Help need with converting Hex string to IEEE format float

Hi all,

Newbie Python programmer here, so please be patient. I have spent all
day googling for an answer to my problem, but everything I try fails to
work (or works from the Interpreter with a set value but not from my
code with dynamic values).

Okay, here is the general gist of the problem. I am using Python to
parse an output file (from a MAK Logger but that is not really
important). Now I have some data that is contained on a line in this
file like:

80 00 00 00

Each of these numbers is a Hex byte making up a four byte (32 bit
Big-Endian) IEEE float. I have read this data into Python using
readlines and then line.split(). This gives me:

['80', '00', '00', '00']

I am then reading these in as:

wib[0] + wib[1] + wib[2] + wib[3] = 8000000 as a string

Now this is the point where I get stuck, I have tried various ways of
implementing the pack/unpack methods of the struct module but with no
luck.

One example I tried was:

wibble = struct.unpack("f", struct.pack("l", long(conv_str, 16)))
OverflowError: long int too large to convert to int

If I follow the examples I have found on the net using a set value of
0x80000000 in them, everything works fine from the interpreter line.

Arrrggggghhhh.

Everything has worked really easily up this point but I am now stuck
completely, I would be grateful for any help people can offer.
Regards

Ian Vincent

re: Help need with converting Hex string to IEEE format float

<i_vincent@hotmail.com> wrote:
[color=blue]
> Newbie Python programmer here, so please be patient. I have spent all
> day googling for an answer to my problem, but everything I try fails to
> work (or works from the Interpreter with a set value but not from my
> code with dynamic values).
>
> Okay, here is the general gist of the problem. I am using Python to
> parse an output file (from a MAK Logger but that is not really
> important). Now I have some data that is contained on a line in this
> file like:
>
> 80 00 00 00
>
> Each of these numbers is a Hex byte making up a four byte (32 bit
> Big-Endian) IEEE float. I have read this data into Python using
> readlines and then line.split(). This gives me:
>
> ['80', '00', '00', '00'][/color]

how about:

# convert to byte string
import struct
s = "".join([chr(int(c, 16)) for c in x])
v = struct.unpack("!f", s)

or

# convert to byte string, via the array module
import array, struct
a = array.array("B", [int(c, 16) for c in x])
v = struct.unpack("!f", )

</F>

re: Help need with converting Hex string to IEEE format float

> # convert to byte string, via the array module[color=blue]
> import array, struct
> a = array.array("B", [int(c, 16) for c in x])
> v = struct.unpack("!f", )[/color]

eh? should be:

# convert to byte string, via the array module
import array, struct
a = array.array("B", [int(c, 16) for c in x])
v = struct.unpack("!f", a)

</F>

re: Help need with converting Hex string to IEEE format float

i_vincent@hotmail.com wrote:
[color=blue]
> Each of these numbers is a Hex byte making up a four byte (32 bit
> Big-Endian) IEEE float. I have read this data into Python using
> readlines and then line.split(). This gives me:
>
> ['80', '00', '00', '00'][/color]


Oh, programmers loves this kind stuff. You should get tons of answers.

##
st = '80 00 00 00'

import binascii
import struct

s = ''.join([binascii.a2b_hex(s) for s in st.split()])
v = struct.unpack("f", s)[0]
print v
##

regards Max M

--

hilsen/regards Max M, Denmark

http://www.mxm.dk/
IT's Mad Science

re: Help need with converting Hex string to IEEE format float

Max M wrote:
[color=blue]
> Oh, programmers loves this kind stuff. You should get tons of answers.[/color]

data = '80 00 00 00'

import Image
v = Image.fromstring("F", (1, 1), data, "hex", "F;32BF").getpixel((0, 0))

</F>

re: Help need with converting Hex string to IEEE format float

On Tue, 14 Dec 2004 16:57:02 +0100, rumours say that "Fredrik Lundh"
<fredrik@pythonware.com> might have written:
[color=blue]
>how about:
>
> # convert to byte string
> import struct
> s = "".join([chr(int(c, 16)) for c in x])
> v = struct.unpack("!f", s)[/color]

I think that the third line in the snippet above could also be:

s = "".join(x).decode("hex")

I am not sure I remember in which version of Python the hex codec was
added, but it is handy.
--
TZOTZIOY, I speak England very best.
"Be strict when sending and tolerant when receiving." (from RFC1958)
I really should keep that in mind when talking with people, actually...

re: Help need with converting Hex string to IEEE format float

Christos TZOTZIOY Georgiou wrote:[color=blue]
> s = "".join(x).decode("hex")
>
> I am not sure I remember in which version of Python the hex codec was
> added, but it is handy.[/color]

Of course, binascii could do this since 2.0 or so, but not
having to import another module *is* nice:
[color=blue][color=green][color=darkred]
>>> 'ff12'.decode('hex')[/color][/color][/color]
'\xff\x12'
[color=blue][color=green][color=darkred]
>>> import binascii
>>> binascii.unhexlify('ff12')[/color][/color][/color]
'\xff\x12'

Thanks for pointing it out, Christos.

re: Help need with converting Hex string to IEEE format float

Max M <maxm@mxm.dk> wrote in news:41bf121e$0$280
$edfadb0f@dread12.news.tele.dk:[color=blue]
>
> ##
> st = '80 00 00 00'
>
> import binascii
> import struct
>
> s = ''.join([binascii.a2b_hex(s) for s in st.split()])
> v = struct.unpack("f", s)[0]
> print v
> ##[/color]

This one worked great for what I was trying to do.

Thanks to everybody for your help.

TTFN

Ian