ok difftime
works...
transform(xxts, lag = difftime(as.POSIXct(BDate, format = "%d/%m/%Y %H:%M"), index(xxts), unit = "hours"))
Question
I am subtracting dates in xts
i.e.
library(xts)
# make data
x <- data.frame(x = 1:4,
BDate = c("1/1/2000 12:00","2/1/2000 12:00","3/1/2000 12:00","4/1/2000 12:00"),
CDate = c("2/1/2000 12:00","3/1/2000 12:00","4/1/2000 12:00","9/1/2000 12:00"),
ADate = c("3/1/2000","4/1/2000","5/1/2000","10/1/2000"),
stringsAsFactors = FALSE)
x$ADate <- as.POSIXct(x$ADate, format = "%d/%m/%Y")
# object we will use
xxts <- xts(x[, 1:3], order.by= x[, 4] )
#### The subtractions
# anwser in days
transform(xxts, lag = as.POSIXct(BDate, format = "%d/%m/%Y %H:%M") - index(xxts))
# asnwer in hours
transform(xxts, lag = as.POSIXct(CDate, format = "%d/%m/%Y %H:%M") - index(xxts))
grep
and regex
and then multiply within an if
clause.I have tried to work through this and went for the grep
regex
apprach but this doesnt even keep the negative sign..
p <- transform(xxts, lag = as.POSIXct(BDate, format = "%d/%m/%Y %H:%M") - index(xxts))
library(stringr)
ind <- grep("days", p$lag)
p$lag[ind] <- as.numeric( str_extract_all(p$lag[ind], "\\(?[0-9,.]+\\)?")) * 24
p$lag
#2000-01-03 2000-01-04 2000-01-05 2000-01-10
# 36 36 36 132
I am convinced there is a more elegant solution...
La solution
ok difftime
works...
transform(xxts, lag = difftime(as.POSIXct(BDate, format = "%d/%m/%Y %H:%M"), index(xxts), unit = "hours"))