How do I vectorize adding array elements based on indices?

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Derek Smith
Derek Smith 2019-9-23
编辑： Stephen23 2019-9-23
% I'm building a form of spiking neural network. Long story, but the reason
% for the inquiry is rather brief, and (I thought) simple.
% Background, if it helps. The network has neurons and synapses. Each synapse
% has a train (capturing all of the spikes) and destination (the index of the
% neuron to which the spike is added). Each neuron has a potential, to which
% input spikes are added (basically, 1s or 0s).
% So, we enter an optimization problem...
% When the program is "fully operational," the element sizes of the potential
% and train/destination can be upwards of 100,000 and 600,000,000,
% respectively! I would like to use vectorization, maybe even GPU Arrays,
% but I can't seem to figure out how to efficiently vectorize without the
% results being wrong!
% take the following arrays:
>> potential = [ 1 , 2 , 3 , 4 , 5 ];
>> train = [ 8 , 7 , 8 , 7 , 8 , 7 , 8 , 7 ];
>> destination= [ 1 , 1 , 1 , 2 , 2 , 2 , 3 , 4 ];
% run the following for-loop
>> for i = 1 : 8
potential( destination(i) ) = potential( destination(i) ) + mod( train(i) , 2 );
end
% arrive at a solution
>> potential
3 3 4 4 5
% try to vectorize: the data is wrong
>> potential(destination) = potential(destination) + mod(train,2)
1 3 3 5 5
% try to use "arrayfun": not even the right size, anymore!
>> potential = arrayfun( @(i) potential(destination(i))+mod(train(i),2), destination)
1 1 1 2 2 2 1 3
  1 个评论
Stephen23
Stephen23 2019-9-23
Your example output values do not match what your code actually produces:
>> potential = [1,2,3,4,5];
>> train = [8,7,8,7,8,7,8,7];
>> destination = [1,1,1,2,2,2,3,4];
>> for k = 1:8, potential(destination(k)) = potential(destination(k)) + mod(train(k),2); end
>> potential
potential =
2 4 3 5 5

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采纳的回答

Stephen23
Stephen23 2019-9-23
编辑：Stephen23 2019-9-23
Using accumarray:
>> potential = [1,2,3,4,5];
>> train = [8,7,8,7,8,7,8,7];
>> destination = [1,1,1,2,2,2,3,4];
>> potential(:) + accumarray(destination(:),mod(train(:),2),[numel(potential),1])
ans =
2
4
3
5
5

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