# Binary representation of float in Python (bits not hex)-ThrowExceptions

Exception or error:

How to get the string as binary IEEE 754 representation of a 32 bit float?

Example

1.00 -> ‘00111111100000000000000000000000’

How to solve:

You can do that with the `struct` package:

``````import struct
def binary(num):
return ''.join(bin(ord(c)).replace('0b', '').rjust(8, '0') 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]

# 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:
``````

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']
'00111111100000000000000000000000'

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

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.

Found another solution using the bitstring module.

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

Output:

``````00111111100000000000000000000000
``````

This problem is more cleanly handled by breaking it into two parts.

The first is to convert the float into an int with the equivalent bit pattern:

``````def float32_bit_pattern(value):
return sum(ord(b) << 8*i for i,b in enumerate(struct.pack('f', value)))
``````

Next convert the int to a string:

``````def int_to_binary(value, bits):
return bin(value).replace('0b', '').rjust(bits, '0')
``````

Now combine them:

``````>>> int_to_binary(float32_bit_pattern(1.0), 32)
'00111111100000000000000000000000'
``````

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))
``````

Piggy-tailing on Dan’s answer with colored version for Python3:

``````import struct

BLUE = "\033[1;34m"
CYAN = "\033[1;36m"
GREEN = "\033[0;32m"
RESET = "\033[0;0m"

def binary(num):
return [bin(c).replace('0b', '').rjust(8, '0') for c in struct.pack('!f', num)]

def binary_str(num):
bits = ''.join(binary(num))
return ''.join([BLUE, bits[:1], GREEN, bits[1:10], CYAN, bits[10:], RESET])

def binary_str_fp16(num):
bits = ''.join(binary(num))
return ''.join([BLUE, bits[:1], GREEN, bits[1:10][-5:], CYAN, bits[10:][:11], RESET])

x = 0.7
print(x, "as fp32:", binary_str(0.7), "as fp16 is sort of:", binary_str_fp16(0.7))
``````

After browsing through lots of similar questions I’ve written something which hopefully does what I wanted.

``````f = 1.00
negative = False
if f < 0:
f = f*-1
negative = True

s = struct.pack('>f', f)
p = struct.unpack('>l', s)[0]
hex_data =  hex(p)

scale = 16
num_of_bits = 32
binrep = bin(int(hex_data, scale))[2:].zfill(num_of_bits)
if negative:
binrep = '1' + binrep[1:]
``````

`binrep` is the result.
Each part will be explained.

``````f = 1.00
negative = False
if f < 0:
f = f*-1
negative = True
``````

Converts the number to a positive if negative, and sets the variable negative to false. The reason for this is that the difference between positive and negative binary representations is just in the first bit, and this was the simpler way than to figure out what goes wrong when doing the whole process with negative numbers.

``````s = struct.pack('>f', f)                          #'?\x80\x00\x00'
p = struct.unpack('>l', s)[0]                     #1065353216
hex_data =  hex(p)                                #'0x3f800000'
``````

`s` is a hex representation of the binary `f`. it is however not in the pretty form i need. Thats where p comes in. It is the int representation of the hex s. And then another conversion to get a pretty hex.

``````scale = 16
num_of_bits = 32
binrep = bin(int(hex_data, scale))[2:].zfill(num_of_bits)
if negative:
binrep = '1' + binrep[1:]
``````

`scale` is the base 16 for the hex. `num_of_bits` is 32, as float is 32 bits, it is used later to fill the additional places with 0 to get to 32. Got the code for `binrep` from this question. If the number was negative, just change the first bit.

I know this is ugly, but i didn’t find a nice way and I needed it fast. Comments are welcome.

This is a little more than was asked, but it was what I needed when I found this entry. This code will give the mantissa, base and sign of the IEEE 754 32 bit float.

``````import ctypes
def binRep(num):
binNum = bin(ctypes.c_uint.from_buffer(ctypes.c_float(num)).value)[2:]
print("bits: " + binNum.rjust(32,"0"))
mantissa = "1" + binNum[-23:]
print("sig (bin): " + mantissa.rjust(24))
mantInt = int(mantissa,2)/2**23
print("sig (float): " + str(mantInt))
base = int(binNum[-31:-23],2)-127
print("base:" + str(base))
sign = 1-2*("1"==binNum[-32:-31].rjust(1,"0"))
print("sign:" + str(sign))
print("recreate:" + str(sign*mantInt*(2**base)))

binRep(-0.75)
``````

output:

``````bits: 10111111010000000000000000000000
sig (bin): 110000000000000000000000
sig (float): 1.5
base:-1
sign:-1
recreate:-0.75
``````

With these two simple functions (Python >=3.6) you can easily convert a float number to binary and vice versa, for IEEE 754 binary64.

``````import struct

def bin2float(b):
''' Convert binary string to a float.

Attributes:
:b: Binary string to transform.
'''
h = int(b, 2).to_bytes(8, byteorder="big")
return struct.unpack('>d', h)[0]

def float2bin(f):
''' Convert float to 64-bit binary string.

Attributes:
:f: Float number to transform.
'''
[d] = struct.unpack(">Q", struct.pack(">d", f))
return f'{d:064b}'
``````

For example:

``````print(float2bin(1.618033988749894))
print(float2bin(3.14159265359))
print(float2bin(5.125))
print(float2bin(13.80))

print(bin2float('0011111111111001111000110111011110011011100101111111010010100100'))
print(bin2float('0100000000001001001000011111101101010100010001000010111011101010'))
print(bin2float('0100000000010100100000000000000000000000000000000000000000000000'))
print(bin2float('0100000000101011100110011001100110011001100110011001100110011010'))
``````

The output is:

``````0011111111111001111000110111011110011011100101111111010010100100
0100000000001001001000011111101101010100010001000010111011101010
0100000000010100100000000000000000000000000000000000000000000000
0100000000101011100110011001100110011001100110011001100110011010
1.618033988749894
3.14159265359
5.125
13.8
``````

I hope you like it, it works perfectly for me.

Several of these answers did not work as written with Python 3, or did not give the correct representation for negative floating point numbers. I found the following to work for me (though this gives 64-bit representation which is what I needed)

``````def float_to_binary_string(f):
def int_to_8bit_binary_string(n):
stg=bin(n).replace('0b','')
fillstg = '0'*(8-len(stg))
return fillstg+stg
return ''.join( int_to_8bit_binary_string(int(b)) for b in struct.pack('>d',f) )
``````

Convert float between 0..1

``````def float_bin(n, places = 3):
if (n < 0 or n > 1):
return "ERROR, n must be in 0..1"

while n > 0:

b = n * 2
if b > 1:
n = b - 1
else:
n = b

``````

You can use the .format for the easiest representation of bits in my opinion:

my code would look something like:

``````def fto32b(flt):
# is given a 32 bit float value and converts it to a binary string
if isinstance(flt,float):
# THE FOLLOWING IS AN EXPANDED REPRESENTATION OF THE ONE LINE RETURN
#   packed = struct.pack('!f',flt) <- get the hex representation in (!)Big Endian format of a (f) Float
#   integers = []
#   for c in packed:
#       integers.append(ord(c))    <- change each entry into an int
#   binaries = []
#   for i in integers:
#       binaries.append("{0:08b}".format(i)) <- get the 8bit binary representation of each int (00100101)
#   binarystring = ''.join(binaries) <- join all the bytes together
#   return binarystring
return ''.join(["{0:08b}".format(i) for i in [ord(c) for c in struct.pack('!f',flt)]])
return None
``````

Output:

``````>>> a = 5.0
'01000000101000000000000000000000'
>>> b = 1.0
'00111111100000000000000000000000'
``````