Question

如何与列表和集合相交以保留列表的顺序？简单的例子：

k=[1,2,3,4]
d={3,2}
d.intersection(k)
[2,3]#this is the ideal result

编辑：速度是这里最重要的因素

Answer 1

您必须使用列表理解，这样可以通过过滤来保持订单：

[i for i in k if i in d]

演示：

>>> k = [1, 2, 3, 4]
>>> d = {2, 3}
>>> [i for i in k if i in d]
[2, 3]

对于Python 3，这是最快的选择;你可以使用list(filter(d.__contains__, l))，但这个简单的小例子比较慢：

>>> from timeit import timeit
>>> def listcomp(k, d):
...     return [i for i in k if i in d]
...
>>> def filtered(k, d):
...     return list(filter(d.__contains__, k))
...
>>> timeit('listcomp(k, d)', 'from __main__ import listcomp, k, d')
0.49590064199946937
>>> timeit('filtered(k, d)', 'from __main__ import filtered, k, d')
0.6533352420010488

当数据集大小增加时，时间会变得更糟：

>>> import random
>>> k = sorted([random.randrange(1000) for _ in range(1000)])
>>> d = {random.randrange(1000) for _ in range(100)}
>>> timeit('listcomp(k, d)', 'from __main__ import listcomp, k, d', number=10000)
0.30027976899873465
>>> timeit('filtered(k, d)', 'from __main__ import filtered, k, d', number=10000)
0.4524774450001132

在Python 2中，filter()是更快的选择，给定足够大的输入：

>>> from timeit import timeit
>>> import random
>>> def listcomp(k, d):
...     return [i for i in k if i in d]
...
>>> def filtered(k, d):
...     return filter(d.__contains__, k)
...
>>> k = [1, 2, 3, 4]
>>> d = {2, 3}
>>> timeit('listcomp(k, d)', 'from __main__ import listcomp, k, d')
0.4015800952911377
>>> timeit('filtered(k, d)', 'from __main__ import filtered, k, d')
0.4407978057861328
>>> k = sorted([random.randrange(1000) for _ in range(1000)])
>>> d = {random.randrange(1000) for _ in range(100)}
>>> timeit('listcomp(k, d)', 'from __main__ import listcomp, k, d', number=10000)
0.4594550132751465
>>> timeit('filtered(k, d)', 'from __main__ import filtered, k, d', number=10000)
0.28088998794555664

我不知道有任何更快的选择; available ordered set implementations都是Pure-python解决方案，而且速度较慢。

如何交叉列表并在python中设置并保留列表的顺序？

1 个答案: