Access a lagged row based on a given row in R

Question

Let's assume my code accesses a certain row. How is it possible to manually access a lagged row. In the example my code picked the row with the date "03.01.2010". Based on this, how can I access for example the following row.

date <- c("01.01.2010","02.01.2010","03.01.2010","04.01.2010","07.01.2010")
ret <- c(1.1,1.2,1.3,1.4,1.5)

mydf <- data.frame(date, ret)
mydf

#         date ret
# 1 01.01.2010 1.1
# 2 02.01.2010 1.2
# 3 03.01.2010 1.3
# 4 04.01.2010 1.4
# 5 07.01.2010 1.5

certaindate <- "03.01.2010"

mydf[mydf$date %in% certaindate, "ret"] # this is an important line in my code and I want to keep it there!

I thought something like

mydf[mydf$date %in% certaindate +1, "ret"]

would do the trick, but it doesn't..

Answer 1

So this works:

mydf[mydf$date %in% certaindate, "ret"] 
# [1] 1.3
mydf[which(mydf$date %in% certaindate)+1,]$ret
# [1] 1.4

The function which(...) returns the indices of all elements which meet the condition. Then you can add 1.

Your code was returning a logical vector (T/F). Adding 1 to that coerces all the logical to 0 or 1, giving a numeric vector of 1 or 2, which was then used to index df.

Answer 2

This is easy to do with dplyr::lag():

library(dplyr)

df <- data.frame(
  date = as.Date(c("2010-01-01", "2010-01-02", "2010-01-03", "2010-01-04", 
    "2010-01-07")),
  ret = c(1.1, 1.2, 1.3, 1.4, 1.5)
)
df
#>         date ret
#> 1 2010-01-01 1.1
#> 2 2010-01-02 1.2
#> 3 2010-01-03 1.3
#> 4 2010-01-04 1.4
#> 5 2010-01-07 1.5

lag(df$date)
#> [1] NA           "2010-01-01" "2010-01-02" "2010-01-03" "2010-01-04"

mutate(df, last_date = lag(date))
#>         date ret  last_date
#> 1 2010-01-01 1.1       <NA>
#> 2 2010-01-02 1.2 2010-01-01
#> 3 2010-01-03 1.3 2010-01-02
#> 4 2010-01-04 1.4 2010-01-03
#> 5 2010-01-07 1.5 2010-01-04