Python array indexing

Question

I am writing a code for doing some integration. But I am stuck at some point(the last two lines). As it says Index Error: Index out of bounds. What I would like to do is- each time r has a value it should write to the empty corresponding index as an increment of 1 so that it can keep track of it. Any help would be appreciated. Here is the code-

from pylab import*
from math import*

dx = 981
dy = 1043
bx = 534.4
by = 109.5

index = zeros(shape=(1,dx+dy))
r=0
r_max=0

for i in xrange(1,dy+1):
    for j in xrange (1,dx+1):
        if i-by > 0:
            theta = 180*atan(abs(j-bx)/(i-by))/pi
            if theta<10:
                r = round(sqrt((j-bx)**2+(i-by)**2))
                if r>0:
                    index[r]+=1

Just a general tip: You should (usually) not use from foo import *. It's better to import foo and then call foo.bar(). For example, I was thrown off by zeros() - where is it defined? You can't tell. If you do import pylab and then call pylab.zeros(), it's much easier to understand what is happening. — Tim Pietzcker
– Tim Pietzcker, Commented Apr 12, 2013 at 7:56
TimPietzcker is absolutely right, this code would start producing weird behaviour if you would add NumPy to it as well (due to the numpy.zeros). — Bas Jansen
– Bas Jansen, Commented Apr 12, 2013 at 8:10

shx2 · Accepted Answer · 2013-04-12 08:02:43Z

2

When you do index = zeros(shape=(1,dx+dy)) you create a 2-dimensional array, with the first axis of size 1. When you do index[r]+=1 you access that first axis of the array, with index r, which can be > 1.

So it looks like what you want is to get rid of the first "useless" dimension, by doing

index = zeros(shape=(dx+dy))

Or, alternatively, indexing the second axis:

index[0, r]+=1

Or,

index[:, r]+=1

answered Apr 12, 2013 at 8:02

shx2

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1 Comment

shx2 Over a year ago

@user2095624 glad I could help. If you find an answer helpful, feel free to accept it

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