spawn_multiple with mpi4py on Windows

[maartenb]

I'm trying to spawn multiple processes from python using mpi4py.

sub_comm = MPI.COMM_SELF.Spawn_multiple(["app.exe"]*nProcess,info=mpiInfo, args = args)

It fails because the root of my intel python distribution (C:\intelpython3\) gets prepended to the command.

[proxy:1:0@xxxx] HYD_spawn (..\windows\src\hydra_spawn.c:231): unable to run process c:\intelpython3/app.exe (error code 2)
[proxy:1:0@xxxx] launch_processes (proxy.c:603): error creating process (error code 2). The system cannot find the file specified.

I tried passing an info object with the "wdir" set to the location of "app.exe", but it didn't help. How can I prevent this?

Thanks,

Maarten

[maartenb · Accepted solution]

OK I solved it. Turning on debug information by setting environment variable I_MPI_HYDRA_DEBUG=on showed that there's another parameter in the MPI Info object called "path", which is set by default to the location of the python distribution. The wdir directory is assumed to be relative to this path. Setting it to the appropriate directory and passing it to Spawn_multiple() solved it.

mpiInfo = MPI.Info.Create()
mpiInfo.Set('path','path/to/set')
mpiInfo.Set('wdir','working/directory')
sub_comm = MPI.COMM_SELF.Spawn_multiple(["app.exe"]*nProcess,info=[mpiInfo]*nProcess, args = args)

[RaeesaM_Intel]

Hi Maarten,

Glad to know that your issue is resolved and thank you for sharing the solution as it will be beneficial for the community.We will no longer monitor this thread and please feel free to raise a new thread if you have any further queries.

Regards,

Raeesa