blocks|key|1463094|text|您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1463095|-math.pow(3,+float(1)/3)|code-block|syntax|javascript|1463096|或者更一般地说：|1463097|if+x+>+0:
++++return+math.pow(x,+float(1)/3)
elif+x+<+0:
++++return+-math.pow(abs(x),+float(1)/3)
else:
++++return+0|1463098|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You could use:

<pre><code>-math.pow(3, float(1)/3)
</code></pre>

Or more generally:

<pre><code>if x &gt; 0:
 return math.pow(x, float(1)/3)
elif x &lt; 0:
 return -math.pow(abs(x), float(1)/3)
else:
 return 0
</code></pre>

blocks|key|1463191|text|math.pow(abs(x),float(1)/3)+*+(1,-1)[x<0]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1463192|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>math.pow(abs(x),float(1)/3) * (1,-1)[x&lt;0]
</code></pre>

blocks|key|1248504|text|接受前面的答案，并将其变成一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1248505|import+math
def+cubic_root(x):
++++return+math.copysign(math.pow(abs(x),+1.0/3.0),+x)|code-block|syntax|javascript|1248506|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Taking the earlier answers and making it into a one-liner:

<pre><code>import math
def cubic_root(x):
 return math.copysign(math.pow(abs(x), 1.0/3.0), x)
</code></pre>

blocks|key|1248598|text|您还可以包装提供cbrt+(立方根)函数的libm库：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1248599|from+ctypes+import+*
libm+=+cdll.LoadLibrary('libm.so.6')
libm.cbrt.restype+=+c_double
libm.cbrt.argtypes+=+[c_double]
libm.cbrt(-8.0)|code-block|syntax|javascript|1248600|给出了预期的|1248601|-2.0|1248602|entityMap^0|8|4|L|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

You can also wrap the <code>libm</code> library that offers a <code>cbrt</code> (cube root) function:

<pre><code>from ctypes import *
libm = cdll.LoadLibrary('libm.so.6')
libm.cbrt.restype = c_double
libm.cbrt.argtypes = [c_double]
libm.cbrt(-8.0)
</code></pre>

gives the expected

<pre><code>-2.0
</code></pre>

blocks|key|1248636|text|负数的立方根恰好是该数的绝对值的立方根的负数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1248637|即x<0的x%5E(1/3)等于(-1)*(%7Cx%7C)%5E(1/3)|1248638|只需将数字设为正数，然后执行立方根。|1248639|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

The cubic root of a negative number is just the negative of the cubic root of the absolute value of that number.

i.e. x^(1/3) for x &lt; 0 is the same as (-1)*(|x|)^(1/3)

Just make your number positive, and then perform cubic root.

blocks|key|1248869|text|numpy的an+inbuilt+cube+root+function+cbrt可以很好地处理负数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1248870|>>>+import+numpy+as+np
>>>+np.cbrt(-8)
-2.0|code-block|syntax|javascript|1248871|这是在版本1.10.0+(发布的2015-10-06)中添加的。|1248872|也适用于numpy+array+/+list输入：|1248873|>>>+np.cbrt([-8,+27])
array([-2.,++3.])|1248874|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy-1.14.0/reference/generated/numpy.cbrt.html^0|0|5|10|4|6|Y|0|0|0|5|6|G|A|0|4|5|A|5|I|4|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1C|8|@$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|1J|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

<code>numpy</code> has <a href="https://docs.scipy.org/doc/numpy-1.14.0/reference/generated/numpy.cbrt.html" rel="nofollow noreferrer">an inbuilt cube root function <code>cbrt</code></a> that handles negative numbers fine:

<pre><code>&gt;&gt;&gt; import numpy as np
&gt;&gt;&gt; np.cbrt(-8)
-2.0
</code></pre>

This was added in version <code>1.10.0</code> (released <code>2015-10-06</code>).

Also works for <code>numpy</code> <code>array</code> / <code>list</code> inputs:

<pre><code>&gt;&gt;&gt; np.cbrt([-8, 27])
array([-2., 3.])
</code></pre>

blocks|key|1248679|text|您可以从scipy.special使用cbrt|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1248680|>>>+from+scipy.special+import+cbrt
>>>+cbrt(-3)
-1.4422495703074083|code-block|syntax|javascript|1248681|这也适用于数组。|1248682|entityMap^0|4|D|J|4|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

You can use <code>cbrt</code> from <code>scipy.special</code>:

<pre><code>&gt;&gt;&gt; from scipy.special import cbrt
&gt;&gt;&gt; cbrt(-3)
-1.4422495703074083
</code></pre>

This also works for arrays.

blocks|key|1463593|text|这也适用于numpy数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1463594|cbrt+=+lambda+n:+n/abs(n)*abs(n)**(1./3)|code-block|syntax|javascript|1463595|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

this works with numpy array as well:

<pre><code>cbrt = lambda n: n/abs(n)*abs(n)**(1./3)
</code></pre>

blocks|key|1248429|text|原始解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1248430|def+cubic_root(nr):
+++if+nr<0:
+++++return+-math.pow(-nr,+float(1)/3)
+++else:
+++++return+math.pow(nr,+float(1)/3)|code-block|syntax|javascript|1248431|可能是大规模的非蟒蛇，但它应该可以工作。|1248432|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Primitive solution:

<pre><code>def cubic_root(nr):
 if nr&lt;0:
 return -math.pow(-nr, float(1)/3)
 else:
 return math.pow(nr, float(1)/3)
</code></pre>

Probably massively non-pythonic, but it should work.

blocks|key|1463514|text|我刚刚遇到了一个非常类似的问题，并从this+forum+post找到了NumPy解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1463515|在nushell中，我们可以使用NumPy+sign和absolute方法来帮助我们。下面是一个对我很有效的例子：|style|CODE|1463516|import+numpy+as+np

x+=+np.array([-81,25])
print+x
#>>>+[-81++25]

xRoot5+=+np.sign(x)+*+np.absolute(x)**(1.0/5.0)+++++
print+xRoot5
#>>>+[-2.40822469++1.90365394]

print+xRoot5**5
#>>>+[-81.++25.]|code-block|syntax|javascript|1463517|因此，回到最初的立方根问题：|1463518|import+numpy+as+np

y+=+-3.
np.sign(y)+*+np.absolute(y)**(1./3.)
#>>>+-1.4422495703074083|1463519|我希望这能帮到你。|1463520|entityMap|0|LINK|mutability|MUTABLE|url|http://scipy-user.10969.n7.nabble.com/numpy-array-root-operation-td1727.html^0|I|F|0|0|M|4|R|8|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|6|7|14|8|@$A|15|B|16|F|G]|$A|17|B|18|F|G]]|9|@]|C|$]]|$1|H|3|I|5|J|7|19|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|1A|8|@]|9|@]|C|$]]|$1|O|3|P|5|J|7|1B|8|@]|9|@]|C|$K|L]]|$1|Q|3|R|5|6|7|1C|8|@]|9|@]|C|$]]|$1|S|3|-4|5|6|7|1D|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]]]

I just had a very similar problem and found the NumPy solution from <a href="http://scipy-user.10969.n7.nabble.com/numpy-array-root-operation-td1727.html" rel="nofollow">this forum post</a>.

In a nushell, we can use of the NumPy <code>sign</code> and <code>absolute</code> methods to help us out. Here is an example that has worked for me:

<pre><code>import numpy as np

x = np.array([-81,25])
print x
#&gt;&gt;&gt; [-81 25]

xRoot5 = np.sign(x) * np.absolute(x)**(1.0/5.0) 
print xRoot5
#&gt;&gt;&gt; [-2.40822469 1.90365394]

print xRoot5**5
#&gt;&gt;&gt; [-81. 25.]
</code></pre>

So going back to the original cube root problem:

<pre><code>import numpy as np

y = -3.
np.sign(y) * np.absolute(y)**(1./3.)
#&gt;&gt;&gt; -1.4422495703074083
</code></pre>

I hope this helps.

blocks|key|1248809|text|对于Python+3中的算术、类似计算器的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1248810|>>>+-3.0**(1/3)
-1.4422495703074083|code-block|syntax|javascript|1248811|或者Python2中的-3.0**(1./3)。|offset|length|style|CODE|1248812|对于x**3+%2B+(0*x**2+%2B+0*x)+%2B+3+=+0的代数解决方案，请使用numpy：|1248813|>>>+p+=+[1,0,0,3]
>>>+numpy.roots(p)
[-3.0%2B0.j++++++++++1.5%2B2.59807621j++1.5-2.59807621j]|1248814|entityMap^0|0|0|B|C|0|2|T|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]]|9|@]|A|$]]|$1|O|3|P|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

For an arithmetic, calculator-like answer in Python 3:

<pre><code>&gt;&gt;&gt; -3.0**(1/3)
-1.4422495703074083
</code></pre>

or <code>-3.0**(1./3)</code> in Python 2.

For the algebraic solution of <code>x**3 + (0*x**2 + 0*x) + 3 = 0</code> use numpy:

<pre><code>&gt;&gt;&gt; p = [1,0,0,3]
&gt;&gt;&gt; numpy.roots(p)
[-3.0+0.j 1.5+2.59807621j 1.5-2.59807621j]
</code></pre>

Can someone help me to find a solution on how to calculate a cubic root of the negative number using python?

<pre><code>&gt;&gt;&gt; math.pow(-3, float(1)/3)
nan
</code></pre>

it does not work. Cubic root of the negative number is negative number. Any solutions?

Cubic root of the negative number on python

Python

有没有人能帮我找出一个用python计算负数的立方根的方法？>>> math.pow(-3, float(1)/3)nan它不起作用。负数的立方根是负数。有什么解决方案吗？

问python上负数的三次根
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回答 11

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