convert for into parfor

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gwur
gwur 2017-9-6
回答： Kris Fedorenko 2017-9-8
Hi,
I've read through some parfor information but I am struggling to understand some of the implementation.
At the moment I am using a for loop. For simplicity it looks something like this:
Itterations=1000;
MyVariable=[10,20];
for i=1:length(MyVariable)
X=MyVariable(i);
Table=[];
counter1=0;
counter2=0;
%nested loop with its own nested loop
for j=1:Itterations
MyInt=*random number generator to produce an integer*
counter1=counter1+MyInt;
Table=[Table;zeros(MyInt,5)];
*do some stuff on the items in the table, which is conditional on MyVariable value*
if *condition is met in the table == something*
for all rows that meet the condition (counting down)
Table(k,:)=[] %remove the row
counter2=counter2+number of rows removed;
end
end
end
display counters' value
end
What I would like to do is use multiple cpu cores, each core working on its own Table using MyVariable. Instead of running the inner nested loop twice (in this case), for each MyVariable value, I would like to run two instances of the nested loop. Is this possible? It would be great if there could be two instances of each variable that would not interact (effectively sandbox each instance of the nested loop) but I get the impression that this is not possible so I am trying to understand how I modify my code so I could use parfor. Help would be very much appreciated.
Thank you.

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回答（1 个）

Kris Fedorenko
Kris Fedorenko 2017-9-8
Hi!
If I understand your question correctly, you would like to parallelize working with each value from MyVariable, i.e. turn your most outer loop (for i=1:length(MyVariable)) into a parfor. I do not see any issues with that, as each iteration of that loop is independent (operates with and changes its own set of variables like Table). For examples of possible issues and fixes for them you can refer to this documentation page: https://www.mathworks.com/help/distcomp/nested-parfor-loops-and-for-loops.html
I would also be careful with your most inner loop (over rows of Table meeting a certain condition). It looks like you are looping over the rows of the Table while also removing rows from the Table. Consider the following example, where we try to remove every other row of A:
A = randn(10, 5);
for k=2:2:10
k
A(k,:) = [];
end
With this for loop we will run into index out of range problem (while also deleting the wrong rows, as the indices would shift as we remove rows). Instead we can accomplish the task like this:
A = randn(10, 5);
k=2:2:10;
A(k,:) = [];
You might have to use logical indexing to determine which rows to remove.
I hope this helps!
Kris

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