问题描述
我想创建一系列长度不一的列表.每个列表将包含相同的元素e
,重复n
次(其中n
=列表的长度).
I want to create a series of lists, all of varying lengths. Each list will contain the same element e
, repeated n
times (where n
= length of the list).
如何创建列表,而不对每个列表使用列表理解[e for number in xrange(n)]
?
How do I create the lists, without using a list comprehension [e for number in xrange(n)]
for each list?
推荐答案
您还可以编写:
[e] * n
您应该注意,例如,如果e是一个空列表,您将得到一个具有n个指向同一列表的引用的列表,而不是n个独立的空列表.
You should note that if e is for example an empty list you get a list with n references to the same list, not n independent empty lists.
性能测试
乍一看,似乎重复是创建具有n个相同元素的列表的最快方法:
At first glance it seems that repeat is the fastest way to create a list with n identical elements:
>>> timeit.timeit('itertools.repeat(0, 10)', 'import itertools', number = 1000000)
0.37095273281943264
>>> timeit.timeit('[0] * 10', 'import itertools', number = 1000000)
0.5577236771712819
但是等等-这不是一个公平的测试...
But wait - it's not a fair test...
>>> itertools.repeat(0, 10)
repeat(0, 10) # Not a list!!!
函数itertools.repeat
实际上并没有创建列表,它只是创建一个对象,如果您愿意,可以使用该对象创建列表!让我们再试一次,但转换为列表:
The function itertools.repeat
doesn't actually create the list, it just creates an object that can be used to create a list if you wish! Let's try that again, but converting to a list:
>>> timeit.timeit('list(itertools.repeat(0, 10))', 'import itertools', number = 1000000)
1.7508119747063233
因此,如果要列表,请使用[e] * n
.如果要延迟生成元素,请使用repeat
.
So if you want a list, use [e] * n
. If you want to generate the elements lazily, use repeat
.
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