loops - Python range with start larger than stop -

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i want extract i indices of vector uniformly t times. instance, if have vector x = [1,2,3,4,5,6,7], i = 3 , t = 5, indices in each time must be:

t = 1; [1,2,3]  t = 2; [4,5,6]  t = 3; [7,1,2]  t = 4; [3,4,5]  t = 5; [6,7,1]

would possible in python range()?

you can use itertools.islice on itertools.cycle. make cycle object iterable, , slice object using window size i:

from itertools import cycle itertools import islice  l = [1,2,3,4,5,6,7] t = 5; = 3  c =  cycle(l)  r = [list(islice(c, i)) _ in range(t)] # range appears here # [[1, 2, 3], [4, 5, 6], [7, 1, 2], [3, 4, 5], [6, 7, 1]]

you can apply different non-negative values of i, , when i greater length of list:

i = 10 r = [list(islice(c, i)) _ in range(t)] print(r) # [[1, 2, 3, 4, 5, 6, 7, 1, 2, 3], [4, 5, 6, 7, 1, 2, 3, 4, 5, 6], [7, 1, 2, 3, 4, 5, 6, 7, 1, 2], [3, 4, 5, 6, 7, 1, 2, 3, 4, 5], [6, 7, 1, 2, 3, 4, 5, 6, 7, 1]]

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