Python - Flipping Binary 1's and 0's in a String

Question

I'm trying to take a binary number in string form and flip the 1's and 0's, that is, change all of the 1's in the string to 0's, and all of the 0's to 1's. I'm new to Python and have been racking my brain for several hours now trying to figure it out.

Answer 1

Amber's answer, while superior, possibly isn't the most clear, so here's a super basic iterative example:

b_string = "1100101"
ib_string = ""

for bit in b_string:
  if bit == "1":
    ib_string += "0"
  else:
    ib_string += "1"

print ib_string

This can be done in much better ways...replacements, comprehensions, but this is an example.

I would learn from the other answers in this question once you understand the basis of this one. This method is slow and painful. For the best performance, as Muhammad Alkarouri pointed out, the string.translate/maketrans combo is the way to go. Right behind it is the comprehension. My code is the slowest by a significant margin.

Answer 2

>>> ''.join('1' if x == '0' else '0' for x in '1000110')
'0111001'

The a for b in c pattern is a generator expression, which produces a series of items based on a different series. In this case, the original series is the characters (since you can iterate over strings in Python, which gives you the characters that make up that string), and the new series is a set of characters with the 0's and 1's flipped.

'1' if x == '0' else '0' is pretty straightforward - it gives us whichever of 1 or 0 isn't x. We do this for each such x in the original set of characters, and then join() them all together (with an empty string '', a.k.a. nothing, in between each item), thus giving us a final string which is all of the opposite characters from the original, combined.

Answer 3

Another way to do it is with string.translate() and string.maketrans()

from string import maketrans
bitString = "10101010100011010"
flippedString = bitString.translate(maketrans("10","01"))

Answer 4

If speed is important:

And you already have the decimal integer representing the binary string, then bit manipulation is slightly faster.

bin((i ^ (2 ** (i.bit_length()+1) - 1)))[3:]

If you are only given the binary string, then use the str.replace method given by @Amy:

s.replace('1', '2').replace('0', '1').replace('2', '0')

I tested the various methods proposed here, and the bit manipulation method, with this gist:

Test Results

i = 129831201;
s = '111101111010001000100100001';

Bit manipulation given decimal int:

bin((i ^ (2 ** (i.bit_length()+1) - 1)))[3:]

1000000 loops, best of 3: 0.647 usec per loop

Bit manipulation given binary string:

bin((int(s, 2) ^ (2**(len(s)+1) - 1)))[3:]

1000000 loops, best of 3: 0.922 usec per loop

Sequential str.replace:

s.replace('1', '2').replace('0', '1').replace('2', '0')

1000000 loops, best of 3: 0.619 usec per loop

str.maketrans:

s.translate(str.maketrans('10', '01'))

1000000 loops, best of 3: 1.16 usec per loop

''.join with dictionary mapping:

flip = {'1':'0', '0':'1'}; ''.join(flip[b] for b in s)

100000 loops, best of 3: 2.78 usec per loop

''.join with conditional:

''.join('1' if b == '0' else '0' for b in s)

100000 loops, best of 3: 2.82 usec per loop

Answer 5

Using a dictionary should be very straightforward.

>>> flip={"1":"0","0":"1"}
>>> s="100011"
>>> import sys
>>> for i in s:
...   sys.stdout.write(flip[i])
...
011100

Answer 6

You've already marked an answer to this, but I haven't seen the method that I would prefer. I'm assuming here that you have a binary number of known length - 8 bits in the examples I am giving.

If you can start with the number as just a number, you can do the following:

myNumber = 0b10010011
myNumberInverted = myNumber ^ 0b11111111

The ^ operator performs a bitwise XOR.

If you really do have to start with a string, you can convert to an integer first and then perform this operation.

myString = '10010011'
myNumber = int(myString, 2)
myNumberInverted = myNumber ^ 0b11111111

Answer 7

http://docs.python.org/library/string.html#string.replace

Replace all the 1's with 2's, then replace the 0's with 1's, finally replacing the 2's with 0's.

"10011".replace("1", "2").replace("0", "1").replace("2", "0")

Answer 8

Along the lines of Amber, but using ASCII arithmetic (for no particular reason). This obviously isn't meant for production code.

''.join(chr(97 - ord(c)) for c in my_bit_string)

48 and 49 are the ASCII (and Unicode) values for '0' and '1' respectively. ord gives you the numeric value for a character, while chr does the reverse.

Answer 9

Some things have changed since this question was answered. Hopefully my update proves useful to someone.

The translate method is the fastest for me. Full test code follows.

Thank you to those who contributed previously to allow me to test this.

# System & Software
# core i7 4790k @ ~ 4.6 GHz   32 GB RAM   Samsung Evo NVME
# Visual Studio 2019 16.3.6

# I do not understand Will's bit manipulation code
# worst times shown in comments
# ordered by speed on my system

import timeit # required to use timeit
import string # Required to call maketrans function.

# https://www.afternerd.com/blog/timeit-multiple-lines/
# if you are trying to time slightly longer pieces of code than a single line
# timeit wants a string


# the syntax  b_string.translate(maketrans("10", "01"))
# presented by Muhammad Alkarouri does not appear to be valid anymore
# i suspect this is due to changes within python versions
a1_bit_flip = """\
a1 = 0b1101010001101111 # accepts the binary input but will store it as an int
a1_bin = bin(a1)[2:]
# creates a string of the binary form of the integer minus the first 2 characters
flip_bin_a1 = a1_bin.translate(str.maketrans("10","01"))"""

trans_time = timeit.timeit(a1_bit_flip, number=100000) # time 100k iterations
print('translate time')
print(trans_time / 100000)
# determine average time of a single iteration ~ 0.6282 us
print('\n')



a2_bit_flip = """\
a2 = 0b1101010001101111
a2_bin = bin(a2)[2:]
a2_bin.replace('1', '2').replace('0', '1').replace('2', '0')"""

replace_time = timeit.timeit(a2_bit_flip, number=100000) # time 100k iterations
print('replace time')
print(replace_time / 100000)
# determine average time of a single iteration ~ 0.7557 us
print('\n')



a3_bit_flip = """\
a3 = 0b1101010001101111
a3_bin = bin(a3)[2:]
bin((int(a3_bin, 2) ^ (2**(len(a3_bin)+1) - 1)))[3:]"""
# I do not understand this line (Will)

bin_bit_time = timeit.timeit(a3_bit_flip, number=100000)
# time 100k iterations
print('bin_bit time')
print(bin_bit_time / 100000)
# determine average time of a single iteration ~ 1.14 us
print('\n')



a4_bit_flip = """\
a4 = 0b1101010001101111
a4_bin = bin(a4)[2:]
bin((i ^ (2 ** (i.bit_length()+1) - 1)))[3:]"""
# I do not understand this line (Will)

int_bit_time = timeit.timeit(a3_bit_flip, number=100000) # time 100k iterations
print('int_bit time')
print(int_bit_time / 100000)
# determine average time of a single iteration ~ 1.14 us
print('\n')



join_bit_flip = """\
a0 = 0b1101010001101111 # accepts the binary input but will store it as an int
a0_bin = bin(a0)[2:]
# creates a string of the binary form of the integer minus the first 2 characters
flip_a0 = "".join('1' if x == '0' else '0' for x in a0_bin)""" # (Amber)

join_time = timeit.timeit(join_bit_flip, number=100000) # time 100k iterations
print('join time')
print(join_time / 100000)
# determine average time of a single iteration ~ 14.511 us
print('\n')