Replacing conditional values with previous values in r - r

Question

I have some data on organism survival as a function of time. The data is constructed using the averages of many replicates for each time point, which can yield a forward time step with an increase in survival. Occasionally, this results in a survivorship greater than 1, which is impossible. How can I conditionally change values greater than 1 to the value preceeding it in the same column?
Here's what the data looks like:
>df
Generation Treatment time lx
1 0 1 0 1
2 0 1 2 1
3 0 1 4 0.970
4 0 1 6 0.952
5 0 1 8 0.924
6 0 1 10 0.913
7 0 1 12 0.895
8 0 1 14 0.729
9 0 2 0 1
10 0 2 2 1
I've tried mutating the column of interest as such, which still yields values above 1:
df1 <- df %>%
group_by(Generation, Treatment) %>%
mutate(lx_diag = as.numeric(lx/lag(lx, default = first(lx)))) %>% #calculate running survival
mutate(lx_diag = if_else(lx_diag > 1.000000, lag(lx_diag), lx_diag)) #substitute values >1 with previous value
>df1
Generation Treatment time lx lx_diag
1 12 1 0 1 1
2 12 1 2 1 1
3 12 1 4 1 1
4 12 1 6 0.996 0.996
5 12 1 8 0.988 0.992
6 12 1 10 0.956 0.968
7 12 1 12 0.884 0.925
8 12 1 14 0.72 0.814
9 12 1 15 0.729 1.01
10 12 1 19 0.76 1.04
I expect the results to look something like:
>df1
Generation Treatment time lx lx_diag
1 12 1 0 1 1
2 12 1 2 1 1
3 12 1 4 1 1
4 12 1 6 0.996 0.996
5 12 1 8 0.988 0.992
6 12 1 10 0.956 0.968
7 12 1 12 0.884 0.925
8 12 1 14 0.72 0.814
9 12 1 15 0.729 0.814
10 12 1 19 0.76 0.814
I know you can conditionally change the values to a specific value (i.e. ifelse with no else), but I haven't found any solutions that can conditionally change a value in a column to the value in the previous row. Any help is appreciated.
EDIT: I realized that mutate and if_else are quite efficient when it comes to converting values. Instead of replacing values in sequence from the first to last, as I would have expected, the commands replace all values at the same time. So in a series of values >1, you will have some left behind. Thus, if you just run the command:
SurvTot1$lx_diag <- if_else(SurvTot1$lx_diag > 1, lag(SurvTot1$lx_diag), SurvTot1$lx_diag)
over again, you can rid of the values >1. Not the most elegant solution, but it works.

Answer 1

This looks like a very ugly solution to me, but I couldn't think of anything else:
df = data.frame(
"Generation" = rep(12,10),
"Treatent" = rep(1,10),
"Time" = c(seq(0,14,by=2),15,19),
"lx_diag" = c(1,1,1,0.996,0.992,0.968,0.925,0.814,1.04,1.04)
)
update_lag = function(x){
k <<- k+1
x
}
k=1
df %>%
mutate(
lx_diag2 = ifelse(lx_diag <=1,update_lag(lx_diag),lag(lx_diag,n=k))
)

Answer 2

Using the data from #Fino, here is my vectorized solution using base R
vals.to.replace <- which(df$lx_diag > 1)
vals.to.substitute <- sapply(vals.to.replace, function(x) tail( df$lx_diag[which(df$lx_diag[1:x] <= 1)], 1) )
df$lx_diag[vals.to.replace] = vals.to.substitute
df
Generation Treatent Time lx_diag
1 12 1 0 1.000
2 12 1 2 1.000
3 12 1 4 1.000
4 12 1 6 0.996
5 12 1 8 0.992
6 12 1 10 0.968
7 12 1 12 0.925
8 12 1 14 0.814
9 12 1 15 0.814
10 12 1 19 0.814