Re: msbin to ieee
On May 6, 6:44 pm, revuesbio [EMAIL PROTECTED] wrote: Hi Does anyone have the python version of the conversion from msbin to ieee? Thank u Not sure if this helps, but I think this thread has the answer; http://groups.google.com/group/comp.lang.python/browse_thread/thread/286d9f6daff9bfab/ce76d5fcd887a47d?lnk=gstq=geskerrettrnum=2#ce76d5fcd887a47d Check out the response from Bengt Richter. His function did the right thing. -- http://mail.python.org/mailman/listinfo/python-list
Re: msbin to ieee
On May 10, 8:48 pm, imageguy [EMAIL PROTECTED] wrote: On May 6, 6:44 pm, revuesbio [EMAIL PROTECTED] wrote: Hi Does anyone have the python version of the conversion from msbin to ieee? Thank u Not sure if this helps, but I think this thread has the answer;http://groups.google.com/group/comp.lang.python/browse_thread/thread/... Check out the response from Bengt Richter. His function did the right thing. Yes, Bengt's function did the right thing on the input for that particular problem, which involved IEEE 64-bit floating point numbers stored in little-endian format. The current problem involves 32-bit MBF (Microsoft Binary/Basic Floating-point/Format) numbers. Different problem. -- http://mail.python.org/mailman/listinfo/python-list
Re: msbin to ieee
Hi, I found something interresting. First, MBF Files come from metastock software but i use another one (MLDownloader) to get quotes and convert them to MBF format probably using functions you've just described (C, borland,...). In final, all my files are created by mldownloader. 2nd, I've tried to modify quotes directly in metastock. And when you read bytes corresponding to zero ... : '\x00\x00\x00\x00' ! cheers, Antoine -- http://mail.python.org/mailman/listinfo/python-list
Re: msbin to ieee
On 7 mai, 23:38, John Machin [EMAIL PROTECTED] wrote:
On May 7, 11:37 pm, revuesbio [EMAIL PROTECTED] wrote:
On 7 mai, 14:56, John Machin [EMAIL PROTECTED] wrote:
On May 7, 10:00 pm, revuesbio [EMAIL PROTECTED] wrote:
On 7 mai, 13:21, John Machin [EMAIL PROTECTED] wrote:
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
well done ! it's exactly what i'm waiting for !!
my code was: from struct import *
x = list(unpack('','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.
pairs of values :
(bytes string, mbf4_as_float(s) result)right
float value
('P\xad\x02\x95', 1070506.0)
1070506.0
('\x00\x00\x00\x02', 5.8774717541114375e-039) 0.0
There is no way that \x00\x00\x00\x02' could represent exactly zero.
What makes you think it does? Rounding?
('\x00\x00\x00\x81', 1.0)
1.0
('\x00\x00\x00\x82', 2.0)
2.0
('[EMAIL PROTECTED]', 3.0)
3.0
('\x00\x00\x00\x83', 4.0)
4.0
('\x00\x00 \x83', 5.0)
5.0
('\xcd\xcc\x0c\x81', 1.100238418579) 1.1
('\xcd\xcc\x0c\x82', 2.200476837158) 2.2
('33S\x82', 3.299523162842) 3.3
('\xcd\xcc\x0c\x83', 4.400953674316) 4.4
It is not apparent whether you regard the output from the function as
correct or not.
4.4 converted to mbf4 format is '\xcd\xcc\x0c\x83' which is
4.400953674316 which is the closest possible mbf4 representation
of 4.4 (difference is 9.5e-008).
The next lower mbf4 value '\xcc\xcc\x0c\x83' is 4.396185302734
(difference is -3.8e-007).
Note that floating-point representation of many decimal fractions is
inherently inexact. print repr(4.4) produces 4.4004
Have you read this:
http://docs.python.org/tut/node16.html
?
If you need decimal-fraction output that matches what somebody typed
into the original software, or saw on the screen, you will need to
know/guess the precision that was involved, and round the numbers
accordingly -- just like the author of the original software would
have needed to do.
['%.*f' % (decplaces, 4.400953674316) for decplaces in range
Re: msbin to ieee
On May 8, 8:15 pm, revuesbio [EMAIL PROTECTED] wrote:
On 7 mai, 23:38, John Machin [EMAIL PROTECTED] wrote:
On May 7, 11:37 pm, revuesbio [EMAIL PROTECTED] wrote:
On 7 mai, 14:56, John Machin [EMAIL PROTECTED] wrote:
On May 7, 10:00 pm, revuesbio [EMAIL PROTECTED] wrote:
On 7 mai, 13:21, John Machin [EMAIL PROTECTED] wrote:
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
well done ! it's exactly what i'm waiting for !!
my code was: from struct import *
x = list(unpack('','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.
pairs of values :
(bytes string, mbf4_as_float(s) result)right
float value
('P\xad\x02\x95', 1070506.0)
1070506.0
('\x00\x00\x00\x02', 5.8774717541114375e-039) 0.0
There is no way that \x00\x00\x00\x02' could represent exactly zero.
What makes you think it does? Rounding?
('\x00\x00\x00\x81', 1.0)
1.0
('\x00\x00\x00\x82', 2.0)
2.0
('[EMAIL PROTECTED]', 3.0)
3.0
('\x00\x00\x00\x83', 4.0)
4.0
('\x00\x00 \x83', 5.0)
5.0
('\xcd\xcc\x0c\x81', 1.100238418579) 1.1
('\xcd\xcc\x0c\x82', 2.200476837158) 2.2
('33S\x82', 3.299523162842) 3.3
('\xcd\xcc\x0c\x83', 4.400953674316) 4.4
It is not apparent whether you regard the output from the function as
correct or not.
4.4 converted to mbf4 format is '\xcd\xcc\x0c\x83' which is
4.400953674316 which is the closest possible mbf4 representation
of 4.4 (difference is 9.5e-008).
The next lower mbf4 value '\xcc\xcc\x0c\x83' is 4.396185302734
(difference is -3.8e-007).
Note that floating-point representation of many decimal fractions is
inherently inexact. print repr(4.4) produces 4.4004
Have you read this:
http://docs.python.org/tut/node16.html
?
If you need decimal-fraction output that matches what somebody typed
into the original software, or saw on the screen, you will need to
know/guess the precision that was involved
Re: msbin to ieee
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). 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' -- http://mail.python.org/mailman/listinfo/python-list
Re: msbin to ieee
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 -- http://mail.python.org/mailman/listinfo/python-list
Re: msbin to ieee
On 7 mai, 13:21, John Machin [EMAIL PROTECTED] wrote:
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
well done ! it's exactly what i'm waiting for !!
my code was:
from struct import *
x = list(unpack('','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])
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.
thank you very much for your script.
A.
--
http://mail.python.org/mailman/listinfo/python-list
Re: msbin to ieee
On 7 mai, 14:56, John Machin [EMAIL PROTECTED] wrote:
On May 7, 10:00 pm, revuesbio [EMAIL PROTECTED] wrote:
On 7 mai, 13:21, John Machin [EMAIL PROTECTED] wrote:
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
well done ! it's exactly what i'm waiting for !!
my code was: from struct import *
x = list(unpack('','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.
pairs of values :
(bytes string, mbf4_as_float(s) result)right
float value
('P\xad\x02\x95', 1070506.0)
1070506.0
('\x00\x00\x00\x02', 5.8774717541114375e-039) 0.0
('\x00\x00\x00\x81', 1.0)
1.0
('\x00\x00\x00\x82', 2.0)
2.0
('[EMAIL PROTECTED]', 3.0)
3.0
('\x00\x00\x00\x83', 4.0)
4.0
('\x00\x00 \x83', 5.0)
5.0
('\xcd\xcc\x0c\x81', 1.100238418579) 1.1
('\xcd\xcc\x0c\x82', 2.200476837158) 2.2
('33S\x82', 3.299523162842) 3.3
('\xcd\xcc\x0c\x83', 4.400953674316) 4.4
('\x00\x00z\x8a', 1000.0)
1000.0
--
http://mail.python.org/mailman/listinfo/python-list
Re: msbin to ieee
On May 7, 10:00 pm, revuesbio [EMAIL PROTECTED] wrote:
On 7 mai, 13:21, John Machin [EMAIL PROTECTED] wrote:
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
well done ! it's exactly what i'm waiting for !!
my code was: from struct import *
x = list(unpack('','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.
--
http://mail.python.org/mailman/listinfo/python-list
Re: msbin to ieee
On May 7, 11:37 pm, revuesbio [EMAIL PROTECTED] wrote:
On 7 mai, 14:56, John Machin [EMAIL PROTECTED] wrote:
On May 7, 10:00 pm, revuesbio [EMAIL PROTECTED] wrote:
On 7 mai, 13:21, John Machin [EMAIL PROTECTED] wrote:
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
well done ! it's exactly what i'm waiting for !!
my code was: from struct import *
x = list(unpack('','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.
pairs of values :
(bytes string, mbf4_as_float(s) result)right
float value
('P\xad\x02\x95', 1070506.0)
1070506.0
('\x00\x00\x00\x02', 5.8774717541114375e-039) 0.0
There is no way that \x00\x00\x00\x02' could represent exactly zero.
What makes you think it does? Rounding?
('\x00\x00\x00\x81', 1.0)
1.0
('\x00\x00\x00\x82', 2.0)
2.0
('[EMAIL PROTECTED]', 3.0)
3.0
('\x00\x00\x00\x83', 4.0)
4.0
('\x00\x00 \x83', 5.0)
5.0
('\xcd\xcc\x0c\x81', 1.100238418579) 1.1
('\xcd\xcc\x0c\x82', 2.200476837158) 2.2
('33S\x82', 3.299523162842) 3.3
('\xcd\xcc\x0c\x83', 4.400953674316) 4.4
It is not apparent whether you regard the output from the function as
correct or not.
4.4 converted to mbf4 format is '\xcd\xcc\x0c\x83' which is
4.400953674316 which is the closest possible mbf4 representation
of 4.4 (difference is 9.5e-008).
The next lower mbf4 value '\xcc\xcc\x0c\x83' is 4.396185302734
(difference is -3.8e-007).
Note that floating-point representation of many decimal fractions is
inherently inexact. print repr(4.4) produces 4.4004
Have you read this:
http://docs.python.org/tut/node16.html
?
If you need decimal-fraction output that matches what somebody typed
into the original software, or saw on the screen, you will need to
know/guess the precision that was involved, and round the numbers
accordingly -- just like the author of the original software would
have needed to do.
['%.*f' % (decplaces, 4.400953674316) for decplaces in range(10)]
['4', '4.4', '4.40', '4.400', '4.4000', '4.4', '4.40',
'4.401', '4.4010', '4.40095']
HTH,
John
--
http://mail.python.org/mailman/listinfo/python-list
msbin to ieee
Hi Does anyone have the python version of the conversion from msbin to ieee? Thank u -- http://mail.python.org/mailman/listinfo/python-list
Re: msbin to ieee
En Sun, 06 May 2007 18:44:07 -0300, revuesbio [EMAIL PROTECTED] escribió: Does anyone have the python version of the conversion from msbin to ieee? I imagine this will be done just once - msbin is a really old format. Instead of coding the conversion in Python, you could: - write a small QuickBasic program using the functions CVSMBF, CVDMBF to do the conversion - download this DLL http://www.microdexterity.com/products.html -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
Re: msbin to ieee
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 -- http://mail.python.org/mailman/listinfo/python-list
