返回列表中每个列表的元素范围

最简单的方法可能是使用列表理解:

newlist = [sublist[:2] for sublist in mylist]

快速回答:

first_two = [sublist[:2] for sublist in mylist]

如果有一个元组列表是可以的,那么一个更快的答案(根据我的测量是2倍):

import operator
map(operator.itemgetter(0, 1), mylist)

t = timeit.Timer("[i[:2] for i in ll]", "ll = [[i, i + 1, i + 2] for i in xrange(1000)]")
t.timeit(10000)
>>> 2.2732808589935303

t2 = timeit.Timer("map(operator.itemgetter(0, 1), ll)", "import operator; ll = [[i, i + 1, i + 2] for i in xrange(1000)]")
t2.timeit(10000)
>>> 1.3041009902954102

使用列表理解。

 newlist = [x[:2] for x in mylist]