Rolling array around

Question

Let's say I have

arr = np.arange(6)
arr
array([0, 1, 2, 3, 4, 5])

and I decide that I want to treat an array "like a circle": When I run out of material at the end, I want to start at index 0 again. That is, I want a convenient way of selecting x elements, starting at index i.

Now, if x == 6, I can simply do

i = 3
np.hstack((arr[i:], arr[:i]))
Out[9]: array([3, 4, 5, 0, 1, 2])

But is there a convenient way of generally doing this, even if x > 6, without having to manually breaking the array apart and thinking through the logic?

For example:

print(roll_array_arround(arr)[2:17])

should return.

array([2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0])

You could write a function roll_array_around(arr, startIndex, endIndex) that returns an array. If you had a function roll_array_around(arr) - i.e., it just took arr as a parameter - then you would theoretically need to return an infinite array. — Ollie
– Ollie, Commented Apr 12, 2018 at 8:09

Eelco Hoogendoorn · Accepted Answer · 2018-04-12 09:32:35Z

2

See mode='wrap' in ndarray.take:

https://docs.scipy.org/doc/numpy/reference/generated/numpy.take.html

Taking your hypothetical function:

print(roll_array_arround(arr)[2:17])

If it is implied that it is a true slice of the original array that you are after, that is not going to happen; a wrapped-around array cannot be expressed as a strided view of the original; so if you seek a function that maps an ndarray to an ndarray, this will necessarily involve a copy of your data.

That is, efficiency-wise, you shouldnt expect to find solution that significantly differs in performance from the expression below.

print(arr.take(np.arange(2,17), mode='wrap'))

edited Apr 12, 2018 at 9:32

answered Apr 12, 2018 at 8:34

Eelco Hoogendoorn

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Comments

Divakar · Accepted Answer · 2018-04-12 08:22:56Z

2

Modulus operation seems like the best fit here -

def rolling_array(n, x, i):
    # n is rolling period
    # x is length of array
    # i is starting number
    return np.mod(np.arange(i,i+x),n)

Sample runs -

In [61]: rolling_array(n=6, x=6, i=3)
Out[61]: array([3, 4, 5, 0, 1, 2])

In [62]: rolling_array(n=6, x=17, i=2)
Out[62]: array([2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0])

answered Apr 12, 2018 at 8:22

Divakar

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Kenstars · Accepted Answer · 2018-04-12 08:29:33Z

1

A solution you can look into would probably be :

from itertools import cycle
list_to_rotate = np.array([1,2,3,4,5])
rotatable_list = cycle(list_to_rotate)

answered Apr 12, 2018 at 8:29

Kenstars

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Anand Chitipothu · Accepted Answer · 2018-04-12 08:11:45Z

-1

You need to roll your array.

>>> x = np.arange(10)
>>> np.roll(x, 2)
array([8, 9, 0, 1, 2, 3, 4, 5, 6, 7])

See numpy documentation for more details.

answered Apr 12, 2018 at 8:11

Anand Chitipothu

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1 Comment

Sraw Over a year ago

I don't think that is what OP means. OP is trying to use an array as an infinite circular queue. Look at OP's example, len(arr) is just 6, but OP wants to get arr[2:17] which definitely exceeds even double length.

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