Converting integer to binary in python

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

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

Just another idea:

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

Shorter way via string interpolation (Python 3.6+):

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

.. 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'

A bit twiddling method…

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

Just use the format function

format(6, "08b")

The general form is

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

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))

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

or

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

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'

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)

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.

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)

You can use just:

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

In my opinion this is the easiest way!

even an easier way

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

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.

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)

    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]