I have a TemporalData
object that has td["ValueDimensions"] =3
. Right now I only have one sample, K=1
. So the following example is the simplest case, dim=3
and K=1
. To begin with, I just want to plot each dimension separately. Should be trivial but it seems I have to twist the td
object into knots to get the time dimension to show correctly…
ts = {{{0.`, {1.333999006891247`, 12.053810602014542`, 87.94618939798545`}}, {0.1`, {1.3514434090695817`, 12.093415930914793`, 86.53455823856495`}}, {0.2`, {1.2484079436257138`, 12.25553483931355`, 92.89743502043808`}}, {0.30000000000000004`, \ {1.3781565261854645`, 12.184908506564101`, 84.45147211241067`}}, {0.4`, {1.3285491537059095`, 12.29338463899116`, 87.1154457958651`}}, {0.5`, {1.353124500190124`, 12.328305046121512`, 85.38069968635773`}}, {0.6000000000000001`, {1.4134624696453986`, 12.326915095624361`, 81.70867228313669`}}, {0.7000000000000001`, {1.195964818586219`, 12.60441567097316`, 94.68144692187659`}}, {0.8`, {1.2736346893937893`, 12.587741552453553`, 89.29573476671187`}}, {0.9`, {1.0705151506735828`, 12.853307642920917`, 103.2366857670525`}}, {1.`, {0.9802798326333646`, 13.00862708606569`, 110.54583685327417`}}}} tds = TemporalData[ts] ListLinePlot[td] (* Pols suggests if "ValuesDimension" returns 3 then this would just work *) ListLinePlot[td["Values", 1]] ListLinePlot[td["Values", 2]] ListLinePlot[td["Values", 3]] (* The following shows on dimension for the one path, but feels wrong on so many levels, e.g. the time dimension has been lost*) tds["ValueList"][[1, All, 1]] // ListLinePlot
Once I can do this, I’d like to generate (using ItoProcess) K paths instead of just one, and then I’d like to plot each of these.
Would appreciate any hints or tips about how to do this elegantly.