fix measure bottlenecks #337
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Above commits may seem strange but I had to fix these for the new RCLL tests |
R/PredictionDataSurv.R
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| ok = inherits(distr, c("VectorDistribution", "Matdist", "Arrdist")) && | ||
| length(keep) > 1 # edge case: Arrdist(1xYxZ) and keep = FALSE | ||
| if (ok) { | ||
| pdata$distr = distr[keep] # we can subset row/samples like this | ||
| } else { | ||
| pdata$distr = base::switch(keep, distr) # one distribution only | ||
| } |
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@bblodfon can you tell me what's happening here?
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I was trying to filter Distribution objects (like the prediction object from surv.parametric) since I noticed we don't do any filtering on those and I needed on the tests to compare RCLL old and new code kind-of. The above is the result of testing some edge cases.
In general you can subset with distr[keep] when you have a Matdist, Arrdist or an VectorDistribution if you have two or more observations. The problem starts when 1 observation is left. distr[keep] with a VectorDistribution leaves a Distribution (e.g. WeibullAFT-something in surv.parametric), a WeightedDiscrete() for Matdist, and an Arrdist of dim 1x... (it doesn't degenerate to a Matdist). And that's all fine, we still use pdata$distr = distr[keep] to do this. But then if you redo some more filtering (on that same specific row index or another to go to 0 observations - a very edge case :), keep will be either TRUE or FALSE, and most of the above are not subsettable or don't work as expected:
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Thanks for asking this - based on the above, we might want to make an Arrdist to a Matdist when only one observation is left by design in distr6?
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@RaphaelS1 Do you have an opinion on this?
| distr = prediction$distr | ||
| if (inherits(distr, c("Matdist", "Arrdist"))) { | ||
| si = diag(distr$survival(true_times)) | ||
| } else { # VectorDistribution or single Distribution, e.g. WeightDisc() | ||
| si = as.numeric(distr$survival(data = matrix(true_times, nrow = 1L))) | ||
| } | ||
| } |
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This one is from using lrn('surv.parametric') and filtering to one observation, p$distr is a WeibullAFT() which is not a VectorDistribution, so diag(distr$survival(true_times)) returned 0 and later NaN in the score
Depends on xoopR/distr6#296