问题
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 a Python equivalent to the smooth.spline function in R, Python SciPy UnivariateSpline vs R smooth.spline, ...))
Therefore I am using rpy2
like described in https://morioh.com/p/eb4151821dc4, but I want to set directly lambda
instead of spar
:
import rpy2.robjects as robjects
r_y = robjects.FloatVector(y_train)
r_x = robjects.FloatVector(x_train)
r_smooth_spline = robjects.r['smooth.spline'] #extract R function# run smoothing function
spline1 = r_smooth_spline(x=r_x, y=r_y, lambda=42)
#alternative: spline1 = r_smooth_spline(x=r_x, y=r_y, spar=0.7) would work fine, but I would like to control lambda dirctly
ySpline=np.array(robjects.r['predict'](spline1,robjects.FloatVector(x_smooth)).rx2('y'))
plt.plot(x_smooth,ySpline)
When I do this the line spline1 = r_smooth_spline(x=r_x, y=r_y, lambda=42)
doesn't work because Python has already a predefined interpretation of lambda
(you can see this from the blue code-highlighting of lambda
) :(
I want lambda
to be interpreted as the smoothing penalty parameter lambda.
If I replace lambda
by spar
I would get a natural cubic spline, but I want to control lambda
directly.
回答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)
回答2:
Perhaps you could use rpy2
's Function.rcall() method when calling smooth.spline
?
import rpy2.robjects as robjects
r_y = robjects.FloatVector(y_train)
r_x = robjects.FloatVector(x_train)
r_smooth_spline = robjects.r['smooth.spline']
args = (('x',r_x), ('y',r_y), ('lambda',42)) # pattern (('argname', value),...)
# import R's "GlobalEnv" to evaluate the function
from rpy2.robjects import globalenv
spline1 = r_smooth_spline.rcall(args, globalenv)
回答3:
You can use Python's **<dict>
in a function call to specify R named arguments that have a name that is not syntactically valid in Python.
See the documentation for more details: https://rpy2.github.io/doc/v3.2.x/html/robjects_functions.html
来源:https://stackoverflow.com/questions/58239011/how-to-use-the-lambda-argument-of-smooth-spline-in-rpy-without-python-interprati