Reversing bits of Python integer

Question

Given a decimal integer (eg. 65), how does one reverse the underlying bits in Python? ie. the following operation:

65 → 01000001 → 10000010 → 130

It seems that this task can be broken down into three steps:

1. Convert the decimal integer to binary representation
2. Reverse the bits
3. Convert back to decimal

Steps #2 and 3 seem pretty straightforward (see this and this SO question related to step #2), but I'm stuck on step #1. The issue with step #1 is retrieving the full decimal representation with filling zeros (ie. 65 = 01000001, not 1000001).

I've searched around, but I can't seem to find anything.

Answer 1

int('{:08b}'.format(n)[::-1], 2)

You can specify any filling length in place of the 8. If you want to get really fancy,

b = '{:0{width}b}'.format(n, width=width)
int(b[::-1], 2)

lets you specify the width programmatically.

Answer 2

If you are after more speed, you can use the technique described in http://leetcode.com/2011/08/reverse-bits.html

def reverse_mask(x):
    x = ((x & 0x55555555) << 1) | ((x & 0xAAAAAAAA) >> 1)
    x = ((x & 0x33333333) << 2) | ((x & 0xCCCCCCCC) >> 2)
    x = ((x & 0x0F0F0F0F) << 4) | ((x & 0xF0F0F0F0) >> 4)
    x = ((x & 0x00FF00FF) << 8) | ((x & 0xFF00FF00) >> 8)
    x = ((x & 0x0000FFFF) << 16) | ((x & 0xFFFF0000) >> 16)
    return x

Answer 3

best way to do is perform bit by bit shifting

def reverse_Bits(n, no_of_bits):
    result = 0
    for i in range(no_of_bits):
        result <<= 1
        result |= n & 1
        n >>= 1
    return result
# for example we reverse 12 i.e 1100 which is 4 bits long
print(reverse_Bits(12,4))

Answer 4

def reverse_bit(num):
    result = 0
    while num:
        result = (result << 1) + (num & 1)
        num >>= 1
    return result

We don't really need to convert the integer into binary, since integers are actually binary in Python.

The reversing idea is like doing the in-space reversing of integers.

def reverse_int(x):
    result = 0
    pos_x = abs(x)
    while pos_x:
        result = result * 10 + pos_x % 10
        pos_x /= 10
    return result if x >= 0 else (-1) * result

For each loop, the original number is dropping the right-most bit(in binary). We get that right-most bit and multiply 2 ( <<1 ) in the next loop when the new bit is added.

Answer 5

There's no need, and no way, to "convert a decimal integer to binary representation". All Python integers are represented as binary; they're just converted to decimal when you print them for convenience.

If you want to follow this solution to the reversal problem, you only need to find appropriate numbits . You can either specify this by hand, or compute the number of bits needed to represent an integer n with n.bit_length() (new in Python 2.7 and 3.1).

However, for 65, that would give you 7, as there's no reason why 65 should require any more bits. (You might want to round up to the nearest multiple of 8...)

Answer 6

You can test the i'th bit of a number by using a shift and mask. For example, bit 6 of 65 is (65 >> 6) & 1 . You can set a bit in a similar way by shifting 1 left the right number of times. These insights gives you code like this (which reverses x in a field of 'n' bits).

def reverse(x, n):
    result = 0
    for i in xrange(n):
        if (x >> i) & 1: result |= 1 << (n - 1 - i)
    return result

print bin(reverse(65, 8))

Answer 7

An inefficient but concise method that works in both Python 2.7 and Python 3:

def bit_reverse(i, n):
    return int(format(i, '0%db' % n)[::-1], 2)

For your example:

>>> bit_reverse(65, 8)
130

Answer 8

Regularly there is the need to apply this operation on array of numbers and not for single number. To increase speed, it's probably better to use NumPy array. There are two solutions.

x1.34 faster than second solution:

import numpy as np
def reverse_bits_faster(x):
  x = np.array(x)
  bits_num = x.dtype.itemsize * 8
  # because bitwise operations may change number of bits in numbers
  one_array = np.array([1], x.dtype)
  # switch bits in-place
  for i in range(int(bits_num / 2)):
    right_bit_mask = (one_array << i)[0]
    left_bit = (x & right_bit_mask) << (bits_num - 1 - i * 2)
    left_bit_mask = (one_array << (bits_num - 1 - i))[0]
    right_bit = (x & left_bit_mask) >> (bits_num - 1 - i * 2)
    moved_bits_mask = left_bit_mask | right_bit_mask
    x = x & (~moved_bits_mask) | left_bit | right_bit
  return x

Slower, but more easy to understand (based on solution proposed by Sudip Ghimire ):

import numpy as np
def reverse_bits(x):
  x = np.array(x)
  bits_num = x.dtype.itemsize * 8
  x_reversed = np.zeros_like(x)
  for i in range(bits_num):
    x_reversed = (x_reversed << 1) | x & 1
    x >>= 1
  return x_reversed

Answer 9

You could also use a Look up table (that can be generated once using methods above):

LUT = [0, 128, 64, 192, 32, 160, 96, 224, 16, 144, 80, 208, 48, 176, 112, 240,
       8, 136, 72, 200, 40, 168, 104, 232, 24, 152, 88, 216, 56, 184, 120,
       248, 4, 132, 68, 196, 36, 164, 100, 228, 20, 148, 84, 212, 52, 180,
       116, 244, 12, 140, 76, 204, 44, 172, 108, 236, 28, 156, 92, 220, 60,
       188, 124, 252, 2, 130, 66, 194, 34, 162, 98, 226, 18, 146, 82, 210, 50,
       178, 114, 242, 10, 138, 74, 202, 42, 170, 106, 234, 26, 154, 90, 218,
       58, 186, 122, 250, 6, 134, 70, 198, 38, 166, 102, 230, 22, 150, 86, 214,
       54, 182, 118, 246, 14, 142, 78, 206, 46, 174, 110, 238, 30, 158, 94,
       222, 62, 190, 126, 254, 1, 129, 65, 193, 33, 161, 97, 225, 17, 145, 81,
       209, 49, 177, 113, 241, 9, 137, 73, 201, 41, 169, 105, 233, 25, 153, 89,
       217, 57, 185, 121, 249, 5, 133, 69, 197, 37, 165, 101, 229, 21, 149, 85,
       213, 53, 181, 117, 245, 13, 141, 77, 205, 45, 173, 109, 237, 29, 157,
       93, 221, 61, 189, 125, 253, 3, 131, 67, 195, 35, 163, 99, 227, 19, 147,
       83, 211, 51, 179, 115, 243, 11, 139, 75, 203, 43, 171, 107, 235, 27,
       155, 91, 219, 59, 187, 123, 251, 7, 135, 71, 199, 39, 167, 103, 231, 23,
       151, 87, 215, 55, 183, 119, 247, 15, 143, 79, 207, 47, 175, 111, 239,
       31, 159, 95, 223, 63, 191, 127, 255]

def reverseBitOrder(uint8):
    return LUT[uint8]

Answer 10

One more way to do it is to loop through the bits from both end and swap each other. This i learned from EPI python book.

i = 0; j = 7
num = 230
print(bin(num))
while i<j:
    # Get the bits from both end iteratively
    if (x>>i)&1 != (x>>j)&1:
        # if the bits don't match swap them by creating a bit mask
        # and XOR it with the number 
        mask = (1<<i) | (1<<j)
        num ^= mask
    i += 1; j -= 1
print(bin(num))

Answer 11

All what you need is numpy

import numpy as np
x = np.uint8(65)
print( np.packbits(np.unpackbits(x, bitorder='little')) )

performance:

py -3 -m timeit "import numpy as np; import timeit; x=np.uint8(65); timeit.timeit(lambda: np.packbits(np.unpackbits(x, bitorder='little')), number=100000)"
1 loop, best of 5: 326 msec per loop

Answer 12

kimstik: You should move the import out of the statement for better performance record.

>>> python -m timeit --setup="import numpy as np; x = np.uint8(65);" --repeat=10 --number=1000 "np.packbits(np.unpackbits(x, bitorder='little'))"
1000 loops, best of 10: 2.37 usec per loop 

Answer 13

bin(x)[:1:-1]

one line, and it automatically goes for the top bit.

>>> x = 0b1011000
>>> bin(x)[:1:-1]
'0001101'
>>> x = 0b100
>>> bin(x)[:1:-1]
'001'

the "0b" on the front of the text-conversion is stripped by the "1" in [:1:-1] which, after the inversion (by -1) has 1 automatically added to it (sigh, range is really weird) before being used as the start point not the end.

you'll need zero-padding on the front to get it a fixed-width reversing but even there [:1:-1] will still do the auto-length-detection