Another possibility for your consideration ![]()
I am using PyPlot backend, but I do not believe as intensely as you are,
and this worked out OK to list the Options about managing a Python installation
pkg> add PyPlot
==THEN I chose==>
pkg> add Conda ## whereupon Julia configured PyCall to use a Julia-specific Python
and so Julia Pkg Handled everything beautifully
Here is why I selected “pkg> Add Conda” :
Another alternative is to configure PyCall to **use a Julia-specific Python **
**distribution via the Conda.jl package (which installs a private Anaconda **
Python distribution), which has the advantage that packages can be installed
and kept up-to-date via Julia. As explained in the PyCall documentation,
set ENV[“PYTHON”]=“” , ## ## Run ONCE THEN Comment out – ENV[“PYTHON”]=“”
run Pkg.build(“PyCall”)
Below are just some notes I scratched down , HTH
(v1.4-EXP3) pkg> add PyCall
- PyCall v1.92.2
build PyCall ## again
julia> using PyCall
[ Info: Precompiling PyCall
(v1.4-EXP3) pkg> build PyCall
Building Conda ─→ ~/.julia/packages/Conda/tJJuN/deps/build.log
Building PyCall → ~/.julia/packages/PyCall/tqyST/deps/build.log
.
and re-launch Julia. Then,
To install the matplotlib module, you can use
pyimport_conda("matplotlib", PKG) ,
where PKG is the Anaconda package that contains the module matplotlib,
.
julia> pyimport_conda("matplotlib", "PKG")
[ Info: Installing matplotlib via the Conda PKG package…