使用幂运算** 0.5比math.sqrt效率低吗? [英] Using exponentiation **0.5 less efficient than math.sqrt?

问题描述

引用"Python编程:计算机科学"

我们本可以扎根 使用幂运算**.使用 math.sqrt效率更高.

We could have taken the square root using exponentiation **. Using math.sqrt is somewhat more efficient.

有点"，但是到什么程度，以及如何?

"Somewhat", but to what extent, and how?

推荐答案

从理论上讲， duffymo的答案是很好的猜测.但是实际上，在我的机器上，它不是更有效:

Theoretically, hammar's answer and duffymo's answer are good guesses. But in practice, on my machine, it's not more efficient:

>>> import timeit
>>> timeit.timeit(stmt='[n ** 0.5 for n in range(100)]', setup='import math', number=10000)
0.15518403053283691
>>> timeit.timeit(stmt='[math.sqrt(n) for n in range(100)]', setup='import math', number=10000)
0.17707490921020508

部分问题是.操作.如果直接将sqrt导入名称空间，则会得到一些改进.

Part of the problem is the . operation. If you import sqrt directly into the namespace, you get a slight improvement.

>>> timeit.timeit(stmt='[sqrt(n) for n in range(100)]', setup='from math import sqrt', number=10000)
0.15312695503234863

那里的关键字:轻微.

进一步的测试表明，随着数量的增加，使用sqrt所获得的收益也会增加.但是仍然不是很多！

Further testing indicates that as the number gets larger, the benefit you get from using sqrt increases. But still not by a lot!

>>> timeit.timeit(stmt='[n ** 0.5 for n in range(1000000)]', setup='import math', number=1)
0.18888211250305176
>>> timeit.timeit(stmt='[math.sqrt(n) for n in range(1000000)]', setup='import math', number=1)
0.18425297737121582
>>> timeit.timeit(stmt='[sqrt(n) for n in range(1000000)]', setup='from math import sqrt', number=1)
0.1571958065032959

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