Converting integer to binary in python
PythonBinaryIntegerPython Problem Overview
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?
Python Solutions
Solution 1 - Python
>>> '{0:08b}'.format(6)
'00000110'
Just to explain the parts of the formatting string:
{}places a variable into a string0takes the variable at argument position 0:adds formatting options for this variable (otherwise it would represent decimal6)08formats the number to eight digits zero-padded on the leftbconverts 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'
Solution 2 - Python
Just another idea:
>>> bin(6)[2:].zfill(8)
'00000110'
Shorter way via string interpolation (Python 3.6+):
>>> f'{6:08b}'
'00000110'
Solution 3 - Python
A bit twiddling method...
>>> bin8 = lambda x : ''.join(reversed( [str((x >> i) & 1) for i in range(8)] ) )
>>> bin8(6)
'00000110'
>>> bin8(-3)
'11111101'
Solution 4 - Python
Just use the format function
format(6, "08b")
The general form is
format(<the_integer>, "<0><width_of_string><format_specifier>")
Solution 5 - Python
eumiro's answer is better, however I'm just posting this for variety:
>>> "%08d" % int(bin(6)[2:])
00000110
Solution 6 - Python
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'
Solution 7 - Python
.. 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'
Solution 8 - Python
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))
Solution 9 - Python
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)
Solution 10 - Python
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.
Solution 11 - Python
('0' * 7 + bin(6)[2:])[-8:]
or
right_side = bin(6)[2:]
'0' * ( 8 - len( right_side )) + right_side
Solution 12 - Python
You can use just:
"{0:b}".format(n)
In my opinion this is the easiest way!
Solution 13 - Python
even an easier way
my_num = 6
print(f'{my_num:b}')
Solution 14 - Python
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)
Solution 15 - Python
The python package https://pypi.org/project/binary-fractions/">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.
Solution 16 - Python
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)