Steven G. Johnson

Works for me on M1 (with Conda python). What python are you trying to link with?

PythonCall has something like this with `@py`.

This works and is not much more code than the proposed syntax: ```jl gaussian_filter = pyimport("scipy.ndimage").gaussian_filter blur = pyimport("scipy.ndimage").gaussian_filter ``` In general, I'm not super excited about imitating Python syntax...

Yes, compiling in the `libpython` into PyCall was something we did to improve load times (#169). Nowadays, it might be possible to revisit this; see e.g. the [PythonCall.jl package](https://github.com/cjdoris/PythonCall.jl) for...

I wonder if it's as simple as inserting the following line [here](https://github.com/JuliaPy/PyCall.jl/blob/2728e20162db3c90198e2828c40e675924c73528/deps/build.jl#L84): ```jl libpython = relpath.(libpython, @__DIR__) ``` Want to give it a try?

I think it's fine to drop 1.6 support going forward.

See also JuliaLang/julia#12644 — I thought Julia had fixed this by using LLVM symbol versioning?

Isn't that correct? We are not running `python.exe`.

Oh, I see, the definition of `sys.executable` in the Python docs is > A string giving the absolute path of the executable binary for the **Python interpreter**, on systems where...

It looks like `sys.executable` is [set from the configuration](https://github.com/python/cpython/blob/595225e86dcc6ea520a584839925a878dce7a9b2/Python/sysmodule.c#L2967), and apparently we can use [this API](https://docs.python.org/3/c-api/init_config.html) to change it when we initialize Python by using `Py_InitializeFromConfig`.