Question

如果我想列出每个传播者的变量id，我怎么能这样做？以下是尝试证明这一想法：

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

Answer 1

在MPI中，每个comm.send(...,dest=x)都应与等级comm.recv(...)进程执行的x匹配。所有消息都可以发送到等级0的进程，进程0必须接收所有这些消息。该操作是称为缩减的集体操作。

通过键入mpirun -np 4 main.py

，可以在4个进程上执行以下代码

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)

请注意，此示例使用集合操作comm.bcast()将结果列表广播到所有进程。有关不同集合操作的mpi4py示例，请参阅here。例如，您受到comm.allreduce()操作的诱惑：

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

mpi4py将值id附加到每个通信器实例的列表中

1 个答案: