Python內建型別效能分析過程例項

阿新 • • 發佈：2020-02-02

這篇文章主要介紹了Python內建型別效能分析過程例項,文中通過示例程式碼介紹的非常詳細，對大家的學習或者工作具有一定的參考學習價值,需要的朋友可以參考下

timeit模組

timeit模組可以用來測試一小段Python程式碼的執行速度。

Timer是測量小段程式碼執行速度的類。

class timeit.Timer(stmt='pass',setup='pass',timer=<timer function>)

stmt引數是要測試的程式碼語句（statment）；
setup引數是執行程式碼時需要的設定；
timer引數是一個定時器函式，與平臺有關。

Timer物件.timeit(number=1000000)

Timer類中測試語句執行速度的物件方法。number引數是測試程式碼時的測試次數，預設為1000000次。方法返回執行程式碼的平均耗時，一個float型別的秒數。

list的操作測試

# -*- 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")

執行結果：

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操作測試

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')

測試pop操作：從結果可以看出，"pop最後一個元素"的效率遠遠高於"pop第一個元素"

可以自行嘗試下list的append(value)和insert(0,value),即一個後面插入和一個前面插入？？？

list內建操作的時間複雜度