I am trying to solve numerically the following integral using NumPy and quad from scipy.integrate. The code is kinda working, but I get spurious notches in the resulting data:
Anybody has an idea why are they occurring and how to get the correct smooth result?
Here is the original code in Python:
#!/usr/bin/env python
import numpy as np
from scipy.integrate import quad
import matplotlib.pyplot as pl
numpts = 300
t_min = -4
t_max = 100
tt = np.linspace(t_min, t_max, numpts)
mean = 0. # ps
fwhm = .05 # ps
def gaussian(x, mean, fwhm):
return 1. / np.sqrt(np.pi) / fwhm * np.exp(-1. * (x - mean)**2 / fwhm**2)
def integrand(t_, t, mean, fwhm):
denum = np.sqrt(t - t_)
r = gaussian(t_, mean, fwhm) / denum
return r
def integrate(t, mean, fwhm, tmin):
return quad(integrand, tmin, t - 1e-9, args=(t, mean, fwhm))[0]
if __name__ == '__main__':
vec_int = np.vectorize(integrate)
y = vec_int(tt, mean, fwhm, tt.min())
fig = pl.figure()
ax = fig.add_subplot(111)
ax.plot(tt, y, 'bo-', mec='none')
ax.set_xlim(-5, 101)
pl.show()
My suspicion would be that an integrable singularity is messing up the inner workings of quad(pack). I'd then try (in this order): use weights="cauchy" in quad; add and subtract the singularity; look at ways of telling quadpack that the integral has a difficult point.
So I solved my problem by switching to the mpmath library and its own integration quad using the tanh-sinh method. I also had to split the integral so that the data is monotonous. The output looks like this:
I am not 100% sure why this works, but it may have to do with the numeric precision and the behaviour of the integration method.
Here's the working code:
#!/usr/bin/env python
import numpy as np
import matplotlib.pyplot as pl
import mpmath as mp
numpts = 3000
t_min = -4
t_max = 100
tt = np.linspace(t_min, t_max, numpts)
mean = mp.mpf('0') # ps
fwhm = mp.mpf('.05') # ps
def gaussian(x, mean, fwhm):
return 1. / mp.sqrt(np.pi) / fwhm * mp.exp(-1. * (x - mean)**2 / fwhm**2)
def integrand(t_, t, mean, fwhm):
denum = np.sqrt(t - t_)
g = gaussian(t_, mean, fwhm)
r = g / denum
return r
def integrate(t, mean, fwhm, tmin):
t = mp.mpf(t)
tmin = mp.mpf(tmin)
# split the integral because it can only handle smooth functions
res = mp.quad(lambda t_: integrand(t_, t, mean, fwhm),
[tmin, mean],
method='tanh-sinh') + \
mp.quad(lambda t_: integrand(t_, t, mean, fwhm),
[mean, t],
method='tanh-sinh')
ans = res
return ans
if __name__ == '__main__':
mp.mp.dps = 15
mp.mp.pretty = True
vec_int = np.vectorize(integrate)
y = vec_int(tt, mean, fwhm, tt.min())
fig = pl.figure()
ax = fig.add_subplot(111)
# make sure we plot the real part
ax.plot(tt, map(mp.re, y), 'bo-', mec='none')
ax.set_xlim(-5, 101)
pl.show()
Related
I'm having some computational problems with the following code:
import numpy as np
from numpy import arange
from scipy.integrate import odeint
import matplotlib.pyplot as plt
from scipy.integrate import quad
import matplotlib as mpl
mpl.rcParams['agg.path.chunksize'] = 10000
# parameters
Ms = 100 #GeV Singlet Mass
Me = 0.511e-3 #Gev Electron Mass
Mp = 1.22e19 #GeV Planck Mass
gs = 106.75 # Entropy dof
H0 = 2.133*(0.7)*1e-42 # GeV Hubble parameter (unused)
gx = 2 # WIMP's dof
g = 100 # total dof
sigmav=[1e-25,1e-11,1e-12] # cross section's order of magnitude
xi=1e-2
xe=1e2
npts=int(1e5)
x = np.linspace(xi, xe, npts)
def fMB(p,x,m):
return np.exp(-x*np.sqrt(1+p*p/(m*m)))*p*p
def neq(x,m):
return (gx/(2*np.pi*np.pi))*quad(fMB, 0, np.inf, args=(x,m))[0]
def neq_nr(x,m):
return 2*(m**2/(2*np.pi*x))**(3/2)*np.exp(-x)
def stot(x):
return (2*np.pi*np.pi/45)*gs*Ms*Ms*Ms/(x*x*x)
def Yeq(x,m):
return neq(x,m)/stot(x)
Yeq2=np.vectorize(Yeq)
def Yeq_nr(x):
return 0.145*(gx/gs)*(x)**(3/2)*np.exp(-x)
def Yeq_r(x):
return 0.278*(3*gx/4)/gs
def Ytot(x):
if np.any(x<=1):
return Yeq_r(x)
else:
return Yeq_nr(x)
def eqd(yl,x,Ms,σv):
'''
Ms [GeV] : Singlet Mass
σv: [1/GeV^2] : ⟨σv⟩
'''
H = 1.67*g**(1/2)*Ms**2/Mp
dyl = -neq(x,Ms)*σv*(yl**2-Yeq(x,Ms)**2)/(x**(-2)*H*x*Yeq(x,Ms)) #occorre ancora dividere per Yeq_nr(x) oppure Yeq(x)
return dyl
y0=1e-15
yl0 = odeint( eqd, y0, x,args=(Ms,sigmav[0]), full_output=True)
yl1 = odeint( eqd, y0, x,args=(Ms,sigmav[1]), full_output=True)
yl2 = odeint( eqd, y0, x,args=(Ms,sigmav[2]), full_output=True)
fig = plt.figure(figsize=(11,8))
plt.loglog(x,yl0[0], label = r'$\langle σ v\rangle = %s {\rm GeV}^{-2}$'%(sigmav[0]))
plt.loglog(x,yl1[0], label = r'$\langle σ v\rangle = %s {\rm GeV}^{-2}$'%(sigmav[1]))
plt.loglog(x,yl2[0], label = r'$\langle σ v\rangle = %s {\rm GeV}^{-2}$'%(sigmav[2]))
plt.loglog(x,Yeq_nr(x), '--', label = '$Y_{EQ}^{nr}$')
plt.loglog(x,Yeq2(x,Ms), '--', label = '$Y_{EQ}$')
plt.ylim(ymax=0.1,ymin=y0)
plt.xlim(xmax=xe,xmin=xi)
plt.xlabel('$x = m_χ/T$', size= 15)
plt.ylabel('$Y$', size= 15)
plt.title('$m_χ = %s$ GeV'%(Ms), size= 15)
plt.legend(loc='best',fontsize=12)
plt.grid(True)
plt.savefig('abundance.jpg',bbox_inches='tight', dpi=150)
In particular, as soon as I use little values of sigmav (ranging from 10^-12 to 10^-25) the solution is well displayed, but making use of bigger values (starting from 10^-11) I obtain problems and I guess is a order of magnitudes problem, but I don't know how to handle it!
Thanks to everyone!
Edit 1:
I'm uploading a plot making use of three different values of sigmav and as you may see the bigger one (1e-10) is showing (I guess) precision problems plot_1
I'm trying to plot the graphs of the following equation in Python.
Solution of the radial differential equation of a 2d quantum ring
The beta parameter is
This was my attempt
import numpy as np
from scipy.special import gamma, genlaguerre
import matplotlib.pyplot as plt
from scipy import exp, sqrt
m = 0.067*9.1*10E-31
R = 5E-9
r = np.linspace(0, 20E-9)
#Definição do parâmetro beta
def beta(gama):
flux = np.linspace(0,1.0)
beta = sqrt((m-flux)**2+(gama**4)/4)
return beta
def Rn(n,gama):
raiz = sqrt((gamma(n+1)/((2**beta(gama)) * gamma(n+beta(gama)+1))))
eval_g = genlaguerre((n,beta(gama)),((gama * r/R)**2/2))
exp_g = exp(-((gama * r/R)**2)/4)
return (1/R) * raiz * (gama * r/R)**beta(gama) * exp_g * eval_g
sol1 = Rn(0,1.5)
sol2 = Rn(0,2.0)
sol3 = Rn(0,2.5)
sol4 = Rn(0,3.0)
fig, ax = plt.subplots()
ax.plot(r/R, sol1, color = 'red', label = '$\gamma$ = 1.5')
ax.plot(r/R, sol2, color = 'green', label = '$\gamma$ = 2.0')
ax.plot(r/R, sol3, color = 'blue', label = '$\gamma$ = 2.5')
ax.plot(r/R, sol4, color = 'black', label = '$\gamma$ = 3.0')
ax.legend()
ax.set_xlabel('R/r')
ax.set_ylabel('$R_0(r)$')
erro using genlaguerre
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
Here the link to the article
I'm not into this topic, but afaics there are at least the following mistakes (I assume setting silently n=0 was on purpose):
In opposite to the posted image of the beta function, you added a division by 4 into your function definition.
Correct version:
def beta(gama):
return np.sqrt((m-flux)**2+(gama**4))
That's no reason for distributed instead of fixed peaks but I tell what I see here.
Product instead of division because of lacking parantheses in definition of amplitude function:
Correct version:
def amplitude(gama):
return np.sqrt(gamma(1)/((2**beta(gama)*gamma(beta(gama)+1))))
In the definition of Rn function there's the introducing 1/R missing.
However, all this unfortunately does not change the peaks to happen at equal x positions...
Sorry! The radial wave function of the article is wrong! I made the calculations and found the correct answer. The following code is correct
import numpy as np
from scipy.special import genlaguerre, gamma
import numpy as np
import matplotlib.pyplot as plt
m = 0 # magnetic number
flux = 0 # Phi in eqn 8
R = 5 # nm
r = np.linspace(0, 6 * R)
rho = r / R
def R0(n, gama):
beta = np.sqrt((m - flux)**2 + gama**4/4)
return (gama/R*
np.sqrt(gamma(n+ 1) / ( 2**beta * gamma(n + beta + 1))) *
(gama * rho)**beta *
np.exp(-gama**2 * rho**2 / 4) *
genlaguerre(n, beta)(gama**2 * rho**2 / 2))
sol1 = R0(0, 1.5)
sol2 = R0(0, 2.0)
sol3 = R0(0, 2.5)
sol4 = R0(0, 3.0)
plt.plot(rho, sol1, rho, sol2, rho, sol3, rho, sol4)
plt.legend(['$\gamma = 1.5$', '$\gamma = 2$', '$\gamma = 2.5$', '$\gamma = 3$'])
plt.ylabel('$R_{0}(r)$')
plt.xlabel('$r/R$')
plt.show()
So far I have managed to find the particular solution to this equation for any given mass and drag coefficient. I have not however found a way to plot the solution or even evaluate the solution for a specific point. I really want to find a way to plot the solution.
from sympy import *
m = float(raw_input('Mass:\n> '))
g = 9.8
k = float(raw_input('Drag Coefficient:\n> '))
f = Function('f')
f1 = g * m
t = Symbol('t')
v = Function('v')
equation = dsolve(f1 - k * v(t) - m * Derivative(v(t)), 0)
C1 = Symbol('C1')
C1_ic = solve(equation.rhs.subs({t:0}),C1)[0]
equation = equation.subs({C1:C1_ic})
For completeness, you may also use Sympy's plot, which is probably more convenient if you want a "quick and dirty" plot.
plot(equation.rhs,(t,0,10))
Import these libraries (seaborn just makes the plots pretty).
from matplotlib import pyplot as plt
import seaborn as sns
import numpy as np
Then tack this onto the end. This will plot time, t, against velocity, v(t).
# make a numpy-ready function from the sympy results
func = lambdify(t, equation.rhs,'numpy')
xvals = np.arange(0,10,.1)
yvals = func(xvals)
# make figure
fig, ax = plt.subplots(1,1,subplot_kw=dict(aspect='equal'))
ax.plot(xvals, yvals)
ax.set_xlabel('t')
ax.set_ylabel('v(t)')
plt.show()
I get a plot like this for a mass of 2 and a drag coefficient of 2.
If I've understood correctly, you want to represent the right hand side of your solution, here's one of the multiple ways to do it:
from sympy import *
import numpy as np
import matplotlib.pyplot as plt
m = float(raw_input('Mass:\n> '))
g = 9.8
k = float(raw_input('Drag Coefficient:\n> '))
f = Function('f')
f1 = g * m
t = Symbol('t')
v = Function('v')
equation = dsolve(f1 - k * v(t) - m * Derivative(v(t)), 0)
C1 = Symbol('C1')
C1_ic = solve(equation.rhs.subs({t: 0}), C1)[0]
equation = equation.subs({C1: C1_ic})
t1 = np.arange(0.0, 50.0, 0.1)
y1 = [equation.subs({t: tt}).rhs for tt in t1]
plt.figure(1)
plt.plot(t1, y1)
plt.show()
I am trying to deblend the emission lines of low resolution spectrum in order to get the gaussian components. This plot represents the kind of data I am using:
After searching a bit, the only option I found was the application of the gauest function from the kmpfit package (http://www.astro.rug.nl/software/kapteyn/kmpfittutorial.html#gauest). I have copied their example but I cannot make it work.
I wonder if anyone could please offer me any alternative to do this or how to correct my code:
import numpy as np
import matplotlib.pyplot as plt
from scipy import optimize
def CurveData():
x = np.array([3963.67285156, 3964.49560547, 3965.31835938, 3966.14111328, 3966.96362305,
3967.78637695, 3968.60913086, 3969.43188477, 3970.25463867, 3971.07714844,
3971.89990234, 3972.72265625, 3973.54541016, 3974.36791992, 3975.19067383])
y = np.array([1.75001533e-16, 2.15520995e-16, 2.85030769e-16, 4.10072843e-16, 7.17558032e-16,
1.27759917e-15, 1.57074192e-15, 1.40802933e-15, 1.45038722e-15, 1.55195653e-15,
1.09280316e-15, 4.96611341e-16, 2.68777266e-16, 1.87075114e-16, 1.64335999e-16])
return x, y
def FindMaxima(xval, yval):
xval = np.asarray(xval)
yval = np.asarray(yval)
sort_idx = np.argsort(xval)
yval = yval[sort_idx]
gradient = np.diff(yval)
maxima = np.diff((gradient > 0).view(np.int8))
ListIndeces = np.concatenate((([0],) if gradient[0] < 0 else ()) + (np.where(maxima == -1)[0] + 1,) + (([len(yval)-1],) if gradient[-1] > 0 else ()))
X_Maxima, Y_Maxima = [], []
for index in ListIndeces:
X_Maxima.append(xval[index])
Y_Maxima.append(yval[index])
return X_Maxima, Y_Maxima
def GaussianMixture_Model(p, x, ZeroLevel):
y = 0.0
N_Comps = int(len(p) / 3)
for i in range(N_Comps):
A, mu, sigma = p[i*3:(i+1)*3]
y += A * np.exp(-(x-mu)*(x-mu)/(2.0*sigma*sigma))
Output = y + ZeroLevel
return Output
def Residuals_GaussianMixture(p, x, y, ZeroLevel):
return GaussianMixture_Model(p, x, ZeroLevel) - y
Wave, Flux = CurveData()
Wave_Maxima, Flux_Maxima = FindMaxima(Wave, Flux)
EmLines_Number = len(Wave_Maxima)
ContinuumLevel = 1.64191e-16
# Define initial values
p_0 = []
for i in range(EmLines_Number):
p_0.append(Flux_Maxima[i])
p_0.append(Wave_Maxima[i])
p_0.append(2.0)
p1, conv = optimize.leastsq(Residuals_GaussianMixture, p_0[:],args=(Wave, Flux, ContinuumLevel))
Fig = plt.figure(figsize = (16, 10))
Axis1 = Fig.add_subplot(111)
Axis1.plot(Wave, Flux, label='Emission line')
Axis1.plot(Wave, GaussianMixture_Model(p1, Wave, ContinuumLevel), 'r', label='Fit with optimize.leastsq')
print p1
Axis1.plot(Wave, GaussianMixture_Model([p1[0],p1[1],p1[2]], Wave, ContinuumLevel), 'g:', label='Gaussian components')
Axis1.plot(Wave, GaussianMixture_Model([p1[3],p1[4],p1[5]], Wave, ContinuumLevel), 'g:')
Axis1.set_xlabel( r'Wavelength $(\AA)$',)
Axis1.set_ylabel('Flux' + r'$(erg\,cm^{-2} s^{-1} \AA^{-1})$')
plt.legend()
plt.show()
A typical simplistic way to fit:
def model(p,x):
A,x1,sig1,B,x2,sig2 = p
return A*np.exp(-(x-x1)**2/sig1**2) + B*np.exp(-(x-x2)**2/sig2**2)
def res(p,x,y):
return model(p,x) - y
from scipy import optimize
p0 = [1e-15,3968,2,1e-15,3972,2]
p1,conv = optimize.leastsq(res,p0[:],args=(x,y))
plot(x,y,'+') # data
#fitted function
plot(arange(3962,3976,0.1),model(p1,arange(3962,3976,0.1)),'-')
Where p0 is your initial guess. By the looks of things, you might want to use Lorentzian functions...
If you use full_output=True, you get all kind of info about the fitting. Also check out curve_fit and the fmin* functions in scipy.optimize. There are plenty of wrappers around these around, but often, like here, it's easier to use them directly.
I am trying to fit some data that are distributed in the time following a rising gaussian curve, and then decaying exponentially.
I have found this example on the web, that is very similar to my case, but I just started to fit with python, and the example seems quite confusing to me.
Nonetheless, I have tryied to adapt the example to my script and data, and in the following is my progress:
#!/usr/bin/env python
import pyfits, os, re, glob, sys
from scipy.optimize import leastsq
from numpy import *
from pylab import *
from scipy import *
from scipy import optimize
import numpy as N
import pylab as P
data=pyfits.open('http://heasarc.gsfc.nasa.gov/docs/swift/results/transients/weak/GX304-1.orbit.lc.fits')
time = data[1].data.field(0)/86400. + data[1].header['MJDREFF'] + data[1].header['MJDREFI']
rate = data[1].data.field(1)
error = data[1].data.field(2)
data.close()
cond = ((time > 56200) & (time < 56220))
time=time[cond]
rate=rate[cond]
error=error[cond]
def expGauss(x, pos, wid, tConst, expMod = 0.5, amp = 1):
expMod *= 1.0
gNorm = amp * N.exp(-0.5*((x-pos)/(wid))**2)
g = expBroaden(gNorm, tConst, expMod)
return g, gNorm
def expBroaden(y, t, expMod):
fy = F.fft(y)
a = N.exp(-1*expMod*time/t)
fa = F.fft(a)
fy1 = fy*fa
yb = (F.ifft(fy1).real)/N.sum(a)
return yb
if __name__ == '__main__':
# Fit the first set
#p[0] -- amplitude, p[1] -- position, p[2] -- width
fitfuncG = lambda p, x: p[0]*N.exp(-0.5*(x-p[1])**2/p[2]**2) # Target function
errfuncG = lambda p, x, y: fitfuncG(p, x) - y # Distance to the target function
p0 = [0.20, 56210, 2.0] # Initial guess for the parameters
p1, success = optimize.leastsq(errfuncG, p0[:], args=(time, rate))
p1G = fitfuncG(p1, time)
# P.plot(rate, 'ro', alpha = 0.4, label = "Gaussian")
# P.plot(p1G, label = 'G-Fit')
def expGauss(x, pos, wid, tConst, expMod = 0.5, amp = 1):
#p[0] -- amplitude, p[1] -- position, p[2] -- width, p[3]--tConst, p[4] -- expMod
fitfuncExpG = lambda p, x: expGauss(x, p[1], p[2], p[3], p[4], p[0])[0]
errfuncExpG = lambda p, x, y: fitfuncExpG(p, x) - y # Distance to the target function
p0a = [0.20, 56210, 2.0] # Initial guess for the parameters
p1a, success = optimize.leastsq(errfuncExpG, p0a[:], args=(time, rate))
p1aG = fitfuncExpG(p1a, time)
print type(rate), type(time), len(rate), len(time)
P.plot(rate, 'go', alpha = 0.4, label = "ExpGaussian")
P.plot(p1aG, label = 'ExpG-Fit')
P.legend()
P.show()
I am sure to have confused the whole thing, so sorry in advance for that, but at this point I don't know how to go further...
The code take the data from the web, so it is directly executable.
At the moment the code runs without any error, but it doesn't produce any plot.
Again, my goal is to fit the data with those two functions, how can I improve my code to do that?
Any suggestion is really appreciated.
Similarly to your other question, here also I would use a trigonometric function to fit this peaK:
The following code works if pasted after your code:
import numpy as np
from scipy.optimize import curve_fit
x = time
den = x.max() - x.min()
x -= x.min()
y_points = rate
def func(x, a1, a2, a3):
return a1*sin(1*pi*x/den)+\
a2*sin(2*pi*x/den)+\
a3*sin(3*pi*x/den)
popt, pcov = curve_fit(func, x, y_points)
y = func(x, *popt)
plot(time,rate)
plot(x,y, color='r', linewidth=2.)
show()