*"...the TWED measure originally proposed by Marteau (2009) seems to consistently outperform all the considered distances..."*

Below is my Octave

*.oct function version of the above linked MATLAB code.*

```
#include octave oct.h
#include octave dmatrix.h
#include limits> // for infinity
#include math.h // for sqrt
DEFUN_DLD ( twed, args, nargout,
"-*- texinfo -*-\n\
@deftypefn {Function File} {} twed (@var{A , timeSA , B , timeSB , lambda, nu})\n\
Calculates the Time Warp Edit Distance between two univariate time series, A and B.\n\
timeSA and timeSB are the time stamps of the respective series, lambda is a penalty\n\
for a deletion operation and nu is an Elasticity parameter - nu >=0 needed for distance measure.\n\
@end deftypefn" )
{
octave_value_list retval_list ;
int nargin = args.length () ;
// check the input arguments
if ( nargin != 6 )
{
error ("Invalid number of arguments. See help twed.") ;
return retval_list ;
}
if ( args(0).length () < 2 )
{
error ("Invalid 1st argument length. Must be >= 2.") ;
return retval_list ;
}
if ( args(1).length () != args(0).length () )
{
error ("Arguments 1 and 2 must be vectors of the same length.") ;
return retval_list ;
}
if ( args(2).length () < 2 )
{
error ("Invalid 3rd argument length. Must be >= 2.") ;
return retval_list ;
}
if ( args(3).length () != args(2).length () )
{
error ("Arguments 3 and 4 must be vectors of the same length.") ;
return retval_list ;
}
if ( args(4).length () > 1 )
{
error ("Argument 5 must a single value for lambda.") ;
return retval_list ;
}
if ( args(5).length () > 1 )
{
error ("Argument 6 must a single value for nu >= 0.") ;
return retval_list ;
}
if ( error_state )
{
error ("Invalid arguments. See help twed.") ;
return retval_list ;
}
// end of input checking
Matrix A_input = args(0).matrix_value () ;
if( A_input.rows() == 1 && A_input.cols() >= 2 ) // is a row matrix, so transpose
{
A_input = A_input.transpose () ;
}
Matrix timeSA_input = args(1).matrix_value () ;
if( timeSA_input.rows() == 1 && timeSA_input.cols() >= 2 ) // is a row matrix, so transpose
{
timeSA_input = timeSA_input.transpose () ;
}
Matrix B_input = args(2).matrix_value () ;
if( B_input.rows() == 1 && B_input.cols() >= 2 ) // is a row matrix, so transpose
{
B_input = B_input.transpose () ;
}
Matrix timeSB_input = args(3).matrix_value () ;
if( timeSB_input.rows() == 1 && timeSB_input.cols() >= 2 ) // is a row matrix, so transpose
{
timeSB_input = timeSB_input.transpose () ;
}
double lambda = args(4).double_value () ;
double nu = args(5).double_value () ;
double inf = std::numeric_limits
```::infinity() ;
Matrix distance ( 1 , 1 ) ; distance.fill ( 0.0 ) ;
double cost ;
// Add padding of zero by using zero-filled distance matrix
Matrix A = distance.stack( A_input ) ;
Matrix timeSA = distance.stack( timeSA_input ) ;
Matrix B = distance.stack( B_input ) ;
Matrix timeSB = distance.stack( timeSB_input ) ;
Matrix DP ( A.rows() , B.rows() ) ; DP.fill ( inf ) ; DP( 0 , 0 ) = 0.0 ;
int n = timeSA.rows () ;
int m = timeSB.rows () ;
// Compute minimal cost
for ( octave_idx_type ii (1) ; ii < n ; ii++ )
{
for ( octave_idx_type jj (1) ; jj < m ; jj++ )
{
// Deletion in A
DP( ii , jj ) = DP(ii-1,jj) + sqrt( ( A(ii-1,0) - A(ii,0) ) * ( A(ii-1,0) - A(ii,0) ) ) + nu * ( timeSA(ii,0) - timeSA(ii-1,0) ) + lambda ;
// Deletion in B
cost = DP(ii,jj-1) + sqrt( ( B(jj-1,0) - B(jj,0) ) * ( B(jj-1,0) - B(jj,0) ) ) + nu * ( timeSB(jj,0) - timeSB(jj-1,0) ) + lambda ;
DP( ii , jj ) = cost < DP( ii , jj ) ? cost : DP( ii , jj ) ;
// Keep data points in both time series
cost = DP(ii-1,jj-1) + sqrt( ( A(ii,0) - B(jj,0) ) * ( A(ii,0) - B(jj,0) ) ) + sqrt( ( A(ii-1,0) - B(jj-1,0) ) * ( A(ii-1,0) - B(jj-1,0) ) ) + nu * ( abs( timeSA(ii,0) - timeSB(jj,0) ) + abs( timeSA(ii-1,0) - timeSB(jj-1,0) ) ) ;
DP( ii , jj ) = cost < DP( ii , jj ) ? cost : DP( ii , jj ) ;
} // end of jj loop
} // end of ii loop
distance( 0 , 0 ) = DP( n - 1 , m - 1 ) ;
retval_list(1) = DP ;
retval_list(0) = distance ;
return retval_list ;
} // end of function

As a quick test I took the example problem from this Cross Validated thread, the applicability I hope being quite obvious to readers:A = [1, 2, 3, 4, 5, 6, 7, 8, 9] ;

B1 = [1, 2, 3, 4, 5, 6, 7, 8, 12] ;

distance1 = twed( A , 1:9 , B1 , 1:9 , 1 , 0.001 )

distance1 = 3

B2 = [0, 3, 2, 5, 4, 7, 6, 9, 8] ;

distance2 = twed( A , 1:9 , B2 , 1:9 , 1 , 0.001 )

distance2 = 17

graphics_toolkit('fltk') ; plot(A,'k','linewidth',2,B1,'b','linewidth',2,B2,'r','linewidth',2);

legend( "A" , "B1" , "B2" ) ;

It can be seen that the twed algorithm correctly picks out B1 as being more like A than B2 (a lower twed distance, with default values for lambda and nu of 1 and 0.001 respectively, taken from the above survey paper) when compared with the simple squared error metric, which gives identical results for both B1 and B2.

More on this in due course.

## 4 comments:

Good Day sir, nice tutorial :) but what will you do next with the shortest distance in order to find if they are the same? Thanks.

Hi ark commands,

There are several things that could be done, and which will probably be the subject of a future blog post. In particular I'm thinking of using the top N matches in the historical record as training examples for machine learning algorithms.

The information provided and the explanation of the concept is really good and one can logically relate to the explanation.

Thank you is very clear but if can any help because i've subject for Dynamic Time Warping Under Limited Warping Path Length, I've difficulty to understand the LDWT algorithm

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