Spawn multiprocessing.Process under different python executable with own path -

i have 2 versions of python (these 2 conda environments)

/path/to/bin-1/python /path/to/bin-2/python 

from 1 version of python want launch function runs in other version using multiprocessing.process object. turns out doable using set_executable method:

ctx = multiprocess.get_context('spawn') ctx.set_executable('/path/to/bin-2/python') 

and indeed can see in fact launch using executable:

def f(q):     import sys     q.put(sys.executable)  if __name__ == '__main__':     import multiprocessing     ctx = multiprocessing.get_context('spawn')     ctx.set_executable('/path/to/bin-2/python')     q = ctx.queue()     proc = ctx.process(target=f, args=(q,))     proc.start()     print(q.get())  $ python foo.py /path/to/bin-2/python 

however path wrong

however when same thing sys.path rather sys.executable find sys.path hosting python process printed out instead, rather sys.path find running /path/to/bin-2/python -c "import sys; print(sys.path)" directly.

i'm used sort of thing if use fork. have expected 'spawn' act same though had entered python interpreter shell.

question

is possible use multiprocessing library run functions , use queues python executable environment have had had started shell?

more broadly, how sys.path populated , different between using multiprocessing in way , launching interpreter directly?