Python creating matrix using if condition on indices : incorrect result

Question

I have the following code where I have been trying to create a tridiagonal matrix x using if-conditions.

#!/usr/bin/env python

# import useful modules
import numpy as np


N=5
x=np.identity(N)
#x=np.zeros((N,N))
print x

# Construct NxN matrix 
for i in range(N):
     for j in range(N):
        if i-j==1:
            x[i][j]=1 
        elif j-1==1:
            x[i][j]=-1
        else:
            x[i][j]=0

            print "i= ",i," j= ",j          


print x

I desire to get

[[ 0.  -1. 0.  0.  0.]
 [ 1.  0. -1.  0.  0.]
 [ 0.  1.  0.  -1  0.]
 [ 0.  0.  1.  0.  -1.]
 [ 0.  0.  0.  1.  0.]]

However, I obtain

[[ 0.  0. -1.  0.  0.]
 [ 1.  0. -1.  0.  0.]
 [ 0.  1. -1.  0.  0.]
 [ 0.  0.  1.  0.  0.]
 [ 0.  0. -1.  1.  0.]]

What's going wrong?

Bonus question : Can I forcefully index from 1 to 5 instead of 0 to 4 in this example, or Python never allows that?

Take a look at http://stackoverflow.com/questions/5842903/block-tridiagonal-matrix-python for a more elegant solution — Tonechas, Jun 29 '16 at 21:19

score 1 · Accepted Answer · answered Jun 29 '16 at 13:45

1

elif j-1==1: should be elif j-i==1:.

And no, lists/arrays etc. are always indexed from 0.

answered Jun 29 '16 at 13:45

Tim Pietzcker

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Thanks. How can I do such a mistake? I'm voting to delete my post. Extremely sorry. – hbaromega Jun 29 '16 at 14:45

Tonechas · Answer 2 · 2021-01-16T15:29:54.347

As for the bonus question, the first element of a sequence in Python has always the index 0. However, if for some particular reason (for example to prevent off-by-one errors) you wish to count the elements of a sequence from a value other than 0, you could use the built-in function enumerate() and set the value of the optional parameter start to fit your needs:

>>> seq = ['a', 'b', 'c']
>>> for count, item in enumerate(seq, start=1):
...     print(count, item)
...
1 a
2 b
3 c

Python creating matrix using if condition on indices : incorrect result

2 Answers2