Log to the base 2 in python

Question

How should I compute log to the base two in python. Eg. I have this equation where I am using log base 2

import math
e = -(t/T)* math.log((t/T)[, 2])
         


        

Answer 1

In python 3 or above, math class has the fallowing functions

import math

math.log2(x)
math.log10(x)
math.log1p(x)

or you can generally use math.log(x, base) for any base you want.

Answer 2

log_base_2(x) = log(x) / log(2)

Answer 3

Try this ,

import math
print(math.log(8,2))  # math.log(number,base) 

Answer 4

If you are on python 3.3 or above then it already has a built-in 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)

Answer 5

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.

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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)))
log2int_fast = math.frexp(x)[1] - 1

• 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 0.5x2⁶. This explains the - 1 above. Also works for 1/32 which is stored as 0.5x2⁻⁴.

• Floors toward negative infinity, so log₂31 computed this way is 4 not 5. log₂(1/17) is -5 not -4.

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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.