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为什么除数很大时除法会更快

2019-10-27 09:57:33  阅读:279  来源: 互联网

标签:bignum time-complexity python time


我想看看除法运算的常数,所以我运行了这段代码

import time

def di(n):
    n/101

i = 10
while i < 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000:
    start = time.clock()
    di(i)
    end = time.clock()
    print("On " + str(i) + " " + str(end-start))
    i *= 10000

这是我得到的输出

On 10 5.98714047756e-06
On 100000 4.7041818038e-06
On 1000000000 2.56591734753e-06
On 10000000000000 2.99357023878e-06
On 100000000000000000 2.99357023878e-06
On 1000000000000000000000 2.99357023879e-06
On 10000000000000000000000000 2.99357023878e-06
On 100000000000000000000000000000 3.42122313003e-06
On 1000000000000000000000000000000000 3.42122313003e-06
On 10000000000000000000000000000000000000 3.84887602128e-06
On 100000000000000000000000000000000000000000 3.42122313003e-06
On 1000000000000000000000000000000000000000000000 3.84887602128e-06
On 10000000000000000000000000000000000000000000000000 1.71061156501e-06
On 100000000000000000000000000000000000000000000000000000 1.71061156502e-06
On 1000000000000000000000000000000000000000000000000000000000 1.71061156501e-06
On 10000000000000000000000000000000000000000000000000000000000000 2.13826445628e-06
On 100000000000000000000000000000000000000000000000000000000000000000 1.71061156502e-06
On 1000000000000000000000000000000000000000000000000000000000000000000000 2.13826445628e-06
On 10000000000000000000000000000000000000000000000000000000000000000000000000 2.13826445628e-06
On 100000000000000000000000000000000000000000000000000000000000000000000000000000 2.13826445626e-06
On 1000000000000000000000000000000000000000000000000000000000000000000000000000000000 2.13826445626e-06

为什么通常以后的除法速度会比前两个最小的时间更快?

解决方法:

Why does the division generally get faster later on as opposed to the time of the first two terms which are the smallest?

实际上不是.如果我将di替换为:

def di(n):
    for i in range(10000000): n / 101

然后得到(我假设您正在使用的Python 3.5):

On 10 0.546889
On 100000 0.545004
On 1000000000 0.5454929999999998
On 10000000000000 0.5519709999999998
On 100000000000000000 1.330797
On 1000000000000000000000 1.31053
On 10000000000000000000000000 1.3393129999999998
On 100000000000000000000000000000 1.3524339999999997
On 1000000000000000000000000000000000 1.3817269999999997
On 10000000000000000000000000000000000000 1.3412670000000002
On 100000000000000000000000000000000000000000 1.3358929999999987
On 1000000000000000000000000000000000000000000000 1.3773859999999996
On 10000000000000000000000000000000000000000000000000 1.3326890000000002
On 100000000000000000000000000000000000000000000000000000 1.3704769999999993
On 1000000000000000000000000000000000000000000000000000000000 1.3235019999999995
On 10000000000000000000000000000000000000000000000000000000000000 1.357647
On 100000000000000000000000000000000000000000000000000000000000000000 1.3341190000000012
On 1000000000000000000000000000000000000000000000000000000000000000000000 1.326544000000002
On 10000000000000000000000000000000000000000000000000000000000000000000000000 1.3671139999999973
On 100000000000000000000000000000000000000000000000000000000000000000000000000000 1.3630120000000012
On 1000000000000000000000000000000000000000000000000000000000000000000000000000000000 1.3600200000000022
On 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000 1.3189189999999975
On 100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 1.3503469999999993

如您所见,大约有两次:一次代表较小的数字,一次代表较大的数字.在Python 3.5中,/进行浮点除法,因此无论数字大小如何,它实际上应该花费相同的时间.

但是,从小到大都有差距.使用以下函数保留语义的Python 2.7也会发生相同的结果:

def di(n):
    for i in xrange(10000000): n / 101.0

在同一台计算机上,我得到:

On 10 0.617427
On 100000 0.61805
On 1000000000 0.6366
On 10000000000000 0.620919
On 100000000000000000 0.616695
On 1000000000000000000000 0.927353
On 10000000000000000000000000 1.007156
On 100000000000000000000000000000 0.98597
On 1000000000000000000000000000000000 0.99258
On 10000000000000000000000000000000000000 0.966753
On 100000000000000000000000000000000000000000 0.992684
On 1000000000000000000000000000000000000000000000 0.991711
On 10000000000000000000000000000000000000000000000000 0.994703
On 100000000000000000000000000000000000000000000000000000 0.978877
On 1000000000000000000000000000000000000000000000000000000000 0.982035
On 10000000000000000000000000000000000000000000000000000000000000 0.973266
On 100000000000000000000000000000000000000000000000000000000000000000 0.977911
On 1000000000000000000000000000000000000000000000000000000000000000000000 0.996857
On 10000000000000000000000000000000000000000000000000000000000000000000000000 0.972555
On 100000000000000000000000000000000000000000000000000000000000000000000000000000 0.985676
On 1000000000000000000000000000000000000000000000000000000000000000000000000000000000 0.987412
On 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000 0.997207
On 100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 0.970129

这与将整数转换为浮点数有关(具体细节尚待确定).当整数超过某个点时,速度会变慢,并且当该点超过该点时,除法运算会花费更长的时间.

标签:bignum,time-complexity,python,time
来源: https://codeday.me/bug/20191027/1943749.html

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