如何加快列表理解 [英] How to speed up list comprehension

问题描述

以下是我的列表:

col = [['red', 'yellow', 'blue', 'red', 'green', 'yellow'],
       ['pink', 'orange', 'brown', 'pink', 'brown']
      ]

我的目标是消除在每个列表中出现一次的项目.

My goal is to eliminate items that appear once in each list.

这是我的代码:

eliminate = [[w for w in c if c.count(w)>1]for c in col]

Output: [['red', 'red', 'yellow','yellow'], ['pink','pink', 'brown','brown']]

该代码对于较小的数据集(如上面的列表)正常工作，但是我的数据集非常大.每个列表最多包含1000个项目.

The code works fine for small dataset such as the list above, however, my dataset is very large. Each list contains up to a 1000 items.

有没有办法使上述代码更快地工作?就像将代码分解成两个或多个for循环一样，因为我的理解是正常的for循环比列表理解要快.

Is there a way to make the above code work faster? like breaking the code down into two or more for-loops, as my understanding is that the normal for-loop is faster than list comprehension.

有什么建议吗?谢谢.

Any suggestions? thanks.

推荐答案

我可以尝试尝试OrderedCounter以避免重复的.count()调用:

I'd have a go at trying an OrderedCounter to avoid the repeated .count() calls:

from collections import OrderedDict, Counter

col=[['red', 'yellow', 'blue', 'red', 'green', 'yellow'],['pink', 'orange', 'brown', 'pink', 'brown']]

class OrderedCounter(Counter, OrderedDict):
    pass

new = [[k for k, v in OrderedCounter(el).iteritems() if v != 1] for el in col]
# [['red', 'yellow'], ['pink', 'brown']]

如果我们只想迭代一次，那么(类似于Martijn的方法-加上少玩组合):

And if we just wish to iterate once, then (similar to Martijn's - plus playing less with sets):

from itertools import count
def unique_plurals(iterable):
    seen = {}
    return [el for el in iterable if next(seen.setdefault(el, count())) == 1]

new = map(unique_plurals, col)

这在指定所需的出现次数方面更加灵活，并且保留一个dict而不是多个set.

This is more flexible in terms of specifying how many occurrences are required, and keeps one dict instead of multiple sets.

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