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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:

  1. 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.

  2. 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

Nam Kang
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2 Answers2

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
Pang
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Cadmium
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    Thanks cadmium I appreciate your help, I think I can go from here – Nam Kang Apr 21 '16 at 21:03
  • @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. – Tonechas Apr 22 '16 at 08:35
  • If i already know that some of these values would go to zero, will there be a way to avoid doing those calculations? – Nam Kang Apr 22 '16 at 19:38
  • In which array are there values that go to zero, `r` or `k2`? – Tonechas Apr 23 '16 at 10:20
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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.

Tonechas
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