R: complete cells below by adding 1 each time - r

Question

I have a dataframe with a column about time and this column contains some NA. I would like to fill these cells with the year before + 1 (if the missing cell is not the beginning of the serie). Here's a reproducible example:
df <- data.frame(x = c("A", "B", "C", "A", "B", "C"),
y = c(2000, NA, NA, 2000, 2001, 2002))
I tried to follow this post
df <- df %>%
complete(y = seq(min(y), max(y), by = "year"))
but I can't find out how to do so. Any idea?
Edit: expected output:
df <- data.frame(x = c("A", "B", "C", "A", "B", "C"),
y = c(2000, 2001, 2002, 2000, 2001, 2002))
Note: I would prefer a dplyr solution.
Note 2 (October 23rd 2019): The three answers so far are good but quite complicated. I'm really surprised that it is not possible to do that simply (for example, having the possibility to add a lag in the fill function would be really useful I think).

Answer 1

This solution is a bit annoying but completely vectorized in dplyr. I doubled your df into a new df2 to try across a couple gapped occurrences.
library(tidyr)
library(dplyr)
df <- data.frame(x = c("A", "B", "C", "A", "B", "C"),
y = c(2000, NA, NA, 2000, 2001, 2002))
df2 <- bind_rows(df, df)
Basically you need to create groups across the blocks with NA. Then you can calculate a within-group cumsum and use fill to drag down the prior value. It is annoying because of all the lines.
df2 %>%
group_by(grp = cumsum(!is.na(y) & lag(is.na(y), default = FALSE))) %>%
mutate(add_year = cumsum(is.na(y))) %>%
fill(y) %>%
mutate(y = y + add_year) %>%
ungroup() %>%
select(-grp, -add_year)

Answer 2

In base you can use ave in combination with cumsum to split your dataset and apply there seq, as you have tried already.
df$y <- ave(df$y, cumsum(!is.na(df$y)), FUN=function(x)
seq(x[1], length.out = length(x)))
identical(df, dfExpected)
#[1] TRUE
df$y
#[1] 2000 2001 2002 2000 2001 2002
In case it starts with NA and you want then to let it start with 2000 you can use replace:
df2$y <-ave(df2$y, cumsum(!is.na(df2$y)), FUN=function(x)
seq(replace(x[1],is.na(x[1]),2000), length.out = length(x)))
identical(df2, dfExpected)
#[1] TRUE
Data:
df <- data.frame(x = c("A", "B", "C", "A", "B", "C"),
y = c(2000, NA, NA, 2000, 2001, 2002))
dfExpected <- data.frame(x = c("A", "B", "C", "A", "B", "C"),
y = c(2000, 2001, 2002, 2000, 2001, 2002))
df2 <- data.frame(x = c("A", "B", "C", "A", "B", "C"),
y = c(NA, NA, NA, 2000, 2001, 2002))

Answer 3

This uses dplyr functions case_when() and lag combined with a while-loop in a custom-function.
Output is as expected, try it out.
library(dplyr)
lag_years <- function(df){
while (anyNA(df$y))
{
df %>%
mutate(y = case_when(is.na(y)&!is.na(lag(y))~lag(y)+1,TRUE~y)) %>%
{.} -> df
}
return(df)
}
lag_years(df) %>%
head()