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What's the opposite function to lag for an R vector/dataframe?

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

like image 682
Gilles Cosyn Avatar asked Feb 10 '15 13:02

Gilles Cosyn


People also ask

What is the opposite of lag in R?

The opposite of lag() function is lead()

What does lead function do in R?

lead: lead Window function: returns the value that is offset rows after the current row, and defaultValue if there is less than offset rows after the current row. For example, an offset of one will return the next row at any given point in the window partition.

What does lag () do in R?

lag lag shifts the times one back. It does not change the values, only the times. Thus lag changes the tsp attribute from c(1, 4, 1) to c(0, 3, 1) . The start time is shifted from 1 to 0, the end time is shifted from 4 to 3 and since shifts do not change the frequency the frequency remains 1.


3 Answers

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).

like image 184
agenis Avatar answered Nov 08 '22 12:11

agenis


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
like image 35
mallet Avatar answered Nov 08 '22 11:11

mallet


Simpler solution:

y = dplyr::lead(y,1)
like image 37
saranrajsp Avatar answered Nov 08 '22 11:11

saranrajsp