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
[AppleScript] 纯文本查看 复制代码 def t1(): l = [] for i in range(1000): l = l + [i] def t2(): l = [] for i in range(1000): l.append(i) def t3():
[AppleScript] 纯文本查看 复制代码 l = [i for i in range(1000)] def t4(): l = list(range(1000))
from timeit import Timer
timer1 = 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
[AppleScript] 纯文本查看 复制代码 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
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