be more accurate.
In general functions will be faster if you apply it to a shorter series. Hence, if speedup is really important you could downsample.
For example, if you have a vector that you want to downsample by a factor 2: (you may need to make sure it fits first)
n = 2;
x = sin(0.01:0.01:pi);
x_downsampled = x(1:n:end)+x(2:n:end);
You will now see that x_downsampled is much smaller (and should thus be easier to process), but will still have the same shape. In your case I think this is sufficient.
To see what you got try:
plot(x)
Now you can simply process x_downsampled and map your solution, for example
f = find(x_downsampled == max(x_downsampled));
location_of_maximum = f * n;
Needless to say this should be done in combination with the most efficient options that the fit function has to offer.
first a little background. I'm a psychology student so my background in coding isn't on par with you guys :-)
My problem is as follow and the most important observation is that curve fitting with 2 different programs gives completly different results for my parameters, altough my graphs stay the same. The main program we have used to fit my longitudinal data is kaleidagraph and this should be seen as kinda the 'golden standard', the program I'm trying to modify is matlab.
I was trying to be smart and wrote some code (a lot at least for me) and the goal of that code was the following:
1. Taking an individual longitudinal datafile
2. curve fitting this data on a non-parametric model using lsqcurvefit
3. obtaining figures and the points where f' and f'' are zero
This all worked well (woohoo :-)) but when I started comparing the function parameters both programs generate there is a huge difference. The kaleidagraph program stays close to it's original starting values. Matlab wanders off and sometimes gets larger by a factor 1000. The graphs stay however more or less the same in both situations and both fit the data well. However it would be lovely if I would know how to make the matlab curve fitting more 'conservative' and more located near it's original starting values.
validFitPersons = true(nbValidPersons,1);
for i=1:nbValidPersons
personalData = data{validPersons(i),3};
personalData = personalData(personalData(:,1)>=minAge,:);
% Fit a specific model for all valid persons
try
opts = optimoptions(#lsqcurvefit, 'Algorithm', 'levenberg-marquardt');
[personalParams,personalRes,personalResidual] = lsqcurvefit(heightModel,initialValues,personalData(:,1),personalData(:,2),[],[],opts);
catch
x=1;
end
Above is a the part of the code i've written to fit the datafiles into a specific model.
Below is an example of a non-parametric model i use with its function parameters.
elseif strcmpi(model,'jpa2')
% y = a.*(1-1/(1+(b_1(t+e))^c_1+(b_2(t+e))^c_2+(b_3(t+e))^c_3))
heightModel = #(params,ages) abs(params(1).*(1-1./(1+(params(2).* (ages+params(8) )).^params(5) +(params(3).* (ages+params(8) )).^params(6) +(params(4) .*(ages+params(8) )).^params(7) )));
modelStrings = {'a','b1','b2','b3','c1','c2','c3','e'};
% Define initial values
if strcmpi('male',gender)
initialValues = [176.76 0.339 0.1199 0.0764 0.42287 2.818 18.52 0.4363];
else
initialValues = [161.92 0.4173 0.1354 0.090 0.540 2.87 14.281 0.3701];
end
I've tried to mimick the curve fitting process in kaleidagraph as good as possible. There I've found they use the levenberg-marquardt algorithm which I've selected. However results still vary and I don't have any more clues about how I can change this.
Some extra adjustments:
The idea for this code was the following:
I'm trying to compare different fitting models (they are designed for this purpose). So what I do is I have 5 models with different parameters and different starting values ( the second part of my code) and next I have the general curve fitting file. Since there are different models it would be interesting if I could put restrictions into how far my starting values could wander off.
Anyone any idea how this could be done?
Anybody willing to help a psychology student?
Cheers
This is a common issue when dealing with non-linear models.
If I were, you, I would try to check if you can remove some parameters from the model in order to simplify it.
If you really want to keep your solution not too far from the initial point, you can use upper bounds and lower bounds for each variable:
x = lsqcurvefit(fun,x0,xdata,ydata,lb,ub)
defines a set of lower and upper bounds on the design variables in x so that the solution is always in the range lb ≤ x ≤ ub.
Cheers
You state:
I'm trying to compare different fitting models (they are designed for
this purpose). So what I do is I have 5 models with different
parameters and different starting values ( the second part of my code)
and next I have the general curve fitting file.
You will presumably compare the statistics from fits with different models, to see whether reductions in the fitting error are unlikely to be due to chance. You may want to rely on that comparison to pick the model that not only fits your data suitably but is also simplest (which is often referred to as the principle of parsimony).
The problem is really with the model you have shown resulting in correlated parameters and therefore overfitting, as mentioned by #David. Again, this should be resolved when you compare different models and find that some do just as well (statistically speaking) even though they involve fewer parameters.
edit
To drive the point home regarding the problem with the choice of model, here are (1) results of a trial fit using simulated data (2) the correlation matrix of the parameters in graphical form:
Note that absolute values of the correlation close to 1 indicate strongly correlated parameters, which is highly undesirable. Note also that the trend in the data is practically linear over a long portion of the dataset, which implies that 2 parameters might suffice over that stretch, so using 8 parameters to describe it seems like overkill.
I have 2 sets of data of float numbers, set A and set B. Both of them are matrices of size 40*40. I would like to find out which set is closer to the normal distribution. I know how to use probplot() in matlab to plot the probability of one set. However, I do not know how to find out the level of the fitness of the distribution is.
In python, when people use problot, a parameter ,R^2, shows how good the distribution of the data is against to the normal distribution. The closer the R^2 value to value 1, the better the fitness is. Thus, I can simply use the function to compare two set of data by their R^2 value. However, because of some machine problem, I can not use the python in my current machine. Is there such parameter or function similar to the R^2 value in matlab ?
Thank you very much,
Fitting a curve or surface to data and obtaining the goodness of fit, i.e., sse, rsquare, dfe, adjrsquare, rmse, can be done using the function fit. More info here...
The approach of #nate (+1) is definitely one possible way of going about this problem. However, the statistician in me is compelled to suggest the following alternative (that does, alas, require the statistics toolbox - but you have this if you have the student version):
Given that your data is Normal (not Multivariate normal), consider using the Jarque-Bera test.
Jarque-Bera tests the null hypothesis that a given dataset is generated by a Normal distribution, versus the alternative that it is generated by some other distribution. If the Jarque-Bera test statistic is less than some critical value, then we fail to reject the null hypothesis.
So how does this help with the goodness-of-fit problem? Well, the larger the test statistic, the more "non-Normal" the data is. The smaller the test statistic, the more "Normal" the data is.
So, assuming you have converted your matrices into two vectors, A and B (each should be 1600 by 1 based on the dimensions you provide in the question), you could do the following:
%# Build sample data
A = randn(1600, 1);
B = rand(1600, 1);
%# Perform JB test
[ANormal, ~, AStat] = jbtest(A);
[BNormal, ~, BStat] = jbtest(B);
%# Display result
if AStat < BStat
disp('A is closer to normal');
else
disp('B is closer to normal');
end
As a little bonus of doing things this way, ANormal and BNormal tell you whether you can reject or fail to reject the null hypothesis that the sample in A or B comes from a normal distribution! Specifically, if ANormal is 1, then you fail to reject the null (ie the test statistic indicates that A is probably drawn from a Normal). If ANormal is 0, then the data in A is probably not generated from a Normal distribution.
CAUTION: The approach I've advocated here is only valid if A and B are the same size, but you've indicated in the question that they are :-)
I'm trying to compute the log of the mean of some very small values. For the current data set, the extreme points are
log_a=-1.6430e+03;
log_b=-3.8278e+03;
So in effect I want to compute (a+b) / 2, or log((a+b)/2) since I know (a+b)/2 is too small to store as a double.
I considered trying to pad everything by a constant, so that instead of storing log_a I'd store log_a+c, but it seems that aand b are far enough apart that in order to pad log_b enough to make exp(log_b+c) computable, I'd end up making exp(log_a+c) too large.
Am I missing some obvious way to go about this computation? As far as I know MATLAB won't let me use anything but double precision, so I'm stumped as to how I can do this simple computation.
EDIT: To clarify: I can compute the exact answer for these specific values. For other runs of the algorithm, the values will be different and might be closer together. So far there have been some good suggestions for approximations; if an exact solution isn't practical, are there any other approximations for more general numbers/magnitudes of values?
Mystical has the right idea but for a more general solution that gives you the log of the arithmetic mean of a vector log_v of numbers already in the log domain use:
max_log = max(log_v);
logsum = max_log + log(sum(exp(log_v-max_log)));
logmean = logsum - log(length(log_v));
This is a common problem in statistical machine learning, so if you do a Google search for logsum.m you'll find a few different versions of MATLAB functions that researchers have written for this purpose. For example, here's a Github link to a version that uses the same calling conventions as sum.
Well, exp(log_b) is so much smaller than exp(log_a) that you can completely ignore that term and still get the correct answer with respect to double-precision:
exp(log_a) = 2.845550077506*10^-714
exp(log_b) = 4.05118588390*10^-1663
If you are actually trying to compute (exp(log_a) + exp(log_b)) / 2, the answer would underflow to zero anyways. So it wouldn't really matter unless you're trying to take another logarithm at the end.
If you're trying compute:
log((exp(log_a) + exp(log_b)) / 2)
Your best bet is to examine the difference between log_a and log_b. If the difference is large, then simply take the final value as equal to the larger term - log(2) since the smaller term will be small enough to completely vanish.
EDIT:
So your final algorithm could look like this:
Check the magnitudes. If abs(log_a - log_b) > 800. Return max(log_a,log(b)) - log(2).
Check either magnitude (they will be close together at this point.). If it is much larger or smaller than 1, add/subtract a constant from both log_a and log_b.
Perform the calculation.
If the values were scaled in step 2. Scale the result back.
EDIT 2:
Here's an even better solution:
if (log_a > log_b)
return log_a + log(1 + exp(log_b - log_a)) - log(2)
else
return log_b + log(1 + exp(log_a - log_b)) - log(2)
This will work if log_a and log_b are not too large or are negative.
Well, if you don't like my previous suggestion of completely changing platforms and are looking for an approximation, why not just use the geometric mean (exp((log_a+log_b)/2) instead?
Use http://wolframalpha.com . For example, as discussed by Mysticial, your calculation of
log(exp(-1.6430e+03) + exp(-3.8278e+03)/2) is approximately equal to log_a. More precisely it equals...
1642.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999288154175193167154874243862288962865800888654829363675488466381404578225092913407982036991983506370017587380105049077722517705727311433458060227246074261903850589008701929721367650576354241270720062760800558681236724831345952032973775644175750495894596292205385323394564549904750849335403418234531787942293155499938538026848481952030717783105220543888597195156662520697417952130625416040662694927878196360733032741542418365527213770518383992577797346467266676866552563022498785887306273550235307330535082355570343750317349638125974233837177558240980392298326807001406291035229026016040567173260205109683449441154277953394697235601979288239733693137185710713089424316870093563207034737497769711306780243623361236030692934897720786516684985651633787662244416960982457075265287065358586526093347161275192566468776617613339812566918101457823704547101340270795298909954224594...