Fill missing sequence values with dplyr - r

Question

I have a data frame with missing values for "SNAP_ID". I'd like to fill in the missing values with floating point values based on a sequence from the previous non-missing value (lag()?). I would really like to achieve this using just dplyr if possible.
Assumptions:
There will never be missing data as the first or last row I'm generating the missing dates based on missing days between a min and max of a data set
There can be multiple gaps in the data set
Current data:
end SNAP_ID
1 2015-06-26 12:59:00 365
2 2015-06-26 13:59:00 366
3 2015-06-27 00:01:00 NA
4 2015-06-27 23:00:00 NA
5 2015-06-28 00:01:00 NA
6 2015-06-28 23:00:00 NA
7 2015-06-29 09:00:00 367
8 2015-06-29 09:59:00 368
What I want to achieve:
end SNAP_ID
1 2015-06-26 12:59:00 365.0
2 2015-06-26 13:59:00 366.0
3 2015-06-27 00:01:00 366.1
4 2015-06-27 23:00:00 366.2
5 2015-06-28 00:01:00 366.3
6 2015-06-28 23:00:00 366.4
7 2015-06-29 09:00:00 367.0
8 2015-06-29 09:59:00 368.0
As a data frame:
df <- structure(list(end = structure(c(1435323540, 1435327140, 1435363260,
1435446000, 1435449660, 1435532400, 1435568400, 1435571940), tzone = "UTC", class = c("POSIXct",
"POSIXt")), SNAP_ID = c(365, 366, NA, NA, NA, NA, 367, 368)), .Names = c("end",
"SNAP_ID"), row.names = c(NA, -8L), class = "data.frame")
This was my attempt at achieving this goal, but it only works for the first missing value:
df %>%
arrange(end) %>%
mutate(SNAP_ID=ifelse(is.na(SNAP_ID),lag(SNAP_ID)+0.1,SNAP_ID))
end SNAP_ID
1 2015-06-26 12:59:00 365.0
2 2015-06-26 13:59:00 366.0
3 2015-06-27 00:01:00 366.1
4 2015-06-27 23:00:00 NA
5 2015-06-28 00:01:00 NA
6 2015-06-28 23:00:00 NA
7 2015-06-29 09:00:00 367.0
8 2015-06-29 09:59:00 368.0
The outstanding answer from #mathematical.coffee below:
df %>%
arrange(end) %>%
group_by(tmp=cumsum(!is.na(SNAP_ID))) %>%
mutate(SNAP_ID=SNAP_ID[1] + 0.1*(0:(length(SNAP_ID)-1))) %>%
ungroup() %>%
select(-tmp)

Answer 1

EDIT: new version works for any number of NA runs.
This one doesn't need zoo, either.
First, notice that tmp=cumsum(!is.na(SNAP_ID)) groups the SNAP_IDs such groups of the same tmp consist of one non-NA value followed by a run of NA values.
Then group by this variable and just add .1 to the first SNAP_ID to fill out the NAs:
df %>%
arrange(end) %>%
group_by(tmp=cumsum(!is.na(SNAP_ID))) %>%
mutate(SNAP_ID=SNAP_ID[1] + 0.1*(0:(length(SNAP_ID)-1)))
end SNAP_ID tmp
1 2015-06-26 12:59:00 365.0 1
2 2015-06-26 13:59:00 366.0 2
3 2015-06-27 00:01:00 366.1 2
4 2015-06-27 23:00:00 366.2 2
5 2015-06-28 00:01:00 366.3 2
6 2015-06-28 23:00:00 366.4 2
7 2015-06-29 09:00:00 367.0 3
8 2015-06-29 09:59:00 368.0 4
Then you can drop the tmp column afterwards (add %>% select(-tmp) to the end).
EDIT: this is the old version which doesn't work for subsequent runs of NAs.
If your aim is to fill each NA with the previous value + 0.1, you can use zoo's na.locf (which fills each NA with the previous value), along with cumsum(is.na(SNAP_ID))*0.1 to add the extra 0.1.
library(zoo)
df %>%
arrange(end) %>%
mutate(SNAP_ID=ifelse(is.na(SNAP_ID),
na.locf(SNAP_ID) + cumsum(is.na(SNAP_ID))*0.1,
SNAP_ID))