算法速度

Question

我试图找到一种最快的方法来计算以下三个函数中一个月的天数。

def f1():
    t=time.time()
    for month in range(1,13):
        for year in range(10001):
            leap_year=(year%4==0 and year%100!=0) or year%400==0
            n=[29,31,28,31,30,31,30,31,31,30,31,30,31][month-2*leap_year*(month==2)]
    return time.time()-t1
def f2():
    t=time.time()
    for month in range(1,13):
        for year in range(10001):
            leap_year=(year%4==0 and year%100!=0) or year%400==0
            n=[31,28,31,30,31,30,31,31,30,31,30,31][month-1]+leap_year*(month==2)
    return time.time()-t
def f3():
    t=time.time()
    for month in range(1,13):
        for year in range(10001):
            leap_year=(year%4==0 and year%100!=0) or year%400==0
            n={1: 31, 2: 28, 3: 31, 4: 30, 5: 31, 6: 30, 7: 31, 8: 31, 9: 30, 10: 31, 11: 30, 12: 31}[month]+leap_year*(month==2)
    return time.time()-t
print(sum(f1() for i in range(100)))
print(sum(f2() for i in range(100)))
print(sum(f3() for i in range(100)))

我期待带字典的算法速度最快。但是，我得到了这个：

3.7855005264282227
3.625513792037964
9.0009183883667

显然，我错了。有人可以解释为什么dict算法速度最慢？它应该花费最少的时间，O（1），并且列表算法应该采用O（n），这与我得到的结果相矛盾。我是编程新手。谢谢！

Answer 1

您每次迭代都会构建一个列表/字典。 如果你只构建一次（因为你还没有改变它），f3（）方法是最快的，列表和字典的访问时间是O（1）。

def f1():
    t=time.time()
    m=[29,31,28,31,30,31,30,31,31,30,31,30,31]
    for month in range(1,13):
        for year in range(10001):
            leap_year=(year%4==0 and year%100!=0) or year%400==0
            n=m[month-2*leap_year*(month==2)]
    return time.time()-t
def f2():
    t=time.time()
    m=[31,28,31,30,31,30,31,31,30,31,30,31]
    for month in range(1,13):
        for year in range(10001):
            leap_year=(year%4==0 and year%100!=0) or year%400==0
            n=m[month-1]+leap_year*(month==2)
    return time.time()-t



def f3():
    t=time.time()
    m={1: 31, 2: 28, 3: 31, 4: 30, 5: 31, 6: 30, 7: 31, 8: 31, 9: 30, 10: 31, 11: 30, 12: 31}
    for month in range(1,13):
        for year in range(10001):
            leap_year=(year%4==0 and year%100!=0) or year%400==0
            n=m[month]+leap_year*(month==2)
    return time.time()-t


print(sum(f1() for i in range(100)))
print(sum(f2() for i in range(100)))
print(sum(f3() for i in range(100)))