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python - Pythonic way to return list of every nth item in a larger list

Say we have a list of numbers from 0 to 1000. Is there a pythonic/efficient way to produce a list of the first and every subsequent 10th item, i.e. [0, 10, 20, 30, ... ]?

Yes, I can do this using a for loop, but I'm wondering if there is a neater way to do this, perhaps even in one line?

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>>> lst = list(range(165))
>>> lst[0::10]
[0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160]

Note that this is around 100 times faster than looping and checking a modulus for each element:

$ python -m timeit -s "lst = list(range(1000))" "lst1 = [x for x in lst if x % 10 == 0]"
1000 loops, best of 3: 525 usec per loop
$ python -m timeit -s "lst = list(range(1000))" "lst1 = lst[0::10]"
100000 loops, best of 3: 4.02 usec per loop

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