import math
print(math.sqr(4))
輸出結(jié)果：2Epsilon = 10e-16
def fab_h(x):
    '''
    求實數(shù)的絕對值
    :param x: R
    '''
    if x >= 0:
        return x
    else:
        return x * -1
def sqrt_h(x, n=2.0):
    '''
    n倍根號下x
    牛頓迭代法
    '''
    val = x
    last = 0.0
    if n == 2.0:
        while (fab_h(val - last) > Epsilon):
            last = val
            val = (val + x / val) / 2
        return val
    while (fab_h(val - last) > Epsilon):
        last = val
        val = ((n-1)*val + x / val**(n-1)) / n
    return val

二，cmath庫

cmath多用于復(fù)數(shù)，負數(shù)的開平方。

# importing cmath library 
import cmath 
# using cmath.sqrt() method 
gfg = cmath.sqrt(3) 
print(gfg)

輸出：

(1.7320508075688772+0j)

三，pow（）函數(shù)

pow（x,y）的意思是返回x的y次方，如pow(x,2)就是返回x的平方，就是x^2,

#!/usr/bin/python
# -*- coding: UTF-8 -*-
import math   # 導入 math 模塊
print "math.pow(100, 2) : ", math.pow(100, 2)
# 使用內(nèi)置，查看輸出結(jié)果區(qū)別
print "pow(100, 2) : ", pow(100, 2)
print "math.pow(100, -2) : ", math.pow(100, -2)
print "math.pow(2, 4) : ", math.pow(2, 4)

輸出：

math.pow(100, 2) : 10000.0
pow(100, 2) : 10000
math.pow(100, -2) : 0.0001
math.pow(2, 4) : 16.0
math.pow(3, 0) : 1.0

python不同開根號速度對比

import time
import math
import numpy as np
def timeit1():
    s = time.time()
    for i in range(750000):
        z=i**.5
    print ("Took %f seconds" % (time.time() - s))
def timeit2(arg=math.sqrt):
    s = time.time()
    for i in range(750000):
        z=arg(i)
    print ("Took %f seconds" % (time.time() - s))
def timeit3(arg=np.sqrt):
    s = time.time()
    for i in range(750000):
        z=arg(i)
    print ("Took %f seconds" % (time.time() - s))
def timeit4():
    s = time.time()
    for i in range(750000):
        z=math.pow(i,.5)
    print ("Took %f seconds" % (time.time() - s))
timeit1()
timeit2()
timeit3()
timeit4()