列表理解與生成器表達式的奇怪時間結果？我正在回答這個問題，我更喜歡這里的生成器表達并使用它，我認為它會更快，因為生成器不需要先創(chuàng)建整個列表：>>> lis=[['a','b','c'],['d','e','f']]>>> 'd' in (y for x in lis for y in x)TrueLevon在他的解決方案中使用了列表理解，>>> lis = [['a','b','c'],['d','e','f']]>>> 'd' in [j for i in mylist for j in i]True但是當我做這些LC的時間結果比生成器快時：~$ python -m timeit -s "lis=[['a','b','c'],['d','e','f']]" "'d' in (y for x in lis for y in x)"    100000 loops, best of 3: 2.36 usec per loop~$ python -m timeit -s "lis=[['a','b','c'],['d','e','f']]" "'d' in [y for x in lis for y in x]"    100000 loops, best of 3: 1.51 usec per loop然后我增加了列表的大小，并再次計時：lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]這次搜索'd'生成器比LC快，但是當我搜索中間元素（11）和最后一個元素然后LC再次擊敗生成器表達式時，我無法理解為什么？~$ python -m timeit -s "lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]" "'d' in (y for x in lis for y in x)"    100000 loops, best of 3: 2.96 usec per loop~$ python -m timeit -s "lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]" "'d' in [y for x in lis for y in x]"    100000 loops, best of 3: 7.4 usec per loop~$ python -m timeit -s "lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]" "11 in [y for x in lis for y in x]"100000 loops, best of 3: 5.61 usec per loop~$ python -m timeit -s "lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]" "11 in (y for x in lis for y in x)"100000 loops, best of 3: 9.76 usec per loop~$ python -m timeit -s "lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]" "18 in (y for x in lis for y in x)"100000 loops, best of 3: 8.94 usec per loop~$ python -m timeit -s "lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]" "18 in [y for x in lis for y in x]"100000 loops, best of 3: 7.13 usec per loop