In python 2.7+ there is a int.bit_length() method:

>>> a = 100
>>> a.bit_length()
7

>>> len(bin(1000))-2
10
>>> len(bin(100))-2
7
>>> len(bin(10))-2
4

Note: will not work for negative numbers, may be need to substract 3 instead of 2

If your Python version has it (≥2.7 for Python 2, ≥3.1 for Python 3), use the bit_length method from the standard library.

Otherwise, len(bin(n))-2 as suggested by YOU is fast (because it s implemented in Python). Note that this returns 1 for 0.

Otherwise, a simple method is to repeatedly divide by 2 (which is a straightforward bit shift), and count how long it takes to reach 0.

def bit_length(n): # return the bit size of a non-negative integer
    bits = 0
    while n >> bits: bits += 1
    return bits

It is significantly faster (at least for large numbers — a quick benchmarks says more than 10 times faster for 1000 digits) to shift by whole words at a time, then go back and work on the bits of the last word.

def bit_length(n): # return the bit size of a non-negative integer
    if n == 0: return 0
    bits = -32
    m = 0
    while n:
        m = n
        n >>= 32; bits += 32
    while m: m >>= 1; bits += 1
    return bits

In my quick benchmark, len(bin(n)) came out significantly faster than even the word-sized chunk version. Although bin(n) builds a string that s discarded immediately, it comes out on top due to having an inner loop that s compiled to machine code. (math.log is even faster, but that s not important since it s wrong.)

Here we can also use slicing.

For positive integers, we d start from 2:

len(bin(1)[2:])
len(bin(5)[2:])
len(bin(10)[2:])
len(bin(100)[2:])
len(bin(1000)[2:])

which would print:

1
3
4
7
10

For negative integers, we d start from 3:

len(bin(-1)[3:])
len(bin(-5)[3:])
len(bin(-10)[3:])
len(bin(-100)[3:])
len(bin(-1000)[3:])

which would print:

1
3
4
7
10

def bitcounter(n):
    return math.floor(math.log(n,2)) + 1

EDIT fixed so that it works with 1

Here is another way:

def number_of_bits(n):
    return len( {:b} .format(n))

Not so efficient I suppose, but doesn t show up in any of the previous answers...

This solution takes advantage of .bit_length() if available, and falls back to len(hex(a)) for older versions of Python. It has the advantage over bin that it creates a smaller temporary string, so it uses less memory.

Please note that it returns 1 for 0, but that s easy to change.

_HEX_BIT_COUNT_MAP = {
     0 : 0,  1 : 1,  2 : 2,  3 : 2,  4 : 3,  5 : 3,  6 : 3,  7 : 3}

def bit_count(a):
  """Returns the number of bits needed to represent abs(a). Returns 1 for 0."""
  if not isinstance(a, (int, long)):
    raise TypeError
  if not a:
    return 1
  # Example: hex(-0xabc) ==  -0xabc .  L  is appended for longs.
  s = hex(a)
  d = len(s)
  if s[-1] ==  L :
    d -= 1
  if s[0] ==  - :
    d -= 4
    c = s[3]
  else:
    d -= 3
    c = s[2]
  return _HEX_BIT_COUNT_MAP.get(c, 4) + (d << 2)


# Use int.bit_length and long.bit_length introduced in Python 2.7 and 3.x.
if getattr(0,  bit_length , None):
  __doc = bit_count.__doc__
  def bit_count(a):
    return a.bit_length() or 1
  bit_count.__doc__ = __doc

assert bit_count(0) == 1
assert bit_count(1) == 1
assert bit_count(2) == 2
assert bit_count(3) == 2
assert bit_count(63) == 6
assert bit_count(64) == 7
assert bit_count(75) == 7
assert bit_count(2047) == 11
assert bit_count(2048) == 12
assert bit_count(-4007) == 12
assert bit_count(4095) == 12
assert bit_count(4096) == 13
assert bit_count(1 << 1203) == 1204
assert bit_count(-(1 << 1203)) == 1204
assert bit_count(1 << 1204) == 1205
assert bit_count(1 << 1205) == 1206
assert bit_count(1 << 1206) == 1207