Convert hex to float

Question

How to convert the following hex string to float (single precision 32-bit) in Python?

\"41973333\" -> 1.88999996185302734375E1

\"41995C29\" -> 1.91700         


        

Answer 1

Slice up the hex strings into 2-character chunks (bytes), make each chunk into the right byte with int formatting, struct.unpack when done. I.e.:

import struct 

testcases = {
"41973333": 1.88999996185302734375E1,
"41995C29": 1.91700000762939453125E1,
"470FC614": 3.6806078125E4,
}

def hex2float(s):
    bins = ''.join(chr(int(s[x:x+2], 16)) for x in range(0, len(s), 2))
    return struct.unpack('>f', bins)[0]

for s in testcases:
  print hex2float(s), testcases[s]

emitting, as desired:

18.8999996185 18.8999996185
19.1700000763 19.1700000763
36806.078125 36806.078125

Answer 2

Gentelmen... Behold:

    class fl:
        def __init__(this, value=0, byte_size=4):

            this.value = value

            if this.value: # speedy check (before performing any calculations)
                Fe=((byte_size*8)-1)//(byte_size+1)+(byte_size>2)*byte_size//2+(byte_size==3)
                Fm,Fb,Fie=(((byte_size*8)-(1+Fe)), ~(~0<<Fe-1), (1<<Fe)-1)

                FS,FE,FM=((this.value>>((byte_size*8)-1))&1,(this.value>>Fm)&Fie,this.value&~(~0 << Fm))
                if FE == Fie: this.value=(float('NaN') if FM!=0 else (float('+inf') if FS else float('-inf')))
                else: this.value=((pow(-1,FS)*(2**(FE-Fb-Fm)*((1<<Fm)+FM))) if FE else pow(-1,FS)*(2**(1-Fb-Fm)*FM))

                del Fe; del Fm; del Fb; del Fie; del FS; del FE; del FM

            else: this.value = 0.0

    print fl( 0x41973333 ).value # >>> 18.899999618530273
    print fl( 0x41995C29 ).value # >>> 19.170000076293945
    print fl( 0x470FC614 ).value # >>> 36806.078125
    print fl( 0x00800000 ).value # >>> 1.1754943508222875e-38 (minimum float value)
    print fl( 0x7F7FFFFF ).value # >>> 340282346638528859811704183484516925440L (maximum float value)
    # looks like I've found a small bug o.o
    # the code still works though (the numbers are properly formatted)
    # the result SHOULD be: 3.4028234663852886e+38 (rounded)
    print fl( 0x3f80000000, 5 ).value # >>> 1.0

sorry for the little ".value" at the end...
this code has been used as a class in my program for nearly 2 years now.
(with a little editing, you can easily make it into a function)

credit to PyTony over at DaniWeb for the code.

unlike non-dynamic computing,
the code is not hard-wired to a fixed float size,
and works with any byte-size.

though I guess we still have a few bugs to work out. XDD
(I'll edit this code later (if I can) with the update)

all is good though for now though...
I havn't had a problem converting 3D game models with it. :)

Answer 3

I recommend using the ctypes module which basically lets you work with low level data types. In your case you could say

from ctypes import *

def convert(s):
    i = int(s, 16)                   # convert from hex to a Python int
    cp = pointer(c_int(i))           # make this into a c integer
    fp = cast(cp, POINTER(c_float))  # cast the int pointer to a float pointer
    return fp.contents.value         # dereference the pointer, get the float

print convert("41973333")    # returns 1.88999996185302734375E1

print convert("41995C29")    # returns 1.91700000762939453125E1

print convert("470FC614")    # returns 3.6806078125E4

I believe that the ctypes module makes sense here, because you're essentially asking how to perform low-level bit casting. Your question is basically, how do I tell Python to take some data and interpret that data as if those exact same bits were a different data type?

In C if you had an int and wanted to interpret its bits as a float, you'd do roughly the same thing, taking a pointer and then casting and dereferencing it:

int i = 0x41973333;
float f = *((float*)&i);

and that's exactly what the Python code using the ctypes library is doing in my example.

Answer 4

In Python 3:

>>> import struct
>>> struct.unpack('!f', bytes.fromhex('41973333'))[0]
18.899999618530273
>>> struct.unpack('!f', bytes.fromhex('41995C29'))[0]
19.170000076293945
>>> struct.unpack('!f', bytes.fromhex('470FC614'))[0]
36806.078125

In Python 2:

>>> import struct
>>> struct.unpack('!f', '41973333'.decode('hex'))[0]
18.899999618530273
>>> struct.unpack('!f', '41995C29'.decode('hex'))[0]
19.170000076293945
>>> struct.unpack('!f', '470FC614'.decode('hex'))[0]
36806.078125

Answer 5

I'm guessing this question relates to this one and you are working with 4 bytes rather than 8 hex digits.

"\x41\x91\x33\x33" is a 4 byte string even though it looks like 16

>>> len("\x41\x91\x33\x33")
4
>>> import struct  
>>> struct.unpack(">fff","\x41\x97\x33\x33\x41\x99\x5C\x29\x47\x0F\xC6\x14")
(18.899999618530273, 19.170000076293945, 36806.078125)

If you do need to deal with the string of hexdigits rather than the actual bytes, you can use struct.pack to convert it, like this

>>> for hx in ["41973333","41995C29","470FC614"]:
...     print(struct.unpack(">f",struct.pack(">i",int(hx,16)))[0])
... 
18.8999996185
19.1700000763
36806.078125