Generating circular shifts / reduced Latin Squares in Python

Question

Was just wondering what\'s the most efficient way of generating all the circular shifts of a list in Python. In either direction. For example, given a list [1, 2, 3, 4

Answer 1

The answer by @Bruno Lenzi does not seem to work:

In [10]: from itertools import cycle

In [11]: x = cycle('ABCD')

In [12]: print [[x.next() for _ in range(4)] for _ in range(4)]
[['A', 'B', 'C', 'D'], ['A', 'B', 'C', 'D'], ['A', 'B', 'C', 'D'], ['A', 'B', 'C', 'D']]

I give a correct version below, however the solution by @f5r5e5d is faster.

In [45]: def use_cycle(a):
    x=cycle(a)
    for _ in a:
        x.next()
        print [x.next() for _ in a]
   ....:         

In [46]: use_cycle([1,2,3,4])
[2, 3, 4, 1]
[3, 4, 1, 2]
[4, 1, 2, 3]
[1, 2, 3, 4]

In [50]: def use_slice(a):
    print [ a[n:] + a[:n] for n in range(len(a)) ]
  ....:     

In [51]: use_slice([1,2,3,4])
[[1, 2, 3, 4], [2, 3, 4, 1], [3, 4, 1, 2], [4, 1, 2, 3]]

In [54]: timeit.timeit('use_cycle([1,2,3,4])','from __main__ import use_cycle',number=100000)
Out[54]: 0.4884989261627197

In [55]: timeit.timeit('use_slice([1,2,3,4])','from __main__ import use_slice',number=100000)
Out[55]: 0.3103291988372803

In [58]: timeit.timeit('use_cycle([1,2,3,4]*100)','from __main__ import use_cycle',number=100)
Out[58]: 2.4427831172943115

In [59]: timeit.timeit('use_slice([1,2,3,4]*100)','from __main__ import use_slice',number=100)
Out[59]: 0.12029695510864258

I removed the print statement in use_cycle and use_slice for timing purposes.