>>> def log2( x ):
... return math.log( x ) / math.log( 2 )
...
>>> log2( 2 )
1.0
>>> log2( 4 )
2.0
>>> log2( 8 )
3.0
>>> log2( 2.4 )
1.2630344058337937
>>>
Try this ,
import math
print(math.log(8,2)) # math.log(number,base)
Don't forget that log[base A] x = log[base B] x / log[base B] A.
So if you only have log
(for natural log) and log10
(for base-10 log), you can use
myLog2Answer = log10(myInput) / log10(2)
math.log2(x)
import math
log2 = math.log(x, 2.0)
log2 = math.log2(x) # python 3.3 or later
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.
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.
http://en.wikipedia.org/wiki/Binary_logarithm
def lg(x, tol=1e-13):
res = 0.0
# Integer part
while x<1:
res -= 1
x *= 2
while x>=2:
res += 1
x /= 2
# Fractional part
fp = 1.0
while fp>=tol:
fp /= 2
x *= x
if x >= 2:
x /= 2
res += fp
return res
Using numpy:
In [1]: import numpy as np
In [2]: np.log2?
Type: function
Base Class: <type 'function'>
String Form: <function log2 at 0x03049030>
Namespace: Interactive
File: c:\python26\lib\site-packages\numpy\lib\ufunclike.py
Definition: np.log2(x, y=None)
Docstring:
Return the base 2 logarithm of the input array, element-wise.
Parameters
----------
x : array_like
Input array.
y : array_like
Optional output array with the same shape as `x`.
Returns
-------
y : ndarray
The logarithm to the base 2 of `x` element-wise.
NaNs are returned where `x` is negative.
See Also
--------
log, log1p, log10
Examples
--------
>>> np.log2([-1, 2, 4])
array([ NaN, 1., 2.])
In [3]: np.log2(8)
Out[3]: 3.0
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)
In python 3 or above, math class has the following 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.
Source: Stackoverflow.com