转换为二进制并在Python中保持前导零

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?

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 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'

>>> '{: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'      

I am using

bin(1)[2:].zfill(8)

will print

'00000001'

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))

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 the 0b prefix)
• 0 pad with zeros
• 10 pad to a total length off 10 (this includes the 2 chars for 0b)
• b use binary representation for the number

Sometimes you just want a simple one liner:

binary = ''.join(['{0:08b}'.format(ord(x)) for x in input])

Python 3

You can use something like this

("{:0%db}"%length).format(num)

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’

You can use zfill:

print str(1).zfill(2) 
print str(10).zfill(2) 
print str(100).zfill(2)

prints:

01
10
100

I like this solution, as it helps not only when outputting the number, but when you need to assign it to a variable… e.g. – x = str(datetime.date.today().month).zfill(2) will return x as ’02’ for the month of feb.