Mlpy - Warping dynamic time dependent on x?

Question

I am trying to get the distance between these two arrays shown below DTW.

I use the python mlpy package which offers

 dist, cost, path = mlpy.dtw_std(y1, y2, dist_only=False) 

I understand that DTW cares about the transition. In addition, as can be seen from the above, mlpy.dtw_std() accepts only 2 1-D arrays. Therefore, I expect that no matter how I left / shift to the right, the dist curves returned by the function should never change.

However, after shifting my green curve a little to the right, the dist returned by mlpy.dtw_std() returned!

Before proceeding: Python mlpy.dwt_std reports dist = 14.014

After switching: Python mlpy.dwt_std reports dist = 38.078 Obviously, since the curves are still these two curves, I do not expect the distances to be different!

Why is this so? Where did it go wrong?

Answer 1

Let me repeat what I understood, please correct me if I make a mistake somewhere. I observe that on both of your stories, the 1D series in blue remains the same, and the green color is stretched. How do you do that you explained this in a message from September 19 to September 13 at 9:36. Your premise is that since (1) DTW “cares” about the time shift and (2) everything you do stretches one time series in length without affecting the y values ​​(conclusion: :) you expect the distance to remain the same.

There is a small missing link between [(1), (2)] and [(Inference)]. Which, the individual distance values ​​corresponding to the displays, will change as the set of signals changes. And this will lead to a difference in calculating the total distance. Build a trajectory of deformation, find the cost price to see it yourself.

Take the simplified case ...

Let a=range(0,101,5) = [0,5,10,15 ... 95, 100]

and b=range(0,101,5) = [0,5,10,15 ... 95, 100].

Now, intuitively, you / I would expect one or one correspondence between two signals (for DTW display), and the distance for all displays should be 0, the signals are identical.

Now, if we do, b=range(0,101,4) = [0,4,8,12 ... 96,100], the DTW mapping between a and b should start at 0, obtained by matching with b 0, and end at 100 mapped to b 100 (boundary restrictions). In addition, since the DTW “takes care” of the time shift, I also expected that 20, 40, 60 and 80 of the two signals would be matched to each other. (I have not tried DTWing these two myself, saying this from intuition, so please check. There are few possibilities for unintuitive deformation, which also takes place, depending on the permissible step patterns / global constraints, but let go with intuitive touches for the moment for ease of understanding / for the sake of simplicity).

For the remaining data points, obviously, the distances corresponding to the display are now non-zero, so the total distance is also non-zero. Our distance / total cost has changed from zero to something other than zero.

Now this was the case when our signals were too simplistic, linearly increasing. Imagine the variables that come into the picture when you have non-monotonous signals of real life, and you need to find a temporary deformation between them. :)

(PS: Please remember to answer in response: D). Thanks.

Answer 2

Obviously, the curves are not the same, so the distance function should not be 0 (otherwise it is not a distance by definition).

What does “relatively large” mean? Perhaps the distance is not infinite, right?

140 points in time, each with a small delta, this still adds to a nonzero number.

The distance from New York to Beijing is approximately 11,018 km. Or 1,101,800,000 mm.

The distance to Alpha Centauri is small, only 4.34 liters. This is the closest other star system for us ...

Compare with the distance to the asymmetrical series; this distance should be much larger.