How do you get a binary float in python?

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'

In this article, we will see how to convert floating-point value to binary. Binary uses two digits, 0 and 1. Also called as base 2 number system Each position in a binary number represents a 0 power of the base (2). Last position in a binary number represents a x power of the base (2).

First, we take the integer part from the floating point value and convert it to binary then take fractional part and convert it to binary form and lastly combining both.

Let’s say we have the following floating point number −

22.625

Convert decimal 22 to binary 10110. Convert decimal 0.625 to binary 0.101. Combine integer and fraction to get the following output −

10110.101

Convert floating to binary by taking user input

Example

def floatoctal_convert(my_number, places = 3):
	my_whole, my_dec = str(my_number).split(".")
	my_whole = int(my_whole)
	my_dec = int (my_dec)
	res = bin(my_whole).lstrip("0b") + "."
	for x in range(places):
		my_whole, my_dec = str((my_decimal_converter(my_dec)) * 8).split(".")
		my_dec = int(my_dec)
		res += my_whole
	return res
def my_decimal_converter(num):
	while num > 1:
		num /= 10
	return num


n = input("Enter floating point value : \n")
p = int(input("Enter the number of decimal places of the result : \n"))
print(floatoctal_convert(n, places = p))

Output

Enter floating point value :
2.34
Enter the number of decimal places of the result :
3
10.256

Updated on 12-Aug-2022 11:54:31

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