blocks|key|242919|text|x的第n根是x%5E(1/n)，因此您可以执行9**(1/2)来查找9的第二个根，例如。通常，您可以将x的第n根计算为：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|242920|x**(1/n)|code-block|syntax|javascript|242921|注意事项：在Python2中，您必须执行1/float(n)或1.0/n，以便结果是float而不是int。有关详细信息，请参阅为什么Python给出了平方根的“错误”答案？|BOLD|242922|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/9595135/why-does-python-give-the-wrong-answer-for-square-root^0|0|1|6|7|L|8|0|0|0|4|K|A|V|5|16|5|1E|3|1S|N|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@$9|14|A|15|B|M]|$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@$9|1E|A|1F|1|1G]]|E|$]]|$1|N|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]]]

nth root of <code>x</code> is <code>x^(1/n)</code>, so you can do <code>9**(1/2)</code> to find the 2nd root of 9, for example. In general, you can compute the nth root of x as:
<pre><code>x**(1/n)
</code></pre>
Note: In Python 2, you had to do <code>1/float(n)</code> or <code>1.0/n</code> so that the result would be a <code>float</code> rather than an <code>int</code>. For more details, see <a href="https://stackoverflow.com/questions/9595135/why-does-python-give-the-wrong-answer-for-square-root">Why does Python give the &quot;wrong&quot; answer for square root?</a>

blocks|key|2409903|text|您还可以使用一些对数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2409904|x第N根|offset|length|style|CODE|2409905|exp(log(x)/n)|code-block|syntax|javascript|2409906|例如：|2409907|>>>+from+math+import+exp,+log
>>>+x+=+8
>>>+n+=+3
>>>+exp(log(x)/n)
2.0|2409908|entityMap^0|0|0|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

You may also use some logarithms:
Nth root of <code>x</code>:
<pre><code>exp(log(x)/n)
</code></pre>
For example:
<pre><code>&gt;&gt;&gt; from math import exp, log
&gt;&gt;&gt; x = 8
&gt;&gt;&gt; n = 3
&gt;&gt;&gt; exp(log(x)/n)
2.0
</code></pre>

blocks|key|2409786|text|还有：x**(n**-1)，它与x**(1/float(n))相同，但比x**(1/float(n))短|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2409787|entityMap^0|3|A|G|F|10|F|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

Also: <code>x**(n**-1)</code>, which is the same but shorter than <code>x**(1/float(n))</code>

blocks|key|243022|text|如果您喜欢在功能上应用此操作，而不是使用infix运算符(+**符号)，则可以将基和指数作为参数传递给pow函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|243023|In+[23]:+(9**(0.5))+==+pow(9,+0.5)
Out[23]:+True|code-block|syntax|javascript|243024|我也喜欢为这是在Python中插入的黑客。寻找新的用途，尽管它更多的是一个有趣的，而不是一个繁重的解决方案。但是，通过以下操作，您可以有效地制作自己的个人符号：|243025|class+Infix:
++++def+__init__(self,+function):
++++++++self.function+=+function
++++def+__ror__(self,+other):
++++++++return+Infix(lambda+x,+self=self,+other=other:+self.function(other,+x))
++++def+__or__(self,+other):
++++++++return+self.function(other)
++++def+__rlshift__(self,+other):
++++++++return+Infix(lambda+x,+self=self,+other=other:+self.function(other,+x))
++++def+__rshift__(self,+other):
++++++++return+self.function(other)
++++def+__call__(self,+value1,+value2):
++++++++return+self.function(value1,+value2)


root_of+=+Infix(lambda+x,y:+y**(1.0/x))

print+2+%7Croot_of%7C+9
3.0|243026|entityMap|0|LINK|mutability|MUTABLE|url|http://code.activestate.com/recipes/384122/^0|U|2|1F|3|0|0|5|G|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|D|@$9|13|A|14|1|15]]|E|$]]|$1|M|3|N|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

If you prefer to apply this operation functionally rather than with an infix operator (the <code>**</code> symbol), you can pass the base and exponent as arguments to the <code>pow</code> function:

<pre><code>In [23]: (9**(0.5)) == pow(9, 0.5)
Out[23]: True
</code></pre>

I am also fond of finding new uses for <a href="http://code.activestate.com/recipes/384122/" rel="noreferrer">this Infix hack in Python</a> although it's more of a fun aside than a heavy-duty solution. But you could effectively make your own personal symbol for this by doing the following:

<pre><code>class Infix:
 def __init__(self, function):
 self.function = function
 def __ror__(self, other):
 return Infix(lambda x, self=self, other=other: self.function(other, x))
 def __or__(self, other):
 return self.function(other)
 def __rlshift__(self, other):
 return Infix(lambda x, self=self, other=other: self.function(other, x))
 def __rshift__(self, other):
 return self.function(other)
 def __call__(self, value1, value2):
 return self.function(value1, value2)


root_of = Infix(lambda x,y: y**(1.0/x))

print 2 |root_of| 9
3.0
</code></pre>

blocks|key|242952|text|的确有。只是**+=)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|242953|任何n个根都是1/n的指数，所以要得到9的平方根，使用9**(1/2)+(或9**0.5)获取立方根，使用9+**+(1/3)+(我们不能用更简单的分数编写)，并得到n根9+**+(1/n)。|242954|还请注意，从Python+3开始，将句点添加到整数使其成为浮点数不再是必要的。说1/3的工作方式实际上是你所期望的，给出了0.333...的结果，而不是零。对于Python的遗留版本，您必须记住使用这个时间段(但也非常想知道为什么要使用编程语言的遗留版本)|242955|entityMap^0|6|2|0|7|3|R|8|12|6|1H|A|2D|A|0|14|3|1P|8|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|K|$]]

There is. It's just <code>**</code> =)

Any nth root is an exponentiation by <code>1/n</code>, so to get the square root of 9, you use <code>9**(1/2)</code> (or <code>9**0.5</code>) to get the cube root, you use <code>9 ** (1/3)</code> (which we can't write with a simpler fraction), and to get the nth root, <code>9 ** (1/n)</code>.

Also note that as of Python 3, adding periods to integers to make them a float is no longer necessary. Saying <code>1/3</code> works the way you would actually expect it to, giving <code>0.333...</code> as result, rather than zero. For legacy versions of Python, you'll have to remember to use that period (but also critically wonder why you're using a legacy version of a programming language)

blocks|key|2409708|text|基本上sqrt(9)等于9%5E5|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2409709|>>>9**.5
3.0|code-block|syntax|javascript|2409710|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|$]]

Basically sqrt(9) is equivalent to 9^.5 

<pre><code>&gt;&gt;&gt;9**.5
3.0
</code></pre>

blocks|key|2409794|text|你应该这么做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2409795|16**(0.5)+#If+you+print+it,+you+get+4,+So+you+can+use+this+formula.|code-block|syntax|javascript|2409796|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|$]]

You should do 

<pre><code>16**(0.5) #If you print it, you get 4, So you can use this formula.
</code></pre>

blocks|key|1331732|text|def+nthrootofm(a,n):
++++return+pow(a,(1/n))
a=81
n=4
q=nthrootofm(a,n)
print(q)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1331733|函数接受两个参数。|unstyled|1331734|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>def nthrootofm(a,n):
 return pow(a,(1/n))
a=81
n=4
q=nthrootofm(a,n)
print(q)
</code></pre>

pow() function takes two parameters .

blocks|key|2409926|text|通过使用libnum库中包含的nroot，您可以在Python中获取数字的第n根：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2409927|from+libnum+import+nroot

nth_root+=+nroot(a,+n)|code-block|syntax|javascript|2409928|这相当于|2409929|📷|atomic|offset|length|2409930|​|2409931|尽管注意，它将返回被截断的nth根。|2409932|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://chart.googleapis.com/chart?cht=tx&chl=%255Csqrt%255Bn%255D%257Ba|imageAlt^0|0|0|0|0|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|13|8|@]|9|@$L|14|M|15|1|16]]|A|$]]|$1|N|3|O|5|6|7|17|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|18|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y|Z|-4]]]]

You can take the nth root of a number in Python by using nroot included in the libnum library:
<pre><code>from libnum import nroot

nth_root = nroot(a, n)
</code></pre>
This would be equivalent to <img src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%5Csqrt%5Bn%5D%7Ba" alt="\sqrt[n]{a}" />
Although note, that it will return the truncated nth root of a.

In maths, if I wish to calculate 3 to the power of 2 then no symbol is required, but I write the 2 small: <code>3²</code>. In Python this operation seems to be represented by the <code>**</code> syntax.
<pre><code>&gt;&gt;&gt; 3**2
9
</code></pre>
If I want to go the other direction and calculate the 2nd root of 9 then in maths I need to use a symbol: <code>2√9 = 3</code>
Is there a short-hand symbol in Python, similar to <code>**</code> that achieves this i.e. <code>2&lt;symbol&gt;9</code>? Or do I need to use the <code>math</code> module?

Is there a short-hand for nth root of x in Python?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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在数学中，如果我想计算3的幂为2，那么不需要符号，但我写了2小：3²。在Python中，这个操作似乎由**语法表示。>>> 3**29如果我想走相反的方向，计算9的第二个根，那么在数学中我需要使用一个符号：2√9 = 3在Python中是否有一个简短的符号，类似于实现这一点的**，即2<symbol>9？还是我需要使用math模块？

问在Python中，x的第n根有短手吗？
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回答 9

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问在Python中，x的第n根有短手吗？
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