CODEWITHSUNDEEP
pythonsympy
user6451264
user6451264

Reputation:

Set default values for sympy lambdify arguments

I know that sympy’s lambdify function can create a lambda with arguments variables and body expression like this:

lambdify(variables, expression)

so

lambdify((Symbol('x'), Symbol('y')) , 'x**2 + y')

produces

lambda x, y: x**2 + y

But how do I give default values to some of the variables? To get something like

lambda x, y=0: x**2 + y

?

I know I can do this with strings like lambdify(('x', 'y=0') , 'x**2 + y'), but I need to work with sympy symbol objects, since I’m extracting them from other sympy expressions.

Upvotes: 3

Views: 910

Answers (1)

asmeurer
asmeurer

Reputation: 91620

The fact that it works with strings is an accident. The way lambdify works is it creates a string form of the expression, then evals the string "lambda %s: (%s)" % (', '.join(map(str, args)), expr) (it's actually a little more complicated than that in general, but that's the basic idea). So if args is ('x', 'y=0'), that ends up being inserted as lambda x, y=0: ....

Right now lambdify doesn't have an option to make the parameters keyword arguments, so your best options are:

  • Create keyword arguments manually using strings (what you already discovered works), or

  • Wrap the lambdified function in a new lambda, like

      lambda x, y=0: lambdify((x, y), expr)(x, y)

    To do it in the general case, you can write some function that creates a keyword argument wrapper with whatever logic you want, like

      from functools import wraps

  def create_wrapped(args, expr):
      func = lambdify(args, expr)
      @wraps(func)
      def wrapped(*_args, **kwargs):
          # Write the logic here to parse _args and kwargs for the arguments as you want them
          return func(*actual_args)
      return wrapped

Upvotes: 1

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