def f(x):
return 1/(1 + (x**2))
from scipy.interpolate import CubicSpline
a = -1
b = 1
n = 5
xArray = np.linspace(a,b,n)
yArray = f(xArray)
x = np.linspace(a,b,nPts)
y = CubicSpline(xArray, yArray, x)
plt.plot(x, y, label="Interpolation, " + str(n) + " points")
Im wondering whats the problem in using cubic spline in this way. The error that I get says there is a wrong dimension?
ValueError: x and y must have same first dimension, but have shapes (101,) and (1,
I see your misunderstanding here roots from misinterpretation of the 'extrapolate' keyword, to quote the documentation of CubicSpline
extrapolate{bool, ‘periodic’, None}, optional
If bool, determines whether to extrapolate to out-of-bounds points
based on first and last intervals, or to return NaNs. If ‘periodic’,
periodic extrapolation is used. If None (default), extrapolate is set
to ‘periodic’ for bc_type='periodic' and to True otherwise.
is a boolean and not the list of points for which you want to interpolate and or extrapolate.
The correct usage is to fit a CubicSpline first and then use it to interpolate or extrapolate
def f(x):
return 1/(1 + (x**2))
from scipy.interpolate import CubicSpline
import numpy as np
import matplotlib.pyplot as plt
a = -1
b = 1
n = 5
xArray = np.linspace(a,b,n)
yArray = f(xArray)
x = np.linspace(a,b,101)
cs = CubicSpline(xArray, yArray, True) # fit a cubic spline
y = cs(x) # interpolate/extrapolate
plt.plot(x, y, label="Interpolation, " + str(n) + " points")
plt.show()
The above code will work
Related
I am trying to model the time evolution of a membrane based on the following code in MATLAB.
The basic outline is that the evolution is based on a differential equation
where j=0,1 and x^0 = x, x^1 = y and x^j(s_i) = x^j_i.
My code is the following.
import numpy as np
from matplotlib import pyplot as plt
R0 = 5 #radius
N = 360 #number of intervals
x0 = 2*np.pi*R0/(N/2) #resting membrane lengths
phi = np.linspace(0,2*np.pi, num=360, dtype=float)
R1 = R0 + 0.5*np.sin(20*phi)
X = R1*np.cos(phi)
Y = R1*np.sin(phi)
L = np.linspace(-1,358, num=360, dtype=int)
R = np.linspace(1,360, num=360,dtype=int) #right and left indexing vectors
R[359] = 0
X = R1*np.cos(phi)
Y = R1*np.sin(phi)
plt.plot(X,Y)
plt.axis("equal")
plt.show()
ds = 1/N
ds2 = ds**2
k = 1/10
w = 10**6
for i in range(0,20000):
lengths = np.sqrt( (X[R]-X)**2 + (Y[R]-Y)**2 )
Ex = k/ds2*(X[R] - 2*X + X[L] - x0*( (X[R]-X)/lengths - (X-X[L])/lengths[L]) )
Ey = k/ds2*(Y[R] - 2*Y + Y[L] - x0*( (Y[R]-Y)/lengths - (Y-Y[L])/lengths[L]) )
X = X + 1/w*Ex
Y = Y + 1/w*Ey
plt.plot(X,Y)
plt.axis("equal")
plt.show()
The model is supposed to devolve into a circular membrane, as below
but this is what mine does
Your definition of x0 is wrong.
In the Matlab code, it is equal to
x0 = 2*pi*R/N/2 # which is pi*R/N
while in your Python code it is
x0 = 2*np.pi*R0/(N/2) # which is 4*np.pi*R0/N
Correcting that, the end result is a circular shape, but with a different radius. I'm assuming that this is because of the reduced number of iterations (20000 instead of 1000000).
Edit:
As expected, using the correct number of iterations results in a plot similar to your expected one.
I have some experiment data. Hereby, I need to fit the following function to determine one of the variable. A Levenberg–Marquardt least-squares algorithm was used in this procedure.
I have used curve fitting option in Igor Pro software. I defined new fit function and tried to define independent and dependent variable.
Nevertheless, I don't know what is the reason that I got the this error:
"The fitting function returned INF for at least one X variable"
My function is :
sin(theta) = -1+2*sqrt(alpha/x)*exp(-beta*(x-alpha)^2)
beta = 1.135e-4;
sin(theta) = [-0.81704 -0.67649 -0.83137 -0.73468 -0.66744 -0.43602 0.45368 0.75802 0.96705 0.99717 ]
x = [72.01 59.99 51.13 45.53 36.15 31.66 30.16 29.01 25.62 23.47 ]
Is there any suggestion to find alpha variable here?
Is there any handy software or program for nonlinear curve fitting?
In gnuplot, it would look like this. The fit is not great, but that's not the "fault" of gnuplot, but apparently this data cannot be fitted with this function very well.
Code:
### nonlinear curve fitting
reset session
$Data <<EOD
72.01 -0.81704
59.99 -0.67649
51.13 -0.83137
45.53 -0.73468
36.15 -0.66744
31.66 -0.43602
30.16 0.45368
29.01 0.75802
25.62 0.96705
23.47 0.99717
EOD
f(x) = -1+2*sqrt(alpha/x)*exp(-beta*(x-alpha)**2)
# initial guessed values
alpha = 25
beta = 1
set fit nolog results
fit f(x) $Data u 1:2 via alpha,beta
plot $Data u 1:2 w lp pt 7, \
f(x) lc rgb "red"
print sprintf("alpha=%g, beta=%g",alpha,beta)
### end of code
Result:
alpha=25.818, beta=0.0195229
If it might be of some use, my equation search on your data turned up a good fit to a standard 4-parameter logistic equation "y = d + (a - d) / (1.0 + pow(x / c, b))" with parameters a = 0.96207949, b = 44.14292256, c = 30.67324939, and d = -0.74830947 yielding RMSE = 0.0565 and R-squared = 0.9943, and I have included code for a Python graphical fitter using this equation.
import numpy, scipy, matplotlib
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
theta = [-0.81704, -0.67649, -0.83137, -0.73468, -0.66744, -0.43602, 0.45368, 0.75802, 0.96705, 0.99717]
x = [72.01, 59.99, 51.13, 45.53, 36.15, 31.66, 30.16, 29.01, 25.62, 23.47]
# rename to match previous example code
xData = numpy.array(x)
yData = numpy.array(theta)
# StandardLogistic4Parameter equation from zunzun.com
def func(x, a, b, c, d):
return d + (a - d) / (1.0 + numpy.power(x / c, b))
# these are the same as the scipy defaults
initialParameters = numpy.array([1.0, 1.0, 1.0, 1.0])
# curve fit the test data
fittedParameters, pcov = curve_fit(func, xData, yData, initialParameters)
modelPredictions = func(xData, *fittedParameters)
absError = modelPredictions - yData
SE = numpy.square(absError) # squared errors
MSE = numpy.mean(SE) # mean squared errors
RMSE = numpy.sqrt(MSE) # Root Mean Squared Error, RMSE
Rsquared = 1.0 - (numpy.var(absError) / numpy.var(yData))
print('Parameters:', fittedParameters)
print('RMSE:', RMSE)
print('R-squared:', Rsquared)
print()
##########################################################
# graphics output section
def ModelAndScatterPlot(graphWidth, graphHeight):
f = plt.figure(figsize=(graphWidth/100.0, graphHeight/100.0), dpi=100)
axes = f.add_subplot(111)
# first the raw data as a scatter plot
axes.plot(xData, yData, 'D')
# create data for the fitted equation plot
xModel = numpy.linspace(min(xData), max(xData))
yModel = func(xModel, *fittedParameters)
# now the model as a line plot
axes.plot(xModel, yModel)
axes.set_xlabel('X Data') # X axis data label
axes.set_ylabel('Y Data') # Y axis data label
plt.show()
plt.close('all') # clean up after using pyplot
graphWidth = 800
graphHeight = 600
ModelAndScatterPlot(graphWidth, graphHeight)
Matlab
I slightly changed the function, -1 changed to -gamma and optimize to find gamma
The code is as follow
ydata = [-0.81704 -0.67649 -0.83137 -0.73468 -0.66744 -0.43602 0.45368...
0.75802 0.96705 0.99717 ];
xdata = [72.01 59.99 51.13 45.53 36.15 31.66 30.16 29.01 25.62 23.47 ];
sin_theta = #(alpha, beta, gamma, xdata) -gamma+2.*sqrt(alpha./xdata).*exp(beta.*(xdata-alpha).^2);
%Fitting function as function of array(x) required by lsqcurvefit
f = #(x,xdata) sin_theta(x(1),x(2), x(3),xdata);
% [alpha, beta, gamma]
x0 = [25, 0, 1] ;
options = optimoptions('lsqcurvefit','Algorithm','levenberg-marquardt', 'FunctionTolerance', 1e-30);
[x,resnorm,residual,exitflag,output] = lsqcurvefit(f,x0,xdata,ydata,[], [], options);
% Accuracy
RMSE = sqrt(sum(residual.^2)/length(residual));
alpha = x(1); beta = x(2); gamma = x(3);
%Plotting data
data = linspace(xdata(1),xdata(end));
plot(xdata,ydata,'ro',data,f(x,data),'b-', 'linewidth', 3)
legend('Data','Fitted exponential')
title('Data and Fitted Curve')
set(gca,'FontSize',20)
Result
alpha = 26.0582, beta = -0.0329, gamma = 0.7881 instead of 1, RMSE = 0.1498
Graph
I have following assembly procedure
nP = small integer ( 1~100 )
nQ = small but larger than nP
nE = large integer ( 10^5~7)
f = #(x) math_expression % for example, sin(pi*x).^4
x = (nQ x nE)
Mass = (nQ x nP)
What I want to construct is
M = (nP x nP x nE) : 3D matrix
by
h = x(nQ,:)-x(1,:);
for k = 1 : nE
M(:,:,k) = Mass'*diag(f(x(:,k))*Mass*h(k);
end
This will be used to construct block diagonal matrix with predefined index matrices
sparse(IM(:),JM(:),M(:),nE*nP,nE*nP,nP*nP*nE);
What I want to do is removing for loop by vectorize
Mass'*diag(f(x(:,k))*Mass*h(k)
I tried to use bsxfun like
assmble = #(dummy,k) Mass'*diag(f(x(:,k))*Mass*h(k)
Mass = bsxfun(assmble,Mass,reshape(1:nE,1,1,nE));
However, matlab said it is not proper way to use bsxfun.
Is there any suggestion?
this is a solution:
i = repmat(1:nQ,1,nE);
j = 1:(nQ*nE);
d = sparse(i,j,f(x(:)));
Mass_d = reshape(Mass'*d,nP,nQ,nE);
Mass_d_t = permute(Mass_d,[2,1,3]);
Mass_d_t_r = reshape(Mass_d_t, nQ ,nP*nE);
M_d_M = Mass'*Mass_d_t_r;
M_d_M_h = bsxfun(#times,M_d_M, repelem(h,nP));
M = reshape(M_d_M_h, nP , nP , nE);
However you may need clear unneeded variables to prevent memory problems
If the object function is
How to code it in python?
I've already coded the normal one:
import numpy as np
import scipy as sp
from scipy.optimize import leastsq
import pylab as pl
m = 9 #the degree of the polynomial
def real_func(x):
return np.sin(2*np.pi*x) #sin(2 pi x)
def fake_func(p, x):
f = np.poly1d(p) #polynomial
return f(x)
def residuals(p, y, x):
return y - fake_func(p, x)
#randomly choose 9 points as x
x = np.linspace(0, 1, 9)
x_show = np.linspace(0, 1, 1000)
y0 = real_func(x)
#add normalize noise
y1 = [np.random.normal(0, 0.1) + y for y in y0]
p0 = np.random.randn(m)
plsq = leastsq(residuals, p0, args=(y1, x))
print 'Fitting Parameters ：', plsq[0]
pl.plot(x_show, real_func(x_show), label='real')
pl.plot(x_show, fake_func(plsq[0], x_show), label='fitted curve')
pl.plot(x, y1, 'bo', label='with noise')
pl.legend()
pl.show()
Since the penalization term is also just quadratic, you could just stack it together with thesquares of the error and use weights 1 for data and lambda for the penalization rows.
scipy.optimize.curvefit does weighted least squares, if you don't want to code it yourself.
How would one go plotting a plane in matlab or matplotlib from a normal vector and a point?
For all the copy/pasters out there, here is similar code for Python using matplotlib:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
point = np.array([1, 2, 3])
normal = np.array([1, 1, 2])
# a plane is a*x+b*y+c*z+d=0
# [a,b,c] is the normal. Thus, we have to calculate
# d and we're set
d = -point.dot(normal)
# create x,y
xx, yy = np.meshgrid(range(10), range(10))
# calculate corresponding z
z = (-normal[0] * xx - normal[1] * yy - d) * 1. /normal[2]
# plot the surface
plt3d = plt.figure().gca(projection='3d')
plt3d.plot_surface(xx, yy, z)
plt.show()
For Matlab:
point = [1,2,3];
normal = [1,1,2];
%# a plane is a*x+b*y+c*z+d=0
%# [a,b,c] is the normal. Thus, we have to calculate
%# d and we're set
d = -point*normal'; %'# dot product for less typing
%# create x,y
[xx,yy]=ndgrid(1:10,1:10);
%# calculate corresponding z
z = (-normal(1)*xx - normal(2)*yy - d)/normal(3);
%# plot the surface
figure
surf(xx,yy,z)
Note: this solution only works as long as normal(3) is not 0. If the plane is parallel to the z-axis, you can rotate the dimensions to keep the same approach:
z = (-normal(3)*xx - normal(1)*yy - d)/normal(2); %% assuming normal(3)==0 and normal(2)~=0
%% plot the surface
figure
surf(xx,yy,z)
%% label the axis to avoid confusion
xlabel('z')
ylabel('x')
zlabel('y')
For copy-pasters wanting a gradient on the surface:
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
import numpy as np
import matplotlib.pyplot as plt
point = np.array([1, 2, 3])
normal = np.array([1, 1, 2])
# a plane is a*x+b*y+c*z+d=0
# [a,b,c] is the normal. Thus, we have to calculate
# d and we're set
d = -point.dot(normal)
# create x,y
xx, yy = np.meshgrid(range(10), range(10))
# calculate corresponding z
z = (-normal[0] * xx - normal[1] * yy - d) * 1. / normal[2]
# plot the surface
plt3d = plt.figure().gca(projection='3d')
Gx, Gy = np.gradient(xx * yy) # gradients with respect to x and y
G = (Gx ** 2 + Gy ** 2) ** .5 # gradient magnitude
N = G / G.max() # normalize 0..1
plt3d.plot_surface(xx, yy, z, rstride=1, cstride=1,
facecolors=cm.jet(N),
linewidth=0, antialiased=False, shade=False
)
plt.show()
The above answers are good enough. One thing to mention is, they are using the same method that calculate the z value for given (x,y). The draw back comes that they meshgrid the plane and the plane in space may vary (only keeping its projection the same). For example, you cannot get a square in 3D space (but a distorted one).
To avoid this, there is a different way by using the rotation. If you first generate data in x-y plane (can be any shape), then rotate it by equal amount ([0 0 1] to your vector) , then you will get what you want. Simply run below code for your reference.
point = [1,2,3];
normal = [1,2,2];
t=(0:10:360)';
circle0=[cosd(t) sind(t) zeros(length(t),1)];
r=vrrotvec2mat(vrrotvec([0 0 1],normal));
circle=circle0*r'+repmat(point,length(circle0),1);
patch(circle(:,1),circle(:,2),circle(:,3),.5);
axis square; grid on;
%add line
line=[point;point+normr(normal)]
hold on;plot3(line(:,1),line(:,2),line(:,3),'LineWidth',5)
It get a circle in 3D:
A cleaner Python example that also works for tricky $z,y,z$ situations,
from mpl_toolkits.mplot3d import axes3d
from matplotlib.patches import Circle, PathPatch
import matplotlib.pyplot as plt
from matplotlib.transforms import Affine2D
from mpl_toolkits.mplot3d import art3d
import numpy as np
def plot_vector(fig, orig, v, color='blue'):
ax = fig.gca(projection='3d')
orig = np.array(orig); v=np.array(v)
ax.quiver(orig[0], orig[1], orig[2], v[0], v[1], v[2],color=color)
ax.set_xlim(0,10);ax.set_ylim(0,10);ax.set_zlim(0,10)
ax = fig.gca(projection='3d')
return fig
def rotation_matrix(d):
sin_angle = np.linalg.norm(d)
if sin_angle == 0:return np.identity(3)
d /= sin_angle
eye = np.eye(3)
ddt = np.outer(d, d)
skew = np.array([[ 0, d[2], -d[1]],
[-d[2], 0, d[0]],
[d[1], -d[0], 0]], dtype=np.float64)
M = ddt + np.sqrt(1 - sin_angle**2) * (eye - ddt) + sin_angle * skew
return M
def pathpatch_2d_to_3d(pathpatch, z, normal):
if type(normal) is str: #Translate strings to normal vectors
index = "xyz".index(normal)
normal = np.roll((1.0,0,0), index)
normal /= np.linalg.norm(normal) #Make sure the vector is normalised
path = pathpatch.get_path() #Get the path and the associated transform
trans = pathpatch.get_patch_transform()
path = trans.transform_path(path) #Apply the transform
pathpatch.__class__ = art3d.PathPatch3D #Change the class
pathpatch._code3d = path.codes #Copy the codes
pathpatch._facecolor3d = pathpatch.get_facecolor #Get the face color
verts = path.vertices #Get the vertices in 2D
d = np.cross(normal, (0, 0, 1)) #Obtain the rotation vector
M = rotation_matrix(d) #Get the rotation matrix
pathpatch._segment3d = np.array([np.dot(M, (x, y, 0)) + (0, 0, z) for x, y in verts])
def pathpatch_translate(pathpatch, delta):
pathpatch._segment3d += delta
def plot_plane(ax, point, normal, size=10, color='y'):
p = Circle((0, 0), size, facecolor = color, alpha = .2)
ax.add_patch(p)
pathpatch_2d_to_3d(p, z=0, normal=normal)
pathpatch_translate(p, (point[0], point[1], point[2]))
o = np.array([5,5,5])
v = np.array([3,3,3])
n = [0.5, 0.5, 0.5]
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure()
ax = fig.gca(projection='3d')
plot_plane(ax, o, n, size=3)
ax.set_xlim(0,10);ax.set_ylim(0,10);ax.set_zlim(0,10)
plt.show()