Is there a short-hand for nth root of x in Python?
Question:
In maths, if I wish to calculate 3 to the power of 2 then no symbol is required, but I write the 2 small: 3². In Python this operation seems to be represented by the ** syntax.
>>> 3**2
9
If I want to go the other direction and calculate the 2nd root of 9 then in maths I need to use a symbol: 2√9 = 3
Is there a short-hand symbol in Python, similar to ** that achieves this i.e. 2<symbol>9? Or do I need to use the math module?
Answers:
nth root of x is x^(1/n), so you can do 9**(1/2) to find the 2nd root of 9, for example. In general, you can compute the nth root of x as:
x**(1/n)
Note: In Python 2, you had to do 1/float(n) or 1.0/n so that the result would be a float rather than an int. For more details, see Why does Python give the "wrong" answer for square root?
Basically sqrt(9) is equivalent to 9^.5
>>>9**.5
3.0
There is. It’s just ** =)
Any nth root is an exponentiation by 1/n, so to get the square root of 9, you use 9**(1/2) (or 9**0.5) to get the cube root, you use 9 ** (1/3) (which we can’t write with a simpler fraction), and to get the nth root, 9 ** (1/n).
Also note that as of Python 3, adding periods to integers to make them a float is no longer necessary. Saying 1/3 works the way you would actually expect it to, giving 0.333... 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)
Also: x**(n**-1), which is the same but shorter than x**(1/float(n))
You should do
16**(0.5) #If you print it, you get 4, So you can use this formula.
If you prefer to apply this operation functionally rather than with an infix operator (the ** symbol), you can pass the base and exponent as arguments to the pow function:
In [23]: (9**(0.5)) == pow(9, 0.5)
Out[23]: True
I am also fond of finding new uses for this Infix hack in Python 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:
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
def nthrootofm(a,n):
return pow(a,(1/n))
a=81
n=4
q=nthrootofm(a,n)
print(q)
pow() function takes two parameters .
You may also use some logarithms:
Nth root of x:
exp(log(x)/n)
For example:
>>> from math import exp, log
>>> x = 8
>>> n = 3
>>> exp(log(x)/n)
2.0
You can take the nth root of a number in Python by using nroot included in the libnum library:
from libnum import nroot
nth_root = nroot(a, n)
This would be equivalent to ![sqrt[n]{a}](https://chart.googleapis.com/chart?cht=tx&chl=%5Csqrt%5Bn%5D%7Ba)
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: 3². In Python this operation seems to be represented by the ** syntax.
>>> 3**2
9
If I want to go the other direction and calculate the 2nd root of 9 then in maths I need to use a symbol: 2√9 = 3
Is there a short-hand symbol in Python, similar to ** that achieves this i.e. 2<symbol>9? Or do I need to use the math module?
nth root of x is x^(1/n), so you can do 9**(1/2) to find the 2nd root of 9, for example. In general, you can compute the nth root of x as:
x**(1/n)
Note: In Python 2, you had to do 1/float(n) or 1.0/n so that the result would be a float rather than an int. For more details, see Why does Python give the "wrong" answer for square root?
Basically sqrt(9) is equivalent to 9^.5
>>>9**.5
3.0
There is. It’s just ** =)
Any nth root is an exponentiation by 1/n, so to get the square root of 9, you use 9**(1/2) (or 9**0.5) to get the cube root, you use 9 ** (1/3) (which we can’t write with a simpler fraction), and to get the nth root, 9 ** (1/n).
Also note that as of Python 3, adding periods to integers to make them a float is no longer necessary. Saying 1/3 works the way you would actually expect it to, giving 0.333... 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)
Also: x**(n**-1), which is the same but shorter than x**(1/float(n))
You should do
16**(0.5) #If you print it, you get 4, So you can use this formula.
If you prefer to apply this operation functionally rather than with an infix operator (the ** symbol), you can pass the base and exponent as arguments to the pow function:
In [23]: (9**(0.5)) == pow(9, 0.5)
Out[23]: True
I am also fond of finding new uses for this Infix hack in Python 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:
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
def nthrootofm(a,n):
return pow(a,(1/n))
a=81
n=4
q=nthrootofm(a,n)
print(q)
pow() function takes two parameters .
You may also use some logarithms:
Nth root of x:
exp(log(x)/n)
For example:
>>> from math import exp, log
>>> x = 8
>>> n = 3
>>> exp(log(x)/n)
2.0
You can take the nth root of a number in Python by using nroot included in the libnum library:
from libnum import nroot
nth_root = nroot(a, n)
This would be equivalent to
Although note, that it will return the truncated nth root of a.