Sure. lambdify has a default set of translations (see
sympy.utilities.lambdify.NUMPY_TRANSLATIONS to see what they are), but
you can override them. The best way is to pass in a list for the
module argument to lambdify, where the first argument is a dictionary
of custom translations, like
In [35]: import numpy
In [36]: lambdify(x, Matrix([[x, 2], [3, 4]]), [{'ImmutableMatrix':
numpy.array}, "numpy"])(1)
Out[36]:
array([[1, 2],
[3, 4]])
(note that ImmutableMatrix should be used because that is what the
matrix gets converted to before it gets lambdified)
As you can imagine, you can use this functionality to translate SymPy
objects into whatever objects or functions you want.
Aaron Meurer
On Tue, Dec 10, 2013 at 4:55 PM, Nicholas Chisholm
<[email protected]> wrote:
> I'm new to sympy (and very impressed with what I can do with it), but have
> been unable to figure something out about converting certain matrix
> expressions into their "lambda function" form for fast evaluation.
>
> I have a sympy matrix containing symbols x, y, and z. If I invoke lambdify
> on this matrix for those symbols, I get a function that returns a numpy
> matrix (great!). Is it also possible to make it return a numpy array (rather
> than a matrix)? The reason I need this is because my matrix is actually a
> rank-3 tensor when the arguments x,y,z are supplied (which are vectors). In
> other words, I want to supply 3 equally sized vectors to my matrix to create
> an m-by-n-by-k numpy array where my original [sympy] matrix was m-by-n.
>
> For example, I could do:
>>>> fn = lambda x,y: np.array( [ [x+y, x-y], [x-y, y-x] ] )
>>>> fn(1,2)
> array([[ 3, -1],
> [-1, 1]]) # Gives back original array shape (2x2)
>>>>
>>>> arr1 = np.array([1,2,3])
>>>> arr2 = np.array([4,5,6])
>>>> fn(arr1, arr2)
> array([[[ 5, 7, 9],
> [-3, -3, -3]],
>
> [[-3, -3, -3],
> [ 3, 3, 3]]])
>>>> fn(arr1, arr2).shape # (2x2x3)
> (2L, 2L, 3L)
>
> I basically want to do the above with a sympy matrix rather than explicitly
> typing the lambda function.
>
> Thanks!
>
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