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'

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 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

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

I like python f-string formatting for a little more complex things like using a parameter in format:

>>> x = 5
>>> n = 8
>>> print(f"{x:0{n}b}")
00000101

Here I print variable x with following formatting: I want it to be left-filled with 0 to have length = n , in b (binary) format. See Format Specification Mini-Language from previous answers for more.

Answer 8

你可以使用这样的东西

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

Answer 9

module Adder(
    input upperBit, lowerBit, c_in,
    output s, c_out)

write gate1, gate2, gate3

xor (gate1, upperBit, lowerBit)
xor (s, gate1, c_in)
and (upperBit, lowerBit)
and (gate1, c_in)
or  (c_out, gate1, gate2)

endmodule

module ful_adder8(
    input [7:0) a, b
    input c_in
    output [7:0) s,
    output c_out)

write [7:0] carry

full_adder fa0(
    a(a[o])
    b(b[0])
    c_in(c_in)
    s(s[0])
    c_out(carry[0]))
full_adder fa0(
    a(a[o])
    b(b[0])
    c_in(c_in)
    s(s[0])
    c_out(carry[0]))
full_adder fa0(
    a(a[o])
    b(b[0])
    c_in(c_in)
    s(s[0])
    c_out(carry[0]))
full_adder fa0(
    a(a[o])
    b(b[0])
    c_in(c_in)
    s(s[0])
    c_out(carry[0]))
full_adder fa0(
    a(a[o])
    b(b[0])
    c_in(c_in)
    s(s[0])
    c_out(carry[0]))
full_adder fa0(
    a(a[o])
    b(b[0])
    c_in(c_in)
    s(s[0])
    c_out(carry[0]))
full_adder fa0(
    a(a[o])
    b(b[0])
    c_in(c_in)
    s(s[0])
    c_out(carry[0]))
full_adder fa0(
    a(a[o])
    b(b[0])
    c_in(c_in)
    s(s[0])
    c_out(carry[0]))

endmodule
test
def split (n):
    return (n&0x1,n&0x2,n&0x4,n&0x8,n&0x10,n&0x20,n&0x40,n&0x80)
def glue (b0,b1,b2,b3,b4,b5,b6,b7,c):
    t = 0
    if b0:
        t += 1
    if b1:
        t += 2
    if b2:
        t += 4
    if b3:
        t += 8
    if b4:
        t += 16
    if b5:
        t += 32
    if b6:
        t += 64
    if b7:
        t += 128
    if c:
        t += 256
    return t


def myadd (a,b):
    (a0,a1,a2,a3,a4,a5,a6,a7) = split(a)
    (b0,b1,b2,b3,b4,b5,b6,b7) = split(b)
    (s0,s1,s2,s3,s4,s5,s6,s7,c) = addEightBits(a0,a1,a2,a3,a4,a5,a6,a7,b0,b1,b2,b3,b4,b5,b6,b7,false)
    return glue (s0,s1,s2,s3,s4,s5,s6,s7,c)

Answer 10

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'

Answer 11

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... eg - x = str(datetime.date.today().month).zfill(2) will return x as '02' for the month of feb.