On Monday, 16 September 2013 20:09:52 UTC+2, Aaron Meurer wrote:
>
> On Mon, Sep 16, 2013 at 5:56 AM, Moritz Beber
> <moritz...@gmail.com<javascript:>>
> wrote:
> > Dear all,
> >
> > My question basically has two parts:
> >
> > 1.) I have a number of (a few thousand) logical expressions each
> consisting
> > of a handful of symbols (in total there are several thousand symbols as
> > well). Currently, I generate a dictionary (let's call it 'big_dict')
> with
> > the symbols and their truth values. Then I loop through each expression
> and
> > evaluate it by calling expr.subs(big_dict). This is painfully slow. I
> looked
> > through the source code for 'subs' a little and saw that it roughly
> > translates the dict into an iterable of old, new pairs which it then
> loops
> > through in order to apply replacements. Is there a more efficient way? I
> > guess what I could code but actually expect an evaluation of a boolean
> > expressions to do is: using the symbols in it, extract the values and
> solve.
>
> You can try using xreplace, or (I belive) subs(simultaneous=True),
> which should only walk the expression tree once.
>
xreplace was the best option, indeed. Here are some timings that might be
informative:
In [4]:
from sympy.abc import a, b, c, x, y, z
In [5]:
expr = sympy.sympify("a & b")
In [6]:
variables = dict(zip(range(100), range(100)))
variables[a] = True
variables[b] = False
In [7]:
%timeit expr.subs(variables)
100 loops, best of 3: 2.22 ms per loop
In [9]:
%timeit expr.subs(variables, simultaneous=True)
100 loops, best of 3: 3.12 ms per loop
In [10]:
%timeit expr.subs([(var, variables[var]) for var in expr.args])
10000 loops, best of 3: 58.2 µs per loop
In [11]:
%timeit expr.subs([(var, variables[var]) for var in expr.args],
simultaneous=True)
1000 loops, best of 3: 957 µs per loop
In [13]:
%timeit expr.xreplace([(var, variables[var]) for var in expr.args])
10000 loops, best of 3: 91.7 µs per loop
In [8]:
%timeit expr.xreplace(variables)
10000 loops, best of 3: 37.2 µs per loop
>
> >
> > 2.) Since I have many such expressions, am I missing a way to solve the
> > whole system?
>
> I'm not clear what you mean by "solve", but look at simplify_logic()
> and satisfiable().
>
Sorry for being unclear. By 'solve' I meant evaluate the expressions as
above with actual truth values. I can't help but thinking, "if only this
were linear algebra" since I basically have a matrix with expressions and
involved symbols and then want to find the result of applying a vector with
a certain conditions. I just don't know if there is a representation that
would allow me to map one onto the other. Either way, I'm happy with the
xreplace.
>
> Aaron Meurer
>
> >
> > Thank you for your insights,
> > Moritz
> >
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Cheers,
Moritz
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