Calculate recovery time to a condition in time series in R - r

Question

I would like to calculate the number of days from the time a condition is not met, to when it is met again, in a time series of daily data in R.
Toy data:
day <- data.frame(
date = seq.POSIXt(
from = ISOdatetime(2017,07,01,0,0,0),
to = ISOdatetime(2017,08,26,0,0,0),
by = "1 day" ))
var <- c(5,6,5,5,0,0,0,0,0,1,1,2,3,3,4,3,4,5,4,5,5,4,5,4,0,1,1,2,3,4,5,5,5,4,4,4,4,5,3,3,3,2,2,2,2,2,2,2,2,2,2,2,2,1,1,0,0)
ts = cbind(day, var)
The condition is var > 3.
I'd like to identify each "recovery" period as the time where var > 0 but <= 3, but only following var going to zero. Then, I'd like the number of days to recovery for each period.
So, for the example data given here, I'd expect this output:
period 1 6
period 2 5
Since var never "recovers" at the end of the dataset, I would either want it not identified as a recovery period, or given a recovery time of 0 days.
I tried this:
ifelse(ts$var >3, 0 ,(ifelse(ts$var>0 & ts$var<4, 1, 0)))
and I think I could pair this if else statement with something that only counts sequential 1s and that would mostly do it. Only problem is that it identifies the end period with the slow drop-off as a “recovery period”, and it shouldn’t. It should only identify periods following a zero as a recovery period.
Here is what this example data look like: plot of var over time. I think it's the minimal data I can provide that show the realistic issues I've had with making counts of data outside of recovery periods.
I need to do this over a long and much more dynamic time series, so an efficient way to do this would be greatly appreciated.

Answer 1

edit
- I don't think this will behave the way you expect it to if var does something like this
[... 0, 1, 2, 1, 0, 2, 4, ...]
But may possibly be adapted to handle this case.
original answer
I haven't tested this much, I'd suggest checking it works with weirder cases (e.g. var is all zeros, starts or ends at a period boundary, other corner cases...)
# ignore zeroes if they precede another zero
s <- which(var == 0 & c(tail(var, -1), NA) != 0)
e <- which(var > 3)
sapply(s, function(x) head(e[e > x], 1) - x)
The approach here is to identify all possible start and end points of periods, then find the first end point that occurs after each start point and taking the difference. A simple loop or maybe even a clever regex could be a good alternative.

Answer 2

Here is an alternative approach which uses the rleid() function from the data.table package to group by contiguous streaks of zero and non-zero values. It then finds the position within each group of the first occurrence of a value > 3:
library(data.table)
setDT(ts)[, if (.GRP > 1) first(which(var > 3)), rleid(var == 0)]
rleid V1
1: 3 6
2: 5 5
The first group is skipped because it is either a streak of zeros or has no preceeding zero value.
This approach works even in the case Callum Webb has described in the edit of his answer:
# append data
var <- c(var, 0,1,2,1,0,2,4)
date = seq.POSIXt(
from = ISOdatetime(2017,07,01,0,0,0),
along.with = var,
by = "1 day" )
ts = data.frame(date, var)
setDT(ts)[, if (.GRP > 1) first(which(var > 3)), rleid(var == 0)]
rleid V1
1: 3 6
2: 5 5
3: 9 2
So, it has recognized that there is a recovery period of 2 days after the final zero.
For the sake of completeness, in case the sequence 0, 1, 2, 1, 0 is considered to include also a recovery period of 3 days length although it has not reached a value greater 3:
setDT(ts)[, if (.GRP > 1) if (all(var %between% c(1, 3))) .N else first(which(var > 3)),
rleid(var == 0)]
rleid V1
1: 3 6
2: 5 5
3: 7 3
4: 9 2
Here all days between two zeros are counted if all values lie between 1 and 3.