I have two solutions to this problem actually, they are both applied below to a test case. The thing is that none of them is perfect: first one only take into account the two end points, the other one can't be made "arbitrarily smooth": there is a limit in the amount of smoothness one can achieve (the one I am showing).
I am sure there is a better solution, that kind-of go from the first solution to the other and all the way to no smoothing at all. It may already be implemented somewhere. Maybe solving a minimization problem with an arbitrary number of splines equidistributed?
Thank you very much for your help
Ps: the seed used is a challenging one
import matplotlib.pyplot as plt
from scipy import interpolate
from scipy.signal import savgol_filter
import numpy as np
import random
def scipy_bspline(cv, n=100, degree=3):
""" Calculate n samples on a bspline
cv : Array ov control vertices
n : Number of samples to return
degree: Curve degree
"""
cv = np.asarray(cv)
count = cv.shape[0]
degree = np.clip(degree,1,count-1)
kv = np.clip(np.arange(count+degree+1)-degree,0,count-degree)
# Return samples
max_param = count - (degree * (1-periodic))
spl = interpolate.BSpline(kv, cv, degree)
return spl(np.linspace(0,max_param,n))
def round_up_to_odd(f):
return np.int(np.ceil(f / 2.) * 2 + 1)
def generateRandomSignal(n=1000, seed=None):
"""
Parameters
----------
n : integer, optional
Number of points in the signal. The default is 1000.
Returns
-------
sig : numpy array
"""
np.random.seed(seed)
print("Seed was:", seed)
steps = np.random.choice(a=[-1, 0, 1], size=(n-1))
roughSig = np.concatenate([np.array([0]), steps]).cumsum(0)
sig = savgol_filter(roughSig, round_up_to_odd(n/10), 6)
return sig
# Generate a random signal to illustrate my point
n = 1000
t = np.linspace(0, 10, n)
seed = 45136. # Challenging seed
sig = generateRandomSignal(n=1000, seed=seed)
sigInit = np.copy(sig)
# Add noise to the signal
mean = 0
std = sig.max()/3.0
num_samples = n/5
idxMin = n/2-100
idxMax = idxMin + num_samples
tCut = t[idxMin+1:idxMax]
noise = np.random.normal(mean, std, size=num_samples-1) + 2*std*np.sin(2.0*np.pi*tCut/0.4)
sig[idxMin+1:idxMax] += noise
# Define filtering range enclosing the noisy area of the signal
idxMin -= 20
idxMax += 20
# Extreme filtering solution
# Spline between first and last points, the points in between have no influence
sigTrim = np.delete(sig, np.arange(idxMin,idxMax))
tTrim = np.delete(t, np.arange(idxMin,idxMax))
f = interpolate.interp1d(tTrim, sigTrim, kind='quadratic')
sigSmooth1 = f(t)
# My attempt. Not bad but not perfect because there is a limit in the maximum
# amount of smoothing we can add (degree=len(tSlice) is the maximum)
# If I could do degree=10*len(tSlice) and converging to the first solution
# I would be done!
sigSlice = sig[idxMin:idxMax]
tSlice = t[idxMin:idxMax]
cv = np.stack((tSlice, sigSlice)).T
p = scipy_bspline(cv, n=len(tSlice), degree=len(tSlice))
tSlice = p.T[0]
sigSliceSmooth = p.T[1]
sigSmooth2 = np.copy(sig)
sigSmooth2[idxMin:idxMax] = sigSliceSmooth
# Plot
plt.figure()
plt.plot(t, sig, label="Signal")
plt.plot(t, sigSmooth1, label="Solution 1")
plt.plot(t, sigSmooth2, label="Solution 2")
plt.plot(t[idxMin:idxMax], sigInit[idxMin:idxMax], label="What I'd want (kind of, smoother will be even better actually)")
plt.plot([t[idxMin],t[idxMax]], [sig[idxMin],sig[idxMax]],"o")
plt.legend()
plt.show()
sys.exit()
Yes, a minimization is a good way to approach this smoothing problem.
Least squares problem
Here is a suggestion for a least squares formulation: let s[0], ..., s[N] denote the N+1 samples of the given signal to smooth, and let L and R be the desired slopes to preserve at the left and right endpoints. Find the smoothed signal u[0], ..., u[N] as the minimizer of
min_u (1/2) sum_n (u[n] - s[n])² + (λ/2) sum_n (u[n+1] - 2 u[n] + u[n-1])²
subject to
s[0] = u[0], s[N] = u[N] (value constraints),
L = u[1] - u[0], R = u[N] - u[N-1] (slope constraints),
where in the minimization objective, the sums are over n = 1, ..., N-1 and λ is a positive parameter controlling the smoothing strength. The first term tries to keep the solution close to the original signal, and the second term penalizes u for bending to encourage a smooth solution.
The slope constraints require that
u[1] = L + u[0] = L + s[0] and u[N-1] = u[N] - R = s[N] - R. So we can consider the minimization as over only the interior samples u[2], ..., u[N-2].
Finding the minimizer
The minimizer satisfies the Euler–Lagrange equations
(u[n] - s[n]) / λ + (u[n+2] - 4 u[n+1] + 6 u[n] - 4 u[n-1] + u[n-2]) = 0
for n = 2, ..., N-2.
An easy way to find an approximate solution is by gradient descent: initialize u = np.copy(s), set u[1] = L + s[0] and u[N-1] = s[N] - R, and do 100 iterations or so of
u[2:-2] -= (0.05 / λ) * (u - s)[2:-2] + np.convolve(u, [1, -4, 6, -4, 1])[4:-4]
But with some more work, it is possible to do better than this by solving the E–L equations directly. For each n, move the known quantities to the right-hand side: s[n] and also the endpoints u[0] = s[0], u[1] = L + s[0], u[N-1] = s[N] - R, u[N] = s[N]. The you will have a linear system "A u = b", and matrix A has rows like
0, ..., 0, 1, -4, (6 + 1/λ), -4, 1, 0, ..., 0.
Finally, solve the linear system to find the smoothed signal u. You could use numpy.linalg.solve to do this if N is not too large, or if N is large, try an iterative method like conjugate gradients.
you can apply a simple smoothing method and plot the smooth curves with different smoothness values to see which one works best.
def smoothing(data, smoothness=0.5):
last = data[0]
new_data = [data[0]]
for datum in data[1:]:
new_value = smoothness * last + (1 - smoothness) * datum
new_data.append(new_value)
last = datum
return new_data
You can plot this curve for multiple values of smoothness and pick the curve which suits your needs. You can also apply this method only on a range of values in the actual curve by defining start and end
I am trying to apply an exponential fit to my data to determine the point at which the value drops by 1/e. When plotted, the fit seems to favor smaller values and does not portray the true relationship.
import numpy as np
import matplotlib
matplotlib.use("TkAgg") # need to set the TkAgg backend explicitly otherwise it introduced a low-level error
from matplotlib import pyplot as plt
import scipy as sc
def autoCorrelation(sample, longTime, temp, plotTau = False ):
# compute empirical autocovariance with lag tau averaged over time longTime
sample.takeTimeStep(timesteps=1500) # 1500 timesteps to let sample reach equilibrium
M = np.zeros(longTime)
for tau in range(longTime):
M[tau] = sample.calcMagnetisation()
sample.takeTimeStep()
M_ave = np.average(M) #time - average
M = (M - M_ave)
autocorrelation = np.correlate(M, M, mode='full')
autocorrelation /= autocorrelation.max() # normalise such that max autocorrelation is 1
autocorrelationArray = autocorrelation[int(len(autocorrelation)/2):]
x = np.arange(0, len(autocorrelationArray), 1)
# apply exponential fit
def exponenial(x, a, b):
return a * np.exp(-b * x)
popt, pcov = curve_fit(exponenial, x, np.absolute(autocorrelationArray)) # array, 2d array
yy = exponenial(x, *popt)
plt.plot(x, np.absolute(autocorrelationArray), 'o', x, yy)
plt.title('Exponential Fit of Magnetisation Autocorrelation against Time for Temperature = ' + str(T) + ' J/k')
plt.xlabel('Time / Number of Iterations ')
plt.ylabel('Magnetisation Autocorrelation')
plt.show()
# prints tau_e value b from exponential a * np.exp(-b * x)
print('tau_e is ' + str(1/popt[1])) # units converted to time steps by taking reciprocal
if __name__ == '__main__':
#plot autocorrelation against time
longTime = 100
temp = [1, 2, 2.3, 2.6, 3, 4]
for T in temp:
magnet = Ising(30, T) # (N, temp)
autoCorrelation(magnet, longTime, temp)
Note: Ising is a class in another .py file containing the functions takeTimeStep and calcMagnetisation.
Expect greater values of tau_e
I would like to use the Fourier transform to find the center of a simulated entity under periodic boundary condition; periodic boundary conditions means, that whenever something exits through one side of the box, it is warped around to appear on the opposite side just like in the classic game asteroids.
So what I have is for each time frame a matrix (Nx3) with N the number of points in xyz. what I want to do is determine the center of that cloud even if it all moved over the periodic boundary and is so to say stuck in between.
My idea for an solution would now be do a (mass weigted) histogram of these points and then perform an FFT on that and use the phase of the first Fourier coefficient to determine where in the box the maximum would be.
as a test case I have used
import numpy as np
Points_x = np.random.randn(10000)
Box_min = -10
Box_max = 10
X = np.linspace( Box_min, Box_max, 100 )
### make a Histogram of the points
Histogram_Points = np.bincount( np.digitize( Points_x, X ), minlength=100 )
### make an artifical shift over the periodic boundary
Histogram_Points = np.r_[ Histogram_Points[45:], Histogram_Points[:45] ]
So now I can use FFT since it expects a periodic function anyways.
## doing fft
F = np.fft.fft(Histogram_Points)
## getting rid of everything but first harmonic
F[2:] = 0.
## back transforming
Fist_harmonic = np.fft.ifft(F)
That way I get a sine wave with its maximum exactly where the maximum of the histogram is.
Now I'd like to extract the position of the maximum not by taking the max function on the sine vector, but somehow it should be retrievable from the first (not the 0th) Fourier coefficient, since that should somehow contain the phase shift of the sine to have its maximum exactly at the maximum of the histogram.
Indeed, plotting
Cos_approx = cos( linspace(0,2*pi,100) * angle(F[1]) )
will give
But I can't figure out how to get the position of the peak from this angle.
Using the FFT is overkill when all you need is one Fourier coefficent. Instead, you can simply compute the dot product of your data with
w = np.exp(-2j*np.pi*np.arange(N) / N)
where N is the number of points. (The time to compute all the Fourier coefficients with the FFT is O(N*log(N)). Computing just one coefficient is O(N).)
Here's a script similar to yours. The data is put in y; the coordinates of the data points are in x.
import numpy as np
N = 100
# x coordinates of the data
xmin = -10
xmax = 10
x = np.linspace(xmin, xmax, N, endpoint=False)
# Generate data in y.
n = 35
y = np.zeros(N)
y[:n] = 1 - np.cos(np.linspace(0, 2*np.pi, n))
y[:n] /= 0.7 + 0.3*np.random.rand(n)
m = 10
y = np.r_[y[m:], y[:m]]
# Compute coefficent 1 of the discrete Fourier transform.
w = np.exp(-2j*np.pi*np.arange(N) / N)
F1 = y.dot(w)
print "F1 =", F1
# Get the angle of F1 (in the interval [0,2*pi]).
angle = np.angle(F1.conj())
if angle < 0:
angle += 2*np.pi
center_x = xmin + (xmax - xmin) * angle / (2*np.pi)
print "center_x = ", center_x
# Create the first sinusoidal mode for the plot.
mode1 = (F1.real * np.cos(2*np.pi*np.arange(N)/N) -
F1.imag*np.sin(2*np.pi*np.arange(N)/N))/np.abs(F1)
import matplotlib.pyplot as plt
plt.clf()
plt.plot(x, y)
plt.plot(x, mode1)
plt.axvline(center_x, color='r', linewidth=1)
plt.show()
This generates the plot:
To answer the question "Why F1.conj()?":
The complex conjugate of F1 is used because of the minus sign in
w = np.exp(-2j*np.pi*np.arange(N) / N) (which I used because it
is a common convention).
Since w can be written
w = np.exp(-2j*np.pi*np.arange(N) / N)
= cos(-2*pi*arange(N)/N) + 1j*sin(-2*pi*arange(N)/N)
= cos(2*pi*arange(N)/N) - 1j*sin(2*pi*arange(N)/N)
the dot product y.dot(w) is basically a projection of y onto
cos(2*pi*arange(N)/N) (the real part of F1) and -sin(2*pi*arange(N)/N)
(the imaginary part of F1). But when we figure out the phase of
the maximum, it is based on the functions cos(...) and sin(...). Taking
the complex conjugate accounts for the opposite sign of the sin()
function. If w = np.exp(2j*np.pi*np.arange(N) / N) were used instead, the
complex conjugate of F1 would not be needed.
You could calculate the circular mean directly on your data.
When calculating the circular mean, your data is mapped to -pi..pi. This mapped data is interpreted as angle to a point on the unit circle. Then the mean value of x and y component is calculated. The next step is to calculate the resulting angle and map it back to the defined "box".
import numpy as np
import matplotlib.pyplot as plt
Points_x = np.random.randn(10000)+1
Box_min = -10
Box_max = 10
Box_width = Box_max - Box_min
#Maps Points to Box_min ... Box_max with periodic boundaries
Points_x = (Points_x%Box_width + Box_min)
#Map Points to -pi..pi
Points_map = (Points_x - Box_min)/Box_width*2*np.pi-np.pi
#Calc circular mean
Pmean_map = np.arctan2(np.sin(Points_map).mean() , np.cos(Points_map).mean())
#Map back
Pmean = (Pmean_map+np.pi)/(2*np.pi) * Box_width + Box_min
#Plotting the result
plt.figure(figsize=(10,3))
plt.subplot(121)
plt.hist(Points_x, 100);
plt.plot([Pmean, Pmean], [0, 1000], c='r', lw=3, alpha=0.5);
plt.subplot(122,aspect='equal')
plt.plot(np.cos(Points_map), np.sin(Points_map), '.');
plt.ylim([-1, 1])
plt.xlim([-1, 1])
plt.grid()
plt.plot([0, np.cos(Pmean_map)], [0, np.sin(Pmean_map)], c='r', lw=3, alpha=0.5);
I have a periodic function of period T and would like to know how to obtain the list of the Fourier coefficients. I tried using fft module from numpy but it seems more dedicated to Fourier transforms than series.
Maybe it a lack of mathematical knowledge, but I can't see how to calculate the Fourier coefficients from fft.
Help and/or examples appreciated.
In the end, the most simple thing (calculating the coefficient with a riemann sum) was the most portable/efficient/robust way to solve my problem:
import numpy as np
def cn(n):
c = y*np.exp(-1j*2*n*np.pi*time/period)
return c.sum()/c.size
def f(x, Nh):
f = np.array([2*cn(i)*np.exp(1j*2*i*np.pi*x/period) for i in range(1,Nh+1)])
return f.sum()
y2 = np.array([f(t,50).real for t in time])
plot(time, y)
plot(time, y2)
gives me:
This is an old question, but since I had to code this, I am posting here the solution that uses the numpy.fft module, that is likely faster than other hand-crafted solutions.
The DFT is the right tool for the job of calculating up to numerical precision the coefficients of the Fourier series of a function, defined as an analytic expression of the argument or as a numerical interpolating function over some discrete points.
This is the implementation, which allows to calculate the real-valued coefficients of the Fourier series, or the complex valued coefficients, by passing an appropriate return_complex:
def fourier_series_coeff_numpy(f, T, N, return_complex=False):
"""Calculates the first 2*N+1 Fourier series coeff. of a periodic function.
Given a periodic, function f(t) with period T, this function returns the
coefficients a0, {a1,a2,...},{b1,b2,...} such that:
f(t) ~= a0/2+ sum_{k=1}^{N} ( a_k*cos(2*pi*k*t/T) + b_k*sin(2*pi*k*t/T) )
If return_complex is set to True, it returns instead the coefficients
{c0,c1,c2,...}
such that:
f(t) ~= sum_{k=-N}^{N} c_k * exp(i*2*pi*k*t/T)
where we define c_{-n} = complex_conjugate(c_{n})
Refer to wikipedia for the relation between the real-valued and complex
valued coeffs at http://en.wikipedia.org/wiki/Fourier_series.
Parameters
----------
f : the periodic function, a callable like f(t)
T : the period of the function f, so that f(0)==f(T)
N_max : the function will return the first N_max + 1 Fourier coeff.
Returns
-------
if return_complex == False, the function returns:
a0 : float
a,b : numpy float arrays describing respectively the cosine and sine coeff.
if return_complex == True, the function returns:
c : numpy 1-dimensional complex-valued array of size N+1
"""
# From Shanon theoreom we must use a sampling freq. larger than the maximum
# frequency you want to catch in the signal.
f_sample = 2 * N
# we also need to use an integer sampling frequency, or the
# points will not be equispaced between 0 and 1. We then add +2 to f_sample
t, dt = np.linspace(0, T, f_sample + 2, endpoint=False, retstep=True)
y = np.fft.rfft(f(t)) / t.size
if return_complex:
return y
else:
y *= 2
return y[0].real, y[1:-1].real, -y[1:-1].imag
This is an example of usage:
from numpy import ones_like, cos, pi, sin, allclose
T = 1.5 # any real number
def f(t):
"""example of periodic function in [0,T]"""
n1, n2, n3 = 1., 4., 7. # in Hz, or nondimensional for the matter.
a0, a1, b4, a7 = 4., 2., -1., -3
return a0 / 2 * ones_like(t) + a1 * cos(2 * pi * n1 * t / T) + b4 * sin(
2 * pi * n2 * t / T) + a7 * cos(2 * pi * n3 * t / T)
N_chosen = 10
a0, a, b = fourier_series_coeff_numpy(f, T, N_chosen)
# we have as expected that
assert allclose(a0, 4)
assert allclose(a, [2, 0, 0, 0, 0, 0, -3, 0, 0, 0])
assert allclose(b, [0, 0, 0, -1, 0, 0, 0, 0, 0, 0])
And the plot of the resulting a0,a1,...,a10,b1,b2,...,b10 coefficients:
This is an optional test for the function, for both modes of operation. You should run this after the example, or define a periodic function f and a period T before running the code.
# #### test that it works with real coefficients:
from numpy import linspace, allclose, cos, sin, ones_like, exp, pi, \
complex64, zeros
def series_real_coeff(a0, a, b, t, T):
"""calculates the Fourier series with period T at times t,
from the real coeff. a0,a,b"""
tmp = ones_like(t) * a0 / 2.
for k, (ak, bk) in enumerate(zip(a, b)):
tmp += ak * cos(2 * pi * (k + 1) * t / T) + bk * sin(
2 * pi * (k + 1) * t / T)
return tmp
t = linspace(0, T, 100)
f_values = f(t)
a0, a, b = fourier_series_coeff_numpy(f, T, 52)
# construct the series:
f_series_values = series_real_coeff(a0, a, b, t, T)
# check that the series and the original function match to numerical precision:
assert allclose(f_series_values, f_values, atol=1e-6)
# #### test similarly that it works with complex coefficients:
def series_complex_coeff(c, t, T):
"""calculates the Fourier series with period T at times t,
from the complex coeff. c"""
tmp = zeros((t.size), dtype=complex64)
for k, ck in enumerate(c):
# sum from 0 to +N
tmp += ck * exp(2j * pi * k * t / T)
# sum from -N to -1
if k != 0:
tmp += ck.conjugate() * exp(-2j * pi * k * t / T)
return tmp.real
f_values = f(t)
c = fourier_series_coeff_numpy(f, T, 7, return_complex=True)
f_series_values = series_complex_coeff(c, t, T)
assert allclose(f_series_values, f_values, atol=1e-6)
Numpy isn't the right tool really to calculate fourier series components, as your data has to be discretely sampled. You really want to use something like Mathematica or should be using fourier transforms.
To roughly do it, let's look at something simple a triangle wave of period 2pi, where we can easily calculate the Fourier coefficients (c_n = -i ((-1)^(n+1))/n for n>0; e.g., c_n = { -i, i/2, -i/3, i/4, -i/5, i/6, ... } for n=1,2,3,4,5,6 (using Sum( c_n exp(i 2 pi n x) ) as Fourier series).
import numpy
x = numpy.arange(0,2*numpy.pi, numpy.pi/1000)
y = (x+numpy.pi/2) % numpy.pi - numpy.pi/2
fourier_trans = numpy.fft.rfft(y)/1000
If you look at the first several Fourier components:
array([ -3.14159265e-03 +0.00000000e+00j,
2.54994550e-16 -1.49956612e-16j,
3.14159265e-03 -9.99996710e-01j,
1.28143395e-16 +2.05163971e-16j,
-3.14159265e-03 +4.99993420e-01j,
5.28320925e-17 -2.74568926e-17j,
3.14159265e-03 -3.33323464e-01j,
7.73558750e-17 -3.41761974e-16j,
-3.14159265e-03 +2.49986840e-01j,
1.73758496e-16 +1.55882418e-17j,
3.14159265e-03 -1.99983550e-01j,
-1.74044469e-16 -1.22437710e-17j,
-3.14159265e-03 +1.66646927e-01j,
-1.02291982e-16 -2.05092972e-16j,
3.14159265e-03 -1.42834113e-01j,
1.96729377e-17 +5.35550532e-17j,
-3.14159265e-03 +1.24973680e-01j,
-7.50516717e-17 +3.33475329e-17j,
3.14159265e-03 -1.11081501e-01j,
-1.27900121e-16 -3.32193126e-17j,
-3.14159265e-03 +9.99670992e-02j,
First neglect the components that are near 0 due to floating point accuracy (~1e-16, as being zero). The more difficult part is to see that the 3.14159 numbers (that arose before we divide by the period of a 1000) should also be recognized as zero, as the function is periodic). So if we neglect those two factors we get:
fourier_trans = [0,0,-i,0,i/2,0,-i/3,0,i/4,0,-i/5,0,-i/6, ...
and you can see the fourier series numbers come up as every other number (I haven't investigated; but I believe the components correspond to [c0, c-1, c1, c-2, c2, ... ]). I'm using conventions according to wiki: http://en.wikipedia.org/wiki/Fourier_series.
Again, I'd suggest using mathematica or a computer algebra system capable of integrating and dealing with continuous functions.
As other answers have mentioned, it seems that what you are looking for is a symbolic computing package, so numpy isn't suitable. If you wish to use a free python-based solution, then either sympy or sage should meet your needs.
Do you have a list of discrete samples of your function, or is your function itself discrete? If so, the Discrete Fourier Transform, calculated using an FFT algorithm, provides the Fourier coefficients directly (see here).
On the other hand, if you have an analytic expression for the function, you probably need a symbolic math solver of some kind.