Re: [Matplotlib-users] Help with converting from scilab

[John H. <jd...@gm...>]

On Fri, Jul 25, 2008 at 10:22 AM, Peter Wesbdell
<fly...@go...> wrote:
> Hello all, my first post here.
>
> I am moving from using scilab to Pylab, can anyone tell me why the two
> following snippets of code produce very different results? BTW. The scilab
> code produces the expected result.

I don't know what scilab does, but the filling of the z array looks
wrong, since python indexing starts at 0, not 1, so you would want to
do

for i in range(0,100):
    for j in range(0,100):

but you rarely want to loop over arrays.  My snippet below shows how
to use meshgrid to use numpy's elementwise operations and avoid loops.
 By only filling starting at 1, you are using the memory in the 0-th
row and column unintialized (np.zeros can be safer than np.empty in
this regard) Also, be careful when defining constants as integers,
since integer division produces integers (3/2=1)

Here is my script - does it produce what you are expecting?


import numpy as np
import matplotlib.pyplot as plt

Lx = 1.
Ly = 1.
n = 2.
m = 2.
f = 100.
w = 2*np.pi*f
t = 1.
A = 2.
Kx = n*np.pi/Lx
Ky = m*np.pi/Ly
x = np.arange(100.)
y = np.arange(100.)
X, Y = np.meshgrid(x, y)
Z = A * np.sin(Kx*X) * np.sin(Ky*Y) * np.exp(1j*w*t)
plt.contourf(Z)
plt.show()



>
> Scilab:
> Lx=1;
> Ly=1;
> n=2;
> m=2;
> f=100;
> w=2*%pi*f;
> t=1;
> A=2;
> Kx=n*%pi/Lx;
> Ky=m*%pi/Ly;
> x=linspace(0,100);
> y=linspace(0,100);
> z=zeros(100,100);
> for i = 1:100
>  for j = 1:100
>   z(i,j) = A * sin(Kx*x(i)) * sin(Ky*y(j)) * %e^(%i*w*t);
>     end
> end
> contour(z)
>
>
> Pylab:
> from pylab import *
> Lx=1
> Ly=1
> n=2
> m=2
> f=100
> w=2*pi*f
> t=1
> A=2
> Kx=n*pi/Lx
> Ky=m*pi/Ly
> x=arange(0,100)
> y=arange(0,100)
> z=empty((100,100))
> for i in range(1,100):
>     for j in range(1,100):
>         z[i, j] = A * sin(Kx*x[i]) * sin(Ky*y[j]) * e**(1j*w*t)
> contourf(z)
> show()
>
> Cheers,
> Pete.
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