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Python - How to resample a 2D shape?

I am writing a python script for some geometrical data manipulation (calculating motion trajectories for a multi-drive industrial machine). Generally, the idea is that there is a given shape (let's say - an ellipse, but it general case it can be any convex shape, defined with a series of 2D points), which is rotated and it's uppermost tangent point must be followed. I don't have a problem with the latter part but I need a little hint with the 2D shape preparation.
Let's say that the ellipse was defined with too little points, for example - 25. (As I said, ultimately this can be any shape, for example a rounded hexagon). To maintain necessary precision I need far more points (let's say - 1000), preferably equally distributed over whole shape or with higher density of points near corners, sharp curves, etc.
I have a few things ringing in my head, I guess that DFT (FFT) would be a good starting point for this resampling, analyzing the scipy.signal.resample() I have found out that there are far more functions in the scipy.signal package which sound promising to me...
What I'm asking for is a suggestion which way I should follow, what tool I should try for this job, which may be the most suitable. Maybe there is a tool meant exactly for what I'm looking for or maybe I'm overthinking this and one of the implementations of FFT like resample() will work just fine (of course, after some adjustments at the starting and ending point of the shape to make sure it's closing without issues)?
Scipy.signal sounds promising, however, as far as I understand, it is meant to work with time series data, not geometrical data - I guess this may cause some problems as my data isn't a function (in a mathematical understanding).
Thanks and best regards!

As far as I understood, what you want is to get an interpolated version of your original data.
The DFT (or FFT) will not achieve this purpose, since it will perform an Fourier Transform (which is not what you want).
Talking theoretically, what you need to interpolate your data is to define a function to calculate the result in the new-data-points.
So, let's say your data contains 5 points, in which one you have a 1D (to simplify) number stored, representing your data, and you want a new array with 10 points, filled with the linear-interpolation of your original data.
Using numpy.interp:
import numpy as np
original_data = [2, 0, 3, 5, 1] # define your data in 1D
new_data_resolution = 0.5 # define new sampling distance (i.e, your x-axis resolution)
interp_data = np.interp(
x = np.arange(0, 5-1+new_data_resolution , new_data_resolution), # new sampling points (new axis)
xp = range(original_data),
fp = original_data
)
# now interp_data contains (5-1) / 0.5 + 1 = 9 points
After this, you will have a (5-1) / new_resolution (which is greater than 5, since new_resolution < 1)-length data, which values will be (in this case) a linear interpolation of your original data.
After you have achieved/understood this example, you can dive in the scipy.interpolate module to get a better understanding in the interpolation functions (my example uses a linear function to get the data in the missing points).
Applying this to n-D dimensional arrays is straight-forward, iterating over each dimension of your data.

Python: Gaussian Copula or invers of cdf

Let's say I have a column x with uniform distributed values.
To these values, I applied a cdf-function.
Now I want to calculate the Gaussian Copula, but I can't find the function in python. I read already, that Gaussian Copula is something like the "inverse of the cdf function".
The reason why I'm doing it comes from this paragraph:
A visual depiction of applying the Gaussian Copula process to normalize
an observation by applying 𝑛 = Phi^-1(𝐹(𝑥)). Calculating 𝐹(𝑥) yields a value 𝑢 ∈ [0, 1]
representing the proportion of shaded area at the left. Then Phi^−1(𝑢) yields a value 𝑛
by matching the shaded area in a Gaussian distribution.
I need your help, does everyone has an idea how to calculate that?
I have 2 ideas so far:
1) gauss = 1/(sqrt(2*pi)*s)*e**(-0.5*(float(x-m)/s)**2)
--> so transform all the values with this to a new value
2) norm.ppf(array,loc,scale)
--> So give the ppf function the mean and the std and the array and it will calculate me the inverse of the CDF... But I doubt #2
The thing is
n.cdf(n.ppf(0.95))
Is not what I want. The idea why I'm doing it, is transforming a not normal/gaussian distribution to a normal distribution.
Like here:
Transform from a non gaussian distribution to a gaussian distribution with Gaussian Copula
Any other ideas or tipps?
Thank you very much :)
EDIT:
I found 2 links which are quite usefull:
1. https://stats.stackexchange.com/questions/197283/how-to-transform-an-arcsine-distribution-to-a-normal-distribution
2. https://stats.stackexchange.com/questions/125648/transformation-chi-squared-to-normal-distribution/125653#125653
In this posts its said that you have to
All the details are in the answer already - you take your random variable, and transform it by its own cdf ..... yielding a uniform result.
Thats true for me. If I take a random distirbution and apply the norm.cdf(data, mean,std) function, I get a uniform distributed cdf
Compare: import pandas as pd
data = [1.5]*7 + [2.5]*2 + [3.5]*8 + [4.5]*3 + [5.5]*1 + [6.5]*8
cdf=n.cdf(data, n.mean(data),n.std(data))
print cdf
But How can I do the
You then transform again, applying the quantile function (inverse cdf) of the desired distribution (in this case by the standard normal quantile function /inverse of the normal cdf, producing a variable with a standard normal distribution).
Because when I use f.e. the norm.ppf function, the values are not reasonable

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.

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).

Scipy - Inverse Sampling Method from custom probability density function

I am trying to perform an inverse sampling from a custom probability density function (PDF). I am just wondering if this even possible, i.e. integrating the PDF, inverting the result and then solving it for a given uniform number. The PDF has the shape f(x, alpha, mean(x))=(1/Gamma(alpha+1)(x))((x*(alpha+1)/mean(x))^(alpha+1))exp(-(alpha+1)*(x/mean(x)) where x > 0. From the shape the only values sub-150 are relevant, and for what I am trying to do the sub-80 values are good enough. Extending the range shouldnt be too hard though.
I have tried to do the inversion method, but only found a numerical way to do the integral, which isnt necessarily helpful considering that I need to invert the function to solve:
u = integral(f(x, alpha, mean(x))dx) from 0 to y, where y is unknown and u is uniform random variable between 0 and 1.
The integral has a gamma function and an incomplete gamma function, so trying to invert it is kind of a mess. Any help is welcome.
Thanks a bunch in advance.
Cheers

Assuming you mean that you're trying to randomly choose values which will be distributed according to your PDF, then yes, it is possible. This is described on Wikipedia as inverse transform sampling. Basically, it's just what you said: integrate the PDF to produce the cumulative distribution (CDF), invert it (which can be done ahead of time), and then choose a random number and run it through the inverted CDF.
If your domain is 0 to positive infinity, your distribution appears to match the gamma distribution which is built into Numpy and Scipy, with theta = 1/alpha and k = alpha+1.