### Python内置类型性能分析 #### timeit模块 timeit模块可以用来测试一小段Python代码的执行速度。 `class timeit.Timer(stmt='pass', setup='pass', timer=<timer function>)` Timer是测量小段代码执行速度的类。 stmt参数是要测试的代码语句（statment）； setup参数是运行代码时需要的设置； timer参数是一个定时器函数，与平台有关。 `timeit.Timer.timeit(number=1000000)` Timer类中测试语句执行速度的对象方法。number参数是测试代码时的测试次数，默认为1000000次。方法返回执行代码的平均耗时，一个float类型的秒数。 **list的操作测试** ~~~ def test1(): l = [] for i in range(1000): l = l + [i] def test2(): l = [] for i in range(1000): l.append(i) def test3(): l = [i for i in range(1000)] def test4(): l = list(range(1000)) from timeit import Timer t1 = Timer("test1()", "from __main__ import test1") print("concat ",t1.timeit(number=1000), "seconds") t2 = Timer("test2()", "from __main__ import test2") print("append ",t2.timeit(number=1000), "seconds") t3 = Timer("test3()", "from __main__ import test3") print("comprehension ",t3.timeit(number=1000), "seconds") t4 = Timer("test4()", "from __main__ import test4") print("list range ",t4.timeit(number=1000), "seconds") # ('concat ', 1.7890608310699463, 'seconds') # ('append ', 0.13796091079711914, 'seconds') # ('comprehension ', 0.05671119689941406, 'seconds') # ('list range ', 0.014147043228149414, 'seconds') ~~~ **pop操作测试** ~~~ x = range(2000000) pop_zero = Timer("x.pop(0)","from __main__ import x") print("pop_zero ",pop_zero.timeit(number=1000), "seconds") x = range(2000000) pop_end = Timer("x.pop()","from __main__ import x") print("pop_end ",pop_end.timeit(number=1000), "seconds") # ('pop_zero ', 1.9101738929748535, 'seconds') # ('pop_end ', 0.00023603439331054688, 'seconds') ~~~ 测试pop操作：从结果可以看出，pop最后一个元素的效率远远高于pop第一个元素 >[success]可以自行尝试下list的append(value)和insert(0,value),即一个后面插入和一个前面插入？？？