How numpy decides number of coefficient in smooth bivariate spline? - python-2.7

I am using SmoothBivariateSpline of numpy library to get a surface fit of a data.
By default degrees of bivariate spline is 3. When i am fitting my data, if the degree is 3, i should get 16 coefficients and 8 knot points in x and y direction respectively. This is happening for one of the spline fit (f1). But i am getting 25 coefficients and 9 knot points for the second fit (f2).
Can someone please tell me why i am getting more number of coefficients?? if you want i can share my data file as well.
here is my code
import numpy as np
# loading the txt file from where i am reading data
L = np.loadtxt('DATA_File_PT.txt')
P = L[:,0] # pressure
T = L[:,1] # temperature
H = L[:,2] # enthalpy
Rho = L[:,3] # density
# fitting Spline for each set (I have P and T as independent variable, H and Rho are dependent)
f1 = interpolate.SmoothBivariateSpline(P, T, H, kx=3, ky=3)
f2 = interpolate.SmoothBivariateSpline(P, T, Rho, kx=3, ky=3)
o1=f1.get_knots()
coeff1 = f1.get_coeffs()
o2=f2.get_knots()
coeff2 = f2.get_coeffs()

Related

How to visualize 2d array of doubles in vtk?

I have 2d array (of size 20x30) of doubles:
double a[20*30];
How to visualize it using VTK? It is extremely difficult to find proper documentation. The closest example which I found is this, however it uses as input 3 unsigned chars which represents color. As I understand I should use vtkScalarsToColors class to somehow map scalars to colors, but I can't figure out how to put everything into single piece of code.
What you probably want to do is to assign scalars to points or cells of a surface or volume mesh. VTK then can take care of the visualization. This is demonstrated in the following example: ScalarBarActor. For the basic usage, follow the Scalars example.
However, you need to provide the suitable mesh yourself to which you want to map the values. From your question, it is not entirely clear what you mean with "how to visualize a 2d array" of values. If you want to assign scalar values in a planar 20x30 grid, you need to first create a surface object (of type vtkPolyData) with triangular or quadrangular cells, and then assign the values to the points of the mesh using surface->GetPointData()->SetScalars(), as demonstrated in the above examples.
Convenient in this case would be the vtkPlaneSource, look here for the corresponding example. The number of grid points you can set by using SetXResolution() or SetYResolution() respectively. (In case this is not clear: vtkPlaneSource inherits vtkPolyDataAlgorithm, to access the underlying vtkPolyData object, use the method GetOutput())
Update: I added sample code that demonstrates the procedure - in python, for better readability.
# This code has been written by normanius under the CC BY-SA 4.0 license.
# License: https://creativecommons.org/licenses/by-sa/4.0/
# Author: normanius: https://stackoverflow.com/users/3388962/normanius
# Date: August 2018
# Reference: https://stackoverflow.com/a/51754466/3388962
import vtk
import numpy as np
###########################################################
# CREATE ARRAY VALUES
###########################################################
# Just create some fancy looking values for z.
n = 100
m = 50
xmin = -1; xmax = 1
ymin = -1; ymax = 1
x = np.linspace(xmin, xmax, n)
y = np.linspace(ymin, ymax, m)
x, y = np.meshgrid(x, y)
x, y = x.flatten(), y.flatten()
z = (x+y)*np.exp(-3.0*(x**2+y**2))
###########################################################
# CREATE PLANE
###########################################################
# Create a planar mesh of quadriliterals with nxm points.
# (SetOrigin and SetPointX only required if the extent
# of the plane should be the same. For the mapping
# of the scalar values, this is not required.)
plane = vtk.vtkPlaneSource()
plane.SetResolution(n-1,m-1)
plane.SetOrigin([xmin,ymin,0]) # Lower left corner
plane.SetPoint1([xmax,ymin,0])
plane.SetPoint2([xmin,ymax,0])
plane.Update()
# Map the values to the planar mesh.
# Assumption: same index i for scalars z[i] and mesh points
nPoints = plane.GetOutput().GetNumberOfPoints()
assert(nPoints == len(z))
# VTK has its own array format. Convert the input
# array (z) to a vtkFloatArray.
scalars = vtk.vtkFloatArray()
scalars.SetNumberOfValues(nPoints)
for i in range(nPoints):
scalars.SetValue(i, z[i])
# Assign the scalar array.
plane.GetOutput().GetPointData().SetScalars(scalars)
###########################################################
# WRITE DATA
###########################################################
writer = vtk.vtkXMLPolyDataWriter()
writer.SetFileName('output.vtp')
writer.SetInputConnection(plane.GetOutputPort())
writer.Write() # => Use for example ParaView to see scalars
###########################################################
# VISUALIZATION
###########################################################
# This is a bit annoying: ensure a proper color-lookup.
colorSeries = vtk.vtkColorSeries()
colorSeries.SetColorScheme(vtk.vtkColorSeries.BREWER_DIVERGING_SPECTRAL_10)
lut = vtk.vtkColorTransferFunction()
lut.SetColorSpaceToHSV()
nColors = colorSeries.GetNumberOfColors()
zMin = np.min(z)
zMax = np.max(z)
for i in range(0, nColors):
color = colorSeries.GetColor(i)
color = [c/255.0 for c in color]
t = zMin + float(zMax - zMin)/(nColors - 1) * i
lut.AddRGBPoint(t, color[0], color[1], color[2])
# Mapper.
mapper = vtk.vtkPolyDataMapper()
mapper.SetInputConnection(plane.GetOutputPort())
mapper.ScalarVisibilityOn()
mapper.SetScalarModeToUsePointData()
mapper.SetLookupTable(lut)
mapper.SetColorModeToMapScalars()
# Actor.
actor = vtk.vtkActor()
actor.SetMapper(mapper)
# Renderer.
renderer = vtk.vtkRenderer()
renderer.SetBackground([0.5]*3)
# Render window and interactor.
renderWindow = vtk.vtkRenderWindow()
renderWindow.SetWindowName('Demo')
renderWindow.AddRenderer(renderer)
renderer.AddActor(actor)
interactor = vtk.vtkRenderWindowInteractor()
interactor.SetRenderWindow(renderWindow)
renderWindow.Render()
interactor.Start()
The result will look similar to this:

Schrodinger equation not evolving properly with time?

I'm writing a code in python to evolve the time-dependent Schrodinger equation using the Crank-Nicolson scheme. I didn't know how to deal with the potential so I looked around and found a way from this question, which I have verified from a couple other sources. According to them, for a harmonic oscillator potential, the C-N scheme gives
AΨn+1=A∗Ψn
where the elements on the main diagonal of A are dj=1+[(iΔt) / (2m(Δx)^2)]+[(iΔt(xj)^2)/4] and the elements on the upper and lower diagonals are a=−iΔt/[4m(Δx)^2]
The way I understand it, I'm supposed to give an initial condition(I've chosen a coherent state) in the form of the matrix Ψn and I need to compute the matrix Ψn+1 , which is the wave function after time Δt. To obtain Ψn+1 for a given step, I'm inverting the matrix A and multiplying it with the matrix A* and then multiplying the result with Ψn. The resulting matrix then becomes Ψn for the next step.
But when I'm doing this, I'm getting an incorrect animation. The wave packet is supposed to oscillate between the boundaries but in my animation, it is barely moving from its initial mean value. I just don't understand what I'm doing wrong. Is my understanding of the problem wrong? Or is it a flaw in my code?Please help! I've posted my code below and the video of my animation here. I'm sorry for the length of the code and the question but it's driving me crazy not knowing what my mistake is.
import numpy as np
import matplotlib.pyplot as plt
L = 30.0
x0 = -5.0
sig = 0.5
dx = 0.5
dt = 0.02
k = 1.0
w=2
K=w**2
a=np.power(K,0.25)
xs = np.arange(-L,L,dx)
nn = len(xs)
mu = k*dt/(dx)**2
dd = 1.0+mu
ee = 1.0-mu
ti = 0.0
tf = 100.0
t = ti
V=np.zeros(len(xs))
u=np.zeros(nn,dtype="complex")
V=K*(xs)**2/2 #harmonic oscillator potential
u=(np.sqrt(a)/1.33)*np.exp(-(a*(xs - x0))**2)+0j #initial condition for wave function
u[0]=0.0 #boundary condition
u[-1] = 0.0 #boundary condition
A = np.zeros((nn-2,nn-2),dtype="complex") #define A
for i in range(nn-3):
A[i,i] = 1+1j*(mu/2+w*dt*xs[i]**2/4)
A[i,i+1] = -1j*mu/4.
A[i+1,i] = -1j*mu/4.
A[nn-3,nn-3] = 1+1j*mu/2+1j*dt*xs[nn-3]**2/4
B = np.zeros((nn-2,nn-2),dtype="complex") #define A*
for i in range(nn-3):
B[i,i] = 1-1j*mu/2-1j*w*dt*xs[i]**2/4
B[i,i+1] = 1j*mu/4.
B[i+1,i] = 1j*mu/4.
B[nn-3,nn-3] = 1-1j*(mu/2)-1j*dt*xs[nn-3]**2/4
X = np.linalg.inv(A) #take inverse of A
plt.ion()
l, = plt.plot(xs,np.abs(u),lw=2,color='blue') #plot initial wave function
T=np.matmul(X,B) #multiply A inverse with A*
while t<tf:
u[1:-1]=np.matmul(T,u[1:-1]) #updating u but leaving the boundary conditions unchanged
l.set_ydata((abs(u))) #update plot with new u
t += dt
plt.pause(0.00001)
After a lot of tinkering, it came down to reducing my step size. That did the job for me- I reduced the step size and the program worked. If anyone is facing the same problem as I am, I recommend playing around with the step sizes. Provided that the rest of the code is fine, this is the only possible area of error.

Object Looking Skewed After Essential Matrix Calculation and Projection

I am trying to calculate an essential and a projection matrix from two images. I will then use them to project a 3D object onto the image. The two images I used are
I picked a few pixel correspondences, and fed that to a SVD based least square mechanism which the books say gives me the essential matrix. I used the code below for this task (code is based mostly on Eric Solem's Programming Computer Vision with Python book):
import scipy.linalg as lin
import pandas as pd
def skew(a):
return np.array([[0,-a[2],a[1]],[a[2],0,-a[0]],[-a[1],a[0],0]])
def essential(x1,x2):
n = x1.shape[1]
A = np.zeros((n,9))
for i in range(n):
A[i] = [ x1[0,i]*x2[0,i], \
x1[0,i]*x2[1,i], \
x1[0,i]*x2[2,i], \
x1[1,i]*x2[0,i], \
x1[1,i]*x2[1,i], \
x1[1,i]*x2[2,i], \
x1[2,i]*x2[0,i], \
x1[2,i]*x2[1,i], \
x1[2,i]*x2[2,i]]
U,S,V = lin.svd(A)
F = V[-1].reshape(3,3)
return F
def compute_P_from_essential(E):
U,S,V = lin.svd(E)
if lin.det(np.dot(U,V))<0: V = -V
E = np.dot(U,np.dot(np.diag([1,1,0]),V))
Z = skew([0,0,-1])
W = np.array([[0,-1,0],[1,0,0],[0,0,1]])
P2 = [np.vstack((np.dot(U,np.dot(W,V)).T,U[:,2])).T,
np.vstack((np.dot(U,np.dot(W,V)).T,-U[:,2])).T,
np.vstack((np.dot(U,np.dot(W.T,V)).T,U[:,2])).T,
np.vstack((np.dot(U,np.dot(W.T,V)).T,-U[:,2])).T]
return P2
points = [ \
[266,163,296,160],[265,237,297,266],\
[76,288,51,340],[135,31,142,4],\
[344,167,371,156],[48,165,71,164],\
[151,68,166,56],[237,26,259,19],\
[226,147,254,140]]
df = pd.DataFrame(points)
df['uno'] = 1.
x1 = np.array(df[[0,1,'uno']].T)
x2 = np.array(df[[2,3,'uno']].T)
print x1
print x2
E = essential(x1,x2)
P = compute_P_from_essential(E)
import pandas as pd
x0 = 3.; y0 = 1.; z0 = 1.
print df.shape
e = 1
cube = [[x0,y0,z0],[x0+e,y0,z0],[x0+e,y0+e,z0],[x0,y0+e,z0],
[x0,y0,z0+e],[x0+e,y0,z0+e],[x0+e,y0+e,z0+e],[x0,y0+e,z0+e]]
cube = pd.DataFrame(cube)
cube['1'] = 1.
xx = np.dot(P[1], cube.T) * 100.
xx[1,:] = 360-xx[1,:]
#xx = xx / xx[2]
print xx[0].shape
plt.plot(xx[0], xx[1],'.')
plt.xlim(0,640)
plt.ylim(0,360)
I calculated the essential matrix, then the projection matrix, then used that to project a 3D cube. The result:
This looks skewed, I am not sure why this happened. Any ideas on how to fix this?
Thanks,
First of all, it looks like you are computing the essential matrix using exactly 9 points. You can do this using only 8 (since scale is a free parameter, you can multiply the essential by a scalar and it will stay the same so you can fix one of the parameters and just use 8 points, but I digress.) However, in practice this is a very bad idea because your 8 points might have poor spatial configuration. So what you want to do is to select N matches (600 for example), and use an algorithm like RANSAC to determine the best Essential matrix. But aside from that, what I'd recommend to debug such applications is this: compute the Fundalental matrix F based on the Essential you just computed. Now you can select a point in image 1 and then display the corresponding epipolar line in the second one. That will help you visually evaluate and thus debug the estimation of the Essential.

Complex cross spectral density

mlab.csd from matplotlib: http://matplotlib.org/api/mlab_api.html#matplotlib.mlab.csd can be used to get real valued cross spectral density. If I want to get the phase information from the spectral density, I need a csd calculation which returns complex values. Is there one ?
This is discussed e.g. in this answer: https://stackoverflow.com/a/29306730/3920342
If you use csd of the mlab library you will get complex values so you can calculate phase angles (and the real valued coherence). In the following code s1 and and s2 contain the two signals (in time domain) to be correlated.
from matplotlib import mlab
# First create power sectral densities for normalization
(ps1, f) = mlab.psd(s1, Fs=1./dt, scale_by_freq=False)
(ps2, f) = mlab.psd(s2, Fs=1./dt, scale_by_freq=False)
plt.plot(f, ps1)
plt.plot(f, ps2)
# Then calculate cross spectral density
(csd, f) = mlab.csd(s1, s2, NFFT=256, Fs=1./dt,sides='default', scale_by_freq=False)
fig = plt.figure()
ax1 = fig.add_subplot(1, 2, 1)
# Normalize cross spectral absolute values by auto power spectral density
ax1.plot(f, np.absolute(csd)**2 / (ps1 * ps2))
ax2 = fig.add_subplot(1, 2, 2)
angle = np.angle(csd, deg=True)
angle[angle<-90] += 360
ax2.plot(f, angle)
# zoom in on frequency with maximum coherence
ax1.set_xlim(9, 11)
ax1.set_ylim(0, 1e-0)
ax1.set_title("Cross spectral density: Coherence")
ax2.set_xlim(9, 11)
ax2.set_ylim(0, 90)
ax2.set_title("Cross spectral density: Phase angle")
Here the real and imaginary(!) part of the cross spectral density:
This code is taken from the question How to use the cross-spectral density to calculate the phase shift of two related signals to create two signals s1 and s2:
"""
Compute the coherence of two signals
"""
import numpy as np
import matplotlib.pyplot as plt
# make a little extra space between the subplots
plt.subplots_adjust(wspace=0.5)
nfft = 256
dt = 0.01
t = np.arange(0, 30, dt)
nse1 = np.random.randn(len(t)) # white noise 1
nse2 = np.random.randn(len(t)) # white noise 2
r = np.exp(-t/0.05)
cnse1 = np.convolve(nse1, r, mode='same')*dt # colored noise 1
cnse2 = np.convolve(nse2, r, mode='same')*dt # colored noise 2
# two signals with a coherent part and a random part
s1 = 0.01*np.sin(2*np.pi*10*t) + cnse1
s2 = 0.01*np.sin(2*np.pi*10*t) + cnse2

Fourier coefficients for NFFT - non uniform fast Fourier transform?

I am trying to use the package pynfft in python 2.7 to do the non-uniform fast Fourier transform (nfft). I have learnt python for only two months, so I have some difficulties.
This is my code:
import numpy as np
from pynfft.nfft import NFFT
#loading data, 104 lines
t_diff, x_diff = np.loadtxt('data/analysis/amplitudes.dat', unpack = True)
N = [13,8]
M = 52
#fourier coefficients
f_hat = np.fft.fft(x_diff)/(2*M)
#instantiation
plan = NFFT(N,M)
#precomputation
x = t_diff
plan.x = x
plan.precompute()
# vector of non uniform samples
f = x_diff[0:M]
#execution
plan.f = f
plan.f_hat = f_hat
f = plan.trafo()
I am basically following the instructions I found in the pynfft tutorial (http://pythonhosted.org/pyNFFT/tutorial.html).
I need the nfft because the time intervals in which my data are taken are not constant (I mean, the first measure is taken at t, the second after dt, the third after dt+dt' with dt' different from dt and so on).
The pynfft package wants the vector of the fourier coefficients ("f_hat") before execution, so I calculated it using numpy.fft, but I am not sure this procedure is correct. Is there another way to do it (maybe with the nfft)?
I would like also to calculate the frquencies; I know that with numpy.fft there is a command: is ther anything like that also for pynfft? I did not find anything in the tutorial.
Thank you for any advice you can give me.
Here is a working example, taken from here:
First we define the function we want to reconstruct, which is the sum of four harmonics:
import numpy as np
import matplotlib.pyplot as plt
np.random.seed(12345)
%pylab inline --no-import-all
# function we want to reconstruct
k=[1,5,10,30] # modulating coefficients
def myf(x,k):
return sum(np.sin(x*k0*(2*np.pi)) for k0 in k)
x=np.linspace(-0.5,0.5,1000) # 'continuous' time/spatial domain; -0.5<x<+0.5
y=myf(x,k) # 'true' underlying trigonometric function
fig=plt.figure(1,(20,5))
ax =fig.add_subplot(111)
ax.plot(x,y,'red')
ax.plot(x,y,'r.')
# we should sample at a rate of >2*~max(k)
M=256 # number of nodes
N=128 # number of Fourier coefficients
nodes =np.random.rand(M)-0.5 # non-uniform oversampling
values=myf(nodes,k) # nodes&values will be used below to reconstruct
# original function using the Solver
ax.plot(nodes,values,'bo')
ax.set_xlim(-0.5,+0.5)
The we initialize and run the Solver:
from pynfft import NFFT, Solver
f = np.empty(M, dtype=np.complex128)
f_hat = np.empty([N,N], dtype=np.complex128)
this_nfft = NFFT(N=[N,N], M=M)
this_nfft.x = np.array([[node_i,0.] for node_i in nodes])
this_nfft.precompute()
this_nfft.f = f
ret2=this_nfft.adjoint()
print this_nfft.M # number of nodes, complex typed
print this_nfft.N # number of Fourier coefficients, complex typed
#print this_nfft.x # nodes in [-0.5, 0.5), float typed
this_solver = Solver(this_nfft)
this_solver.y = values # '''right hand side, samples.'''
#this_solver.f_hat_iter = f_hat # assign arbitrary initial solution guess, default is 0
this_solver.before_loop() # initialize solver internals
while not np.all(this_solver.r_iter < 1e-2):
this_solver.loop_one_step()
Finally, we display the frequencies:
import matplotlib.pyplot as plt
fig=plt.figure(1,(20,5))
ax =fig.add_subplot(111)
foo=[ np.abs( this_solver.f_hat_iter[i][0])**2 for i in range(len(this_solver.f_hat_iter) ) ]
ax.plot(np.abs(np.arange(-N/2,+N/2,1)),foo)
cheers