Nested FOR loop indexing issue - Python

Question

I am having trouble with my indices in a nested for loop. Python is spitting out an index error telling me my index is out of bounds.

Below are my code and subsequent error:

from math import *
import numpy as np
from scipy import integrate
import matplotlib.pyplot as plt
import os

croot = 1
ctip = 1
span = 1
thetaroot = 0
thetatip = 0
a0root = 0.11
a0tip = 0.11
alpha = 0
alpha0root = -2.5
alpha0tip = -2.5
thetaroot = thetaroot * arctan(1.) / 45.
thetatip = thetatip * arctan(1.) / 45.
alpha = alpha * arctan(1.) / 45.
alpha0root = alpha0root * arctan(1.) / 45.
alpha0tip = alpha0tip * arctan(1.) / 45.
n = 10
theta = np.empty(n, dtype = object)
y = np.empty(n, dtype = object)
c = np.empty(n, dtype = object)
cl = np.empty(n, dtype = object)
alp = np.empty(n, dtype = object)
a = np.empty(n, dtype = object)
rhs = np.empty(n, dtype = object)
b = np.empty(n, dtype = object)
a = np.empty(n, dtype = object)
rhs = rhs[:,None]
b = b[:,None]
a = a[:,None]
#
# Define properties at n span stations
#
pi = 4. * arctan(1.)
for i in range(0,n):
    theta[i] = i * pi / (2. * n)
    y[i] = span * 0.5 * cos(theta[i])
    c[i] = croot + (ctip - croot) * y[i] * 2. / span
    alp[i] = alpha + thetaroot - (alpha0root + (alpha0tip - alpha0root + thetaroot - thetatip) * y[i] * 2. / span)
    a[i] = a0root + (a0tip - a0root) * y[i] * 2. / span

pi = 4. * arctan(1.)
# Set up 2n x 2n system of equations for A1, A3 , ... A2n-1
for j in range(0,n):
    mu = c[j] * a[j] / (4. * span); print('mu=',mu)
    rhs[j,0] = alp[j] * sin(theta[j]) * c[j] * a[j] / (4 * span)
    for i in range(0,n):
        l = 2 * i - 1
        b[j,i] = sin(l * theta[j]) * (mu * l + sin(theta[j]))

I then receive the error:

---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-70-b5bd95e69bb5> in <module>()
     45     for i in range(0,n):
     46         l = 2 * i - 1
---> 47         b[j,i] = sin(l * theta[j]) * (mu * l + sin(theta[j]))
     48 
     49 

IndexError: index 1 is out of bounds for axis 1 with size 1

How can I effectively call out both indices? In MATLAB, b(j,i) is the normal syntax.

Any help is appreciated, thanks!

is this numpy - in which case where do you import the relevant libraries — Tony Suffolk 66
– Tony Suffolk 66, Commented Dec 15, 2014 at 23:17
Apologies, I've updated my script with the imported libraries. Numpy is being used. — Doug Fox
– Doug Fox, Commented Dec 15, 2014 at 23:19
b is (n,1). If something close to this worked in MATLAB its because that language expands an array if indexed beyond the current size. In numpy you have to initialize it to the correct final shape. — hpaulj
– hpaulj, Commented Dec 15, 2014 at 23:34

BingsF · Accepted Answer · 2014-12-15 23:23:31Z

1

The call b = np.empty(n, dtype = object) with n=10 makes a one-dimensional array, but you are indexing into it (b[j,i]) as though it is a two-dimensional array.

To initialize a 10 by 10 array, you would call b = np.empty([n, n], dtype = object).

EDIT: I didn't notice this assignment: b = b[:,None] That creates this:

>>> [[None]
 [None]
 [None]
 [None]
 [None]
 [None]
 [None]
 [None]
 [None]
 [None]]

which is a two-dimensional array, but trying to index past the first element of the inner array (containing just None) is causing your error.

edited Dec 15, 2014 at 23:23

answered Dec 15, 2014 at 23:17

BingsF

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9 Comments

Doug Fox Over a year ago

I'm assuming you mean b = np.empty([n, n], dtype = object) ??

BingsF Over a year ago

Oops, yes I did mean that.

Doug Fox Over a year ago

So, since it is trying to index past the first element, what would you suggest as a viable alternative for the nested loop?

BingsF Over a year ago

My suggestion was create b as b = np.empty([n, n], dtype = object), then the nested loop should work. (and remove b = b[:,None])

Doug Fox Over a year ago

In doing that, I receive an error stating 'too many indeces'. It points to the line containing rhs[j,i] = alp[j] * sin(theta[j]) * c[j] * a[j] / (4 * span)

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