Two different numpy arrays are being assigned the same values when only one array is being referenced

Question

I'm trying to write some code to carry out the Jacobi method for solving linear equations (I realise my method is not the most efficient way to do this but I am trying to figure out why it's not working).

I have tried to debug the problem and noticed the following issue. The code finishes after 2 iterations because on the second iteration on line 32 when xnew[i] is assigned a new value, the same value is also assigned to x[i], even though x[i] is not referenced. Why is this happening on the second iteration and not the first time the for loop is run and is there a way to fix this?

Thanks in advance

import numpy as np

A = np.array(
    [[0.93, 0.24, 0],
     [0.04, 0.54, 0.26],
     [1, 1, 1]])

b = np.array([[6.0], [2.0], [10.0]])

n , m  = np.shape(A) 

x = np.zeros(shape=(n,1))

xnew = np.zeros(shape=(n,1))

iterlimit = 100 
tol = 0.0000001 


for  iteration in range(iterlimit):

    convergence = True
    for i in range(n):
        sum=0
        for j in range(n):
            if j != i:
                sum = sum + (A[i,j] * x[j])
        
        #on second iteration (iteration =1) below line begins to
        #assign x[i] the same values as it assigns xnew[i] causing the 
        #convergence check below to not run and results in a premature break
        xnew[i] = 1/A[i,i] * (b[i] - sum) 
                     
        if abs(xnew[i]-x[i]) > tol:
            convergence = False
            
    if convergence:
        break
    
    x = xnew

print("Iteration:", iteration+1)
print("Solution:")
print(np.matrix(xnew))

`x = xnew` doesn't make a copy of the array. Both variables refer to the same array. — Barmar, Oct 29 '20 at 00:18
`x = xnew` assigns a reference to the same array to both variables. — Michael Butscher, Oct 29 '20 at 00:20
Does this answer your question? [Does Python make a copy of objects on assignment?](https://stackoverflow.com/questions/2438938/does-python-make-a-copy-of-objects-on-assignment) — Carcigenicate, Oct 29 '20 at 00:21
Thanks a lot guys! Would the best way to copy the values in xnew to x be numpy.copyto(x,xnew)? — dinusha, Oct 29 '20 at 00:27

score 2 · Accepted Answer · answered Oct 29 '20 at 00:28

2

x = xnew

This line assigns xnew to x. Not the contents of xnew, but the array itself. So after your first iteration, x and xnew reference the same array in memory.

Try instead x[:] = xnew[:]

answered Oct 29 '20 at 00:28

Michael

2,344
6
12

Thanks! Does this rule of x = xnew assigning both variables to the same thing in memory only apply to objects in python? Because it doesn't apply to simple variables right? For example if x and xnew were integers instead of arrays, this would be fine. – dinusha Oct 29 '20 at 00:40
@user157691 This generally applies to all Python variables. If `x` and `xnew` were integers, assigning `x = xnew` would cause both variables to point to the same integer in memory (and you can check this by looking at the `id(x)` and `id(xnew)`, or checking whether `x is xnew` is true, before and after the assignment). If you then ran a function on that integer that altered its value somehow (but without using the python assignment operator `=`, since that creates a new integer in memory) then both variables would appear to be pointing to the same value still. – Michael Oct 29 '20 at 00:50

score 1 · Answer 2 · answered Oct 29 '20 at 00:29

The problem is here : x = xnew

In fact, this way you are referencing to the same object now. So obviously when you will change xnew , you are changing the object represented by xnew which is now also referenced by x.

So to just make a copy you can write :

x = xnew[:]
OR
x = xnew.copy()

Two different numpy arrays are being assigned the same values when only one array is being referenced

2 Answers2