Convert int to binary python

In order to convert an integer to a binary, I have used this code :

>>> bin[6]  
'0b110'

and when to erase the '0b', I use this :

>>> bin[6][2:]  
'110'

What can I do if I want to show 6 as 00000110 instead of 110?

just-Luka

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asked May 2, 2012 at 9:31

1

>>> '{0:08b}'.format[6]
'00000110'

Just to explain the parts of the formatting string:

• {} places a variable into a string
• 0 takes the variable at argument position 0
• : adds formatting options for this variable [otherwise it would represent decimal 6]
• 08 formats the number to eight digits zero-padded on the left
• b converts the number to its binary representation

If you're using a version of Python 3.6 or above, you can also use f-strings:

>>> f'{6:08b}'
'00000110'

TrebledJ

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answered May 2, 2012 at 9:32

eumiroeumiro

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5

Just another idea:

>>> bin[6][2:].zfill[8]
'00000110'

Shorter way via string interpolation [Python 3.6+]:

>>> f'{6:08b}'
'00000110'

answered May 2, 2012 at 9:37

mshsayemmshsayem

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3

A bit twiddling method...

>>> bin8 = lambda x : ''.join[reversed[ [str[[x >> i] & 1] for i in range[8]] ] ]
>>> bin8[6]
'00000110'
>>> bin8[-3]
'11111101'

marbel82

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answered May 2, 2012 at 10:07

sobelsobel

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3

Just use the format function

format[6, "08b"]

The general form is

format[, ""]

answered Apr 7, 2016 at 20:16

theOnetheOne

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1

eumiro's answer is better, however I'm just posting this for variety:

>>> "%08d" % int[bin[6][2:]]
00000110

answered May 2, 2012 at 9:35

jedwardsjedwards

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0

numpy.binary_repr[num, width=None] has a magic width argument

Relevant examples from the documentation linked above:

>>> np.binary_repr[3, width=4]
'0011'

The two’s complement is returned when the input number is negative and width is specified:

>>> np.binary_repr[-3, width=5]
'11101'

answered May 1, 2019 at 2:11

Tom HaleTom Hale

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.. or if you're not sure it should always be 8 digits, you can pass it as a parameter:

>>> '%0*d' % [8, int[bin[6][2:]]]
'00000110'

answered May 2, 2012 at 9:47

thebjornthebjorn

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Going Old School always works

def intoBinary[number]:
binarynumber=""
if [number!=0]:
    while [number>=1]:
        if [number %2==0]:
            binarynumber=binarynumber+"0"
            number=number/2
        else:
            binarynumber=binarynumber+"1"
            number=[number-1]/2

else:
    binarynumber="0"

return "".join[reversed[binarynumber]]

answered Jun 8, 2018 at 16:40

2

The best way is to specify the format.

format[a, 'b']

returns the binary value of a in string format.

To convert a binary string back to integer, use int[] function.

int['110', 2]

returns integer value of binary string.

answered Apr 27, 2020 at 3:18

PranjalyaPranjalya

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1

Assuming you want to parse the number of digits used to represent from a variable which is not always constant, a good way will be to use numpy.binary.

could be useful when you apply binary to power sets

import numpy as np
np.binary_repr[6, width=8]

answered Mar 11, 2020 at 13:31

amaama

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even an easier way

my_num = 6
print[f'{my_num:b}']

answered Sep 11, 2020 at 2:33

Raad AltaieRaad Altaie

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['0' * 7 + bin[6][2:]][-8:]

or

right_side = bin[6][2:]
'0' * [ 8 - len[ right_side ]] + right_side

eyllanesc

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answered Sep 5, 2018 at 0:07

1

You can use just:

"{0:b}".format[n]

In my opinion this is the easiest way!

answered Aug 11, 2020 at 17:21

LehaLeha

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def int_to_bin[num, fill]:
    bin_result = ''

    def int_to_binary[number]:
        nonlocal bin_result
        if number > 1:
            int_to_binary[number // 2]
        bin_result = bin_result + str[number % 2]

    int_to_binary[num]
    return bin_result.zfill[fill]

answered May 18, 2020 at 9:41

The python package Binary Fractions has a full implementation of binaries as well as binary fractions. You can do your operation as follows:

from binary_fractions import Binary
b = Binary[6] # creates a binary fraction string
b.lfill[8] # fills to length 8

This package has many other methods for manipulating binary strings with full precision.

answered Jul 16, 2021 at 14:49

Simple code with recursion:

 def bin[n,number=['']]:
   if n==0:
     return[number]
   else:
     number=str[n%2]+number
     n=n//2
     return bin[n,number]

answered Dec 2, 2021 at 22:11

1

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