Develop Reference

Python - changing color of 3d plot regarding the heat equation

I need to solve the heat equation using finite difference method in Python. However my main problem is not the solving equation but rather plotting the result. See below for my code.
import numpy as np
import matplotlib.pyplot as plt
max_iter_time = 100
interval_length = 1
D = 20
dt = .001
delta_x = .01
u = np.empty((max_iter_time,round(interval_length/delta_x)))
interval = np.linspace(0,1,round(interval_length/delta_x))
interior = np.sin(interval*2*np.pi)
u_start = 0
u = np.array([interior]*max_iter_time)
u[:,0]=u_start
for i in range(0, max_iter_time-1,1):
u[:,round(interval_length/delta_x)-1]=u[:,round(interval_length/delta_x)-2]
def fdm1d(u):
for t in range(0, max_iter_time-1, 1):
for i in range(0, max_iter_time-1, 1):
u[t + 1, i] = u[t][i] + D*dt * (u[t][i+1] + u[t][i-1] + u[t][i] - 2*u[t][i])
return u
u2 = []
u1 = fdm1d(u)
print(u1.shape)
ax = plt.axes(projection='3d')
for t in range(0, max_iter_time-1, 1):
for i in range(0, max_iter_time-1, 1):
u2.append([t,i,u1[t][i]])
xpoints = []
ypoints = []
zpoints =[]
for i in range(len(u2)-1):
xpoints.append(u2[i][0])
ypoints.append(u2[i][1])
zpoints.append(u2[i][2])
ax.scatter(xpoints,ypoints,zpoints,cmap='inferno')
plt.show()
My question is about the plot itself - when I compile the code the plot that shows looks like this:
plot
In the code I have specified the color of this but nothing has changed. Could someone adivse on this? What do I do wrong here? Is there any better way to plot this?

I am getting two plots for one data set in python

I am working through example 8.1 titled Euler's Method from Mark Newman's book Computational Physics. I rewrote the example as a method with Numpy arrays but when I plot it I get two plots on the same figure not sure how to correct it. Also is there better way to convert my 2 1D arrays into 1 2D array to use for plotting in Matplotlib, thanks.
Newman's example :
from math import sin
from numpy import arange
from pylab import plot,xlabel,ylabel,show
def f(x,t):
return -x**3 + sin(t)
a = 0.0 # Start of the interval
b = 10.0 # End of the interval
N = 1000 # Number of steps
h = (b-a)/N # Size of a single step
x = 0.0 # Initial condition
tpoints = arange(a,b,h)
xpoints = []
for t in tpoints:
xpoints.append(x)
x += h*f(x,t)
plot(tpoints,xpoints)
xlabel("t")
ylabel("x(t)")
show()
My modifications:
from pylab import plot,show,xlabel,ylabel
from numpy import linspace,exp,sin,zeros,vstack,column_stack
def f(x,t):
return (-x**(3) + sin(t))
def Euler(f,x0,a,b):
N=1000
h = (b-a)/N
t = linspace(a,b,N)
x = zeros(N,float)
y = x0
for i in range(N):
x[i] = y
y += h*f(x[i],t[i])
return column_stack((t,x)) #vstack((t,x)).T
plot(Euler(f,0.0,0.0,10.0))
xlabel("t")
ylabel("x(t)")
show()

The reason you get two lines is that t as well as x are plotted against their index, instead of x plotted against t
I don't see why you'd want to stack the two arrays. Just keep then separate, which will also solve the problem of the two plots.
The following works fine.
import numpy as np
import matplotlib.pyplot as plt
f = lambda x,t: -x**3 + np.sin(t)
def Euler(f,x0,a,b):
N=1000
h = (b-a)/N
t = np.linspace(a,b,N)
x = np.zeros(N,float)
y = x0
for i in range(N):
x[i] = y
y += h*f(x[i],t[i])
return t,x
t,x = Euler(f,0.0,0.0,10.0)
plt.plot(t,x)
plt.xlabel("t")
plt.ylabel("x(t)")
plt.show()

Find an Inverse for the Exponential Integral function

I have a program where I have to find x.
But I have to use the special function Ei - the exponential integral, and x is inside the argument of Ei.
So Python isn't recognizing it.
ei(mx) = te^r + ei(c)
Here the RHS is a constant alongwith m.
I want to find the value of x, and then append it to a list. But Python isn't able to do this.
from scipy import special
import matplotlib
import matplotlib.pyplot as plt
import numpy as np
Y = []
X = np.arange(0,10,.1)
for i in X:
y = scipy.special.expi(i)
Y.append(y)
N_0 = 2
t_f = 100
r = 2
K = 100
N_t = [N_0,]
t = np.arange(0,100,1)
for i in t:
l = i*e**r + scipy.special.expi(r*N_t[i]/K)
N_t.append(l)
plt.plot(X,Y)
plt.plot(t,N_t)
plt.show

I've corrected some mistakes in your code to give the following. You should compare this with your code line by line.
from scipy.special import expi
import matplotlib
import matplotlib.pyplot as plt
import numpy as np
Y = []
X = np.arange(0,10,.1)
for i in X:
y = expi(i)
Y.append(y)
N_0 = 2
t_f = 100
r = 2
K = 100
N_t = [N_0,]
t = np.arange(0,100,1)
for i in t:
l = i*np.exp(r) + expi(r*N_t[i]/K)
N_t.append(l)
plt.plot(X,Y)
plt.plot(t,N_t)
plt.show()
However, there is still one possible flaw that I notice and can't resolve. You plot X and t together in the same graph at the end yet X ranges over 0 to 10 and t ranges over 0 to 100. Is this what you intended?
Also matplotlib complains that the lengths of the vectors supplied to it in the second call to plot are not the same.

Python plotting data generated from for loop inside for loop

I am trying to convert this code from Matlab to Python:
x(1) = 0.1;
j = 0;
for z = 2.8:0.0011:3.9
j = j+1 %Gives progress of calculation
zz(j) = z;
for n = 1:200
x(n+1) = z*x(n)*(1 - x(n));
xn(n,j) = x(n);
end
end
h = plot(zz,xn(100:200,:),'r.');
set(h,'Markersize',3);
and so far I have got this:
import numpy as np
import matplotlib.pyplot as plt
x = []
x.append(0.1)
xn = []
j = 0
z_range = np.arange(2.8, 3.9, 0.0011)
n_range = range(0,200,1)
plt.figure()
for zz in z_range:
j = j+1
print j # Gives progress of calculation
for n in n_range:
w = zz * x[n] * (1.0-x[n])
x.append(zz * x[n] * (1.0-x[n]))
xn.append(w)
x = np.array(x)
xn = np.array(xn)
xn_matrix = xn.reshape((z_range.size, len(n_range)))
xn_mat = xn_matrix.T
plt.figure()
#for i in z_range:
# plt.plot(z_range, xn_mat[0:i], 'r.')
plt.show()
I'm not sure if this is the best way to convert the for loops from Matlab into Python, and I seem to have problems with plotting the result. The x(n+1) = z*x(n)*(1 - x(n)); and xn(n,j) = x(n); lines in Matlab are bugging me, so could someone please explain if there is a more efficient way of writing this in Python?

import numpy as np
import matplotlib.pyplot as plt
x = 0.1
# preallocate xn
xn = np.zeros([1001, 200])
# linspace is better for a non-integer step
zz = np.linspace(2.8, 3.9, 1001)
# use enumerate instead of counting iterations
for j,z in enumerate(zz):
print(j)
for n in range(200):
# use tuple unpacking so old values of x are unneeded
xn[j,n], x = x, z*x*(1 - x)
plt.plot(zz, xn[:, 100:], 'r.')
plt.show()

Speeding up the code using numpy

I'm new to python, and I have this code for calculating the potential inside a 1x1 box using fourier series, but a part of it is going way too slow (marked in the code below).
If someone could help me with this, I suspect I could've done something with the numpy library, but I'm not that familiar with it.
import matplotlib.pyplot as plt
import pylab
import sys
from matplotlib import rc
rc('text', usetex=False)
rc('font', family = 'serif')
#One of the boundary conditions for the potential.
def func1(x,n):
V_c = 1
V_0 = V_c * np.sin(n*np.pi*x)
return V_0*np.sin(n*np.pi*x)
#To calculate the potential inside a box:
def v(x,y):
n = 1;
sum = 0;
nmax = 20;
while n < nmax:
[C_n, err] = quad(func1, 0, 1, args=(n), );
sum = sum + 2*(C_n/np.sinh(np.pi*n)*np.sin(n*np.pi*x)*np.sinh(n*np.pi*y));
n = n + 1;
return sum;
def main(argv):
x_axis = np.linspace(0,1,100)
y_axis = np.linspace(0,1,100)
V_0 = np.zeros(100)
V_1 = np.zeros(100)
n = 4;
#Plotter for V0 = v_c * sin () x
for i in range(100):
V_0[i] = V_0_1(i/100, n)
plt.plot(x_axis, V_0)
plt.xlabel('x/L')
plt.ylabel('V_0')
plt.title('V_0(x) = sin(m*pi*x/L), n = 4')
plt.show()
#Plot for V_0 = V_c(1-(x-1/2)^4)
for i in range(100):
V_1[i] = V_0_2(i/100)
plt.figure()
plt.plot(x_axis, V_1)
plt.xlabel('x/L')
plt.ylabel('V_0')
plt.title('V_0(x) = 1- (x/L - 1/2)^4)')
#plt.legend()
plt.show()
#Plot V(x/L,y/L) on the boundary:
V_0_Y = np.zeros(100)
V_1_Y = np.zeros(100)
V_X_0 = np.zeros(100)
V_X_1 = np.zeros(100)
for i in range(100):
V_0_Y[i] = v(0, i/100)
V_1_Y[i] = v(1, i/100)
V_X_0[i] = v(i/100, 0)
V_X_1[i] = v(i/100, 1)
# V(x/L = 0, y/L):
plt.figure()
plt.plot(x_axis, V_0_Y)
plt.title('V(x/L = 0, y/L)')
plt.show()
# V(x/L = 1, y/L):
plt.figure()
plt.plot(x_axis, V_1_Y)
plt.title('V(x/L = 1, y/L)')
plt.show()
# V(x/L, y/L = 0):
plt.figure()
plt.plot(x_axis, V_X_0)
plt.title('V(x/L, y/L = 0)')
plt.show()
# V(x/L, y/L = 1):
plt.figure()
plt.plot(x_axis, V_X_1)
plt.title('V(x/L, y/L = 1)')
plt.show()
#Plot V(x,y)
#######
# This is where the code is way too slow, it takes like 10 minutes when n in v(x,y) is 20.
#######
V = np.zeros(10000).reshape((100,100))
for i in range(100):
for j in range(100):
V[i,j] = v(j/100, i/100)
plt.figure()
plt.contour(x_axis, y_axis, V, 50)
plt.savefig('V_1')
plt.show()
if __name__ == "__main__":
main(sys.argv[1:])

You can find how to use FFT/DFT in this document :
Discretized continuous Fourier transform with numpy
Also, regarding your V matrix, there are many ways to improve the execution speed. One is to make sure you use Python 3, or xrange() instead of range() if you a are still in Python 2.. I usually put these lines in my Python code, to allow it to run evenly wether I use Python 3. or 2.*
# Don't want to generate huge lists in memory... use standard range for Python 3.*
range = xrange if isinstance(range(2),
list) else range
Then, instead of re-computing j/100 and i/100, you can precompute these values and put them in an array; knowing that a division is much more costly than a multiplication ! Something like :
ratios = np.arange(100) / 100
V = np.zeros(10000).reshape((100,100))
j = 0
while j < 100:
i = 0
while i < 100:
V[i,j] = v(values[j], values[i])
i += 1
j += 1
Well, anyway, this is rather cosmetic and will not save your life; and you still need to call the function v()...
Then, you can use weave :
http://docs.scipy.org/doc/scipy-0.14.0/reference/tutorial/weave.html
Or write all your pure computation/loop code in C, compile it and generate a module which you can call from Python.

You should look into numpy's broadcasting tricks and vectorization (several references, one of the first good links that pops up is from Matlab but it is just as applicable to numpy - can anyone recommend me a good numpy link in the comments that I might point other users to in the future?).
What I saw in your code (once you remove all the unnecessary bits like plots and unused functions), is that you are essentially doing this:
from __future__ import division
from scipy.integrate import quad
import numpy as np
import matplotlib.pyplot as plt
def func1(x,n):
return 1*np.sin(n*np.pi*x)**2
def v(x,y):
n = 1;
sum = 0;
nmax = 20;
while n < nmax:
[C_n, err] = quad(func1, 0, 1, args=(n), );
sum = sum + 2*(C_n/np.sinh(np.pi*n)*np.sin(n*np.pi*x)*np.sinh(n*np.pi*y));
n = n + 1;
return sum;
def main():
x_axis = np.linspace(0,1,100)
y_axis = np.linspace(0,1,100)
#######
# This is where the code is way too slow, it takes like 10 minutes when n in v(x,y) is 20.
#######
V = np.zeros(10000).reshape((100,100))
for i in range(100):
for j in range(100):
V[i,j] = v(j/100, i/100)
plt.figure()
plt.contour(x_axis, y_axis, V, 50)
plt.show()
if __name__ == "__main__":
main()
If you look carefully (you could use a profiler too), you'll see that you're integrating your function func1 (which I'll rename into the integrand) about 20 times for each element in the 100x100 array V. However, the integrand doesn't change! So you can already bring it out of your loop. If you do that, and use broadcasting tricks, you could end up with something like this:
import numpy as np
from scipy.integrate import quad
import matplotlib.pyplot as plt
def integrand(x,n):
return 1*np.sin(n*np.pi*x)**2
sine_order = np.arange(1,20).reshape(-1,1,1) # Make an array along the third dimension
integration_results = np.empty_like(sine_order, dtype=np.float)
for enu, order in enumerate(sine_order):
integration_results[enu] = quad(integrand, 0, 1, args=(order,))[0]
y,x = np.ogrid[0:1:.01, 0:1:.01]
term = integration_results / np.sinh(np.pi * sine_order) * np.sin(sine_order * np.pi * x) * np.sinh(sine_order * np.pi * y)
# This is the key: you have a 3D matrix here and with this summation,
# you're basically squashing the entire 3D structure into a flat, 2D
# representation. This 'squashing' is done by means of a sum.
V = 2*np.sum(term, axis=0)
x_axis = np.linspace(0,1,100)
y_axis = np.linspace(0,1,100)
plt.figure()
plt.contour(x_axis, y_axis, V, 50)
plt.show()
which runs in less than a second on my system.
Broadcasting becomes much more understandable if you take pen&paper and draw out the vectors that you are "broadcasting" as if you were constructing a building, from basic Tetris-blocks.
These two versions are functionally the same, but one is completely vectorized, while the other uses python for-loops. As a new user to python and numpy, I definitely recommend reading through the broadcasting basics. Good luck!