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[BUG] Installation issue: const PyCall = Base.require
#287
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const PyCall = Base.require
One thing to try, which is a bit of a hack but might fix it — manually install and build PyCall in the base Julia environment: julia -e 'using Pkg; Pkg.install("PyCall"); Pkg.build("PyCall")' Run this line in bash in any directory, doesn’t matter which. (Make sure to have the same Python available on your path as if you were using This should add PyCall to the base env which might fix this. Let me know if this works. If not I would then try the solutions in #257 Cheers, |
Perhaps this is the same issue ...
Here's the test script:
Environment:
|
In my environment, PyCharm was invoking julia in:
instead of from my conda virtual environment:
The julia environment in: /Users/davidlaxer/.julia/juliaup/julia-1.9.0+0.x64.apple.darwin14/bin/julia did not have 'pycall' installed. |
Hm, this might be a different issue. Did you install PySR with pip instead of conda? |
Look like: $ python setup.py develop
|
Okay I think that's a different issue. If using conda's Julia then you likely would also need to use conda's PySR (https://github.com/conda-forge/pysr-feedstock) so it gets the paths correctly. |
Presumably fixed by #535, so closing. |
Describe the bug
I am using Python installed in a spack environment and want to use PySR (the following works with the system-python). So I installed Julia, added it to my
PATH
and then I do the installation instructions, which fail:Curiously it works calling the
julia
-library before:(I can run the example in this REPL)
Closing that REPL and trying to run the example again, I arrive at:
This is weird, do you have any idea what could be the problem? Should I raise this with the Pycall-Developers?
Version (please include the following information):
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