how to get growth rate of a (time series) data?
On Wed, Jun 19, 2013 at 7:04 AM, Yanyuan Zhu <yyz at tongji.edu.cn> wrote:
Hello all, now I'm trying to switch from Excel to R to deal with the data, and as a newbie i got the problem as follows. suppose I have a data named "test" test<- data.frame(year=c(1996:2011), Y=c(74163.6,81658.5,86531.6,91125.0,98749.0,109028.0,120475.6,136613.4,160956.6,187423.5,222712.5,266599.2,315974.6,348775.1,402816.5,465731.3)) in which Y means the GDP of a country If i want to get Delta Y = Y(t)-Y(t-1) , i could use diff() in R diff(test$Y) but what if i want to get gY=(Y(t)-Y(t-1))/Y(t-1)? seems diff(test$Y)/(test$Y)[-1] doesnt work ...
As already mentioned, that R expression gives (Y[t] - Y[t-1]) / Yt[t] whereas you want Y <- test$Y n <- length(Y) diff(Y) / Y[-n] or you might want to use a time series class for simpler manipluations: Using ts class: Y.ts <- ts(test[, 2], start = test[1,1]) Ydiff.ts <- diff(Y.ts) / lag(Y.ts, - 1) or, using tis class which is frequently used for equally spaced eoconomic series: Y.tis <- tis(test[, 2], start = test[1,1], freq = 1) Ydiff.tis <- diff(Y.tis) / lag(Y.tis, - 1) or using zoo class which, in addition, supports non-equally spaced series and also allows for two different approaches here: library(zoo) Y.z <- read.zoo(test, FUN = identity) Ydiff.z <- diff(Y.z) / lag(Y.z, - 1) This also works with zoo: Ydiff.z <- diff(Y.z, arith = FALSE) - 1 -- Statistics & Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com