Python内置类型性能分析过程实例
这篇文章主要介绍了Python内置类型性能分析过程实例,文中通过示例代码介绍的非常详细,对大家的学习或者工作具有一定的参考学习价值,需要的朋友可以参考下
timeit模块
timeit模块可以用来测试一小段Python代码的执行速度。
Timer是测量小段代码执行速度的类。
class timeit.Timer(stmt='pass', setup='pass', timer=<timer function>)
- 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。
Timer对象.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
# -*- coding:utf-8 -*- import timeit def t2(): li = [] for i in range(10000): li.insert(0, i) def t0(): li = [] for i in range(10000): li.extend([i]) def t1(): li = [] for i in range(10000): li.append(i) def t3(): li = [] for i in range(10000): li += [i] def t3_1(): li = [] for i in range(10000): li = li + [i] def t4(): li = [ i for i in range(10000)] def t5(): li = list(range(10000)) timer2 = timeit.Timer(stmt="t2()", setup="from __main__ import t2") print("insert", timer2.timeit(number=1000), "seconds") timer0 = timeit.Timer(stmt="t0()", setup="from __main__ import t0") print("extend", timer0.timeit(number=1000), "seconds") timer1 = timeit.Timer(stmt="t1()", setup="from __main__ import t1") print("append", timer1.timeit(number=1000), "seconds") timer3 = timeit.Timer(stmt="t3()", setup="from __main__ import t3") print("+=", timer3.timeit(number=1000), "seconds") timer3_1 = timeit.Timer(stmt="t3_1()", setup="from __main__ import t3_1") print("+加法", timer3_1.timeit(number=1000), "seconds") timer4 = timeit.Timer(stmt="t4()", setup="from __main__ import t4") print("[i for i in range()]", timer4.timeit(number=1000), "seconds") timer5 = timeit.Timer(stmt="t5()", setup="from __main__ import t5") print("list", timer5.timeit(number=1000), "seconds")
执行结果: insert 18.678989517 seconds extend 1.022223395000001 seconds append 0.6755100029999994 seconds += 0.773258104 seconds +加法 126.929554195 seconds [i for i in range()] 0.36483252799999377 seconds list 0.19607099800001038 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
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