[R] how to get growth rate of a (time series) data?

Gabor Grothendieck ggrothendieck at gmail.com
Wed Jun 19 14:11:47 CEST 2013

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:

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

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email: ggrothendieck at gmail.com

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