sympy.Function that returns a function

[Nico]

So far, I've always defined functions as

```

class f(sympy.Function):

     pass

```

which I could then conveniently use as

```

x = sympy.Symbol('x')

y = f(x)

```

Nice.

Now, I would like to define a function that itself returns a function, aka, an operator. It should be used as

```

y = op(u)(x)

```

With the naïve definition above, I'm getting

```

TypeError: 'op' object is not callable

```

Any hints?

Cheers,

Nico

[Aaron Meurer]

It depends if you want op(u) to also be usable in expressions. If you
do, the simplest way is to make op be a Function, and define __call__
on it so that it returns an object representing op(u)(x).

Defining that other object is the tricky part. I think something like

class AppliedOp(Expr):
def __new__(cls, op, x):
obj = Expr.__new__(cls, op, x)
return obj

@property
def func(self):
return self._args[0]

@property
def args(self):
return self._args[1:]

and op is

class op(Function):
def __call__(self, x):
return AppliedOp(self, x)

You'll also want to define some printing methods on AppliedOp so that
it prints as op(u)(x).

Aaron Meurer

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[Nico]

Thanks Anton for the tip!
I'll see what'll work best for me. (As a dirty workaround, I now defined the function to take two arguments, e.g., op(u, x).)

Cheers,

Nico

[Aaron Meurer]

If you don't need op(u) as a distinct expression that is way easier.
You can also define custom printing on it to make it display however
you want.

Aaron Meurer

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