How do I calculate square root in Python?
Pregunta
Why does Python give the "wrong" answer?
x = 16
sqrt = x**(.5)
returns 4
sqrt = x**(1/2)
returns 1
Yes, I know import math
and use sqrt
. But I'm looking for an answer to the above.
Solución
sqrt=x**(1/2)
is doing integer division. 1/2 == 0
.
So you're computing x(1/2) in the first instance, x(0) in the second.
So it's not wrong, it's the right answer to a different question.
Otros consejos
You have to write: sqrt = x**(1/2.0)
, otherwise an integer division is performed and the expression 1/2
returns 0
.
This behavior is "normal" in Python 2.x, whereas in Python 3.x 1/2
evaluates to 0.5
. If you want your Python 2.x code to behave like 3.x w.r.t. division write from __future__ import division
- then 1/2
will evaluate to 0.5
and for backwards compatibility, 1//2
will evaluate to 0
.
And for the record, the preferred way to calculate a square root is this:
import math
math.sqrt(x)
import math
math.sqrt( x )
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.
/
performs an integer division in Python 2:
>>> 1/2
0
If one of the numbers is a float, it works as expected:
>>> 1.0/2
0.5
>>> 16**(1.0/2)
4.0
What you're seeing is integer division. To get floating point division by default,
from __future__ import division
Or, you could convert 1 or 2 of 1/2 into a floating point value.
sqrt = x**(1.0/2)
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 (num)
and you have a guess of its square root (estimate)
. Estimate can be any number bigger than 0, but a number that makes sense shortens the recursive call depth significantly.
new_estimate = (estimate + num / estimate) / 2
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.
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)
For example we need to find 30's square root. We know that the result is between 5 and 6.
newtons_method(30,5)
number is 30 and estimate is 5. The result from each recursive calls are:
5.5
5.477272727272727
5.4772255752546215
5.477225575051661
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().
Perhaps a simple way to remember: add a dot after the numerator (or denominator)
16 ** (1. / 2) # 4
289 ** (1. / 2) # 17
27 ** (1. / 3) # 3
You can use NumPy to calculate square roots of arrays:
import numpy as np
np.sqrt([1, 4, 9])
I hope the below mentioned code will answer your question.
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)