blocks|key|415878|text|您必须编写：sqrt+=+x**(1/2.0)，否则将执行整数除法，并且表达式1/2将返回0。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|415879|这种行为在Python2.x中是“正常的”，而在Python3.x中1/2的计算结果是0.5。如果您希望Python+2.x代码的行为像3.x+w.r.t。部门写入from+__future__+import+division+-然后1/2将评估为0.5，为了向后兼容，1//2将评估为0。|415880|根据记录，计算平方根的首选方法是：|415881|import+math
math.sqrt(x)|code-block|syntax|javascript|415882|entityMap^0|6|H|13|3|19|1|0|Y|3|17|3|2A|V|39|3|3G|3|3R|4|3Z|1|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]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1C|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|1D|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|P|$]]

You have to write: <code>sqrt = x**(1/2.0)</code>, otherwise an integer division is performed and the expression <code>1/2</code> returns <code>0</code>.

This behavior is "normal" in Python 2.x, whereas in Python 3.x <code>1/2</code> evaluates to <code>0.5</code>. If you want your Python 2.x code to behave like 3.x w.r.t. division write <code>from __future__ import division</code> - then <code>1/2</code> will evaluate to <code>0.5</code> and for backwards compatibility, <code>1//2</code> will evaluate to <code>0</code>.

And for the record, the preferred way to calculate a square root is this:

<pre><code>import math
math.sqrt(x)
</code></pre>

blocks|key|631213|text|import+math
math.sqrt(+x+)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|631214|这是对答案链的一个微不足道的补充。然而，由于这个主题是非常常见的谷歌点击，这值得添加，我认为。|unstyled|631215|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>import math
math.sqrt( x )
</code></pre>

It is a trivial addition to the answer chain. However since the Subject is very common google hit, this deserves to be added, I believe.

blocks|key|415827|text|/在Python2中执行整数除法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|415828|>>>+1/2
0|code-block|syntax|javascript|415829|如果其中一个数字是浮点数，它将按预期工作：|415830|>>>+1.0/2
0.5
>>>+16**(1.0/2)
4.0|415831|entityMap^0|0|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

<code>/</code> performs an integer division in Python 2:

<pre><code>&gt;&gt;&gt; 1/2
0
</code></pre>

If one of the numbers is a float, it works as expected:

<pre><code>&gt;&gt;&gt; 1.0/2
0.5
&gt;&gt;&gt; 16**(1.0/2)
4.0
</code></pre>

blocks|key|631062|text|你看到的是整数除法。要在默认情况下获得浮点除法，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|631063|from+__future__+import+division|code-block|syntax|javascript|631064|或者，您可以将1/2的1或2转换为浮点值。|631065|sqrt+=+x**(1.0/2)|631066|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|$]]

What you're seeing is integer division. To get floating point division by default, 

<pre><code>from __future__ import division
</code></pre>

Or, you could convert 1 or 2 of 1/2 into a floating point value.

<pre><code>sqrt = x**(1.0/2)
</code></pre>

blocks|key|416027|text|这可能有点晚了，但计算平方根的最简单、最准确的方法是牛顿方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|416028|你有一个数字，你想要计算它的平方根(num)，并且你有一个关于它的平方根(estimate)的猜测。Estimate可以是大于0的任何数字，但有意义的数字会显著缩短递归调用深度。|offset|length|style|CODE|416029|new_estimate+=+(estimate+%2B+num+/+estimate)+/+2|code-block|syntax|javascript|416030|此行使用这两个参数计算更精确的估计值。您可以将new_estimate值传递给函数并计算另一个比前一个更精确的new_estimate，也可以像这样定义递归函数。|416031|def+newtons_method(num,+estimate):
++++#+Computing+a+new_estimate
++++new_estimate+=+(estimate+%2B+num+/+estimate)+/+2
++++print(new_estimate)
++++#+Base+Case:+Comparing+our+estimate+with+built-in+functions+value
++++if+new_estimate+==+math.sqrt(num):
++++++++return+True
++++else:
++++++++return+newtons_method(num,+new_estimate)|416032|例如，我们需要找到30的平方根。我们知道结果在5到6之间。|416033|newtons_method(30,5)|416034|数字为30，估计为5。每次递归调用的结果为：|416035|5.5
5.477272727272727
5.4772255752546215
5.477225575051661|416036|最后的结果是数的平方根的最精确的计算。它与内置函数math.sqrt()的值相同。|416037|entityMap^0|0|H|5|10|A|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@$D|14|E|15|F|G]|$D|16|E|17|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|18|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|1A|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|T|5|J|7|1C|8|@]|9|@]|A|$K|L]]|$1|U|3|V|5|6|7|1D|8|@]|9|@]|A|$]]|$1|W|3|X|5|J|7|1E|8|@]|9|@]|A|$K|L]]|$1|Y|3|Z|5|6|7|1F|8|@]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|11|$]]

This might be a little late to answer but most simple and accurate way to compute square root is newton's method.

You have a number which you want to compute its square root <code>(num)</code> and you have a guess of its square root <code>(estimate)</code>. Estimate can be any number bigger than 0, but a number that makes sense shortens the recursive call depth significantly.

<pre><code>new_estimate = (estimate + num / estimate) / 2
</code></pre>

This line computes a more accurate estimate with those 2 parameters. You can pass new_estimate value to the function and compute another new_estimate which is more accurate than the previous one or you can make a recursive function definition like this.

<pre><code>def newtons_method(num, estimate):
 # Computing a new_estimate
 new_estimate = (estimate + num / estimate) / 2
 print(new_estimate)
 # Base Case: Comparing our estimate with built-in functions value
 if new_estimate == math.sqrt(num):
 return True
 else:
 return newtons_method(num, new_estimate)
</code></pre>

For example we need to find 30's square root. We know that the result is between 5 and 6. 

<pre><code>newtons_method(30,5)
</code></pre>

number is 30 and estimate is 5. The result from each recursive calls are:

<pre><code>5.5
5.477272727272727
5.4772255752546215
5.477225575051661
</code></pre>

The last result is the most accurate computation of the square root of number. It is the same value as the built-in function math.sqrt().

blocks|key|631244|text|也许是一种简单的记忆方法:在分子(或分母)后面加一个点|type|unstyled|depth|inlineStyleRanges|entityRanges|data|631245|16+**+(1.+/+2)+++#+4
289+**+(1.+/+2)++#+17
27+**+(1.+/+3)+++#+3|code-block|syntax|javascript|631246|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|$]]

Perhaps a simple way to remember: add a dot after the numerator (or denominator)

<pre><code>16 ** (1. / 2) # 4
289 ** (1. / 2) # 17
27 ** (1. / 3) # 3
</code></pre>

blocks|key|416202|text|如果你想用计算器实际做的方式来做，那就使用巴比伦技巧。对here和here进行了说明。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|416203|假设您想要计算2的平方根：|416204|a=2

a1+=+(a/2)%2B1
b1+=+a/a1
aminus1+=+a1
bminus1+=+b1


while+(aminus1-bminus1+>+0):
++++an+=+0.5+*+(aminus1+%2B+bminus1)
++++bn+=+a+/+an
++++aminus1+=+an
++++bminus1+=+bn
++++print(an,bn,an-bn)|code-block|syntax|javascript|416205|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/werowe/HypatiaAcademy/blob/master/algorithms/squareRoot.ipynb|1|https://www.youtube.com/watch?v=6u6oOUiia2Q&feature=youtu.be^0|S|4|0|X|4|1|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]|$A|Y|B|Z|1|10]]|C|$]]|$1|D|3|E|5|6|7|11|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|12|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

If you want to do it the way the calculator actually does it, use the Babylonian technique. It is explained <a href="https://github.com/werowe/HypatiaAcademy/blob/master/algorithms/squareRoot.ipynb" rel="nofollow noreferrer">here</a> and <a href="https://www.youtube.com/watch?v=6u6oOUiia2Q&amp;feature=youtu.be" rel="nofollow noreferrer">here</a>.
Suppose you want to calculate the square root of 2:
<pre><code>a=2

a1 = (a/2)+1
b1 = a/a1
aminus1 = a1
bminus1 = b1


while (aminus1-bminus1 &gt; 0):
 an = 0.5 * (aminus1 + bminus1)
 bn = a / an
 aminus1 = an
 bminus1 = bn
 print(an,bn,an-bn)
</code></pre>

blocks|key|631099|text|您可以使用NumPy来计算数组的平方根：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|631100|+import+numpy+as+np
+np.sqrt([1,+4,+9])|code-block|syntax|javascript|631101|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 can use NumPy to calculate square roots of arrays:

<pre><code> import numpy as np
 np.sqrt([1, 4, 9])
</code></pre>

blocks|key|631128|text|我希望下面提到的代码能回答你的问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|631129|def+root(x,a):
++++y+=+1+/+a
++++y+=+float(y)
++++print+y
++++z+=+x+**+y
++++print+z

base+=+input("Please+input+the+base+value:")
power+=+float(input("Please+input+the+root+value:"))


root(base,power)+|code-block|syntax|javascript|631130|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|$]]

I hope the below mentioned code will answer your question.

<pre><code>def root(x,a):
 y = 1 / a
 y = float(y)
 print y
 z = x ** y
 print z

base = input("Please input the base value:")
power = float(input("Please input the root value:"))


root(base,power) 
</code></pre>

Why does Python give the &quot;wrong&quot; answer?
<pre><code>x = 16

sqrt = x**(.5) #returns 4
sqrt = x**(1/2) #returns 1
</code></pre>
Yes, I know <code>import math</code> and use <code>sqrt</code>. But I'm looking for an answer to the above.

How do I calculate square root in Python?

Python

为什么Python会给出“错误”的答案？x = 16sqrt = x**(.5)  #returns 4sqrt = x**(1/2) #returns 1是的，我知道import math并使用sqrt。但我正在寻找上述问题的答案。

问如何在Python中计算平方根？
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