I have pairs of series of timed measurements I want to align. I have annotators judging and marking ambiguous events in a signal and want to optimally match their times and events. The input is four columns: the event onset times and labels for one annotator, and the times and labels for the other annotator. For example (as rows):

```
annotator_1_times: .34, .39, .50, .68, .88
annotator_1_label: A, X, Q, L, Z
annotator_2_times: .33, .41, .67, .90
annotator_2_label: A, X, L, X
```

Annotators don't necessarily have the same number of events when they interpret the same signal. Annotators in general are expected to have similar but nonidentical labels, and similar but nonidentical times.

How this should be done depends on some sort of cost function -- something that will decide how "bad" it is for a time to be off a certain amount, and for two labels to disagree.

A desirable output in my example case:

```
annotator_1_times: .34, .39, .50, .68, .88
annotator_1_label: A, X, Q, L, Z
annotator_2_times: .33, .41, [], .67, .90 <-note gap inserted
annotator_2_label: A, X, [], L, X
```

Stuff I would then do post hoc:

```
time_mismatch_dif: .01, .02, XX, .01, .02 <- for computing agreemt
label_mismatches_: 0, 0, ADD, 0, SUBST <- for computing agreemt
```

The hard part is to know where to insert the gaps.

If need be, I can do just the numerical alignments and separately just the label alignments and then merge them somehow. I know there are character alignment algorithms (e.g. in genetics) and there must be time-series alignment algorithms.

Any suggestions welcome.

Answer:

Your problem is very similar to Levenshtein distance problem, and you can adapt the same algorithm there.

Firstly define your cost function.

Then, run a dynamic programming on a quadratic table: for each `i`

and `j`

calculate `ans[i][j]`

, that is the 'cost of alignment' of first `i`

events from the first annotator and first `j`

events from the second. This can be done in three ways:

- either you align
`i`

and`j`

, then`ans[i][j]`

becomes`ans[i-1][j-1] + costAlignment(a[i],b[j])`

- either you align
`i`

and 'gap', then`ans[i][j]`

becomes`ans[i-1][j] + costGap(a[i])`

- either you align
`j`

and 'gap', then`ans[i][j]`

becomes`ans[i][j-1] + costGap(b[j])`

You should choose minimum of three options.

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