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Which is faster in Python: x**.5 or math.sqrt(x)?

Asked 2008-11-29T01:24:09.293
255

I've been wondering this for some time. As the title say, which is faster, the actual function or simply raising to the half power?

UPDATE

This is not a matter of premature optimization. This is simply a question of how the underlying code actually works. What is the theory of how Python code works?

I sent Guido van Rossum an email cause I really wanted to know the differences in these methods.

My email:

There are at least 3 ways to do a square root in Python: math.sqrt, the '**' operator and pow(x,.5). I'm just curious as to the differences in the implementation of each of these. When it comes to efficiency which is better?

His response:

pow and ** are equivalent; math.sqrt doesn't work for complex numbers, and links to the C sqrt() function. As to which one is faster, I have no idea...

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4

Someone commented about the "fast Newton-Raphson square root" from Quake 3... I implemented it with ctypes, but it's super slow in comparison to the native versions. I'm going to try a few optimizations and alternate implementations.

from ctypes import c_float, c_long, byref, POINTER, cast

def sqrt(num):
 xhalf = 0.5*num
 x = c_float(num)
 i = cast(byref(x), POINTER(c_long)).contents.value
 i = c_long(0x5f375a86 - (i>>1))
 x = cast(byref(i), POINTER(c_float)).contents.value

 x = x*(1.5-xhalf*x*x)
 x = x*(1.5-xhalf*x*x)
 return x * num

Here's another method using struct, comes out about 3.6x faster than the ctypes version, but still 1/10 the speed of C.

from struct import pack, unpack

def sqrt_struct(num):
 xhalf = 0.5*num
 i = unpack('L', pack('f', 28.0))[0]
 i = 0x5f375a86 - (i>>1)
 x = unpack('f', pack('L', i))[0]

 x = x*(1.5-xhalf*x*x)
 x = x*(1.5-xhalf*x*x)
 return x * num
answered 2010-04-23T05:27:31.410

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