Using lag in mutate() on a arranged data - r

Question

I am working on a data set which is similar to
data <-tribble(
~id, ~ dates, ~days_prior,
1,20190101, NA,
1,NA, 15,
1,NA, 20,
2, 20190103, NA,
2,NA, 3,
2,NA, 4)
I have the first date for each ID and I am trying to calculate the next date by adding days_prior to the previous date. I am using the lag function to refer to the previous date.
df<- df%>% mutate(dates = as.Date(ymd(dates)), days_prior =as.integer(days_prior))
df<-df %>% mutate(dates =
as.Date(ifelse(is.na(days_prior),dates,days_prior+lag(dates)),
origin="1970-01-01"))
This works but only for the next row as you can see attached data.
What am I doing wrong? I would like all the dates to be calculated by mutate(). What different approach should I take to calculate this.

Answer 1

I don't really see how lag would help here; unless I misunderstood here is an option using tidyr::fill
data %>%
group_by(id) %>%
mutate(dates = as.Date(ymd(dates))) %>%
fill(dates) %>%
mutate(dates = dates + if_else(is.na(days_prior), 0L, as.integer(days_prior))) %>%
ungroup()
## A tibble: 6 x 3
# id dates days_prior
# <dbl> <date> <dbl>
#1 1 2019-01-01 NA
#2 1 2019-01-16 15
#3 1 2019-01-21 20
#4 2 2019-01-03 NA
#5 2 2019-01-06 3
#6 2 2019-01-07 4
Or a slight variation, replacing the NA entries in days_prior with 0
data %>%
group_by(id) %>%
mutate(
dates = as.Date(ymd(dates)),
days_prior = replace(days_prior, is.na(days_prior), 0)) %>%
fill(dates) %>%
mutate(dates = dates + as.integer(days_prior)) %>%
ungroup()
Update
In response to your clarifications in the comments, here is what you can do
data %>%
group_by(id) %>%
mutate(
dates = as.Date(ymd(dates)),
days_prior = replace(days_prior, is.na(days_prior), 0)) %>%
fill(dates) %>%
mutate(dates = dates + cumsum(days_prior)) %>%
ungroup()
## A tibble: 6 x 3
# id dates days_prior
# <dbl> <date> <dbl>
#1 1 2019-01-01 0
#2 1 2019-01-16 15
#3 1 2019-02-05 20
#4 2 2019-01-03 0
#5 2 2019-01-06 3
#6 2 2019-01-10 4

Answer 2

You can use the na.locf from the zoo package to fill in the last observed date before adding the prior days.
library("tidyverse")
library("zoo")
data %>%
# Fill in NA dates with the previous non-NA date
# The `locf` stands for "last observation carried forward"
# Fill in NA days_prior with 0
mutate(dates = zoo::na.locf(dates),
days_prior = replace_na(days_prior, 0)) %>%
mutate(dates = lubridate::ymd(dates) + days_prior)
This solution makes two assumptions:
The rows are sorted by id. You can get around this assumption with a group_by(id) followed by an ungroup() statement as shows in the solution by Maurits Evers.
For each id, the row with the observed date is first in the group. This needs to be true in any case with either na.locf and fill because both functions fill in NAs using the previous non-NA entry.
If you don't want to make any assumptions about the ordering, you can sort the rows at the start with data %>% arrange(id, dates).