Question :
I’m trying to convert an integer to binary using the bin() function in Python. However, it always removes the leading zeros, which I actually need, such that the result is always 8-bit:
Example:
bin(1) -> 0b1
# What I would like:
bin(1) -> 0b00000001
Is there a way of doing this?
Answer #1:
Use the format()
function:
>>> format(14, '#010b')
'0b00001110'
The format()
function simply formats the input following the Format Specification mini language. The #
makes the format include the 0b
prefix, and the 010
size formats the output to fit in 10 characters width, with 0
padding; 2 characters for the 0b
prefix, the other 8 for the binary digits.
This is the most compact and direct option.
If you are putting the result in a larger string, use an formatted string literal (3.6+) or use str.format()
and put the second argument for the format()
function after the colon of the placeholder {:..}
:
>>> value = 14
>>> f'The produced output, in binary, is: {value:#010b}'
'The produced output, in binary, is: 0b00001110'
>>> 'The produced output, in binary, is: {:#010b}'.format(value)
'The produced output, in binary, is: 0b00001110'
As it happens, even for just formatting a single value (so without putting the result in a larger string), using a formatted string literal is faster than using format()
:
>>> import timeit
>>> timeit.timeit("f_(v, '#010b')", "v = 14; f_ = format") # use a local for performance
0.40298633499332936
>>> timeit.timeit("f'{v:#010b}'", "v = 14")
0.2850222919951193
But I’d use that only if performance in a tight loop matters, as format(...)
communicates the intent better.
If you did not want the 0b
prefix, simply drop the #
and adjust the length of the field:
>>> format(14, '08b')
'00001110'
Answer #2:
>>> '{:08b}'.format(1)
'00000001'
See: Format Specification Mini-Language
Note for Python 2.6 or older, you cannot omit the positional argument identifier before :
, so use
>>> '{0:08b}'.format(1)
'00000001'
Answer #3:
I am using
bin(1)[2:].zfill(8)
will print
'00000001'
Answer #4:
You can use the string formatting mini language:
def binary(num, pre='0b', length=8, spacer=0):
return '{0}{{:{1}>{2}}}'.format(pre, spacer, length).format(bin(num)[2:])
Demo:
print binary(1)
Output:
'0b00000001'
EDIT:
based on @Martijn Pieters idea
def binary(num, length=8):
return format(num, '#0{}b'.format(length + 2))
Answer #5:
When using Python >= 3.6
, the cleanest way is to use f-strings with string formatting:
>>> var = 23
>>> f"{var:#010b}"
'0b00010111'
Explanation:
var
the variable to format:
everything after this is the format specifier#
use the alternative form (adds the0b
prefix)0
pad with zeros10
pad to a total length off 10 (this includes the 2 chars for0b
)b
use binary representation for the number
Answer #6:
Sometimes you just want a simple one liner:
binary = ''.join(['{0:08b}'.format(ord(x)) for x in input])
Python 3
Answer #7:
You can use something like this
("{:0%db}"%length).format(num)
Answer #8:
you can use rjust string method of python
syntax:
string.rjust(length, fillchar)
fillchar is optional
and for your Question you acn write like this
'0b'+ '1'.rjust(8,'0)
so it wil be ‘0b00000001’