On May 7, 6:18 pm, revuesbio <[EMAIL PROTECTED]> wrote:
> On 7 mai, 03:52, John Machin <[EMAIL PROTECTED]> wrote:
>
> > On May 7, 7:44 am, revuesbio <[EMAIL PROTECTED]> 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
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