Creating a callable with numexpr

Question

I'm doing some symbolic math with sympy, then generating a Python lambda function using eval and sympy's lambdastr utility. Here's a simplified example of what I mean:

import sympy
import numpy as np
from sympy.utilities.lambdify import lambdastr

# simple example expression (my use-case is more complex)
expr = sympy.S('b*sqrt(a) - a**2')
a, b = sorted(expr.free_symbols, key=lambda s: s.name)

func = eval(lambdastr((a,b), expr), dict(sqrt=np.sqrt))

# call func on some numpy arrays
foo, bar = np.random.random((2, 4))
print func(foo, bar)

This works, but I don't like the use of eval, and sympy doesn't necessarily generate computationally-efficient code. Instead, I'd like to use numexpr, which seems perfect for this use-case:

import numexpr
print numexpr.evaluate(str(expr), local_dict=dict(a=foo, b=bar))

The only problem is that I'd like to generate a callable (like the func lambda), instead of calling numexpr.evaluate every time. Is this possible?

I'm not sure I'm following with the ask here ... But can't you just wrap the `numexpr.evaluate` call in a `lambda` function? — mgilson, Apr 22 '15 at 19:31
I suppose I could, but would that incur the cost of re-parsing the expression string at every evaluation? — perimosocordiae, Apr 22 '15 at 19:32
Can you [`lambdify`](http://docs.sympy.org/latest/modules/utilities/lambdify.html)? — miradulo, Apr 22 '15 at 19:33
Oh, wow, I completely missed `lambdify`. If you make that an answer, I'll accept it. Thanks! — perimosocordiae, Apr 22 '15 at 19:35
numexpr expressions are cached, so reuse is fast: https://github.com/pydata/numexpr/wiki/Numexpr-Users-Guide#usage-notes — mgilson, Apr 22 '15 at 19:36

miradulo · Accepted Answer · 2015-04-22T20:10:16.160

2

You can make use of the lambdify module, which allows you to transform SymPy expressions to lambda functions for efficient calculation. This is nice in its ability to return a function with an attached implementation.

Lambdifying your own function could look something like this:

func = lambdify((a,b),expr, dict(sqrt=np.sqrt))

edited Apr 22 '15 at 20:10

answered Apr 22 '15 at 20:04

miradulo

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Works great, though you don't need the parens around `expr`. Thanks! – perimosocordiae Apr 22 '15 at 20:09

Creating a callable with numexpr

1 Answers1