基於SSM專案使用pagination分頁外掛實現分頁查詢案例程式碼
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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
dict內建操作的時間複雜度
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