Here is my first steps within the NumPy world.
As a matter of fact the target is plotting below 2-D function as a 3-D mesh:
N = \frac{n}{2\sigma\sqrt{\pi}}\exp^{-\frac{n^{2}x^{2}}{4\sigma^{2}}}
That could been done as a piece a cake in Matlab with below snippet:
[x,n] = meshgrid(0:0.1:20, 1:1:100);
mu = 0;
sigma = sqrt(2)./n;
f = normcdf(x,mu,sigma);
mesh(x,n,f);
But the bloody result is ugly enough to drive me trying Python capabilities to generate scientific plots.
I searched something and found that the primary steps to hit above mark in Pyhton might be acquired by below snippet:
from matplotlib.patches import Polygon
import numpy as np
from scipy.integrate import quad
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
sigma = 1
def integrand(x,n):
return (n/(2*sigma*np.sqrt(np.pi)))*np.exp(-(n**2*x**2)/(4*sigma**2))
t = np.linespace(0, 20, 0.01)
n = np.linespace(1, 100, 1)
lower_bound = -100000000000000000000 #-inf
upper_bound = t
tt, nn = np.meshgrid(t,n)
real_integral = quad(integrand(tt,nn), lower_bound, upper_bound)
Axes3D.plot_trisurf(real_integral, tt,nn)
Edit: With due attention to more investigations on Greg's advices, above code is the most updated snippet.
Here is the generated exception:
RuntimeError: infinity comparisons don't work for you
It is seemingly referring to the quad call...
Would you please helping me to handle this integrating-plotting problem?!...
Best
Just a few hints to get you in the right direction.
numpy.meshgrid can do the same as MatLABs function:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.meshgrid.html
When you have x and n you can do math just like in matlab:
sigma = numpy.sqrt(2)/n
(in python multiplication/division is default index by index - no dot needed)
scipy has a lot more advanced functions, see for example How to calculate cumulative normal distribution in Python for a 1D case.
For plotting you can use matplotlibs pcolormesh:
import matplotlib.pyplot as plt
plt.pcolormesh(x,n,real_integral)
Hope this helps until someone can give you a more detailed answer.
Related
I'm trying to do a guassian fit for some experimental data but I keep running into error after error. I've followed a few different threads online but either the fit isn't good (it's just a horizontal line) or the code just won't run. I'm following this code from another thread. Below is my code.
I apologize if my code seems a bit messy. There are some bits from other attempts when I tried making it work. Hence the "astropy" import.
import math as m
import matplotlib.pyplot as plt
import numpy as np
from scipy import optimize as opt
import pandas as pd
import statistics as stats
from astropy import modeling
def gaus(x,a,x0,sigma, offset):
return a*m.exp(-(x-x0)**2/(2*sigma**2)) + offset
# Python program to get average of a list
def Average(lst):
return sum(lst) / len(lst)
wavelengths = [391.719, 391.984, 392.248, 392.512, 392.777, 393.041, 393.306, 393.57, 393.835, 394.099, 391.719, 391.455, 391.19, 390.926, 390.661, 390.396]
intensities = [511.85, 1105.85, 1631.85, 1119.85, 213.85, 36.85, 10.85, 6.85, 13.85, 7.85, 511.85, 200.85, 80.85, 53.85, 14.85, 24.85]
n=sum(intensities)
mean = sum(wavelengths*intensities)/n
sigma = m.sqrt(sum(intensities*(wavelengths-mean)**2)/n)
def gaus(x,a,x0,sigma):
return a*m.exp(-(x-x0)**2/(2*sigma**2))
popt,pcov = opt.curve_fit(gaus,wavelengths,intensities,p0=[1,mean,sigma])
print(popt)
plt.scatter(wavelengths, intensities)
plt.title("Helium Spectral Line Peak 1")
plt.xlabel("Wavelength (nm)")
plt.ylabel("Intensity (a.u.)")
plt.show()
Thanks to the kind user, my curve seems to be working more reasonably well. However, one of the points seems to be back connecting to an earlier point? Screenshot below:
There are two problems with your code. The first is that you are performing vector operation on list which gives you the first error in the line mean = sum(wavelengths*intensities)/n. Therefore, you should use np.array instead. The second is that you take math.exp on python list which again throws an error as it takes a real number, so you should use np.exp here instead.
The following code solves your problem:
import matplotlib.pyplot as plt
import numpy as np
from scipy import optimize as opt
wavelengths = [391.719, 391.984, 392.248, 392.512, 392.777, 393.041,
393.306, 393.57, 393.835, 394.099, 391.719, 391.455,
391.19, 390.926, 390.661, 390.396]
intensities = [511.85, 1105.85, 1631.85, 1119.85, 213.85, 36.85, 10.85, 6.85,
13.85, 7.85, 511.85, 200.85, 80.85, 53.85, 14.85, 24.85]
wavelengths_new = np.array(wavelengths)
intensities_new = np.array(intensities)
n=sum(intensities)
mean = sum(wavelengths_new*intensities_new)/n
sigma = np.sqrt(sum(intensities_new*(wavelengths_new-mean)**2)/n)
def gaus(x,a,x0,sigma):
return a*np.exp(-(x-x0)**2/(2*sigma**2))
popt,pcov = opt.curve_fit(gaus,wavelengths_new,intensities_new,p0=[1,mean,sigma])
print(popt)
plt.scatter(wavelengths_new, intensities_new, label="data")
plt.plot(wavelengths_new, gaus(wavelengths_new, *popt), label="fit")
plt.title("Helium Spectral Line Peak 1")
plt.xlabel("Wavelength (nm)")
plt.ylabel("Intensity (a.u.)")
plt.show()
I need to solve a 2nd order ODE which I have decoupled into two first order ODEs. I've tried solving it using solve_ivp, but it doesn't seem to provide the solution I expect. I have provided the code below.
import numpy as np
import matplotlib.pyplot as plt
from scipy.misc import derivative
from scipy.integrate import solve_ivp
from matplotlib import rc
rc('text', usetex = True)
V0 = 2*10**-10
A = 0.130383
f = 0.129576
phi_i_USR2 = 6.1
phi_Ni_USR2 = 1.2
def V_USR2(phi):
return V0*(np.tanh(phi/np.sqrt(6)) + A*np.sin(1/f*np.tanh(phi/np.sqrt(6))))**2
def V_phi_USR2(phi):
return derivative(V_USR2, phi)
N = np.linspace(0,66,1000)
def USR2(N, X):
phi, g = X
return [g, (g**2/2 - 3)*(V_phi_USR2(phi)/V_USR2(phi) + g)]
X0 = [phi_i_USR2, phi_Ni_USR2]
sol = solve_ivp(USR2, (0,66), X0, method = 'LSODA', t_eval = N)
phi_USR2 = sol.y[0]
phi_N_USR2 = sol.y[1]
N_USR2 = sol.t
plt.plot(phi_USR2, phi_N_USR2)
plt.xlabel("$\phi$")
plt.ylabel("$\phi'$")
plt.title("Phase plot for USR2")
plt.show()
solve_ivp gives me the following plot:
The problem is that there is supposed to be an oscillation near the origin, which is not well captured by solve_ivp. However, for the same equation and initial conditions, Mathematica gives me exactly what I want:
I want the same plot in Python as well. I tried various methods in solve_ivp such as RK45, LSODA, Radau and BDF, but all of them show the same problem (LSODA tries to depict an oscillation but fails, but the other methods don't even move past the point where the oscillation starts). It would be great if someone can shed light on the problem. Thanks in advance.
I have been thinking about it for a long time, but I don't find out what the problem is. Hope you can help me, Thank you.
F(s) Gaussian function
F(s)=1/(√2π s) e^(-(w-μ)^2/(2s^2 ))
Code:
import numpy as np
from matplotlib import pyplot as plt
from math import pi
from scipy.fft import fft
def F_S(w, mu, sig):
return (np.exp(-np.power(w-mu, 2)/(2 * np.power(sig, 2))))/(np.power(2*pi, 0.5)*sig)
w=np.linspace(-5,5,100)
plt.plot(w, np.real(np.fft.fft(F_S(w, 0, 1))))
plt.show()
Result:
As was mentioned before you want the absolute value, not the real part.
A minimal example, showing the the re/im , abs/phase spectra.
import numpy as np
import matplotlib.pyplot as p
%matplotlib inline
n=1001 # add 1 to keep the interval a round number when using linspace
t = np.linspace(-5, 5, n ) # presumed to be time
dt=t[1]-t[0] # time resolution
print(f'sampling every {dt:.3f} sec , so at {1/dt:.1f} Sa/sec, max. freq will be {1/2/dt:.1f} Hz')
y = np.exp(-(t**2)/0.01) # signal in time
fr= np.fft.fftshift(np.fft.fftfreq(n, dt)) # shift helps with sorting the frequencies for better plotting
ft=np.fft.fftshift(np.fft.fft(y)) # fftshift only necessary for plotting in sequence
p.figure(figsize=(20,12))
p.subplot(231)
p.plot(t,y,'.-')
p.xlabel('time (secs)')
p.title('signal in time')
p.subplot(232)
p.plot(fr,np.abs(ft), '.-',lw=0.3)
p.xlabel('freq (Hz)')
p.title('spectrum, abs');
p.subplot(233)
p.plot(fr,np.real(ft), '.-',lw=0.3)
p.xlabel('freq (Hz)')
p.title('spectrum, real');
p.subplot(235)
p.plot(fr,np.angle(ft), '.-', lw=0.3)
p.xlabel('freq (Hz)')
p.title('spectrum, phase');
p.subplot(236)
p.plot(fr,np.imag(ft), '.-',lw=0.3)
p.xlabel('freq (Hz)')
p.title('spectrum, imag');
you have to change from time scale to frequency scale
When you make a FFT you will get the simetric tranformation, i.e, mirror of the positive to negative curve. Usually, you only will look at the positive side.
Also, you should take care with sample rate, as FFT is designed to transform time domain input to frequency domain, the time, or sample rate, of input info matters. So add timestep in np.fft.fftfreq(n, d=timestep) for your sample rate.
If you simple want to make a fft of normal dist signal, here is another question with it and some good explanations on why are you geting this behavior:
Fourier transform of a Gaussian is not a Gaussian, but thats wrong! - Python
There are two mistakes in your code:
Don't take the real part, take the absoulte value when plotting.
From the docs:
If A = fft(a, n), then A[0] contains the zero-frequency term (the mean
of the signal), which is always purely real for real inputs. Then
A[1:n/2] contains the positive-frequency terms, and A[n/2+1:] contains
the negative-frequency terms, in order of decreasingly negative
frequency.
You can rearrange the elements with np.fft.fftshift.
The working code:
import numpy as np
from matplotlib import pyplot as plt
from math import pi
from scipy.fftpack import fft, fftshift
def F_S(w, mu, sig):
return (np.exp(-np.power(w-mu, 2)/(2 * np.power(sig, 2))))/(np.power(2*pi, 0.5)*sig)
w=np.linspace(-5,5,100)
plt.plot(w, fftshift(np.abs(np.fft.fft(F_S(w, 0, 1)))))
plt.show()
Also, you might want to consider scaling the x axis too.
I need help getting the right plot for tau as a function of c (conversion).
Here's my code:
from __future__ import division, print_function
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from math import *
from scipy.optimize import root
def tau(c,T = 298):
ci = 2
pw = 1000
Cp = 4
k0 = 0.01
e = 1000
Hr = (-3*10**4)
a = np.linspace(0,10000,10000)
t = np.zeros(len(a))
for i in range(len(a)):
t[i] = (ci * k0 * exp(e/298))/(ci - c[i])
plt.plot(t,c)
plt.show()
print(tau(np.linspace(0,1,10000)))
My plot is not correct.
How do I modify my equation to get the following kind of plot?
It's difficult to say why your plot of conversion is incorrect without knowing the original problem statement. Can you please provide more information? Your plot of conversion does increase with residence time (tau), which it should do. It looks like you've defined some constants that are unused, such as the enthalpy of reaction (Hr), pw, and the temperature. At any rate, it is apparent that your question is not about python or plotting, but rather about how to correctly solve a chemical engineering reactor design problem.
I intend for part of a program I'm writing to automatically generate Gaussian distributions of various statistics over multiple raw text sources, however I'm having some issues generating the graphs as per the guide at:
python pylab plot normal distribution
The general gist of the plot code is as follows.
import numpy as np
import matplotlib.mlab as mlab
import matplotlib.pyplot as pyplot
meanAverage = 222.89219487179491 # typical value calculated beforehand
standardDeviation = 3.8857889432054091 # typical value calculated beforehand
x = np.linspace(-3,3,100)
pyplot.plot(x,mlab.normpdf(x,meanAverage,standardDeviation))
pyplot.show()
All it does is produce a rather flat looking and useless y = 0 line!
Can anyone see what the problem is here?
Cheers.
If you read documentation of matplotlib.mlab.normpdf, this function is deprycated and you should use scipy.stats.norm.pdf instead.
Deprecated since version 2.2: scipy.stats.norm.pdf
And because your distribution mean is about 222, you should use np.linspace(200, 220, 100).
So your code will look like:
import numpy as np
from scipy.stats import norm
import matplotlib.pyplot as pyplot
meanAverage = 222.89219487179491 # typical value calculated beforehand
standardDeviation = 3.8857889432054091 # typical value calculated beforehand
x = np.linspace(200, 220, 100)
pyplot.plot(x, norm.pdf(x, meanAverage, standardDeviation))
pyplot.show()
It looks like you made a few small but significant errors. You either are choosing your x vector wrong or you swapped your stddev and mean. Since your mean is at 222, you probably want your x vector in this area, maybe something like 150 to 300. This way you get all the good stuff, right now you are looking at -3 to 3 which is at the tail of the distribution. Hope that helps.
I see that, for the *args which are sending meanAverage, standardDeviation, the correct thing to be sent is:
mu : a numdims array of means of a
sigma : a numdims array of atandard deviation of a
Does this help?