Python内置类型性能分析
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 function>)
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
def t1():
l = []
for i in range(1000):
l = l + [i]
def t2():
l = []
for i in range(1000):
l.append(i)
def t3():
l = [i for i in range(1000)]
def t4():
l = list(range(1000))
from timeit import Timer
timer1 = Timer("t1()", "from __main__ import t1")
print("concat ",timer1.timeit(number=1000), "seconds")
timer2 = Timer("t2()", "from __main__ import t2")
print("append ",timer2.timeit(number=1000), "seconds")
timer3 = Timer("t3()", "from __main__ import t3")
print("comprehension ",timer3.timeit(number=1000), "seconds")
timer4 = Timer("t4()", "from __main__ import t4")
print("list range ",timer4.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
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
