Binary representation of float in Python (bits not hex)

Solution 1:

You can do that with the struct package:

import struct
def binary(num):
    return ''.join('{:0>8b}'.format(c) for c in struct.pack('!f', num))

That packs it as a network byte-ordered float, and then converts each of the resulting bytes into an 8-bit binary representation and concatenates them out:

>>> binary(1)
'00111111100000000000000000000000'

Edit: There was a request to expand the explanation. I'll expand this using intermediate variables to comment each step.

def binary(num):
    # Struct can provide us with the float packed into bytes. The '!' ensures that
    # it's in network byte order (big-endian) and the 'f' says that it should be
    # packed as a float. Alternatively, for double-precision, you could use 'd'.
    packed = struct.pack('!f', num)
    print 'Packed: %s' % repr(packed)

    # For each character in the returned string, we'll turn it into its corresponding
    # integer code point
    # 
    # [62, 163, 215, 10] = [ord(c) for c in '>\xa3\xd7\n']
    integers = [ord(c) for c in packed]
    print 'Integers: %s' % integers

    # For each integer, we'll convert it to its binary representation.
    binaries = [bin(i) for i in integers]
    print 'Binaries: %s' % binaries

    # Now strip off the '0b' from each of these
    stripped_binaries = [s.replace('0b', '') for s in binaries]
    print 'Stripped: %s' % stripped_binaries

    # Pad each byte's binary representation's with 0's to make sure it has all 8 bits:
    #
    # ['00111110', '10100011', '11010111', '00001010']
    padded = [s.rjust(8, '0') for s in stripped_binaries]
    print 'Padded: %s' % padded

    # At this point, we have each of the bytes for the network byte ordered float
    # in an array as binary strings. Now we just concatenate them to get the total
    # representation of the float:
    return ''.join(padded)

And the result for a few examples:

>>> binary(1)
Packed: '?\x80\x00\x00'
Integers: [63, 128, 0, 0]
Binaries: ['0b111111', '0b10000000', '0b0', '0b0']
Stripped: ['111111', '10000000', '0', '0']
Padded: ['00111111', '10000000', '00000000', '00000000']
'00111111100000000000000000000000'

>>> binary(0.32)
Packed: '>\xa3\xd7\n'
Integers: [62, 163, 215, 10]
Binaries: ['0b111110', '0b10100011', '0b11010111', '0b1010']
Stripped: ['111110', '10100011', '11010111', '1010']
Padded: ['00111110', '10100011', '11010111', '00001010']
'00111110101000111101011100001010'

Solution 2:

Here's an ugly one ...

>>> import struct
>>> bin(struct.unpack('!i',struct.pack('!f',1.0))[0])
'0b111111100000000000000000000000'

Basically, I just used the struct module to convert the float to an int ...

Here's a slightly better one using ctypes:

>>> import ctypes
>>> bin(ctypes.c_uint.from_buffer(ctypes.c_float(1.0)).value)
'0b111111100000000000000000000000'

Basically, I construct a float and use the same memory location, but I tag it as a c_uint. The c_uint's value is a python integer which you can use the builtin bin function on.

Solution 3:

Found another solution using the bitstring module.

import bitstring
f1 = bitstring.BitArray(float=1.0, length=32)
print(f1.bin)

Output:

00111111100000000000000000000000

Solution 4:

For the sake of completeness, you can achieve this with numpy using:

f = 1.00
int32bits = np.asarray(f, dtype=np.float32).view(np.int32).item()  # item() optional

You can then print this, with padding, using the b format specifier

print('{:032b}'.format(int32bits))