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

Question

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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Here is a better implementation of an  every 10th item  list comprehension  that does not use the list contents as part of the membership test    gt  gt  gt  l   range 165   gt  gt  gt    item for i item in enumerate l  if i 10  0    0  10  20  30  40  50  60  70  80  90  100  110  120  130  140  150  160   gt  gt  gt  l   list  ABCDEFGHIJKLMNOPQRSTUVWXYZ    gt  gt  gt    item for i item in enumerate l  if i 10  0     A    K    U     But this is still far slower than just using list slicing

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source list   10  is the most obvious  but this doesn t work for any iterable and is not memory efficient for large lists  itertools islice source sequence  0  None  10  works for any iterable and is memory-efficient  but probably is not the fastest solution for large list and big step   source list i  for i in xrange 0  len source list   10

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newlist   oldlist   10    This picks out every 10th element of the list

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gt  gt  gt  lst   list range 165    gt  gt  gt  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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Why not just use a step parameter of range function as well to get   l   range 0  1000  10    For comparison  on my machine   H   gt python -m timeit -s  l   range 1000    l1    x for x in l if x   10    0   10000 loops  best of 3  90 8 usec per loop H   gt python -m timeit -s  l   range 1000    l1   l 0  10   1000000 loops  best of 3  0 861 usec per loop H   gt python -m timeit -s  l   range 0  1000  10   100000000 loops  best of 3  0 0172 usec per loop

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From manual  s i j k   slice of s from i to j with step k  li   range 100  sub   li 0  10    gt  gt  gt  sub  0  10  20  30  40  50  60  70  80  90

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You can use the slice operator like this      l    1 2 3 4 5  l2   l   2    get subsequent 2nd item

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existing list   range 0  1001  filtered list    i for i in existing list if i   10    0

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List comprehensions are exactly made for that   smaller list    x for x in range 100001  if x   10    0    You can get more info about them in the python official documentation  http   docs python org tutorial datastructures html list-comprehensions

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