CODEWITHSUNDEEP
scipyjuliainterpolationpycall
Mohammad Saad
Mohammad Saad

Reputation: 2005

Cannot assign custom radial for scipy rbf imported in julia through pycall?

I have imported scipy.interpolate.Rbf from python to Julia 1.6 using PyCall. It works, but when I am trying to assign the radial function value to multiquadric it doesn't allow me to do so, due to syntax issue. For example:

using PyCall
interpolate = pyimport("scipy.interpolate")

x = [1,2,3,4,5]
y = [1,2,3,4,5]
z = [1,2,3,4,5]

rbf = interpolate.Rbf(x,y,z, function='multiquadric', smoothness=1.0)

for the above example I am getting this error:

LoadError: syntax: unexpected "="
Stacktrace:
 [1] top-level scope

This is due to the use of function variable. May I know what can be the way around this error, so that i can assign the multiquadric radial for the rbf.

Look forward to the suggestions.

Thanks!!

Upvotes: 2

Views: 57

Answers (1)

Przemyslaw Szufel
Przemyslaw Szufel

Reputation: 42214

Your problem is that function is a reserved keyword in Julia. Additionally, note that 'string' for strings is OK in Python but in Julia you have "string".

The easiest workaround in your case is py"" string macro:

julia> py"$interpolate.Rbf($x,$y,$z, function='multiquadric', smoothness=1.0)"
PyObject <scipy.interpolate.rbf.Rbf object at 0x000000006424F0A0>

Now suppose you actually need at some point to pass function= parameter? It is not easy because function is a reserved keyword in Julia. What you could do is to pack it into a named tuple using Symbol notation:

myparam = NamedTuple{(:function,)}(("multiquadric",))

Now if you do:

some_f(x,y; myparam...)

than myparam would get unpacked to a keyword parameter.

Upvotes: 2

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