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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