返回较大列表中第n个项目的列表的Python方法

问题:返回较大列表中第n个项目的列表的Python方法

假设我们有一个从0到1000的数字列表。是否有一种pythonic /高效的方法来生成第一个以及随后的第10个项目的列表,即[0, 10, 20, 30, ... ]

是的,我可以使用for循环来执行此操作,但是我想知道是否有更整洁的方法可以执行此操作,甚至在一行中也可以?

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?


回答 0

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

请注意,这比循环并检查每个元素的模量快约100倍:

$ 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
>>> 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

回答 1

  1. source_list[::10] 是最明显的,但这对任何可迭代的方法都无效,并且对于大列表而言内存效率不高。
  2. itertools.islice(source_sequence, 0, None, 10) 适用于任何可迭代且内存有效的方法,但对于大型列表和大型步骤而言,可能不是最快的解决方案。
  3. (source_list[i] for i in xrange(0, len(source_list), 10))
  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))

回答 2

您可以像这样使用slice运算符:

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

You can use the slice operator like this:

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

回答 3

从手册: 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]

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]

回答 4

newlist = oldlist[::10]

这将选择列表中的第10个元素。

newlist = oldlist[::10]

This picks out every 10th element of the list.


回答 5

为什么不只使用range函数的step参数来获得:

l = range(0, 1000, 10)

为了进行比较,在我的机器上:

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

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

回答 6

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

回答 7

这是“每10个项目”列表理解的一种更好的实现,它不使用列表内容作为成员资格测试的一部分:

>>> 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']

但这仍然比仅使用列表切片要慢得多。

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.


回答 8

列表理解正是为此而做出的:

smaller_list = [x for x in range(100001) if x % 10 == 0]

您可以在python官方文档中获取有关它们的更多信息:http : //docs.python.org/tutorial/datastructures.html#list-comprehensions

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