I need help setting up matrices to solve using Gaussian elimination in Python

Question

Apologies in advance to those who has to read through my poor coding skill

The objective of this coding is to first develop a 17x17 matrix and solve for the 17 unknowns using methods presented in linear algebra.

The part I am having the most difficulty is:

implementing 2 counters i and j, where the value of i will increase once the value of j reaches its limit and goes back to 0 again.
Lastly, being able to insert new values to a single array for later manipulation. I tried using np.insert, np.hstack, np.vstack, np.append, etc could not work it.

So i can generate matrix that looks like

x11 x12 x13....x1j 
x21 .......... x2j
xi1............xij

here is some attempt

import numpy as np
import math as mt
r=[2,2.8,3.2,3.5,3.7,3.8,3.8,3.8,3.8,3.8,3.8,3.8,3.7,3.5,3.2,2.8,2]
n=np.linspace(1,17,17)
m=np.linspace(1,17,17)
i=0
k=np.array([])
l=1
k2=[]
while i <=18:
    for j in range(17):
        h1=mt.sqrt(r[i]**2+(l*(n[i]-m[j])+l/2)**2)
        h2=mt.sqrt(r[i]**2+(l*(n[i]-m[j])-l/2)**2)
        h=h1-h2    
        k2.append(h)
        i=i+1

I am trying to obtain stokes' stream function in axially symmetrical flow for those who are interested,

I will appreciate any type of feedback, please guide me in the right direction

Pang · Accepted Answer · 2018-10-29 07:35:59Z

1

Your code suffers from two mistakes. The first that in Python, you start counting from zero; you may think of your matrix as having 17 rows, 1 to 17, but Python sees it as going from 0 to 16. The second is that when working with numpy, you should build your array first, and then insert your calculated values. There's a good explanation of why here:(How do I create an empty array/matrix in NumPy?).

I made r an array for consistency's sake, and I inserted the calculated values into k2. I'm not sure k was for.

import numpy as np
import math as mt

r=np.array([2,2.8,3.2,3.5,3.7,3.8,3.8,3.8,3.8,3.8,3.8,3.8,3.7,3.5,3.2,2.8,2])
n=np.linspace(1,17,17)
m=np.linspace(1,17,17)
l=1

k2 = np.empty(shape=(17,17))
i=0
j=0
while i <=16:
    while j<=16: 
        h1=mt.sqrt(r[i]**2+(l*(n[i]-m[j])+l/2)**2)
        h2=mt.sqrt(r[i]**2+(l*(n[i]-m[j])-l/2)**2)
        h=np.array(h1-h2)    
        k2[i,j]= h
        j+=1
    j=0    
    i+=1

edited Oct 29, 2018 at 7:35

Pang

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answered Apr 21, 2016 at 18:38

Cadmium

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4 Comments

Nam Kang Over a year ago

Thanks cadmium I appreciate your help, I think I can go from here

Tonechas Over a year ago

@NamKang: The use of magic constants such as 17 is considered as bad programming practice. If you change the size of r by adding or removing some elements, the code above won't work properly unless you manually replace the value 17 by the new size of r (the same applies to the value 16 which appears in the while loops). That's why I introduced the variable R in my solution.

Nam Kang Over a year ago

If i already know that some of these values would go to zero, will there be a way to avoid doing those calculations?

Tonechas Over a year ago

In which array are there values that go to zero, r or k2?

Tonechas · Accepted Answer · 2016-04-22 13:29:46Z

The code below is a vectorized solution to your problem:

import numpy as np

r = np.asarray([2,2.8,3.2,3.5,3.7,3.8,3.8,3.8,3.8,3.8,3.8,3.8,3.7,3.5,3.2,2.8,2])
l = 1

R = r.size
n, m = np.mgrid[1:R+1, 1:R+1]

h1 = np.sqrt(r[:, np.newaxis]**2 + (l*(n-m) + l/2.)**2)
h2 = np.sqrt(r[:, np.newaxis]**2 + (l*(n-m) - l/2.)**2)
k2 = h1 - h2

The result k2 is a 2-dimensional array rather than a vector:

>>> np.set_printoptions(precision=1)
>>> k2
array([[ 0. , -0.4, -0.7, -0.8, -0.9, -0.9, -0.9, -1. , -1. , -1. , -1. , -1. , -1. , -1. , -1. , -1. , -1. ],
       [ 0.3,  0. , -0.3, -0.6, -0.7, -0.8, -0.9, -0.9, -0.9, -0.9, -1. , -1. , -1. , -1. , -1. , -1. , -1. ],
       [ 0.5,  0.3,  0. , -0.3, -0.5, -0.7, -0.8, -0.8, -0.9, -0.9, -0.9, -0.9, -1. , -1. , -1. , -1. , -1. ], 
       [ 0.6,  0.5,  0.3,  0. , -0.3, -0.5, -0.6, -0.8, -0.8, -0.9, -0.9, -0.9, -0.9, -0.9, -1. , -1. , -1. ],
       [ 0.7,  0.6,  0.5,  0.3,  0. , -0.3, -0.5, -0.6, -0.7, -0.8, -0.9, -0.9, -0.9, -0.9, -0.9, -0.9, -1. ],
       [ 0.8,  0.7,  0.6,  0.5,  0.3,  0. , -0.3, -0.5, -0.6, -0.7, -0.8, -0.8, -0.9, -0.9, -0.9, -0.9, -0.9],
       [ 0.8,  0.8,  0.7,  0.6,  0.5,  0.3,  0. , -0.3, -0.5, -0.6, -0.7, -0.8, -0.8, -0.9, -0.9, -0.9, -0.9],
       [ 0.9,  0.8,  0.8,  0.7,  0.6,  0.5,  0.3,  0. , -0.3, -0.5, -0.6, -0.7, -0.8, -0.8, -0.9, -0.9, -0.9],
       [ 0.9,  0.9,  0.8,  0.8,  0.7,  0.6,  0.5,  0.3,  0. , -0.3, -0.5, -0.6, -0.7, -0.8, -0.8, -0.9, -0.9],
       [ 0.9,  0.9,  0.9,  0.8,  0.8,  0.7,  0.6,  0.5,  0.3,  0. , -0.3, -0.5, -0.6, -0.7, -0.8, -0.8, -0.9],
       [ 0.9,  0.9,  0.9,  0.9,  0.8,  0.8,  0.7,  0.6,  0.5,  0.3,  0. , -0.3, -0.5, -0.6, -0.7, -0.8, -0.8],
       [ 0.9,  0.9,  0.9,  0.9,  0.9,  0.8,  0.8,  0.7,  0.6,  0.5,  0.3,  0. , -0.3, -0.5, -0.6, -0.7, -0.8],
       [ 1. ,  0.9,  0.9,  0.9,  0.9,  0.9,  0.9,  0.8,  0.7,  0.6,  0.5,  0.3,  0. , -0.3, -0.5, -0.6, -0.7],
       [ 1. ,  1. ,  1. ,  0.9,  0.9,  0.9,  0.9,  0.9,  0.8,  0.8,  0.6,  0.5,  0.3,  0. , -0.3, -0.5, -0.6],
       [ 1. ,  1. ,  1. ,  1. ,  1. ,  0.9,  0.9,  0.9,  0.9,  0.8,  0.8,  0.7,  0.5,  0.3,  0. , -0.3, -0.5],
       [ 1. ,  1. ,  1. ,  1. ,  1. ,  1. ,  1. ,  0.9,  0.9,  0.9,  0.9,  0.8,  0.7,  0.6,  0.3,  0. , -0.3],
       [ 1. ,  1. ,  1. ,  1. ,  1. ,  1. ,  1. ,  1. ,  1. ,  1. ,  0.9,  0.9,  0.9,  0.8,  0.7,  0.4,  0. ]])

Hopefully this is the result you were looking for.

Notice that in order to save space, only one decimal digit is displayed.

You may find it helpful to have a look on the description of the function mgrid and the object newaxis in Numpy's documentation to figure out how this code works.

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