Improve performance of my code

Question

Suraj Shetiya 2017-4-21

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回答： Greg Dionne 2017-4-24

I was working on time warping on sign language and it seems to take about an hour to classify around a 1000 objects. I have heard that it takes 5 mins to run for other friends of mine.

Note that this is a assignement, which I have solved but I need to know the source of my performance hit.

The ASL data set and their format is as below.

-------------------------------------------------
object ID: 1121
class label: 8
sign meaning: 4
dominant hand trajectory:
   -0.488893      0.470786
   -0.488893      0.470786
   -0.488893      0.470786
   -0.488893      0.470786
   -0.488893      0.470786
   -0.466259      0.470786
   -0.466259      0.470786
   -0.466259      0.470786
   -0.466259      0.470786
   -0.466259      0.470786
   -0.466259      0.470786
   -0.466259      0.470786
   -0.466259      0.470786
   -0.466259      0.470786
   -0.466259      0.470786
-------------------------------------------------

There are multiple such objects and I have read each object into a struct. So, I have a vector of these structs which are created in the format below.

struct('ID', current_id, 'label', current_class_label, 'sign', current_sign, 'data', current_data)

The algorithm for time warping is a dynamic programming solution that calculates the overall cost for the match. The matlab code that corresponds to this algorithm is as below.

function cost = cost_calc(pattern1, pattern2)
    new_mat = zeros(size(pattern1,1), size(pattern2,1));
    new_mat(1,1) = distance_2d( pattern1(1, :), pattern2(1, :));
    for i = 2:size(pattern2,1)
        new_mat(1,i) = new_mat(1, i-1) + distance_2d( pattern1(1, :), pattern2(i, :));
    end
    for i = 2:size(pattern1,1)
        new_mat(i,1) = new_mat(i-1, 1) + distance_2d( pattern1(i, :), pattern2(1, :));
        for j = 2:size(pattern2,1)
            new_mat(i,j) = min([new_mat(i-1,j), new_mat(i-1, j-1), new_mat(i, j-1)]) + distance_2d( pattern1(i, :), pattern2(j, :));
        end
    end
    cost = new_mat(end,end);
end

Below is the code for Euclidean distance, that I am using.

function d = distance_2d(p1, p2)
    d = norm(p1 - p2);
end

The call the cost_calc function is happening as below. obj being a test object.

distances = arrayfun(@(x) cost_calc(obj.data, x.data), train);

When I run the profiler on my code, I can find that distance_2d which finds the Euclidean distance, seems to take 2-2.5s for every test object, the min call seems to be taking a little above 2s while the overall DTW method seems to take around 4.5s for every test object.

Please note that I cant post the entire code here as it is against the rules of my university.

Any pointers ?

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Answer 1

Eamonn 2017-4-21

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Five minutes!!?

You can do this in under 0.5 seconds.

Use the UCR suite, google UCR Suite DTW

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Suraj Shetiya 2017-4-21

Thanks for the link I will use that to implement my DTW.

But, can you point me to why my code runs slow so that I get to know what to use and what not for speed ?

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Answer 2

Greg Dionne 2017-4-24

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在 MATLAB Online 中打开

There is a version of DTW in the Signal Processing Toolbox that can accommodate multivariate data, it also has a version for searching via time-stretching with normalization (FINDSIGNAL).

Since it looks like you're implementing the "classic" unconstrained algorithm, you could first compute a point-wise distance matrix first (call it "D") where D(i,j) = norm(x(:,i)-y(:,j)). Note that I've changed the order in which your data appears to keep elements that you expect to access together in the first index -- MATLAB is column-major in its ordering (not row-major), so you'll see this convention frequently in routines that use MATLAB.

Something like:

function d = getDistanceMatrix(x, y) 
d = zeros(size(x,2),size(y,2)); 
  for c=1:size(x,1); 
    [u,v] = meshgrid(y(c,:),x(c,:)); 
    d = d + conj(u-v).*(u-v); 
  end 
  d = sqrt(d); 
end

Then you could feed "D" into a routine that computes the cumulative distance matrix, "C". The benefit of splitting out the distance computation from the cumulative computation is now (with a little bit of arithmetic) you can use purely vector operations to do this along each column.

Something like:

function C = cumulativeDistance(D) 
m = size(D,1); 
n = size(D,2); 
C=NaN(size(D));
Dcum = cumsum(D); 
C(:,1) = Dcum(:,1); 
for j=2:n 
  prevmin = [C(1,j-1); min(C(1:end-1,j-1),C(2:end,j-1))]; 
  delta = prevmin - [0; Dcum(1:end-1,j)]; 
  C(:,j) = cummin(delta)+Dcum(:,j); 
end