返回较大列表中每个第 n 个项目的列表的 Pythonic 方法

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, ie [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?

Answer 1

>>> 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

Answer 2

1. source_list[::10] is the most obvious, but this doesn't work for any iterable and is not memory efficient for large lists.
2. 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.
3. (source_list[i] for i in xrange(0, len(source_list), 10))

Answer 3

You can use the slice operator like this:

l = [1,2,3,4,5]
l2 = l[::2] # get subsequent 2nd item

Answer 4

From manual: s[i:j:k] slice of s from i to j with step k

li = range(100)
sub = li[0::10]

>>> sub
[0, 10, 20, 30, 40, 50, 60, 70, 80, 90]

Answer 5

newlist = oldlist[::10]

这将选出列表的每 10 个元素。

Answer 6

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:\>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:\>python -m timeit -s "l = range(1000)" "l1 = l[0::10]"
1000000 loops, best of 3: 0.861 usec per loop
H:\>python -m timeit -s "l = range(0, 1000, 10)"
100000000 loops, best of 3: 0.0172 usec per loop

Answer 7

existing_list = range(0, 1001)
filtered_list = [i for i in existing_list if i % 10 == 0]

Answer 8

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:

>>> l = range(165)
>>> [ 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]
>>> l = list("ABCDEFGHIJKLMNOPQRSTUVWXYZ")
>>> [ 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.

Answer 9

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