Python内置方法的时间复杂度测试分析

timeit模块

timeit模块可以用来测试一小段Python代码的执行速度。

class timeit.Timer(stmt=‘pass’, setup=‘pass’, timer=)

Timer是测量小段代码执行速度的类。

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。

timeit.Timer.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

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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操作测试

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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