How can I find the peak curvature of a spline fitted using scipy? (Actually, peak second differential would be enough)
I have calculated the tck values as follows, using my 1d xs and ys vectors:
tck = splrep(xs, ys, s=0)
I know I can evaluate the second differential at any x of my choice:
ddy = splev([x], tck, 2)
So I could loop over many values of x, calculate the curvature and take the maximum. But I would prefer to interpret the values in tck to get the coefficients of the individual cubic functions, and thus calculate the peak curvature directly. However, tck appears rather opaque - how can I extract the cubic function coefficients from it?
Just use the der keyword argument on splev function:
ddy = splev(X, tck, der=2)
and preferrably don't loop over many values of x, instead make a Nx1 array X containing every value you want to evaluate, so as to get back an array of values instead of individual values you'll have to put in a sequence anyway.
Also, it is extremely adviseable to PLOT your results as a way to debug it. If plots make sense, things are most likely working (and, if not, they surely are NOT working) as you expect.
EDIT: in case the interpolation using X gives just an approximate value and you want the TRUE maximum, you can use parabolic interpolation of the three points that define the maximum (the local interpolated maximum and its neighbors), considering the spline is locally smooth:
def parabolic_interpolation(p1, p2, p3):
x1, y1 = p1
x2, y2 = p2
x3, y3 = p3
denom = (x1-x2)*(x1-x3)*(x2-x3);
a = (x3*(y2-y1)+x2*(y1-y3)+x1*(y3-y2))/denom
b = (x3*x3*(y1-y2)+x2*x2*(y3-y1)+x1*x1*(y2-y3))/denom
c = (x2*x3*(x2-x3)*y1+x3*x1*(x3-x1)*y2+x1*x2*(x1-x2)*y3)/denom
xv = -b/(2*a)
yv = c-b**2/(4*a)
return (xv, yv) # coordinates of the vertex
Hope this helps!
Related
I am trying to evaluate the convolution of two continuous functions using scipy.signal.fftconvolve. The scenario of the code is as following:
I am trying to approximate the following double integral:
, i.e. in a region C_1(x',y'), representing a circle of radius 1 centered at (x', y'). This can be approximated by the following integral:
where the function K is chosen as a continuous integrable function, say, exp(-x^2-y^2), the shape of which is approximately that of a circle of radius 1. If I take a function K'(x,y)=K(-x,-y), then the integral is exactly a convolution of the two functions:
So I try to discretize these two functions into arrays and then carry out convolution.
The following code will be written in Julia and the fftconvolve function will be imported using PyCall.jl.
using PyCall
using Interpolations
r = 1
xc = -10:0.05:10
yc = -10:0.05:10
K(x, y) = exp(-(x^2+y^2)/r^2)
rho(x, y) = x^2+y^3 # Try some arbitrary function
ss = pyimport("scipy.signal") # Import scipy.signal module from Python
a = [rho(x,y) for x in xc, y in yc]
b = [K(-x,-y) for x in xc, y in yc]
c = ss.fftconvolve(a,b,mode="same") # zero-paddings beyond boundary, unimportant since rho is near zero beyond the boundary anyway
c_unscaled = interpolate(c', BSpline(Cubic(Line(OnCell()))))
# Adjoint because the array comprehension switched x and y, then interpolate the array
c_scaled = Interpolations.scale(c_unscaled, xc, yc) # Scale the interpolated function w.r.t. xc, yc
print(c_scaled(0.0,0.0)) # The result of the integral for (x', y') = (0, 0)
The result is 628.3185307178969, while the result from numerical integration is 0.785398. What is the problem here?
You could probably try to use scipy.signal.convolve which will convolve two N-dimensional arrays, but not by using Fast Fourier Transform.
It uses a direct method to calculate a convolution. Here, I mean that the convolution is determined directly from sums.
So you could maybe try to replace the line where you calculate c with this one:
c = ss.convolve(a,b,mode="same", method='direct')
I'm trying to implement the following formula in python for X and Y points
I have tried following approach
def f(c):
"""This function computes the curvature of the leaf."""
tt = c
n = (tt[0]*tt[3] - tt[1]*tt[2])
d = (tt[0]**2 + tt[1]**2)
k = n/d
R = 1/k # Radius of Curvature
return R
There is something incorrect as it is not giving me correct result. I think I'm making some mistake while computing derivatives in first two lines. How can I fix that?
Here are some of the points which are in a data frame:
pts = pd.DataFrame({'x': x, 'y': y})
x y
0.089631 97.710199
0.089831 97.904541
0.090030 98.099313
0.090229 98.294513
0.090428 98.490142
0.090627 98.686200
0.090827 98.882687
0.091026 99.079602
0.091225 99.276947
0.091424 99.474720
0.091623 99.672922
0.091822 99.871553
0.092022 100.070613
0.092221 100.270102
0.092420 100.470020
0.092619 100.670366
0.092818 100.871142
0.093017 101.072346
0.093217 101.273979
0.093416 101.476041
0.093615 101.678532
0.093814 101.881451
0.094013 102.084800
0.094213 102.288577
pts_x = np.gradient(x_c, t) # first derivatives
pts_y = np.gradient(y_c, t)
pts_xx = np.gradient(pts_x, t) # second derivatives
pts_yy = np.gradient(pts_y, t)
After getting the derivatives I am putting the derivatives x_prim, x_prim_prim, y_prim, y_prim_prim in another dataframe using the following code:
d = pd.DataFrame({'x_prim': pts_x, 'y_prim': pts_y, 'x_prim_prim': pts_xx, 'y_prim_prim':pts_yy})
after having everything in the data frame I am calling function for each row of the data frame to get curvature at that point using following code:
# Getting the curvature at each point
for i in range(len(d)):
temp = d.iloc[i]
c_temp = f(temp)
curv.append(c_temp)
You do not specify exactly what the structure of the parameter pts is. But it seems that it is a two-dimensional array where each row has two values x and y and the rows are the points in your curve. That itself is problematic, since the documentation is not quite clear on what exactly is returned in such a case.
But you clearly are not getting the derivatives of x or y. If you supply only one array to np.gradient then numpy assumes that the points are evenly spaced with a distance of one. But that is probably not the case. The meaning of x' in your formula is the derivative of x with respect to t, the parameter variable for the curve (which is separate from the parameters to the computer functions). But you never supply the values of t to numpy. The values of t must be the second parameter passed to the gradient function.
So to get your derivatives, split the x, y, and t values into separate one-dimensional arrays--lets call them x and y and t. Then get your first and second derivatives with
pts_x = np.gradient(x, t) # first derivatives
pts_y = np.gradient(y, t)
pts_xx = np.gradient(pts_x, t) # second derivatives
pts_yy = np.gradient(pts_y, t)
Then continue from there. You no longer need the t values to calculate the curvatures, which is the point of the formula you are using. Note that gradient is not really designed to calculate the second derivatives, and it absolutely should not be used to calculate third or higher-order derivatives. More complex formulas are needed for those. Numpy's gradient uses "second order accurate central differences" which are pretty good for the first derivative, poor for the second derivative, and worthless for higher-order derivatives.
I think your problem is that x and y are arrays of double values.
The array x is the independent variable; I'd expect it to be sorted into ascending order. If I evaluate y[i], I expect to get the value of the curve at x[i].
When you call that numpy function you get an array of derivative values that are the same shape as the (x, y) arrays. If there are n pairs from (x, y), then
y'[i] gives the value of the first derivative of y w.r.t. x at x[i];
y''[i] gives the value of the second derivative of y w.r.t. x at x[i].
The curvature k will also be an array with n points:
k[i] = abs(x'[i]*y''[i] -y'[i]*x''[i])/(x'[i]**2 + y'[i]**2)**1.5
Think of x and y as both being functions of a parameter t. x' = dx/dt, etc. This means curvature k is also a function of that parameter t.
I like to have a well understood closed form solution available when I program a solution.
y(x) = sin(x) for 0 <= x <= pi
y'(x) = cos(x)
y''(x) = -sin(x)
k = sin(x)/(1+(cos(x))**2)**1.5
Now you have a nice formula for curvature as a function of x.
If you want to parameterize it, use
x(t) = pi*t for 0 <= t <= 1
x'(t) = pi
x''(t) = 0
See if you can plot those and make your Python solution match it.
I have a data set data[xi,yi,zi] I'd like to plot (with interpolation values). With scipy.interpolate, everything looks almost perfect however the interpolation is generating some values beyond the bounds of the input data. for example suppose zi is bound by 0 < zi < 1, the rbf interpolatation seems to be returning interpolated values out of the bounds (e.g. >1); here my simplified attempt:
N=100
data=[xi yi zi]
xis = np.linspace(xi.min(), xi.max(), N)
yis = np.linspace(yi.min(), yi.max(), N)
XI, YI = np.meshgrid(xis,yis)
rbf = scipy.interpolate.Rbf(xi, yi, zi, function='linear')
ZI=rbf(XI,YI)
print ZI.max()
->1.01357328514
Is there a way to pass limits to Rbf and let it know not to go past zi.max() and zi.min() ?
Interpolation with radial basis functions may result in values above the maximum and below the minimum of the given data values. (Illustration). This is a mathematical feature of the method, one cannot pass in an option to disable it. Two possible solutions:
use np.clip to clip the interpolant between the min-max values of the data, when plotting it.
use piecewise linear interpolation instead (scipy.interpolate.LinearNDInterpolator), which is guaranteed to respect the minimum and maximum of data values.
I am using the following code to fit a 3D line on a cloud of 3D points.
I am using a least squares method of lmfit to minimize.
I need to add weights to different points, but do not know how to do it when using an array (and not a scalar) distance output. The problem when using scalar is that the it is not as good as when using an array. I assume because of a larger number of variables.
So the question is - is there a way to add weights for each element of array to the minimizer?
Using something like Nelder w/scaral input does not perform the 3D fit well.
from lmfit import minimize, Parameters, Parameter,report_fit,fit_report, Minimizer, printfuncs
import numpy as np
#parameters
params = Parameters()
params.add('y1', value= 0)
params.add('x0', value= 129)
params.add('x1', value= 0)
params.add('y0', value= 129)
params.add('y1', value= 0)
#function calculating point-line distance
def fun(params,x,y,z):
x0 = params['x0'].value; x1 = params['x1'].value; y0 = params['y0'].value; y1 = params['y1'].value
distance = []
#parametric equations
v0 = np.array([x0, y0, 0])
v1 = np.array([x0+x1,y0+ y1, 1])
#for loop over all the 3D points to calculate distance
for point in range(len(x)):
p = np.array([x[point], y[point], z[point]])
distance.append(np.linalg.norm(np.cross(v0-p,v0-v1)))
return distance
result = minimize(fun, params,args=(x,y,z))
print(fit_report(result))
theta = np.arccos(1/ np.sqrt(result.params['x1']*result.params['x1']+result.params['y1']*result.params['y1']+1))
I suppose that you read the documentation on how to write an objective function (this link as provided in one of your other questions)? If so, which part of the description there is not clear and/or what did you try and didn't work?
In your example you generate an array called distance, which effectively is the same as (model-data) in the documentation. What you want to add here is eps, which will scale the residual by the uncertainty in the data.
I am trying to find higher order derivatives of a dataset (x,y). x and y are 1D arrays of length N.
Let's say I generate them as :
xder0=np.linspace(0,10,1000)
yder0=np.sin(xder0)
I define the derivative function which takes in 2 array (x,y) and returns (x1, y1) where y1 is the derivative calculated at each index as : (y[i+1]-y[i])/(x[i+1]-x[i]). x1 is just the mean of x[i+1] and x[i]
Here is the function that does it:
def deriv(x,y):
delx =np.zeros((len(x)-1), dtype=np.longdouble)
ydiff=np.zeros((len(x)-1), dtype=np.longdouble)
for i in range(len(x)-1):
delx[i] =(x[i+1]+x[i])/2.0
ydiff[i] =(y[i+1]-y[i])/(x[i+1]-x[i])
return delx, ydiff
Now to calculate the first derivative, I call this function as:
xder1, yder1 = deriv(xder0, yder0)
Similarly for second derivative, I call this function giving first derivatives as input:
xder2, yder2 = deriv(xder1, yder1)
And it goes on:
xder3, yder3 = deriv(xder2, yder2)
xder4, yder4 = deriv(xder3, yder3)
xder5, yder5 = deriv(xder4, yder4)
xder6, yder6 = deriv(xder5, yder5)
xder7, yder7 = deriv(xder6, yder6)
xder8, yder8 = deriv(xder7, yder7)
xder9, yder9 = deriv(xder8, yder8)
Something peculiar happens after I reach order 7. The 7th order becomes very noisy! Earlier derivatives are all either sine or cos functions as expected. However 7th order is a noisy sine. And hence all derivatives after that blow up.
Any idea what is going on?
This is a well known stability issue with numerical interpolation using equally-spaced points. Read the answers at http://math.stackexchange.com.
To overcome this problem you have to use non-equally-spaced points, like the roots of Lagendre polynomial. The instability occurs due to the unavailability of information at the boundaries, thus more concentration of points at the boundaries is required, as per the roots of say Lagendre polynomials or others with similar properties such as Chebyshev polynomial.