logarithm – Log to the base 2 in python
logarithm – Log to the base 2 in python
Its good to know that
but also know that
math.log
takes an optional second argument which allows you to specify the base:
In [22]: import math
In [23]: math.log?
Type: builtin_function_or_method
Base Class: <type builtin_function_or_method>
String Form: <builtin function log>
Namespace: Interactive
Docstring:
log(x[, base]) > the logarithm of x to the given base.
If the base not specified, returns the natural logarithm (base e) of x.
In [25]: math.log(8,2)
Out[25]: 3.0
Depends on whether the input or output is int
or float
.
assert 5.392317422778761 == math.log2(42.0)
assert 5.392317422778761 == math.log(42.0, 2.0)
assert 5 == math.frexp(42.0)[1]  1
assert 5 == (42).bit_length()  1
float → float math.log2(x)
import math
log2 = math.log(x, 2.0)
log2 = math.log2(x) # python 3.3 or later
 Thanks @akashchandrakar and @unutbu.
float → int math.frexp(x)
If all you need is the integer part of log base 2 of a floating point number, extracting the exponent is pretty efficient:
log2int_slow = int(math.floor(math.log(x, 2.0))) # these give the
log2int_fast = math.frexp(x)[1]  1 # same result

Python frexp() calls the C function frexp() which just grabs and tweaks the exponent.

Python frexp() returns a tuple (mantissa, exponent). So
[1]
gets the exponent part. 
For integral powers of 2 the exponent is one more than you might expect. For example 32 is stored as 0.5×2⁶. This explains the
 1
above. Also works for 1/32 which is stored as 0.5×2⁻⁴. 
Floors toward negative infinity, so log₂31 computed this way is 4 not 5. log₂(1/17) is 5 not 4.
int → int x.bit_length()
If both input and output are integers, this native integer method could be very efficient:
log2int_faster = x.bit_length()  1

 1
because 2ⁿ requires n+1 bits. Works for very large integers, e.g.2**10000
. 
Floors toward negative infinity, so log₂31 computed this way is 4 not 5.
logarithm – Log to the base 2 in python
If you are on python 3.3 or above then it already has a builtin function for computing log2(x)
import math
finds log base2 of x
answer = math.log2(x)
If you are on older version of python then you can do like this
import math
finds log base2 of x
answer = math.log(x)/math.log(2)