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])
```

It's 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: <built-in 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
```

`math.log2(x)`

```
import math
log2 = math.log(x, 2.0)
log2 = math.log2(x) # python 3.4 or later
```

- Thanks @akashchandrakar and @unutbu.

`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 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 is 4 not 5. log₂(1/17) is -5 not -4.

Licensed under: CC-BY-SA with attribution

Not affiliated with: Stack Overflow