What's the opposite function to lag for an R v

Question

I have a problem dealing with time series in R.

#--------------read data

wb = loadWorkbook("Countries_Europe_Prices.xlsx") 
df = readWorksheet(wb, sheet="Sheet2")

x <- df$Year
y <- df$Index1

y <- lag(y, 1, na.pad = TRUE)
cbind(x, y)

It gives me the following output:

        x     y
 [1,] 1974    NA
 [2,] 1975  50.8
 [3,] 1976  51.9
 [4,] 1977  54.8
 [5,] 1978  58.8
 [6,] 1979  64.0
 [7,] 1980  68.8
 [8,] 1981  73.6
 [9,] 1982  74.3
[10,] 1983  74.5
[11,] 1984  72.9
[12,] 1985  72.1
[13,] 1986  72.3
[14,] 1987  71.7
[15,] 1988  72.9
[16,] 1989  75.3
[17,] 1990  81.2
[18,] 1991  84.3
[19,] 1992  87.2
[20,] 1993  90.1

But I want the first value in y to be 50.8 and so forth. In other words, I want to get a negative lag. I don't get it, how can I do it?

My problem is very similar to this problem, but however I cannot solve it. I guess I still do not understand the solution(s)...

Basic lag in R vector/dataframe

Answer 1

The opposite of lag() function is lead()

Answer 2

Simpler solution:

y = dplyr::lead(y,1)

Answer 3

How about the built-in 'lead' function? (from the dplyr package) Doesn't it do exactly the job of Ahmed's function?

cbind(x, lead(y, 1))

If you want to be able to calculate either positive or negative lags in the same function, i suggest a 'shorter' version of his 'shift' function:

shift = function(x, lag) {
  require(dplyr)
  switch(sign(lag)/2+1.5, lead(x, abs(lag)), lag(x, abs(lag)))
}

What it does is creating 2 cases, one with lag the other with lead, and chooses one case depending on the sign of your lag (the +1.5 is a trick to transform a {-1, +1} into a {1, 2} alternative).

Answer 4

There is an easier way of doing this which I have captured fully from this link. What I will do here is explaining what should you do in steps:

First create the following function by running the following code:

shift<-function(x,shift_by){
    stopifnot(is.numeric(shift_by))
    stopifnot(is.numeric(x))

    if (length(shift_by)>1)
        return(sapply(shift_by,shift, x=x))

    out<-NULL
    abs_shift_by=abs(shift_by)
    if (shift_by > 0 )
        out<-c(tail(x,-abs_shift_by),rep(NA,abs_shift_by))
    else if (shift_by < 0 )
        out<-c(rep(NA,abs_shift_by), head(x,-abs_shift_by))
    else
        out<-x
    out
}

This will create a function called shift with two arguments; one is the vector you need to operate its lag/lead and the other is number of lags/leads you need.

Example:

Suppose you have the following vector:

x<-seq(1:10)

x
 [1]  1  2  3  4  5  6  7  8  9 10

if you need x's first order lag

shift(x,-1)
[1] NA  1  2  3  4  5  6  7  8  9 

if you need x's first order lead (negative lag)

shift(x,1)
[1]  2  3  4  5  6  7  8  9 10 NA