Alvaro Joao Alvaro Joao - 2 months ago 8
Python Question

Python how cyclic fetch a pre-fixed number of elements in array

I'm trying to make a function that will always return me a pre-fixed number of elements from an array which will be larger than the pre-fixed number:

def getElements(i,arr,size=10):
return cyclic array return


where
i
stands for index of array to fetch and
arr
represent the array of all elements:

Example:



a = [0,1,2,3,4,5,6,7,8,9,10,11]
b = getElements(9,a)
>> b
>> [9,10,11,0,1,2,3,4,5,6]
b = getElements(1,a)
>> b
>> [1,2,3,4,5,6,7,8,9,10]


where
i = 9
and the array return the
[9:11]+[0:7]
to complete 10 elements with
i = 1
don't need to cyclic the array just get
[1:11]


thanks for the help

Initial code (not working):



def getElements(i,arr,size=10):
total = len(arr)
start = i%total
end = start+size
return arr[start:end]

#not working cos not being cyclic


EDIT:



I can't make any
import
for this script

Answer

You could return

array[i: i + size] + array[: max(0, i + size - len(array))]

For example

In [144]: array = list(range(10))

In [145]: array
Out[145]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [146]: i, size = 1, 10

In [147]: array[i: i + size] + array[: max(0, i + size - len(array))]
Out[147]: [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]

In [148]: i, size = 2, 3

In [149]: array[i: i + size] + array[: max(0, i + size - len(array))]
Out[149]: [2, 3, 4]

In [150]: i, size = 5, 9

In [151]: array[i: i + size] + array[: max(0, i + size - len(array))]
Out[151]: [5, 6, 7, 8, 9, 0, 1, 2, 3]