msbin to ieee

[John Machin]

On May 7, 10:00 pm, revuesbio <revues... at gmail.com> wrote:
> On 7 mai, 13:21, John Machin <sjmac... at lexicon.net> wrote:
>
>
>
> > On May 7, 6:18 pm, revuesbio <revues... at gmail.com> wrote:
>
> > > On 7 mai, 03:52, John Machin <sjmac... at lexicon.net> wrote:
>
> > > > On May 7, 7:44 am, revuesbio <revues... at gmail.com> wrote:
>
> > > > > Hi
> > > > > Does anyone have the python version of the conversion from msbin to
> > > > > ieee?
> > > > > Thank u
>
> > > > Yes, Google has it. Google is your friend. Ask Google. It will lead
> > > > you to such as:
>
> > > >http://mail.python.org/pipermail/python-list/2005-August/337817.html
>
> > > > HTH,
> > > > John
>
> > > Thank you,
>
> > > I've already read it but the problem is always present. this script is
> > > for double precision MBF format ( 8 bytes).
>
> > It would have been somewhat more helpful had you said what you had
> > done so far,  even posted your code ...
>
> > > I try to adapt this script for single precision MBF format ( 4 bytes)
> > > but i don't find the right float value.
>
> > > for example : 'P\xad\x02\x95' will return '0.00024924660101532936'
>
> > If you know what the *correct* value is, you might like to consider
> > shifting left by log2(correct_value/erroneous_value) :-)
>
> > Do you have any known correct pairs of (mbf4 string, decimal_float
> > value)? My attempt is below -- this is based on a couple of
> > descriptive sources that my friend Google found, with no test data. I
> > believe the correct answer for the above input is 1070506.0 i.e. you
> > are out by a factor of 2 ** 32
>
> > def mbf4_as_float(s):
> >     m0, m1, m2, m3 = [ord(c) for c in s]
> >     exponent = m3
> >     if not exponent:
> >         return 0.0
> >     sign = m2 & 0x80
> >     m2 |= 0x80
> >     mant = (((m2 << 8) | m1) << 8) | m0
> >     adj = 24 + 128
> >     num = mant * 2.0 ** (exponent - adj)
> >     if sign:
> >         return -num
> >     return num
>
> > HTH,
> > John
>
> well done ! it's exactly what i'm waiting for !!
>
> my code was:>>> from struct import *
> >>> x = list(unpack('BBBB','P\xad\x02\x95'))
> >>> x
> [80, 173, 2, 149]
> >>> def conversion1(bytes):
>
> b=bytes[:]
> sign = bytes[-2] & 0x80
> b[-2] |= 0x80
> exp = bytes[-1] - 0x80 - 56
> acc = 0L
> for i,byte in enumerate(b[:-1]):
> acc |= (long(byte)<<(i*8))
> return (float(acc)*2.0**exp)*((1.,-1.)[sign!=0])

Apart from the 2**32 problem, the above doesn't handle *any* of the
2**24 different representations of zero. Try feeding \0\0\0\0' to it
and see what you get.

>
> >>> conversion1(x)
>
> 0.00024924660101532936
>
> this script come from google groups but i don't understand bit-string
> manipulation (I'm a  newbie). informations about bit-string
> manipulation with python is too poor on the net.

The basic operations (and, or, exclusive-or, shift) are not specific
to any language. Several  languages share the same notation (& | ^ <<
>>), having inherited it from C.

>
> thank you very much for your script.

Don't thank me, publish some known correct pairs of values so that we
can verify that it's not just accidentally correct for 1 pair of
values.