Simple Automatic Classification of the (R-->R) Functions - python

Given data values of some real-time physical process (e.g. network traffic) it is to find a name of the function which "at best" matches with the data.
I have a set of functions of type y=f(t) where y and t are real:
funcs = set([cos, tan, exp, log])
and a list of data values:
vals = [59874.141, 192754.791, 342413.392, 1102604.284, 3299017.372]
What is the simplest possible method to find a function from given set which will generate the very similar values?
PS: t is increasing starting from some positive value by almost-equal intervals

Just write the error ( quadratic sum of error at each point for instance ) for each function of the set and choose the function giving the minimum.
But you should still fit each function before choosing

Scipy has functions for fitting data, but they use polynomes or splines. You can use one of Gauß' many discoveries, the method of least squares to fit other functions.

I would try an approach based on fitting too. For each of the four test functions (f1-4 see below), the values of a and b that minimizes the squared error.
f1(t) = a*cos(b*t)
f2(t) = a*tan(b*t)
f3(t) = a*exp(b*t)
f4(t) = a*log(b*t)
After fitting the squared error of the four functions can be used for evaluating the fit goodness (low values means a good fit).
If fitting is not allowed at all, the four functions can be divided into two distinct subgroups, repeating functions (cos and tan) and strict increasing functions (exp and log).
Strict increasing functions can be identified by checking if all the given values are increasing throughout the measuring interval.
In pseudo code an algorithm could be structured like
if(vals are strictly increasing):
% Exp or log
if(increasing more rapidly at the end of the interval):
% exp detected
else:
% log detected
else:
% tan or cos
if(large changes in vals over a short period is detected):
% tan detected
else:
% cos detected
Be aware that this method is not that stable and will be easy to trick into faulty conclusions.

See Curve Fitting

Related

stats.rv_continuous slow when when using custom pdf

Ultimately I am trying to visualise the copula between two PDFs which are estimated from data (both via a KDE). Suppose, for one of the KDEs, I have discrete x,y data sorted in a tuple called data. I need to generate random variables with this distribution in order to perform the probability integral transform (and ultimately to obtain the uniform distribution). My methodology to generate random variables is as follows:
import scipy.stats as st
from scipy import interpolate, integrate
pdf1 = interpolate.interp1d(data[0], data[1])
class pdf1_class(st.rv_continuous):
def _pdf(self,x):
return pdf1(x)
pdf1_rv = pdf1_class(a = data[0][0], b= data[0][-1], name = 'pdf1_class')
pdf1_samples = pdf1_rv.rvs(size=10000)
However, this method is extremely slow. I also get the following warnings:
IntegrationWarning: The maximum number of subdivisions (50) has been achieved.
If increasing the limit yields no improvement it is advised to analyze
the integrand in order to determine the difficulties. If the position of a
local difficulty can be determined (singularity, discontinuity) one will
probably gain from splitting up the interval and calling the integrator
on the subranges. Perhaps a special-purpose integrator should be used.
warnings.warn(msg, IntegrationWarning)
IntegrationWarning: The occurrence of roundoff error is detected, which prevents
the requested tolerance from being achieved. The error may be
underestimated.
warnings.warn(msg, IntegrationWarning)
Is there a better way to generate the random variables?
As per suggestion by #unutbu I implemented _cdf and _ppf, which makes the calculation of 10000 samples instantaneous. To do this I added the following to the above code:
discrete_cdf1 = integrate.cumtrapz(y=data[1], x = data[0])
cdf1 = interpolate.interp1d(data[0][1:], discrete_cdf1)
ppf1 = interpolate.interp1d(discerete_cdf1, data[0][:-1])
I then add the following two methods to pdf1_class
def _cdf(self,x):
return cdf1(x)
def _ppf(self,x):
return ppf1(x)

Python circle fitting to data points less sensitive to random noise

I have a set of measured radii (t+epsilon+error) at an equally spaced angles.
The model is circle of radius (R) with center at (r, Alpha) with added small noise and some random error values which are much bigger than noise.
The problem is to find the center of the circle model (r,Alpha) and the radius of the circle (R). But it should not be too much sensitive to random error (in below data points at 7 and 14).
Some radii could be missing therefore the simple mean would not work here.
I tried least square optimization but it significantly reacts on error.
Is there a way to optimize least deltas but not the least squares of delta in Python?
Model:
n=36
R=100
r=10
Alpha=2*Pi/6
Data points:
[95.85, 92.66, 94.14, 90.56, 88.08, 87.63, 88.12, 152.92, 90.75, 90.73, 93.93, 92.66, 92.67, 97.24, 65.40, 97.67, 103.66, 104.43, 105.25, 106.17, 105.01, 108.52, 109.33, 108.17, 107.10, 106.93, 111.25, 109.99, 107.23, 107.18, 108.30, 101.81, 99.47, 97.97, 96.05, 95.29]
It seems like your main problem here is going to be removing outliers. There are a couple of ways to do this, but for your application, your best bet is to probably just to remove items based on their distance from the median (Since the median is much less sensitive to outliers than the mean.)
If you're using numpy that would looks like this:
def remove_outliers(data_points, margin=1.5):
nd = np.abs(data_points - np.median(data_points))
s = nd/np.median(nd)
return data_points[s<margin]
After which you should run least squares.
If you're not using numpy you can do something similar with native python lists:
def median(points):
return sorted(points)[len(points)/2] # evaluates to an int in python2
def remove_outliers(data_points, margin=1.5):
m = median(data_points)
centered_points = [abs(point - m) for point in data_points]
centered_median = median(centered_points)
ratios = [datum/centered_median for datum in centered_points]
return [point for i, point in enumerate(data_points) if ratios[i]>margin]
If you're looking to just not count outliers as highly you can just calculate the mean of your dataset, which is just a linear equivalent of the least-squares optimization.
If you're looking for something a little better I might suggest throwing your data through some kind of low pass filter, but I don't think that's really needed here.
A low-pass filter would probably be the best, which you can do as follows: (Note, alpha is a number you will have to fiddle with to get your desired output.)
def low_pass(data, alpha):
new_data = [data[0]]
for i in range(1, len(data)):
new_data.append(alpha * data[i] + (1 - alpha) * new_data[i-1])
return new_data
At which point your least squares optimization should work fine.
Replying to your final question
Is there a way to optimize least deltas but not the least squares of delta in Python?
Yes, pick an optimization method (for example downhill simplex implemented in scipy.optimize.fmin) and use the sum of absolute deviations as a merit function. Your dataset is small, I suppose that any general purpose optimization method will converge quickly. (In case of non-linear least squares fitting it is also possible to use general purpose optimization algorithm, but it's more common to use the Levenberg-Marquardt algorithm which minimizes sums of squares.)
If you are interested when minimizing absolute deviations instead of squares has theoretical justification see Numerical Recipes, chapter Robust Estimation.
From practical side, the sum of absolute deviations may not have unique minimum.
In the trivial case of two points, say, (0,5) and (1,9) and constant function y=a, any value of a between 5 and 9 gives the same sum (4). There is no such problem when deviations are squared.
If minimizing absolute deviations would not work, you may consider heuristic procedure to identify and remove outliers. Such as RANSAC or ROUT.

polyfit refining: setting polynomial to be always possitive

I am trying to fit a polynomial to my data, e.g.
import scipy as sp
x = [1,6,9,17,23,28]
y = [6.1, 7.52324, 5.71, 5.86105, 6.3, 5.2]
and say I know the degree of polynomial (e.g.: 3), then I just use scipy.polyfit method to get the polynomial of a given degree:
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
fittedModelFunction = sp.polyfit(x, y, 3)
func = sp.poly1d(fittedModelFunction)
++++++++++++++++++++++++++++++
QUESTIONS: ++++++++++++++++++++++++++++++
1) How can I tell in addition that the resulting function func must be always positive (i.e. f(x) >= 0 for any x)?
2) How can I further define a constraint (e.g. number of (local) min and max points, etc.) in order to get a better fitting?
Is there smth like this:
http://mail.scipy.org/pipermail/scipy-user/2007-July/013138.html
but more accurate?
Always Positve
I haven't been able to find a scipy reference that determines if a function is positive-definite, but an indirect way would be to find the all the roots - Scipy Roots - of the function and inspect the limits near those roots. There are a few cases to consider:
No roots at all
Pick any x and evaluate the function. Since the function does not cross the x-axis because of a lack of roots, any positive result will indicate the function is positive!
Finite number of roots
This is probably the most likely case. You would have to inspect the limits before and after each root - Scipy Limits. You would have to specify your own minimum acceptable delta for the limit however. I haven't seen a 2-sided limit method provided by Scipy, but it looks simple enough to make your own.
from sympy import limit
// f: function, v: variable to limit, p: point, d: delta
// returns two limit values
def twoSidedLimit(f, v, p, d):
return limit(f, v, p-d), limit(f, v, p+d)
Infinite roots
I don't think that polyfit would generate an oscillating function, but this is something to consider. I don't know how to handle this with the method I have already offered... Um, hope it does not happen?
Constraints
The only built-in form of constraints seems to be limited to the optimize library of SciPy. A crude way to enforce constraints for polyfit would be to get the function from polyfit, generate a vector of values for various x, and try to select values from the vector that violate the constraint. If you try to use filter, map, or lambda it may be slow with large vectors since python's filter makes a copy of the list/vector being filtered. I can't really help in this regard.

(Python) Estimating regression parameter confidence intervals with scikits bootstrap

I've just started to try out a nice bootstrapping package available through scikits:
https://github.com/cgevans/scikits-bootstrap
but I've encountered a problem when trying to estimate confidence intervals for the correlation coefficient from linear regression. The confidence intervals returned lie completely outside the range of the original statistic.
Here is the code:
import numpy as np
from scipy import stats
import bootstrap as boot
np.random.seed(0)
x = np.arange(10)
y = 10 + 1.5*x + 2*np.random.randn(10)
r0 = stats.linregress(x, y)[2]
def my_function(y):
return stats.linregress(x, y)[2]
ci = boot.ci(y, statfunction=my_function, alpha=0.05, n_samples=1000, method='pi')
This yields a result of ci = [-0.605, 0.644], but the original statistic is r0=0.894.
I've tried this in R and it seems to work fine there: the ci straddles r0 as expected.
Please help!
Could you provide your R code? I'd be interested in knowing how this is dealt with in R.
The problem here is that you're only passing y to boot.ci, but every time it runs my_function, it uses the entire, original x (note the lack of x input to my_function). Bootstrapping applies the statistic function to resampled data, so if you're applying your statistic function using the original x and a sample of y, you're going to have a nonsensical result. This is why the BCA method doesn't work at all, actually: it can't apply your statistic function to jackknife samples, which don't have the same number of elements.
Samples are taken along axis 0 (rows), so if you want to pass multiple 1D arrays to your statistic function, you can use multiple columns: xy = vstack((x,y)).T would work, and then use a statfunction that takes data from those columns:
def my_function(xysample):
return stats.linregress(xysample[:,0], xysample[:,1])[2]
Alternatively, if you wanted to avoid messing with your data at all, you could define a function that operates on indexes, and then just pass indexes to boot.ci:
def my_function2(i):
return stats.linregress(x[i], y[i])[2]
boot.ci(np.arange(len(x)), statfunction=my_function2, alpha=0.05, n_samples=1000, method='pi')
Note that in either of these cases, BCA works, so you may as well use method='bca' unless you really do want to use percentage intervals; BCA is pretty much always better.
I do realize that both of these methods are less than ideal. Honestly, I've never had a need to pass multiple arrays like this to my statfunction, and the majority of people are likely using mean as their statfunction. I think the best idea here may be to allow lists of equally-size[0] arrays to be passed, eg, boot.ci([x,y],...), and then sample all of those at the same time and pass them all to the statfunction as separate arguments. In that case, you could just have a my_function(x,y). I'll see if I can do this, but if you can show me your R code, that would be great, as I'd like to see if there is a better way of dealing with this.
Update:
In the most recent version of scikits.bootstrap (v0.3.1), a tuple of arrays can be provided, and samples from them will be passed as separate arguments to statfunction. Additionally, statfunction can provide array output, and confidence intervals will be calculated for each point in the output. Thus, this is now very easy to do. The following will give confidence intervals for every output of linregress:
cis = boot.ci( (x,y), statfunction=stats.linregress )
cis[:,2] in this case will be the desired confidence interval.

autocorrelation function of time series data with numpy

I have been trying to calculate an autocorrelation function, as defined in statistical mechanics, using numpy. Most of the documentation I found is relative to functions like correlate and convolve. However, for a given random variable x these functions just seem to calculate the sum
ACF(dt) = sum_{t=0}^T [(x(t)*x(t+dt)]
instead of the average
ACF(dt) = mean[x(t)*x(t+dt)]
so in fact for calculating an autocorrelation function one would need to do something like:
acf = np.correlate(x,x,mode='full')
acf_half = acf[acf.size / 2:]
ldata = len(acf)
acf = np.array([x/(ldata-i) for i,x in enumerate(acf_half)])
Of course we would need to subtract mean(x)**2 from the resulting acf to be correct.
Can anyone confirm that this is correct?
Generally speaking, the autocorrelation, correlation, etc. is the sum (integral). Sometimes it is normalized, but not averaged in the sense as you've written above. This is because they are defined in terms of the mathematical convolution operation, which is simply the integral that you've written as a sum above.
The brackets at the stat mech page indicate a thermal average, which is an ensemble or time average over the 'experiment' taking place many times at many different states at some temperature. This (the finite temperature) causes the fluctuations that give rise to the 'statistical' nature of the problem, and cause the decay of the correlation (loss of long range order). This simply means that you should find the autocorrelation of several datasets, and average those together, but do not take the mean of the function.
As far as I can tell, your code is attempting to weigh the correlation at dt by the length of the overlap length dt, but I do not believe that this is correct.
With respect to the subtraction of <s>2, that's in the case of the spin model, where <s> would be the mean spin (magnetization), so I believe you are correct in that you should use mean(x)**2.
As a side-note, I would suggest using mode='same' instead of 'full' so that the domain of your correlation matches the domain of your input without having to look at just one-half of the output (here the output is symmetric, so it doesn't really make a difference).

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