Python内置类型性能分析过程实例
时间：2020-12-07 16:56:09 | 栏目：Python代码 | 点击：次
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
