How to use the lambda argument of smooth.spline in RPy WITHOUT Python interprating it as lambda

Question

I want to use the natural cubic smoothing splines smooth.spline from R in Python (like som many others want as well (Python natural smoothing splines, Is there

Answer 1

This little trick will work around the specific problem you're having, by allowing you to write "lambda" in a string.

kwargs = {"x": r_x, "y": r_y, "lambda":  42}
spline1 = r_smooth_spline(**kwargs)

In the general case, you can pass around argument containers easily with tuples and dicts.

# as normal
f = function("foo", "bar", my_kwarg="my_value")

# the same call using argument containers
args = ("foo", "bar")
kwargs = {"my_kwarg": "my_value"}
f = function(*args, **kwargs)