tapply(dat$`lagged Date`, INDEX = dat$Location, FUN = diff(dat$`lagged Date`)) - r

Question

Can someone explain me why this is not working?
tapply(dat$`lagged Date`, INDEX = dat$Location, FUN = diff(dat$`lagged Date`))
I receive the following error:
Error in match.fun(FUN) : 'diff(dat$lagged Date)' is not a
function, character or symbol
structure(list(`lagged Date` = structure(c(1466306880, 1466307060,
1466307240, 1466307420, 1466307600, 1466307780), class = c("POSIXct",
"POSIXt"), tzone = "UTC"), Location = c(309, 309, 309, 309, 309,
309), Duration = c(0, 0, 0, 0, 0, 0), Latitude = c(53.50205667,
53.501915, 53.50183667, 53.50178833, 53.50184, 53.50186167),
Longitude = c(-3.354733333, -3.354096667, -3.353838333, -3.353673333,
-3.353711667, -3.353741667), `Number of Records` = c(1, 1,
1, 1, 1, 1), Speed = c(0.9, 0, 0, 0, 0, 0), `Sum of Var` = c(38,
38, 38, 38, 38, 38), check = c(0, 0, 0, 0, 0, 0)), .Names = c("lagged Date",
"Location", "Duration", "Latitude", "Longitude", "Number of Records",
"Speed", "Sum of Var", "check"), row.names = c(NA, -6L), class = c("tbl_df",
"tbl", "data.frame"))
thank you!

Answer 1

I'm not sure what you want to achieve, but using only diff as the FUN part works and produces this output:
tapply(dat$`lagged Date`, INDEX = dat$Location, FUN = diff)
$`309`
Time differences in mins
[1] 3 3 3 3 3
If you want to convert the output into hours, you can do that by selecting only the values of the difftime-list object and convert those:
as.numeric(tapply(dat$`lagged Date`, INDEX = dat$Location, FUN = diff)[[1]], units = "hours")
Output then looks like this:
[1] 0.05 0.05 0.05 0.05 0.05