Circular array indices in Python

Question

I\'m looking to get an array (or matrix) that is circular, such that

Let:

 a = [1,2,3]

Then I would like

a[0] = 1
a         


        

Answer 1

Having such functionality is not good for your code. Instead write a generator function which generates you round robin values.

numbers = [1, 2, 3]

def returnNumber():
    """
    A circular array for yielding list members repeatedly 
    """
    index = -1
    while True:
        index += 1
        yield slangWords[index % len(numbers)]

# Now you can use this generator
numberGenerator = returnNumber()
numberGenerator.next() # returns 1 
numberGenerator.next() # returns 2
numberGenerator.next() # returns 3
numberGenerator.next() # returns 1
numberGenerator.next() # returns 2

Answer 2

You can use modulo operator, like this

print a[3 % len(a)] 

If you don't want to use modulo operator like this, you need to subclass list and implement __getitem__, yourself.

class CustomList(list):
    def __getitem__(self, index):
        return super(CustomList, self).__getitem__(index % len(self))

a = CustomList([1, 2, 3])
for index in xrange(5):
    print index, a[index]

Output

0 1
1 2
2 3
3 1
4 2

If you want to do the same with Numpy Arrays, you can do like this

import numpy as np

class CustomArray(np.ndarray):
    def __new__(cls, *args, **kwargs):
        return np.asarray(args[0]).view(cls)

    def __getitem__(self, index):
        return np.ndarray.__getitem__(self, index % len(self))

a = CustomArray([1, 2, 3])
for index in xrange(5):
    print a[index]

More information about Subclassing Numpy Arrays can be found here (Thanks to JonClements)