CODEWITHSUNDEEP
pythonmpigilmpi4pymultiple-processes
kilojoules
kilojoules

Reputation: 10093

mpi4py append value id to list for each communicator instance

If I want to list each communicator's id of a variable, how could I do that? Below is an attempt to demonstrate this idea:

from mpi4py import MPI
comm = MPI.COMM_WORLD

obj = "I am an example. My ID is unique to each communicator."
mpi_id = 'rank %i has id %s'%(comm.rank, str(id(obj)))
comm.send(mpi_id, tag=11, dtest=comm.rank)

mpi_id_list = []
for i in range(comm.size):
    mpi_id_list.append( comm.recv(source=i, tag=11))

print mpi_id_list

Upvotes: 0

Views: 1243

Answers (1)

francis
francis

Reputation: 9817

In MPI, each comm.send(...,dest=x) should be matched by a comm.recv(...) executed by the process of rank x. All messages can be sent to the process of rank 0 and the process 0 must receive all these messages. This operation is a collective operation called a reduction.

The following code can be executed on 4 processes by typing mpirun -np 4 main.py

from mpi4py import MPI
comm = MPI.COMM_WORLD

obj = "I am an example. My ID is unique to each communicator."
mpi_id = 'rank %i has id %s'%(comm.rank, str(id(obj)))
comm.send(mpi_id, tag=11, dest=0)

mpi_id_list = []
if comm.rank==0:
   mpi_id_list = []
   for i in range(comm.size):
      mpi_id_list.append( comm.recv(source=i, tag=11))

   print mpi_id_list

#broadcasting the list
mpi_id_list = comm.bcast(mpi_id_list, root=0)

#now, the list is the same on all processes.
print "rank "+str(comm.rank)+" has list "+str(mpi_id_list)

Notice that this example makes use of the collective operation comm.bcast() to broadcast the resulting list to all processes. See https://mpi4py.scipy.org/docs/usrman/tutorial.html for mpi4py examples of different collective operations. For instance, you be tempted by the comm.allreduce() operation:

list=comm.allreduce([mpi_id])
print list

Upvotes: 2

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