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?

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16 Answers

>>> '{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'
5

Just another idea:

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

Shorter way via string interpolation (Python 3.6+):

>>> f'{6:08b}' '00000110'
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'
3

Just use the format function

format(6, "08b")

The general form is 

format(<the_integer>, "<0><width_of_string><format_specifier>")
1

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

>>> "%08d" % int(bin(6)[2:]) 00000110
1

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'

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

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

or

right_side = bin(6)[2:] '0' * ( 8 - len( right_side )) + right_side
1

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}') 
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)

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)
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