Precision of variables saved using O.save

Hi,

Type of saved variables obey to what is coded in YADE_CLASS_BASE_DOC_ATTRS_CTOR_..() block in hpp files. E.g. the radius of a sphere is a "Real" (= double ? you should check yourself to what typedef this "Real" corresponds to in Yade) according to https://github.com/yade/trunk/blob/master/pkg/common/Sphere.hpp#L13

But, I do not know if this answers actually to your question..

Jerome

Hi,
Precision depends on which format you save to. If you save in binary (filename.yade) it will have the maximum precision.
If you save in xml there is a conversion to string, and in that case I'm not sure what happens. You could check that in boost::serialization.
I hope it helps.
Bruno

Hello,

you can also easily try it yourself:

import random; r=random.random
O.bodies.append(sphere((r(),r(),r()),1))
base = '/tmp/a.'
for ext in 'yade','xml':
print O.bodies[0].state.pos[0]
O.save(base+ext)
O.reset()
O.load(base+ext)
print O.bodies[0].state.pos[0]

my output:
0.872136848902
0.872136848902
0.872136848902
0.872136848902

it means that in both .yade and .xml format, the numbers (double precision
numbers) are saved in maximum possible precision. Do you have any problems
with O.save/O.load?

cheers
Jan

2015-01-21 9:41 GMT+01:00 Bruno Chareyre <
<email address hidden>>:

> Question #260626 on Yade changed:
> https://answers.launchpad.net/yade/+question/260626
>
> Bruno Chareyre proposed the following answer:
> Hi,
> Precision depends on which format you save to. If you save in binary
> (filename.yade) it will have the maximum precision.
> If you save in xml there is a conversion to string, and in that case I'm
> not sure what happens. You could check that in boost::serialization.
> I hope it helps.
> Bruno
>
> --
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@Jan
The fact that two floats are converted to the same string with "print" does not imply that they are equal. Before printing a number, python is rounding and truncating.

Real number representation is still a mystery to me, I know just enough to know that it is never as simple as one expects.
For instance, this is not a bug:
>>> x=0.1; y=0.2
>>> x+y
0.30000000000000004
>>> print x+y
0.3

This being said, your conclusion may be right (I'm still not 100% sure) even though the proof was flawn, since a variant of your script with "all" digits displayed gives this:
0.2017354310946590789654919717577286064624786376953125
0.2017354310946590789654919717577286064624786376953125
0.2017354310946590789654919717577286064624786376953125
0.2017354310946590789654919717577286064624786376953125
___
import random; r=random.random
from decimal import Decimal #see https://docs.python.org/2/tutorial/floatingpoint.html
O.bodies.append(sphere((r(),r(),r()),1))
base = '/tmp/a.'
for ext in 'yade','xml':
 print Decimal(O.bodies[0].state.pos[0])
 O.save(base+ext)
 O.reset()
 O.load(base+ext)
 print Decimal(O.bodies[0].state.pos[0])

See the hyperlink in #4
;-)

Thanks Jan Stránský, that solved my question.