>>> import collections
>>> a = list(range(1000000))
>>> a[100] = 1 #稍微改变一下列表
#方法一
>>> b = filter(lambda x: a.count(x) > 1, a)
#方法二
>>> d = filter(lambda x: x[1] != 1,collections.Counter(a).items())
为什么方法一要比方法二慢得多呢?
方法一中的count()函数要O(n^2)的时间复杂度。
方法二加速的原因是什么呢?到底是怎么实现的?(值得深究)
帮助文档:
Dict subclass for counting hashable items. Sometimes called a bag
or multiset. Elements are stored as dictionary keys and their counts
are stored as dictionary values.
原来如此。