survmean: broken use of unique(DT)

[WetRobot]

Got this kind of error today:

 Error in `[.data.table`(unique(x, by = key(x)), x[[survScale]] < TF$tol,  : 
  i evaluates to a logical vector length 75953 but there are 75598 rows. Recycling of logical i is no longer allowed as it hides more bugs than is worth the rare convenience. Explicitly use rep(...,length=.N) if you really need to recycle. 

8.
stop("i evaluates to a logical vector length ", length(i), " but there are ", 
    nrow(x), " rows. Recycling of logical i is no longer allowed as it hides more bugs than is worth the rare convenience. Explicitly use rep(...,length=.N) if you really need to recycle.") 
7.
`[.data.table`(unique(x, by = key(x)), x[[survScale]] < TF$tol, 
    ) 
6.
NextMethod(x) 
5.
`[.Lexis`(unique(x, by = key(x)), x[[survScale]] < TF$tol, ) 
4.
unique(x, by = key(x))[x[[survScale]] < TF$tol, ] 
3.
survmean(fot ~ sex + agegr, data = x, breaks = BL, e1.breaks = eBL, 
    r = RSR, pophaz = pm_obs, e1.pophaz = pm_pred, surv.method = "hazard", 
    verbose = T) 
2.
FUN(X[[i]], ...) 
1.
lapply(sites_all, function(site_no) {
    sub <- in.sites.DT(x, site_no)
    if (site_no %in% sites_m) {
        sub[sub] <- x[sub, sex == 1L] ... 

so unique(x, by = key(x))[x[[survScale]] < TF$tol, ] has been used incorrectly. Most likely this
appeared because there were several rows in data by lex.id.

[WetRobot]

It worked with only one row per lex.id, so that was the culprit.