On 09/20/2013 01:31 AM, Aaron Meurer wrote:
On Tue, Sep 17, 2013 at 3:41 AM, Moritz Beber <[email protected]> wrote:
On Monday, 16 September 2013 20:09:52 UTC+2, Aaron Meurer wrote:
On Mon, Sep 16, 2013 at 5:56 AM, Moritz Beber <[email protected]> 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
I don't understand why this one is so slow, especially when the very
similar version below is fast.
Aaron Meurer
The reason is that subs iterates through all key, value pairs in the
dictionary to check for substitutions whereas the code below only
retrieves the two values of the symbols in expr.args from the otherwise
rather large dictionary.
In [10]:
%timeit expr.subs([(var, variables[var]) for var in expr.args])
10000 loops, best of 3: 58.2 盜 per loop
In [11]:
%timeit expr.subs([(var, variables[var]) for var in expr.args],
simultaneous=True)
1000 loops, best of 3: 957 盜 per loop
In [13]:
%timeit expr.xreplace([(var, variables[var]) for var in expr.args])
10000 loops, best of 3: 91.7 盜 per loop
In [8]:
%timeit expr.xreplace(variables)
10000 loops, best of 3: 37.2 盜 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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