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revuesbio
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Hi
Does anyone have the python version of the conversion from msbin to ieee? Thank u 




revuesbio 


 
Gabriel Genellina
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En Sun, 06 May 2007 18:44:07 0300, revuesbio <(EMail Removed)>
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 




Gabriel Genellina 


 
John Machin
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On May 7, 7:44 am, revuesbio <(EMail Removed)> 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/pyt...st/337817.html HTH, John 




John Machin 
revuesbio
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On 7 mai, 03:52, John Machin <(EMail Removed)> wrote:
> On May 7, 7:44 am, revuesbio <(EMail Removed)> 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/pyt...st/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' 




revuesbio 
John Machin
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On May 7, 6:18 pm, revuesbio <(EMail Removed)> wrote:
> On 7 mai, 03:52, John Machin <(EMail Removed)> wrote: > > > On May 7, 7:44 am, revuesbio <(EMail Removed)> 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/pyt...st/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 <<  m1) <<  m0 adj = 24 + 128 num = mant * 2.0 ** (exponent  adj) if sign: return num return num HTH, John 




John Machin 
revuesbio
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On 7 mai, 13:21, John Machin <(EMail Removed)> wrote:
> On May 7, 6:18 pm, revuesbio <(EMail Removed)> wrote: > > > > > On 7 mai, 03:52, John Machin <(EMail Removed)> wrote: > > > > On May 7, 7:44 am, revuesbio <(EMail Removed)> 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/pyt...st/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 <<  m1) <<  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*) 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 bitstring manipulation (I'm a newbie). informations about bitstring manipulation with python is too poor on the net. thank you very much for your script. A. 




revuesbio 
John Machin
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On May 7, 10:00 pm, revuesbio <(EMail Removed)> wrote:
> On 7 mai, 13:21, John Machin <(EMail Removed)> wrote: > > > > > On May 7, 6:18 pm, revuesbio <(EMail Removed)> wrote: > > > > On 7 mai, 03:52, John Machin <(EMail Removed)> wrote: > > > > > On May 7, 7:44 am, revuesbio <(EMail Removed)> 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/pyt...st/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 <<  m1) <<  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*) > 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 bitstring > manipulation (I'm a newbie). informations about bitstring > manipulation with python is too poor on the net. The basic operations (and, or, exclusiveor, 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. 




John Machin 
revuesbio
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On 7 mai, 14:56, John Machin <(EMail Removed)> wrote:
> On May 7, 10:00 pm, revuesbio <(EMail Removed)> wrote: > > > > > On 7 mai, 13:21, John Machin <(EMail Removed)> wrote: > > > > On May 7, 6:18 pm, revuesbio <(EMail Removed)> wrote: > > > > > On 7 mai, 03:52, John Machin <(EMail Removed)> wrote: > > > > > > On May 7, 7:44 am, revuesbio <(EMail Removed)> 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/pyt...st/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 <<  m1) <<  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*) > > 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 bitstring > > manipulation (I'm a newbie). informations about bitstring > > manipulation with python is too poor on the net. > > The basic operations (and, or, exclusiveor, 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', 5.8774717541114375e039) 0.0 ('\x00\x00\x00\x81', 1.0) 1.0 ('\x00\x00\x00\x82', 2.0) 2.0 ('\x00\x00@\x82', 3.0) 3.0 ('\x00\x00\x00\x83', 4.0) 4.0 ('\x00\x00 \x83', 5.0) 5.0 ('\xcd\xcc\x0c\x81', 1.1000000238418579) 1.1 ('\xcd\xcc\x0c\x82', 2.200000047683715 2.2 ('33S\x82', 3.2999999523162842) 3.3 ('\xcd\xcc\x0c\x83', 4.4000000953674316) 4.4 ('\x00\x00z\x8a', 1000.0) 1000.0 




revuesbio 
John Machin
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On May 7, 11:37 pm, revuesbio <(EMail Removed)> wrote:
> On 7 mai, 14:56, John Machin <(EMail Removed)> wrote: > > > > > On May 7, 10:00 pm, revuesbio <(EMail Removed)> wrote: > > > > On 7 mai, 13:21, John Machin <(EMail Removed)> wrote: > > > > > On May 7, 6:18 pm, revuesbio <(EMail Removed)> wrote: > > > > > > On 7 mai, 03:52, John Machin <(EMail Removed)> wrote: > > > > > > > On May 7, 7:44 am, revuesbio <(EMail Removed)> 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/pyt...st/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 <<  m1) <<  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*) > > > 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 bitstring > > > manipulation (I'm a newbie). informations about bitstring > > > manipulation with python is too poor on the net. > > > The basic operations (and, or, exclusiveor, 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', 5.8774717541114375e039) 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 > ('\x00\x00@\x82', 3.0) > 3.0 > ('\x00\x00\x00\x83', 4.0) > 4.0 > ('\x00\x00 \x83', 5.0) > 5.0 > ('\xcd\xcc\x0c\x81', 1.1000000238418579) 1.1 > ('\xcd\xcc\x0c\x82', 2.200000047683715 2.2 > ('33S\x82', 3.2999999523162842) 3.3 > ('\xcd\xcc\x0c\x83', 4.4000000953674316) 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 4.4000000953674316 which is the closest possible mbf4 representation of 4.4 (difference is 9.5e00. The next lower mbf4 value '\xcc\xcc\x0c\x83' is 4.3999996185302734 (difference is 3.8e007). Note that floatingpoint representation of many decimal fractions is inherently inexact. print repr(4.4) produces 4.4000000000000004 Have you read this: http://docs.python.org/tut/node16.html ? If you need decimalfraction 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, 4.4000000953674316) for decplaces in range(10)] ['4', '4.4', '4.40', '4.400', '4.4000', '4.40000', '4.400000', '4.4000001', '4.40000010', '4.400000095'] HTH, John 




John Machin 
revuesbio
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On 7 mai, 23:38, John Machin <(EMail Removed)> wrote:
> On May 7, 11:37 pm, revuesbio <(EMail Removed)> wrote: > > > > > On 7 mai, 14:56, John Machin <(EMail Removed)> wrote: > > > > On May 7, 10:00 pm, revuesbio <(EMail Removed)> wrote: > > > > > On 7 mai, 13:21, John Machin <(EMail Removed)> wrote: > > > > > > On May 7, 6:18 pm, revuesbio <(EMail Removed)> wrote: > > > > > > > On 7 mai, 03:52, John Machin <(EMail Removed)> wrote: > > > > > > > > On May 7, 7:44 am, revuesbio <(EMail Removed)> 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/pyt...st/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 <<  m1) <<  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*) > > > > 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 bitstring > > > > manipulation (I'm a newbie). informations about bitstring > > > > manipulation with python is too poor on the net. > > > > The basic operations (and, or, exclusiveor, 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', 5.8774717541114375e039) 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 > > ('\x00\x00@\x82', 3.0) > > 3.0 > > ('\x00\x00\x00\x83', 4.0) > > 4.0 > > ('\x00\x00 \x83', 5.0) > > 5.0 > > ('\xcd\xcc\x0c\x81', 1.1000000238418579) 1.1 > > ('\xcd\xcc\x0c\x82', 2.200000047683715 2.2 > > ('33S\x82', 3.2999999523162842) 3.3 > > ('\xcd\xcc\x0c\x83', 4.4000000953674316) 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 > 4.4000000953674316 which is the closest possible mbf4 representation > of 4.4 (difference is 9.5e00. > > The next lower mbf4 value '\xcc\xcc\x0c\x83' is 4.3999996185302734 > (difference is 3.8e007). > > Note that floatingpoint representation of many decimal fractions is > inherently inexact. print repr(4.4) produces 4.4000000000000004 > > Have you read this: > http://docs.python.org/tut/node16.html > ? > > If you need decimalfraction 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, 4.4000000953674316) for decplaces in range(10)] > > ['4', '4.4', '4.40', '4.400', '4.4000', '4.40000', '4.400000', > '4.4000001', '4.40000010', '4.400000095'] > > HTH, > John another couples and round number corresponding to the right value ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('0\x8b\x01\x95', 1061222.0, '1061222.000') ('\xb8\x1e=\x83', 5.9099998474121094, '5.910') (')\\O\x83', 6.4800000190734863, '6.480') ('\x9a\x99A\x83', 6.0500001907348633, '6.050') ('\x00\x00P\x83', 6.5, '6.500') ('8BY\x95', 1779783.0, '1779783.000') ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('\xe0\xa0\x02\x95', 1070108.0, '1070108.000') ('33{\x83', 7.8499999046325684, '7.850') ('q=z\x83', 7.820000171661377, '7.820') ('33s\x83', 7.5999999046325684, '7.600') (')\\\x7f\x83', 7.9800000190734863, '7.980') ('\x00\x9aX\x92', 221800.0, '221800.000') ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('0\xa1\x02\x95', 1070118.0, '1070118.000') ('\x85\xebq\x83', 7.559999942779541, '7.560') ('\x14\xaeo\x83', 7.4899997711181641, '7.490') ('\xcd\xccT\x83', 6.6500000953674316, '6.650') ('\x00\x00p\x83', 7.5, '7.500') ('\x00\xa4N\x92', 211600.0, '211600.000') ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('\x90\xa1\x02\x95', 1070130.0, '1070130.000') ('\xaeGa\x83', 7.0399999618530273, '7.040') ('\xc3\xf5p\x83', 7.5300002098083496, '7.530') ('\x8f\xc2e\x83', 7.179999828338623, '7.180') ('H\xe1b\x83', 7.0900001525878906, '7.090') ('\xc0\xe27\x93', 376598.0, '376598.000') ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('\x08\xa4\x02\x95', 1070209.0, '1070209.000') ('\x9a\x99a\x83', 7.0500001907348633, '7.050') ('\xd7\xa3x\x83', 7.7699999809265137, '7.770') ('H\xe1r\x83', 7.5900001525878906, '7.590') ('{\x14v\x83', 7.690000057220459, '7.690') ('\x80.W\x93', 440692.0, '440692.000') ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('h\xa4\x02\x95', 1070221.0, '1070221.000') ('\x8f\xc2\x01\x84', 8.1099996566772461, '8.110') ('=\n\x03\x84', 8.1899995803833008, '8.190') ('\xcd\xcc\x00\x84', 8.0500001907348633, '8.050') ('ffv\x83', 7.6999998092651367, '7.700') ('\x80X\x1a\x94', 632200.0, '632200.000') ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('\x08\xa7\x02\x95', 1070305.0, '1070305.000') ('33s\x83', 7.5999999046325684, '7.600') ('q=r\x83', 7.570000171661377, '7.570') ('\\\x8fj\x83', 7.3299999237060547, '7.330') ('33k\x83', 7.3499999046325684, '7.350') ('\xc0a\r\x94', 579100.0, '579100.000') ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('X\xa7\x02\x95', 1070315.0, '1070315.000') ('\xcd\xcc\x83', 7.9000000953674316, '7.900') ('q=z\x83', 7.820000171661377, '7.820') ('\x00\x00p\x83', 7.5, '7.500') ('\x00\x00p\x83', 7.5, '7.500') ('\x00\x1b7\x92', 187500.0, '187500.000') ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('\xb8\xa7\x02\x95', 1070327.0, '1070327.000') ('{\x14~\x83', 7.940000057220459, '7.940') ('\xcd\xcc\x04\x84', 8.3000001907348633, '8.300') ('\xe1z\x00\x84', 8.0299997329711914, '8.030') ('\xcd\xcc\x10\x84', 9.0500001907348633, '9.050') ('\x00R\x00\x95', 1051200.0, '1051200.000') ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('P\xaa\x02\x95', 1070410.0, '1070410.000') ('R\xb8\x1e\x84', 9.9200000762939453, '9.920') ('\xd7\xa3\x1c\x84', 9.7899999618530273, '9.790') ('\x85\xeb\x19\x84', 9.619999885559082, '9.620') ('\x9a\x99\x19\x84', 9.6000003814697266, '9.600') ('\x98\x1c\x0c\x95', 1147795.0, '1147795.000') ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('\xa0\xaa\x02\x95', 1070420.0, '1070420.000') ('=\n\x0f\x84', 8.9399995803833008, '8.940') ('ff\x0e\x84', 8.8999996185302734, '8.900') ('\xe1z\x0c\x84', 8.7799997329711914, '8.780') ('\x1f\x85\x0f\x84', 8.9700002670288086, '8.970') ('\x00\x1d&\x92', 170100.0, '170100.000') ('\x00\x00\x00\x02', 5.8774717541114375e039, '0.000') ('8\xad\x02\x95', 1070503.0, '1070503.000') ('\xf6(\x0c\x84', 8.7600002288818359, '8.760') ('\xe1z\x14\x84', 9.2799997329711914, '9.280') all is ok. thank u 




revuesbio 


 
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