python - Convert to binary and keep leading zeros

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'