# Frequently asked questions

## General

*How do I choose a step pattern?*

This question has been raised on Stack Overflow; see
here,
here
and
here.
A good first guess is `symmetric2`

(the default), i.e.

```
g[i,j] = min(
g[i-1,j-1] + 2 * d[i ,j ] ,
g[i ,j-1] + d[i ,j ] ,
g[i-1,j ] + d[i ,j ] ,
)
```

*What's all the fuss about normalization? What's wrong with the *`symmetric1`

recursion I found in Wikipedia/in another implementation?

`symmetric1`

recursion I found in Wikipedia/in another implementation?An alignment computed with a non-normalizable step pattern has two serious drawbacks:

- It cannot be meaningfully normalized by timeseries length. Hence, longer timeseries have naturally higher distances, in turn making comparisons impossible.
- It favors diagonal steps, therefore it is not robust: two paths differing for a small local change (eg. horizontal+vertical step rather than diagonal) have very different costs.

This is discussed in section 3.2 of the JSS paper, section 4.2 of the AIIM paper, section 4.7 of Rabiner and Juang's Fundamentals of speech recognition book, and elsewhere. Make sure you familiarize yourself with those references.

TLDR: just stick to the default `symmetric2`

recursion and use the
value of `normalizedDistance`

.

*What about *derivative* dynamic time warping?*

Use the command
`diff`

to preprocess
the timeseries.

*Why do changes in *`dist.method`

appear to have no effect?

`dist.method`

appear to have no effect?Because it only makes a difference when aligning *multivariate*
timeseries. It specifies the "pointwise" or local distance used
(before the alignment) between the query feature *vector* at time *i*,
`query[i,]`

and the reference feature *vector* at time *j*, `ref[j,]`

. Most distance functions coincide with the Euclidean distance in the
one-dimensional case. Note the following:

```
r<-matrix(runif(10),5) # A 2-D timeseries of length 5
s<-matrix(runif(10),5) # Ditto
myMethod<-"Manhattan" # Or anything else
al1<-dtw(r,s,dist.method=myMethod) # Passing the two inputs
al2<-dtw(proxy::dist(r,s,method=myMethod)) # Equivalent, passing the distance matrix
all.equal(al1,al2)
```

*Can the time/memory requirements be relaxed?*

The first thing you should try is to set the `distance.only=TRUE`

parameter, which skips backtracing and some object copies. Second,
consider downsampling the input timeseries.

*What is the relation between *`dist`

and `dtw`

?

`dist`

and `dtw`

?There are two *very different*, *totally unrelated* uses for `dist`

.
This is explained at length in the paper, but let's summarize.

- If you have
**two multivariate**timeseries, you can feed them to`dist`

to obtain a*local distance matrix*. You then pass this matrix to dtw(). This is equivalent to passing the two matrices to the dtw() function and specifying a`dist.method`

(see also the next question). - If you have
**many univariate**timeseries, instead of iterating over all pairs and applying dtw() to each, you may feed the lot (arranged as a matrix) to`dist`

with`method="DTW"`

. In this case your code does NOT explicitly call dtw(). This is equivalent to iterating over all pairs; it is also equivalent to using the`dtwDist`

convenience function.

## Clustering

*Can I use the DTW distance to cluster timeseries?*

Of course. You need to start with a dissimilarity matrix, i.e. a
matrix holding in *i,j* the DTW distance between timeseries *i* and
*j*. This matrix is fed to the clustering functions. Obtaining the
dissimilarity matrix is done differently depending on whether your
timeseries are univariate or or multivariate: see the next questions.

*How do I cluster univariate timeseries of homogeneous length?*

Arrange the timeseries (single-variate) in a matrix *as rows*. Make
sure you use a symmetric pattern. See
dtwDist.

*How do I cluster *multiple* *multivariate* timeseries?*

You have to handle the loop yourself. Assuming you have data arranged
as `x[time,component,series]`

, pseudocode would be:

```
for (i in 1:N) {
for (j in 1:N) {
result[i,j] <- dtw( dist(x[,,i],x[,,j]),
distance.only=T )$normalizedDistance
```

*Can I compute a DTW-based dissimilarity matrix out of timeseries of different lengths?*

Either loop over the inputs yourself, or pad with NAs and use the following code:

```
dtwOmitNA <-function (x,y)
{
a<-na.omit(x)
b<-na.omit(y)
return(dtw(a,b,distance.only=TRUE)$normalizedDistance)
}
## create a new entry in the registry with two aliases
pr_DB$set_entry(FUN = dtwOmitNA, names = c("dtwOmitNA"))
d<-dist(dataset, method = "dtwOmitNA")
```

## Non-discoveries

*I've discovered a multidimensional/multivariate version of the DTW algorithm! Shall it be included in the package?*

Alas, most likely you haven't. DTW had been "multidimensional" since
its conception. Local distances are computed between *N*-dimensional
vectors; feature vectors have been extensively used in speech
recognition since the '70s (see e.g. things like MFCC, RASTA,
cepstrum, etc). Don't worry: several other people have "rediscovered"
multivariate DTW already. The *dtw* package supports the numerous
types of multi-dimensional local distances that the
proxy
package does, as explained in section 3.6 of the paper in
JSS.

*I've discovered a realtime/early detection version of the DTW algorithm!*

Alas, most likely you haven't. A natural solution for real-time
recognition of timeseries is "unconstrained DTW", which relaxes one or
both endpoint boundary conditions. To my knowledge, the algorithm was
published as early as 1978 by Rabiner, Rosenberg, and
Levinson under the name
UE2-1: see e.g. the mini-review in (Tormene and Giorgino,
2008). Feel also free
to learn about the clever algorithms or expositions by Sakurai et al.
(2007); Latecki
(2007);
Mori et al. (2006);
Smith-Waterman
(1981);
Rabiner and Schmidt
(1980); etc. Open-ended
alignments (at one or both ends) are available in the *dtw* package,
as described in section 3.5 of the JSS
paper.

*I've discovered a bug in your backtrack algorithm!*

Alas, most likely you haven't. You may be doing backtracking via steepest descent. It's not the correct way to do it. Here's a counterexample:

```
> library(dtw)
> dm<-matrix(10,4,4)+diag(rep(1,4))
> al<-dtw(dm,k=T,step=symmetric2)
> al$localCostMatrix
[,1] [,2] [,3] [,4]
[1,] *11* *10* 10 10
[2,] 10 11 *10* 10
[3,] 10 10 11 *10*
[4,] 10 10 10 *11*
> al$costMatrix
[,1] [,2] [,3] [,4]
[1,] >11< 21 31 41
[2,] 21 >32< 41 51
[3,] 31 41 >52< 61
[4,] 41 51 61 >72<
```

The sum of costs along my warping path (above) is (starting from [1,1]) 11+10+2*10+2*10+11 = 72 which is correct (=g[4,4]) . If you follow your backtracking "steepest descent" algorithm (red), you get the diagonal 11+2*11+2*11+2*11=77 which is wrong.