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 t1():           l   =   []          for i   in  range(1000):                        l   =   l   +   [i] def t2():           l   =   []          for i   in  range(1000):                        l.append(i) def t3():

l   =   [i  for i   in  range(1000)] def    t4():           l   =   list(range(1000))from    timeit  import  Timertimer1  =   Timer("t1()",   "from   __main__    import  t1") print("concat  ",timer1.timeit(number=1000),   "seconds") timer2   =   Timer("t2()",   "from   __main__    import  t2") print("append  ",timer2.timeit(number=1000),   "seconds") timer3   =   Timer("t3()",   "from   __main__    import  t3") print("comprehension   ",timer3.timeit(number=1000),   "seconds") timer4   =   Timer("t4()",   "from   __main__    import  t4") print("list    range   ",timer4.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

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？？？

list内置操作的时间复杂度

dict内置操作的时间复杂度