The aim of this post is to properly understand Numerical Fourier Transform on Python or Matlab with an example in which the Analytical Fourier Transform is well known. For this purpose I choose the rectangular function, the analytical expression of it and its Fourier Transform are reported here
https://en.wikipedia.org/wiki/Rectangular_function
Here the code in Matlab
x = -3 : 0.01 : 3;
y = zeros(length(x));
y(200:400) = 1;
ffty = fft(y);
ffty = fftshift(ffty);
plot(real(ffty))
And here the code in Python
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(-3, 3, 0.01)
y = np.zeros(len(x))
y[200:400] = 1
ffty = np.fft.fft(y)
ffty = np.fft.fftshift(ffty)
plt.plot(np.real(ffty))
In both the two programming langueages I have the some result with the some problems:
First of all the fourier transfrom is not real as expected, moreover even choosing the real part, the solution does not looks like the analytical solution: in fact the first plot reported here is as it should be at least in shape and the second plot is what I get from my calculations.
Is there anyone who could suggest me how to analytically calculate the fourier transform of the rectangular function?
There are two problems in your Matlab code:
First, y = zeros(length(x)); should be y = zeros(1,length(x));. Currently you create a square matrix, not a vector.
Second, the DFT (or FFT) will be real and symmetric if y is. Your y should be symmetric, and that means with respect to 0. So, instead of y(200:400) = 1; use y(1:100) = 1; y(end-98:end) = 1;. Recall that the DFT is like the Fourier series of a signal from which your input is just one period, and the first sample corresponds to time instant 0.
So:
x = -3 : 0.01 : 3;
y = zeros(1,length(x));
y(1:100) = 1; y(end-98:end) = 1;
ffty = fft(y);
ffty = fftshift(ffty);
plot(ffty)
gives
>> isreal(ffty)
ans =
1
The code in Python is
import matplotlib.pyplot as plt
import numpy as np
x = np.arange(-3, 3, 0.01)
y = np.zeros(len(x))
y[200:400] = 1
yShift = np.fft.fftshift(y)
fftyShift = np.fft.fft(yShift)
ffty = np.fft.fftshift(fftyShift)
plt.plot(ffty)
plt.show()
Related
i have to often normalize the result of circular-convolution, so I 'borrowed'-and-modfied the following routine :
def fft_normalize(x):# Normalize a vector x in complex domain.
c = np.fft.rfft(x)
# Look at real and image as if they were real
ri = np.vstack([c.real, c.imag])
# Normalize magnitude of each complex/real pair
norm = np.linalg.norm(ri, axis=0)
if np.any(norm==0): norm[norm == 0] = np.float64(1e-308) #!fixme
ri= np.divide(ri,norm)
c_proj = ri[0,:] + 1j * ri[1,:]
rv = np.fft.irfft(c_proj, n=x.shape[-1])
return rv
def fft_convolution(a, b):
return np.fft.irfft(np.fft.rfft(a) * np.fft.rfft(b))
so that I do this :
fft_normalize(fft_convolution(a,b))
i see in the numpy docs there is a 'norm' option. Is this equivalent to what i'm doing ?
And if yes, which option should I use ?
def fft_convolution2(a, b):
return np.fft.irfft(np.fft.rfft(a) * np.fft.rfft(b), norm='ortho')
When I test it it behaves better when I do fft_normalize()
Second, i had to add this line, but it does not seem, right. any ideas ?
if np.any(norm==0): norm[norm == 0] = np.float64(1e-308) #!fixme
As a side note, if you know !! numpy docs mentions that they promote float32 to float64 and that scipy.fftpack does not !!
Would fftpack be faster ! because scipy says fftpack is obsolete and there is no info on the new scipy ?
#cris are you sugesting i do it this way :
def fft_normalize(x):# Normalize a vector x in complex domain.
c = np.fft.rfft(x)
ri = np.vstack([c.real, c.imag])
norm = np.abs(c)
if np.any(norm==0): norm[norm == 0] = MIN #!fixme
ri= np.divide(ri,norm)
c_proj = ri[0,:] + 1j * ri[1,:]
rv = np.fft.irfft(c_proj, n=x.shape[-1])
return rv
The norm argument to the FFT functions in NumPy determine whether the transform result is multiplied by 1, 1/N or 1/sqrt(N), with N the number of samples in the array. Normally, the inverse transform is normalized by dividing by N, and the forward transform is not. Specifying “ortho” here causes both transforms to be normalized by 1/sqrt(2). Specifying “forward” causes only the forward transform to be normalized by 1/N.
These normalizations are very different from the one you apply, where you normalize each element in the frequency domain.
I'm trying to code a demonstration of compressed sensing for my final year project but am getting poor image reconstruction when using the Lasso algorithm. I've relied on the following as a reference: http://www.pyrunner.com/weblog/2016/05/26/compressed-sensing-python/
However my code has some differences:
I use scikit-learn to perform a lasso optimisation (basis pursuit) as opposed to using cvxpy to perform an l_1 minimisation with an equality constraint as in the article.
I construct psi differently/more simply, testing seems to show that it's correct.
I use a different package to read and write the image.
import numpy as np
import scipy.fftpack as spfft
import scipy.ndimage as spimg
import imageio
from sklearn.linear_model import Lasso
x_orig = imageio.imread('gt40.jpg', pilmode='L') # read in grayscale
x = spimg.zoom(x_orig, 0.2) #zoom for speed
ny,nx = x.shape
k = round(nx * ny * 0.5) #50% sample
ri = np.random.choice(nx * ny, k, replace=False)
y = x.T.flat[ri] #y is the measured sample
# y = np.expand_dims(y, axis=1) ---- this doesn't seem to make a difference, was presumably required with cvxpy
psi = spfft.idct(np.identity(nx*ny), norm='ortho', axis=0) #my construction of psi
# psi = np.kron(
# spfft.idct(np.identity(nx), norm='ortho', axis=0),
# spfft.idct(np.identity(ny), norm='ortho', axis=0)
# )
# psi = 2*np.random.random_sample((nx*ny,nx*ny)) - 1
theta = psi[ri,:] #equivalent to phi*psi
lasso = Lasso(alpha=0.001, max_iter=10000)
lasso.fit(theta, y)
s = np.array(lasso.coef_)
x_recovered = psi#s
x_recovered = x_recovered.reshape(nx, ny).T
x_recovered_final = x_recovered.astype('uint8') #recovered image is float64 and has negative values..
imageio.imwrite('gt40_recovered.jpg', x_recovered_final)
Unfortunately I'm not allowed to post images yet so here is a link to the original zoomed image, the image recovered with lasso and the image recovered with cvxpy (described later):
https://imgur.com/a/LROSug6
As you can see not only is the recovery poor but the image completely corrupted - the colours seem to be negative and the detail from the 50% sample lost. I think I've managed to track down the problem to the Lasso regression - it returns a vector that, when inverse transformed, has values that are not necessarily in the 0-255 range as expected for the image. So the conversion to from dtype float64 to uint8 is rather random (e.g. -55 becomes 255-55=200).
Following this I tried swapping out lasso for the same optimisation as in the article (minimising the l_1 norm subject to theta*s=y using cvxpy):
import cvxpy as cvx
x_orig = imageio.imread('gt40.jpg', pilmode='L') # read in grayscale
x = spimg.zoom(x_orig, 0.2)
ny,nx = x.shape
k = round(nx * ny * 0.5)
ri = np.random.choice(nx * ny, k, replace=False)
y = x.T.flat[ri]
psi = spfft.idct(np.identity(nx*ny), norm='ortho', axis=0)
theta = psi[ri,:] #equivalent to phi*psi
#NEW CODE STARTS:
vx = cvx.Variable(nx * ny)
objective = cvx.Minimize(cvx.norm(vx, 1))
constraints = [theta#vx == y]
prob = cvx.Problem(objective, constraints)
result = prob.solve(verbose=True)
s = np.array(vx.value).squeeze()
x_recovered = psi#s
x_recovered = x_recovered.reshape(nx, ny).T
x_recovered_final = x_recovered.astype('uint8')
imageio.imwrite('gt40_recovered_altopt.jpg', x_recovered_final)
This took nearly 6 hours but finally I got a somewhat satisfactory result. However I would like to perform a demonstration of lasso if possible. Any help in getting the lasso to return appropriate values or somehow converting its result appropriately would be very much appreciated.
I've been trying to create a 2D map of blobs of matter (Gaussian random field) using a variance I have calculated. This variance is a 2D array. I have tried using numpy.random.normal since it allows for a 2D input of the variance, but it doesn't really create a map with the trend I expect from the input parameters. One of the important input constants lambda_c should manifest itself as the physical size (diameter) of the blobs. However, when I change my lambda_c, the size of the blobs does not change if at all. For example, if I set lambda_c = 40 parsecs, the map needs blobs that are 40 parsecs in diameter. A MWE to produce the map using my variance:
import numpy as np
import random
import matplotlib.pyplot as plt
from matplotlib.pyplot import show, plot
import scipy.integrate as integrate
from scipy.interpolate import RectBivariateSpline
n = 300
c = 3e8
G = 6.67e-11
M_sun = 1.989e30
pc = 3.086e16 # parsec
Dds = 1097.07889283e6*pc
Ds = 1726.62069147e6*pc
Dd = 1259e6*pc
FOV_arcsec_original = 5.
FOV_arcmin = FOV_arcsec_original/60.
pix2rad = ((FOV_arcmin/60.)/float(n))*np.pi/180.
rad2pix = 1./pix2rad
x_pix = np.linspace(-FOV_arcsec_original/2/pix2rad/180.*np.pi/3600.,FOV_arcsec_original/2/pix2rad/180.*np.pi/3600.,n)
y_pix = np.linspace(-FOV_arcsec_original/2/pix2rad/180.*np.pi/3600.,FOV_arcsec_original/2/pix2rad/180.*np.pi/3600.,n)
X_pix,Y_pix = np.meshgrid(x_pix,y_pix)
conc = 10.
M = 1e13*M_sun
r_s = 18*1e3*pc
lambda_c = 40*pc ### The important parameter that doesn't seem to manifest itself in the map when changed
rho_s = M/((4*np.pi*r_s**3)*(np.log(1+conc) - (conc/(1+conc))))
sigma_crit = (c**2*Ds)/(4*np.pi*G*Dd*Dds)
k_s = rho_s*r_s/sigma_crit
theta_s = r_s/Dd
Renorm = (4*G/c**2)*(Dds/(Dd*Ds))
#### Here I just interpolate and zoom into my field of view to get better resolutions
A = np.sqrt(X_pix**2 + Y_pix**2)*pix2rad/theta_s
A_1 = A[100:200,0:100]
n_x = n_y = 100
FOV_arcsec_x = FOV_arcsec_original*(100./300)
FOV_arcmin_x = FOV_arcsec_x/60.
pix2rad_x = ((FOV_arcmin_x/60.)/float(n_x))*np.pi/180.
rad2pix_x = 1./pix2rad_x
FOV_arcsec_y = FOV_arcsec_original*(100./300)
FOV_arcmin_y = FOV_arcsec_y/60.
pix2rad_y = ((FOV_arcmin_y/60.)/float(n_y))*np.pi/180.
rad2pix_y = 1./pix2rad_y
x1 = np.linspace(-FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,n_x)
y1 = np.linspace(-FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,n_y)
X1,Y1 = np.meshgrid(x1,y1)
n_x_2 = 500
n_y_2 = 500
x2 = np.linspace(-FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,n_x_2)
y2 = np.linspace(-FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,n_y_2)
X2,Y2 = np.meshgrid(x2,y2)
interp_spline = RectBivariateSpline(y1,x1,A_1)
A_2 = interp_spline(y2,x2)
A_3 = A_2[50:450,0:400]
n_x_3 = n_y_3 = 400
FOV_arcsec_x = FOV_arcsec_original*(100./300)*400./500.
FOV_arcmin_x = FOV_arcsec_x/60.
pix2rad_x = ((FOV_arcmin_x/60.)/float(n_x_3))*np.pi/180.
rad2pix_x = 1./pix2rad_x
FOV_arcsec_y = FOV_arcsec_original*(100./300)*400./500.
FOV_arcmin_y = FOV_arcsec_y/60.
pix2rad_y = ((FOV_arcmin_y/60.)/float(n_y_3))*np.pi/180.
rad2pix_y = 1./pix2rad_y
x3 = np.linspace(-FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,n_x_3)
y3 = np.linspace(-FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,n_y_3)
X3,Y3 = np.meshgrid(x3,y3)
n_x_4 = 1000
n_y_4 = 1000
x4 = np.linspace(-FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,n_x_4)
y4 = np.linspace(-FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,n_y_4)
X4,Y4 = np.meshgrid(x4,y4)
interp_spline = RectBivariateSpline(y3,x3,A_3)
A_4 = interp_spline(y4,x4)
############### Function to calculate variance
variance = np.zeros((len(A_4),len(A_4)))
def variance_fluctuations(x):
for i in xrange(len(x)):
for j in xrange(len(x)):
if x[j][i] < 1.:
variance[j][i] = (k_s**2)*(lambda_c/r_s)*((np.pi/x[j][i]) - (1./(x[j][i]**2 -1)**3.)*(((6.*x[j][i]**4. - 17.*x[j][i]**2. + 26)/3.)+ (((2.*x[j][i]**6. - 7.*x[j][i]**4. + 8.*x[j][i]**2. - 8)*np.arccosh(1./x[j][i]))/(np.sqrt(1-x[j][i]**2.)))))
elif x[j][i] > 1.:
variance[j][i] = (k_s**2)*(lambda_c/r_s)*((np.pi/x[j][i]) - (1./(x[j][i]**2 -1)**3.)*(((6.*x[j][i]**4. - 17.*x[j][i]**2. + 26)/3.)+ (((2.*x[j][i]**6. - 7.*x[j][i]**4. + 8.*x[j][i]**2. - 8)*np.arccos(1./x[j][i]))/(np.sqrt(x[j][i]**2.-1)))))
variance_fluctuations(A_4)
#### Creating the map
mean = 0
delta_kappa = np.random.normal(0,variance,A_4.shape)
xfinal = np.linspace(-FOV_arcsec_x*np.pi/180./3600.*Dd/pc/2,FOV_arcsec_x*np.pi/180./3600.*Dd/pc/2,1000)
yfinal = np.linspace(-FOV_arcsec_x*np.pi/180./3600.*Dd/pc/2,FOV_arcsec_x*np.pi/180./3600.*Dd/pc/2,1000)
Xfinal, Yfinal = np.meshgrid(xfinal,yfinal)
plt.contourf(Xfinal,Yfinal,delta_kappa,100)
plt.show()
The map looks like this, with the density of blobs increasing towards the right. However, the size of the blobs don't change and the map looks virtually the same whether I use lambda_c = 40*pc or lambda_c = 400*pc.
I'm wondering if the np.random.normal function isn't really doing what I expect it to do? I feel like the pixel scale of the map and the way samples are drawn make no link to the size of the blobs. Maybe there is a better way to create the map using the variance, would appreciate any insight.
I expect the map to look something like this , the blob sizes change based on the input parameters for my variance :
This is quite a well visited problem in (surprise surprise) astronomy and cosmology.
You could use lenstool: https://lenstools.readthedocs.io/en/latest/examples/gaussian_random_field.html
You could also try here:
https://andrewwalker.github.io/statefultransitions/post/gaussian-fields
Not to mention:
https://github.com/bsciolla/gaussian-random-fields
I am not reproducing code here because all credit goes to the above authors. However, they did just all come right out a google search :/
Easiest of all is probably a python module FyeldGenerator, apparently designed for this exact purpose:
https://github.com/cphyc/FyeldGenerator
So (adapted from github example):
pip install FyeldGenerator
from FyeldGenerator import generate_field
from matplotlib import use
use('Agg')
import matplotlib.pyplot as plt
import numpy as np
plt.figure()
# Helper that generates power-law power spectrum
def Pkgen(n):
def Pk(k):
return np.power(k, -n)
return Pk
# Draw samples from a normal distribution
def distrib(shape):
a = np.random.normal(loc=0, scale=1, size=shape)
b = np.random.normal(loc=0, scale=1, size=shape)
return a + 1j * b
shape = (512, 512)
field = generate_field(distrib, Pkgen(2), shape)
plt.imshow(field, cmap='jet')
plt.savefig('field.png',dpi=400)
plt.close())
This gives:
Looks pretty straightforward to me :)
PS: FoV implied a telescope observation of the gaussian random field :)
A completely different and much quicker way may be just to blur the delta_kappa array with gaussian filter. Try adjusting sigma parameter to alter the blobs size.
from scipy.ndimage.filters import gaussian_filter
dk_gf = gaussian_filter(delta_kappa, sigma=20)
Xfinal, Yfinal = np.meshgrid(xfinal,yfinal)
plt.contourf(Xfinal,Yfinal,dk_ma,100, cmap='jet')
plt.show();
this is image with sigma=20
this is image with sigma=2.5
ThunderFlash, try this code to draw the map:
# function to produce blobs:
from scipy.stats import multivariate_normal
def blob (positions, mean=(0,0), var=1):
cov = [[var,0],[0,var]]
return multivariate_normal(mean, cov).pdf(positions)
"""
now prepare for blobs generation.
note that I use less dense grid to pick blobs centers (regulated by `step`)
this makes blobs more pronounced and saves calculation time.
use this part instead of your code section below comment #### Creating the map
"""
delta_kappa = np.random.normal(0,variance,A_4.shape) # same
step = 10 #
dk2 = delta_kappa[::step,::step] # taking every 10th element
x2, y2 = xfinal[::step],yfinal[::step]
field = np.dstack((Xfinal,Yfinal))
print (field.shape, dk2.shape, x2.shape, y2.shape)
>> (1000, 1000, 2), (100, 100), (100,), (100,)
result = np.zeros(field.shape[:2])
for x in range (len(x2)):
for y in range (len(y2)):
res2 = blob(field, mean = (x2[x], y2[y]), var=10000)*dk2[x,y]
result += res2
# the cycle above took over 20 minutes on Ryzen 2700X. It could be accelerated by vectorization presumably.
plt.contourf(Xfinal,Yfinal,result,100)
plt.show()
you may want to play with var parameter in blob() to smoothen the image and with step to make it more compressed.
Here is the image that I got using your code (somehow axes are flipped and more dense areas on the top):
I understand that most FFT/IFFT routines have an error floor. I was expecting NumPy's FFT to have an error floor in the same orders as FFTW (say 1e-15), but the following experiment shows errors in the order of 1e-5.
Consider calculating the IDFT of a box. It is well-known that the result is the sinc-like Dirichlet kernel. But that is not what I get from numpy.fft.irfft. In fact even the first sample that should simply equal the width of the box divided by the number of FFT points is off by an amount around 4e-5 as the following example shows:
import numpy as np
import matplotlib.pyplot as plt
from scipy.special import diric
N = 40960
K = 513
X = np.ones(K, dtype=np.complex)
x = np.fft.irfft(X, N)
print("x[0] = %g: expected %g - error = %g" % (x[0], (2*K+1)/N, x[0]-(2*K+1)/N))
# expected IDFT of a box is Dirichlet function (see
# https://en.wikipedia.org/wiki/Discrete_Fourier_transform#Some_discrete_Fourier_transform_pairs)
y = diric(2*np.pi*np.arange(N)/N, 2*K+1) * (2*K+1) / N
plt.figure()
plt.plot(x[:1024] - y[:1024])
plt.title('error')
plt.show(block=True)
It looks like the error is of sinusoidal form:
Has anybody experience same issue? Am I misunderstanding something about the NumPy's FFT pack or it is just not accurate?
Update
Here is the equivalent of part of the script in Octave:
N = 40960;
K = 513;
X = zeros(1, N);
X(1:K) = 1;
X(N-K:N) = 1;
x = ifft(X);
fprintf("x[0] = %g, expected = %g - error = %g\n", x(1), (2*K+1)/N, x(1)-(2*K+1)/N);
The error on x[0] is practically zero in Octave. (I did not check other samples because I am not aware of equivalent of diric function in Octave.)
Thanks to MarkDickinson, I realized that my math was wrong. The correct comparison would be carried out by:
import numpy as np
import matplotlib.pyplot as plt
from scipy.special import diric
N = 40960
K = 513
X = np.ones(K+1, dtype=np.complex)
x = np.fft.irfft(X, N)
print("x[0] = %g: expected %g - error = %g" % (x[0], (2*K+1)/N, x[0]-(2*K+1)/N))
# expected IDFT of a box is Dirichlet function (see
# https://en.wikipedia.org/wiki/Discrete_Fourier_transform#Some_discrete_Fourier_transform_pairs)
y = diric(2*np.pi*np.arange(N)/N, 2*K+1) * (2*K+1) / N
plt.figure()
plt.plot(x[:1024] - y[:1024])
plt.title('error')
plt.show(block=True)
that shows irfft is accurate. Here is the error plot:
Numpy is correct, my math was incorrect. I am sorry for posting this misleading question. I don't know what is the standard procedure in these cases. Should I delete my question or leave it here with this answer? I just don't want it to be undermining NumPy or challanging its accuracy (as this was clearly a false alarm).
I am trying to fit a quadratic function to some data, and I'm trying to do this without using numpy's polyfit function.
Mathematically I tried to follow this website https://neutrium.net/mathematics/least-squares-fitting-of-a-polynomial/ but somehow I don't think that I'm doing it right. If anyone could assist me that would be great, or If you could suggest another way to do it that would also be awesome.
What I've tried so far:
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
ones = np.ones(3)
A = np.array( ((0,1),(1,1),(2,1)))
xfeature = A.T[0]
squaredfeature = A.T[0] ** 2
b = np.array( (1,2,0), ndmin=2 ).T
b = b.reshape(3)
features = np.concatenate((np.vstack(ones), np.vstack(xfeature), np.vstack(squaredfeature)), axis = 1)
featuresc = features.copy()
print(features)
m_det = np.linalg.det(features)
print(m_det)
determinants = []
for i in range(3):
featuresc.T[i] = b
print(featuresc)
det = np.linalg.det(featuresc)
determinants.append(det)
print(det)
featuresc = features.copy()
determinants = determinants / m_det
print(determinants)
plt.scatter(A.T[0],b)
u = np.linspace(0,3,100)
plt.plot(u, u**2*determinants[2] + u*determinants[1] + determinants[0] )
p2 = np.polyfit(A.T[0],b,2)
plt.plot(u, np.polyval(p2,u), 'b--')
plt.show()
As you can see my curve doesn't compare well to nnumpy's polyfit curve.
Update:
I went through my code and removed all the stupid mistakes and now it works, when I try to fit it over 3 points, but I have no idea how to fit over more than three points.
This is the new code:
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
ones = np.ones(3)
A = np.array( ((0,1),(1,1),(2,1)))
xfeature = A.T[0]
squaredfeature = A.T[0] ** 2
b = np.array( (1,2,0), ndmin=2 ).T
b = b.reshape(3)
features = np.concatenate((np.vstack(ones), np.vstack(xfeature), np.vstack(squaredfeature)), axis = 1)
featuresc = features.copy()
print(features)
m_det = np.linalg.det(features)
print(m_det)
determinants = []
for i in range(3):
featuresc.T[i] = b
print(featuresc)
det = np.linalg.det(featuresc)
determinants.append(det)
print(det)
featuresc = features.copy()
determinants = determinants / m_det
print(determinants)
plt.scatter(A.T[0],b)
u = np.linspace(0,3,100)
plt.plot(u, u**2*determinants[2] + u*determinants[1] + determinants[0] )
p2 = np.polyfit(A.T[0],b,2)
plt.plot(u, np.polyval(p2,u), 'r--')
plt.show()
Instead using Cramer's Rule, actually solve the system using least squares. Remember that Cramer's Rule will only work if the total number of points you have equals the desired order of polynomial plus 1.
If you don't have this, then Cramer's Rule will not work as you're trying to find an exact solution to the problem. If you have more points, the method is unsuitable as we will create an overdetermined system of equations.
To adapt this to more points, numpy.linalg.lstsq would be a better fit as it solves the solution to the Ax = b by computing the vector x that minimizes the Euclidean norm using the matrix A. Therefore, remove the y values from the last column of the features matrix and solve for the coefficients and use numpy.linalg.lstsq to solve for the coefficients:
import numpy as np
import matplotlib.pyplot as plt
ones = np.ones(4)
xfeature = np.asarray([0,1,2,3])
squaredfeature = xfeature ** 2
b = np.asarray([1,2,0,3])
features = np.concatenate((np.vstack(ones),np.vstack(xfeature),np.vstack(squaredfeature)), axis = 1) # Change - remove the y values
determinants = np.linalg.lstsq(features, b)[0] # Change - use least squares
plt.scatter(xfeature,b)
u = np.linspace(0,3,100)
plt.plot(u, u**2*determinants[2] + u*determinants[1] + determinants[0] )
plt.show()
I get this plot now, which matches what the dashed curve is in your graph, also matching what numpy.polyfit gives you: