izaakm · January 8, 2021 14:20
diff --git a/mpi4py_trivial.py b/mpi4py_trivial.py
 #Suppose you have a collection of tasks, which in this example I'll assume is just running a function f.
 #If these tasks are completely separate and independent the most then you can parallelize them easily.
 #In this gist I'll show the simplest possible way to do this using mpi4py.
 #There are better ways to do this, in particular if the tasks vary significantly in time taken to run.

 import mpi4py.MPI

 def f(i):
  "A fake task - in this case let just open a file and write a number to it"
  #open file with name based on task number
  f=open("%d.txt"%i,"w")
  #write some info to it
  f.write("%d * 10 = %d\n"%(i, 10*i))
  #close the file
  f.close()

 #A list of all the tasks to do.  In your case you will probably build this task list in a more complex way.
 #You don't even need to build it in advance for this approach to work
 task_list = range(100)

 #main program loop.  This is the unparallelized verion, for comparison
 for task in task_list:
  f(task)
  
 #And now moving on the parallel version
  
 #mpi4py has the notion of a "communicator" - a collection of processors
 #all operating together, usually on the same program.  Each processor 
 #in the communicator is identified by a number, its rank,  We'll use that
 #number to split the tasks

 #find out which number processor this particular instance is,
 #and how many there are in total
 rank = mpi4py.MPI.COMM_WORLD.Get_rank()
 size = mpi4py.MPI.COMM_WORLD.Get_size()

 #parallelized version
 #the enumerate function gives us a number i in addition
 #to the task.  (In this specific case i is the same as task!  But that's
 #not true usually)
 for i,task in enumerate(task_list):
  #This is how we split up the jobs.
  #The % sign is a modulus, and the "continue" means
  #"skip the rest of this bit and go to the next time
  #through the loop"
  # If we had e.g. 4 processors, this would mean
  # that proc zero did tasks 0, 4, 8, 12, 16, ...
  # and proc one did tasks 1, 5, 9, 13, 17, ...
  # and do on.
  if i%size!=rank: continue
  print "Task number %d (%d) being done by processor %d of %d" % (i, task, rank, size)
  f(task)
	#Suppose you have a collection of tasks, which in this example I'll assume is just running a function f.
	#If these tasks are completely separate and independent the most then you can parallelize them easily.
	#In this gist I'll show the simplest possible way to do this using mpi4py.
	#There are better ways to do this, in particular if the tasks vary significantly in time taken to run.

	import mpi4py.MPI

	def f(i):
	"A fake task - in this case let just open a file and write a number to it"
	#open file with name based on task number
	f=open("%d.txt"%i,"w")
	#write some info to it
	f.write("%d * 10 = %d\n"%(i, 10*i))
	#close the file
	f.close()

	#A list of all the tasks to do. In your case you will probably build this task list in a more complex way.
	#You don't even need to build it in advance for this approach to work
	task_list = range(100)

	#main program loop. This is the unparallelized verion, for comparison
	for task in task_list:
	f(task)

	#And now moving on the parallel version

	#mpi4py has the notion of a "communicator" - a collection of processors
	#all operating together, usually on the same program. Each processor
	#in the communicator is identified by a number, its rank, We'll use that
	#number to split the tasks

	#find out which number processor this particular instance is,
	#and how many there are in total
	rank = mpi4py.MPI.COMM_WORLD.Get_rank()
	size = mpi4py.MPI.COMM_WORLD.Get_size()

	#parallelized version
	#the enumerate function gives us a number i in addition
	#to the task. (In this specific case i is the same as task! But that's
	#not true usually)
	for i,task in enumerate(task_list):
	#This is how we split up the jobs.
	#The % sign is a modulus, and the "continue" means
	#"skip the rest of this bit and go to the next time
	#through the loop"
	# If we had e.g. 4 processors, this would mean
	# that proc zero did tasks 0, 4, 8, 12, 16, ...
	# and proc one did tasks 1, 5, 9, 13, 17, ...
	# and do on.
	if i%size!=rank: continue
	print "Task number %d (%d) being done by processor %d of %d" % (i, task, rank, size)
	f(task)