python python上负数的三次方根
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Cubic root of the negative number on python
提问by Vladimir Prudnikov
Can someone help me to find a solution on how to calculate a cubic root of the negative number using python?
有人可以帮我找到有关如何使用 python 计算负数三次根的解决方案吗？
>>> math.pow(-3, float(1)/3)
nan
it does not work. Cubic root of the negative number is negative number. Any solutions?
这是行不通的。负数的三次方根是负数。任何解决方案？
采纳答案by DrAl
You could use:
你可以使用：
-math.pow(3, float(1)/3)
Or more generally:
或更一般地：
if x > 0:
return math.pow(x, float(1)/3)
elif x < 0:
return -math.pow(abs(x), float(1)/3)
else:
return 0
回答by Andrew Walker
A simple use of De Moivre's formula, is sufficient to show that the cube root of a value, regardless of sign, is a multi-valued function. That means, for any input value, there will be three solutions. Most of the solutions presented to far only return the principle root. A solution that returns all valid roots, and explicitly tests for non-complex special cases, is shown below.
De Moivre 公式的简单使用，足以表明一个值的立方根，无论符号如何，都是一个多值函数。这意味着，对于任何输入值，都会有三种解决方案。提出的大多数解决方案只返回主根。返回所有有效根并显式测试非复杂特殊情况的解决方案如下所示。
import numpy
import math
def cuberoot( z ):
z = complex(z)
x = z.real
y = z.imag
mag = abs(z)
arg = math.atan2(y,x)
return [ mag**(1./3) * numpy.exp( 1j*(arg+2*n*math.pi)/3 ) for n in range(1,4) ]
Edit:As requested, in cases where it is inappropriate to have dependency on numpy, the following code does the same thing.
编辑：根据要求，在不适合依赖 numpy 的情况下，以下代码执行相同的操作。
def cuberoot( z ):
z = complex(z)
x = z.real
y = z.imag
mag = abs(z)
arg = math.atan2(y,x)
resMag = mag**(1./3)
resArg = [ (arg+2*math.pi*n)/3. for n in range(1,4) ]
return [ resMag*(math.cos(a) + math.sin(a)*1j) for a in resArg ]
回答by David
math.pow(abs(x),float(1)/3) * (1,-1)[x<0]
回答by user9876
Taking the earlier answers and making it into a one-liner:
将较早的答案制成单行：
import math
def cubic_root(x):
return math.copysign(math.pow(abs(x), 1.0/3.0), x)
回答by tom10
You can get the complete (all n roots) and more general (any sign, any power) solution using:
您可以使用以下方法获得完整的（所有 n 个根）和更通用（任何符号、任何幂）的解决方案：
import cmath
x, t = -3., 3 # x**(1/t)
a = cmath.exp((1./t)*cmath.log(x))
p = cmath.exp(1j*2*cmath.pi*(1./t))
r = [a*(p**i) for i in range(t)]
Explanation: a is using the equation xu= exp(u*log(x)). This solution will then be one of the roots, and to get the others, rotate it in the complex plane by a (full rotation)/t.
解释：a 使用方程 x u= exp(u*log(x))。该解将成为其中一个根，为了得到其他根，将它在复平面中旋转（完整旋转）/ t。
回答by Otto Allmendinger
You can also wrap the libmlibrary that offers a cbrt(cube root) function:
您还可以包装libm提供cbrt（立方体根）函数的库：
from ctypes import *
libm = cdll.LoadLibrary('libm.so.6')
libm.cbrt.restype = c_double
libm.cbrt.argtypes = [c_double]
libm.cbrt(-8.0)
gives the expected
给出预期
-2.0
回答by Kirk Broadhurst
The cubic root of a negative number is just the negative of the cubic root of the absolute value of that number.
负数的三次方根就是该数绝对值三次方根的负数。
i.e. x^(1/3) for x < 0 is the same as (-1)*(|x|)^(1/3)
即 x^(1/3) for x < 0 与 (-1)*(|x|)^(1/3) 相同
Just make your number positive, and then perform cubic root.
只需使您的数字为正，然后执行三次方根。
回答by Matěj ?míd
You can use cbrtfrom scipy.special:
您可以使用cbrt从scipy.special：
>>> from scipy.special import cbrt
>>> cbrt(-3)
-1.4422495703074083
This also works for arrays.
这也适用于数组。
回答by kidnakyo
this works with numpy array as well:
这也适用于 numpy 数组：
cbrt = lambda n: n/abs(n)*abs(n)**(1./3)
回答by ijoseph
numpyhas an inbuilt cube root function cbrtthat handles negative numbers fine:
numpy有一个内置的立方根函数cbrt，可以很好地处理负数：
>>> import numpy as np
>>> np.cbrt(-8)
-2.0
This was added in version 1.10.0(released 2015-10-06).
这是在版本中添加的1.10.0（已发布2015-10-06）。
Also works for numpyarray/ listinputs:
也适用于numpyarray/list输入：
>>> np.cbrt([-8, 27])
array([-2., 3.])
