### Python內置類型性能分析 #### timeit模塊 timeit模塊可以用來測試一小段Python代碼的執行速度。 `class timeit.Timer(stmt='pass', setup='pass', timer=<timer function>)` Timer是測量小段代碼執行速度的類。 stmt參數是要測試的代碼語句（statment）； setup參數是運行代碼時需要的設置； timer參數是一個定時器函數，與平臺有關。 `timeit.Timer.timeit(number=1000000)` Timer類中測試語句執行速度的對象方法。number參數是測試代碼時的測試次數，默認為1000000次。方法返回執行代碼的平均耗時，一個float類型的秒數。 **list的操作測試** ~~~ def test1(): l = [] for i in range(1000): l = l + [i] def test2(): l = [] for i in range(1000): l.append(i) def test3(): l = [i for i in range(1000)] def test4(): l = list(range(1000)) from timeit import Timer t1 = Timer("test1()", "from __main__ import test1") print("concat ",t1.timeit(number=1000), "seconds") t2 = Timer("test2()", "from __main__ import test2") print("append ",t2.timeit(number=1000), "seconds") t3 = Timer("test3()", "from __main__ import test3") print("comprehension ",t3.timeit(number=1000), "seconds") t4 = Timer("test4()", "from __main__ import test4") print("list range ",t4.timeit(number=1000), "seconds") # ('concat ', 1.7890608310699463, 'seconds') # ('append ', 0.13796091079711914, 'seconds') # ('comprehension ', 0.05671119689941406, 'seconds') # ('list range ', 0.014147043228149414, '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最后一個元素的效率遠遠高于pop第一個元素 >[success]可以自行嘗試下list的append(value)和insert(0,value),即一個后面插入和一個前面插入？？？