Re: [sympy] How to equip a Symbol with an additional attribute?
On 05/19/2015 07:48 AM, Joachim Durchholz wrote: An external dictionary isn't ideal but better than this, at least from the maintainer's perspective. Maybe we can give you better help if you describe your use case in some more detail. At some point I create symbols from which I know, that they are derivatives and I want to store their order. Using the classes Function and Derivative is no real option because I want the symbols to behave as symbols on most occasions. Ah. This sounds like a design problem in Derivative to me that we should fix. Can you post what you did, what you'd have needed to get out, and what SymPy gave you? I think strongest argument for me was that the string representation of the expressions gets too long for Functions and Derivatives. This reduces my chance to see any structure. Second argument is that I have a (own) library for modeling differential forms which internally uses Symbol and does not straight forwardly work with Function instances. E.g. x.name - 'x' x(t).name - AttributeError However, this discussions make me think, if using Functions would not be the better alternative In general, I think there are situations, where it might be useful to store some algorithm-specific extra information to some symbols. For algorithms of SymPy itself, I think the algorithms should be fixed :-) It would be useful for coding external algorithms, but it does come with ramifications (most importantly the potential for namespace conflicts); from a maintenance perpective, it's far better if such an algorithm does its own data management. OK. I can perfectly live with this. -- You received this message because you are subscribed to the Google Groups sympy group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at http://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/555B0556.2040908%40gmx.de. For more options, visit https://groups.google.com/d/optout.
Re: [sympy] How to equip a Symbol with an additional attribute?
On 05/19/2015 02:39 AM, Aaron Meurer wrote:
On Mon, May 18, 2015 at 12:29 PM, Joachim Durchholz j...@durchholz.org
wrote:
Am 18.05.2015 um 15:56 schrieb Carsten Knoll:
I want to equip an Symbol with an additional attribute to store some
specific information right in place.
For 'normal' Python classes it is no problem to dynamically create an
additional attribute for an already existing instance.
However, for sympy Symbols if I try
x = Symbol('x')
x.k = 0
I get the error
AttributeError: 'Symbol' object has no attribute 'k'
That happens because we use __slots__ for performance reasons.
However, storing attributes in that way is a bad idea in general, because
you risk name conflicts with future versions of SymPy. (That's also the
reason why using subclasses to extend a library tends to break over time.)
I'm not so sure about this.
The real reason that storing an attribute on x is a bad idea is that
SymPy implicitly assumes that it can replace one expression with the
same expression if the two are equal, so, e.g., if you had
x1 = Symbol('x')
x2 = Symbol('x')
x1.k = 0
and supposing it worked, along with x1 == x2 giving True, then SymPy
would assume that it has the freedom at any point to swap x1 with x2
and visa versa (and indeed, it does do this quite a bit with the
cache).
Thanks for pointing that out. I will add it to my tests.
I think what you really want to do is create a subclass of Symbol that
stores the attributes in .args, so that they become part of the
equality of the object. Joachim's alternative 1 (store the
information separately) also sounds like a good one (and probably much
simpler to implement).
Subclassing and adding attributes in .args indeed seems more complicated.
I will stick with the external dict solution.
Regards,
Carsten.
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[sympy] Bug with solve
Hello,
I tried to find the roots of : 2*sqrt(x)*sqrt(x**3 - x**2 + 1) + 2*x**3 -
x**2 + x*sqrt(x**3 - x**2 + 1) + 2*x**1.5 + 2*x**2.5*sqrt(x**3 - x**2 + 1)
- 3*x**2.5 - 4*x**3.5*sqrt(x**3 - x**2 + 1) - 2*x**3.5 + 5*x**4.5 -
3*x**5.5 - 1 = 0.
x = symbols('x', real=True)
f = Lambda(x, 2*sqrt(x)*sqrt(x**3 - x**2 + 1) + 2*x**3 - x**2 + x*sqrt(x**3
- x**2 + 1) + 2*x**1.5 + 2*x**2.5*sqrt(x**3 - x**2 + 1) - 3*x**2.5 -
4*x**3.5*sqrt(x**3 - x**2 + 1) - 2*x**3.5 + 5*x**4.5 - 3*x**5.5 - 1)
Input : solve(f(x), x)
Output : [0.149910458676117, 0.865324593883105, 1.00,
1.24896947842683, 2.56371129870321, 3.02429035798029]
But only 0.149910458676117 and 1 are really roots :
f(0.865324593883105) = 0.735269457206734
f(1.24896947842683) = -4.09736628717860
f(2.56371129870321) = -509.709822051607
f(3.02429035798029) = -1361.51149028959
Arnaud,
Python 3.4.2 (v3.4.2:ab2c023a9432, Oct 6 2014, 22:15:05) [MSC v.1600 32
bit (Intel)] on win32
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Re: [sympy] Bug with solve
Hi,
I have : Python 3.4.2 (v3.4.2:ab2c023a9432, Oct 6 2014, 22:15:05) [MSC
v.1600 32 bit (Intel)] on win32
Le mardi 19 mai 2015 18:25:33 UTC+2, Aaron Meurer a écrit :
What version of SymPy is this? On master I get
In [1]: x = symbols('x', real=True)
In [2]: f = Lambda(x, 2*sqrt(x)*sqrt(x**3 - x**2 + 1) + 2*x**3 - x**2
+ x*sqrt(x**3 - x**2 + 1) + 2*x**1.5 + 2*x**2.5*sqrt(x**3 - x**2 + 1)
- 3*x**2.5 - 4*x**3.5*sqrt(x**3 - x**2 + 1) - 2*x**3.5 + 5*x**4.5 -
3*x**5.5 - 1)
In [3]: solve(f(x), x)
Out[3]: [0.149910458676117, 1.0]
(note that it takes a while)
Aaron Meurer
On Tue, May 19, 2015 at 6:29 AM, Arnaud Usciati rai...@gmail.com
javascript: wrote:
Hello,
I tried to find the roots of : 2*sqrt(x)*sqrt(x**3 - x**2 + 1) + 2*x**3
-
x**2 + x*sqrt(x**3 - x**2 + 1) + 2*x**1.5 + 2*x**2.5*sqrt(x**3 - x**2 +
1) -
3*x**2.5 - 4*x**3.5*sqrt(x**3 - x**2 + 1) - 2*x**3.5 + 5*x**4.5 -
3*x**5.5 -
1 = 0.
x = symbols('x', real=True)
f = Lambda(x, 2*sqrt(x)*sqrt(x**3 - x**2 + 1) + 2*x**3 - x**2 +
x*sqrt(x**3
- x**2 + 1) + 2*x**1.5 + 2*x**2.5*sqrt(x**3 - x**2 + 1) - 3*x**2.5 -
4*x**3.5*sqrt(x**3 - x**2 + 1) - 2*x**3.5 + 5*x**4.5 - 3*x**5.5 - 1)
Input : solve(f(x), x)
Output : [0.149910458676117, 0.865324593883105, 1.00,
1.24896947842683, 2.56371129870321, 3.02429035798029]
But only 0.149910458676117 and 1 are really roots :
f(0.865324593883105) = 0.735269457206734
f(1.24896947842683) = -4.09736628717860
f(2.56371129870321) = -509.709822051607
f(3.02429035798029) = -1361.51149028959
Arnaud,
Python 3.4.2 (v3.4.2:ab2c023a9432, Oct 6 2014, 22:15:05) [MSC v.1600 32
bit
(Intel)] on win32
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Re: [sympy] Bug with solve
You gave Python version not sympy.__version__ -- You received this message because you are subscribed to the Google Groups sympy group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at http://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/d05d4cdb-11d4-4474-8543-a76bcf4e0018%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.
Re: [sympy] Bug with solve
Right, I have sympy-0.7.6.win32. -- You received this message because you are subscribed to the Google Groups sympy group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at http://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/f0619d17-cdc0-47dd-858c-6a6415d212d3%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.
Re: [sympy] How to equip a Symbol with an additional attribute?
Good points. So I conclude that it would be nice if somebody came up with a way to allow user-defined attributes without running into namespace conflicts. Oh, and the mechanism should be easy to use for the casual user. I have some ideas in that direction, but I'd like to see what others think first. Independently of that, I'd like to see somebody check whether the performace impact of __slots__ is still relevant; this may have changed with newer Python versions and I'd like to have some contemporary figures so we know what the actual trade-off is. -- You received this message because you are subscribed to the Google Groups sympy group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at http://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/555B7F6B.10002%40durchholz.org. For more options, visit https://groups.google.com/d/optout.
Re: [sympy] How to equip a Symbol with an additional attribute?
On Tue, May 19, 2015 at 4:41 AM, Carsten Knoll carstenkn...@gmx.de wrote: On 05/19/2015 07:48 AM, Joachim Durchholz wrote: An external dictionary isn't ideal but better than this, at least from the maintainer's perspective. Maybe we can give you better help if you describe your use case in some more detail. At some point I create symbols from which I know, that they are derivatives and I want to store their order. Using the classes Function and Derivative is no real option because I want the symbols to behave as symbols on most occasions. Ah. This sounds like a design problem in Derivative to me that we should fix. Can you post what you did, what you'd have needed to get out, and what SymPy gave you? I think strongest argument for me was that the string representation of the expressions gets too long for Functions and Derivatives. This reduces my chance to see any structure. Second argument is that I have a (own) library for modeling differential forms which internally uses Symbol and does not straight forwardly work with Function instances. E.g. x.name - 'x' x(t).name - AttributeError That's because x(t) is not really an instance of x. Symbol implements __call__ so that x(t) returns a Function object (a completely different class). Aaron Meurer However, this discussions make me think, if using Functions would not be the better alternative In general, I think there are situations, where it might be useful to store some algorithm-specific extra information to some symbols. For algorithms of SymPy itself, I think the algorithms should be fixed :-) It would be useful for coding external algorithms, but it does come with ramifications (most importantly the potential for namespace conflicts); from a maintenance perpective, it's far better if such an algorithm does its own data management. OK. I can perfectly live with this. -- You received this message because you are subscribed to the Google Groups sympy group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at http://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/555B0556.2040908%40gmx.de. For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups sympy group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at http://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/CAKgW%3D6Kq9AqMk1O97Of%2BPN_NUJEwf2obaCbYa4fHo45PdbFOWg%40mail.gmail.com. For more options, visit https://groups.google.com/d/optout.
[sympy] Magnitude of a complex number
Am I misunderstanding something here (using master): $ isympy Couldn't locate IPython. Having IPython installed is greatly recommended. ... z = (4+3*I)/(3-4*I) z 4 + 3⋅ⅈ ─── 3 - 4⋅ⅈ abs(z) │ 1 │ 5⋅│───│ │3 - 4⋅ⅈ│ simplify(abs(z)) │ 1 │ 5⋅│───│ │3 - 4⋅ⅈ│ abs(z).evalf() 1.00 abs(z) ** 2 2 │ 1 │ 25⋅│───│ │3 - 4⋅ⅈ│ z * z.conjugate() (4 - 3⋅ⅈ)⋅(4 + 3⋅ⅈ) ─── (3 - 4⋅ⅈ)⋅(3 + 4⋅ⅈ) simplify(z * z.conjugate()) 1 Why doesn't abs(z) simplify (or should I open an issue)? -- Oscar -- You received this message because you are subscribed to the Google Groups sympy group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at http://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/CAHVvXxS6DNZR8dPJ5AZ96V7MKtZQrvug_Pjs2%3DLejVOP6e73aw%40mail.gmail.com. For more options, visit https://groups.google.com/d/optout.
Re: [sympy] theano and sympy
Theano is doing floating point computations whereas SymPy does the
computation using arbitrary precision.
Jason
moorepants.info
+01 530-601-9791
On Tue, May 19, 2015 at 3:14 PM, drazi...@gmail.com wrote:
Hi,
I was experimenting with the following code:
import sympy
from sympy.abc import x
expr = sympy.exp(-x ** 2)
from sympy.printing.theanocode import theano_function
fn_theano = theano_function([x], [expr], dims={x: 1}, dtypes={x:
'float64'})
print fn_theano([29.]),sympy.exp(-29.**2)
I got
[ 0.] 5.73245586032578e-366
My question is why fn_theano does not give the same value?
Thank you,
Costas
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Re: [sympy] Bug with substituting derivatives
Well, that is exactly the problem, and what I think is a bugit should
not work like that..
Den tisdag 19 maj 2015 kl. 22:00:46 UTC+2 skrev Alexander Lindsay:
Not to hijack this post, but why does subs(phi(z,t).diff(z),...) work
when phi(z,t).diff(t,2) does not contain any derivatives of phi with
respect to z? It seems like this substitution is saying that
phi(z,t).diff(z) = phi(z,t)
I'm new to sympy, so I apologize if this is a stupid question.
Alex
On 05/19/2015 12:16 PM, Jonathan Lindgren wrote:
I recently updated from sympy 0.7.4 (I tihnk) to 0.7.6 and now I have some
very strange behaviour with subs. The following code
from sympy.abc import phi
import sympy as sp
z=sp.Symbol('z')
t=sp.Symbol('t')
sp.pprint((phi(z,t).diff(t,2)).subs(phi(z,t).diff(z),sp.Symbol('b')(z,t)).expand())
gives me the output
2
∂
───(b(z, t))
2
∂t
but I would expect the output
2
∂
───(φ(z, t))
2
∂t
This was working perfectly in my previous version of sympy.
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[sympy] theano and sympy
Hi,
I was experimenting with the following code:
import sympy
from sympy.abc import x
expr = sympy.exp(-x ** 2)
from sympy.printing.theanocode import theano_function
fn_theano = theano_function([x], [expr], dims={x: 1}, dtypes={x:
'float64'})
print fn_theano([29.]),sympy.exp(-29.**2)
I got
[ 0.] 5.73245586032578e-366
My question is why fn_theano does not give the same value?
Thank you,
Costas
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Re: [sympy] Bug with substituting derivatives
Not to hijack this post, but why does subs(phi(z,t).diff(z),...) work
when phi(z,t).diff(t,2) does not contain any derivatives of phi with
respect to z? It seems like this substitution is saying that
phi(z,t).diff(z) = phi(z,t)
I'm new to sympy, so I apologize if this is a stupid question.
Alex
On 05/19/2015 12:16 PM, Jonathan Lindgren wrote:
I recently updated from sympy 0.7.4 (I tihnk) to 0.7.6 and now I have
some very strange behaviour with subs. The following code
from sympy.abc import phi
import sympy as sp
z=sp.Symbol('z')
t=sp.Symbol('t')
sp.pprint((phi(z,t).diff(t,2)).subs(phi(z,t).diff(z),sp.Symbol('b')(z,t)).expand())
gives me the output
2
∂
───(b(z, t))
2
∂t
but I would expect the output
2
∂
───(φ(z, t))
2
∂t
This was working perfectly in my previous version of sympy.
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Re: [sympy] Bug with solve
What version of SymPy is this? On master I get
In [1]: x = symbols('x', real=True)
In [2]: f = Lambda(x, 2*sqrt(x)*sqrt(x**3 - x**2 + 1) + 2*x**3 - x**2
+ x*sqrt(x**3 - x**2 + 1) + 2*x**1.5 + 2*x**2.5*sqrt(x**3 - x**2 + 1)
- 3*x**2.5 - 4*x**3.5*sqrt(x**3 - x**2 + 1) - 2*x**3.5 + 5*x**4.5 -
3*x**5.5 - 1)
In [3]: solve(f(x), x)
Out[3]: [0.149910458676117, 1.0]
(note that it takes a while)
Aaron Meurer
On Tue, May 19, 2015 at 6:29 AM, Arnaud Usciati rait...@gmail.com wrote:
Hello,
I tried to find the roots of : 2*sqrt(x)*sqrt(x**3 - x**2 + 1) + 2*x**3 -
x**2 + x*sqrt(x**3 - x**2 + 1) + 2*x**1.5 + 2*x**2.5*sqrt(x**3 - x**2 + 1) -
3*x**2.5 - 4*x**3.5*sqrt(x**3 - x**2 + 1) - 2*x**3.5 + 5*x**4.5 - 3*x**5.5 -
1 = 0.
x = symbols('x', real=True)
f = Lambda(x, 2*sqrt(x)*sqrt(x**3 - x**2 + 1) + 2*x**3 - x**2 + x*sqrt(x**3
- x**2 + 1) + 2*x**1.5 + 2*x**2.5*sqrt(x**3 - x**2 + 1) - 3*x**2.5 -
4*x**3.5*sqrt(x**3 - x**2 + 1) - 2*x**3.5 + 5*x**4.5 - 3*x**5.5 - 1)
Input : solve(f(x), x)
Output : [0.149910458676117, 0.865324593883105, 1.00,
1.24896947842683, 2.56371129870321, 3.02429035798029]
But only 0.149910458676117 and 1 are really roots :
f(0.865324593883105) = 0.735269457206734
f(1.24896947842683) = -4.09736628717860
f(2.56371129870321) = -509.709822051607
f(3.02429035798029) = -1361.51149028959
Arnaud,
Python 3.4.2 (v3.4.2:ab2c023a9432, Oct 6 2014, 22:15:05) [MSC v.1600 32 bit
(Intel)] on win32
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Re: [sympy] How to equip a Symbol with an additional attribute?
Hi--
This seems as good a place as any to mention something I'd been trying to
do some time ago where I ran into the same battle versus __slots__.
I wanted to add a description attribute to my Symbols for documentation
purposes. I use Sympy almost exclusively inside of the IPython notebook
for engineering work, and wind up with long sheets of derivations,
substitutions, dead end attempts, etc. I'm also trying to build up a
personal library of formulas that I can import.
The problem I run in to is eventually, either due to complexity or time, I
can't keep track of what all my symbols mean. I like to keep the names
themselves short so the symbolic output is easily understandable, but that
relies on really strong conventions and a good memory to interpret-- which
aren't always present... =/
One solution was to put the description into the text of the notebook,
which I often do, but then I have to find the place where that Symbol was
defined to locate the descriptive text-- and then need to scroll back to
the active cell to edit, then back and forth.
The solution I was after was to add a x.desc string to the Symbol, so I
could just ask the Symbol itself what the hell I intended it to mean. I
then could modify Symbol to pretty print F_{B}: the force of the bat. If
I got really energetic, I'd change Expr to pretty print:
F_{B}=m_{b}*a_{b}
Newton's Second Law of Motion where:
F_{B} is the force of the bat
m_{b} is the mass of the ball
a_{b} is the acceleration of the ball
For all the maintenance reasons you describe, these are changes that should
happen in the Sympy library itself, but it's easier for me to experiment in
my own sandbox without upsetting the core code and the __slots__ prevents
this. There's also the chance that I can't convince enough people this
should be a core part of the library, or to implement it in the way I'd
like, and then I'm stuck.
Maintenance isn't quite as big an issue for me as I'm not building long
standing applications-- I'm generating documents. Sure, I'd like those to
remain viable for as long as possible so I can edit them in the future, but
it's not the same has having deployed a large codebase. If I have to go
through and change all my references from .desc to .desc2 for my notebook
to run, it's not the end of the world.
Storing all of this in an outside data structure does have the potential
advantage in that I could then look up a symbol if I knew what it meant but
not its symbolic representation:
SymbolIndex['accel ball']
or find all the formulas I have involving the ball:
ExpressionIIndex['ball']
ExpressionIndex[a_{b}]
but that would have required some deeper hacking into the Symbol creation
and destruction mechanisms.
Anyway, probably wandered off topic a bit, but I just wanted to put a vote
in for user attributes and the possible elimination of __slots__ if that
isn't a significant performance hit.
On Monday, May 18, 2015 at 10:29:42 AM UTC-7, Joachim Durchholz wrote:
Am 18.05.2015 um 15:56 schrieb Carsten Knoll:
I want to equip an Symbol with an additional attribute to store some
specific information right in place.
For 'normal' Python classes it is no problem to dynamically create an
additional attribute for an already existing instance.
However, for sympy Symbols if I try
x = Symbol('x')
x.k = 0
I get the error
AttributeError: 'Symbol' object has no attribute 'k'
That happens because we use __slots__ for performance reasons.
However, storing attributes in that way is a bad idea in general,
because you risk name conflicts with future versions of SymPy. (That's
also the reason why using subclasses to extend a library tends to break
over time.)
Alternative 1: You could set up a dict with a Symbol-to-attributes
mapping: Symbols are hashable (that's a guaranteed property).
Alternative 2: We could add a `userdict` attribute in Symbol, so that
SymPy users can store additional attributes there.
I'm not too happy with either alternative.
Maybe we can give you better help if you describe your use case in some
more detail.
Regards,
Jo
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[sympy] Bug with substituting derivatives
I recently updated from sympy 0.7.4 (I tihnk) to 0.7.6 and now I have some
very strange behaviour with subs. The following code
from sympy.abc import phi
import sympy as sp
z=sp.Symbol('z')
t=sp.Symbol('t')
sp.pprint((phi(z,t).diff(t,2)).subs(phi(z,t).diff(z),sp.Symbol('b')(z,t)).expand())
gives me the output
2
∂
───(b(z, t))
2
∂t
but I would expect the output
2
∂
───(φ(z, t))
2
∂t
This was working perfectly in my previous version of sympy.
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