>>> '{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 decimal6)
• 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'

Just another idea:

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

Shorter way viastring interpolation (Python 3.6+):

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

Just use the format function

format(6, "08b")

The general form is

format(<the_integer>, "<0><width_of_string><format_specifier>")

A bit twiddling method...

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

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'

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

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

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.

.. 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'

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)/2else:    binarynumber="0"return "".join(reversed(binarynumber))

An even an easier way:

my_num = 6print(f'{my_num:b}')

You can use just:

"{0:b}".format(n)

In my opinion this is the easiest way!

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 npnp.binary_repr(6, width=8)

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

or

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

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)

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)

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

from binary_fractions import Binaryb = Binary(6) # Creates a binary fraction stringb.lfill(8) # Fills to length 8

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

    def convertToBinary(self, n):        result=""        if n==0:            return 0        while n>0:            r=n%2            result+=str(r)            n=int(n/2)        if n%2==0:            result+="0"        return result[::-1]

• python
• binary
• integer