Create a new lagging column data.frame

Question

I do not understand how to create a new column "lagged" in data.frame. My current data is collected at the end of the data. One of the programs I need to send suggests that she collected the first one in the morning, so I need to mash column 2 into 1 row. The code I wrote just returns the same data.

How can I do it right?

Thanks.

D8 = structure(list(Date = structure(c(14396, 14397, 14398, 14399, 14400, 14403, 14404, 14405, 14406, 14407, 14410, 14411, 14412, 14413, 14414, 14417, 14418, 14419, 14420, 14421, 14424, 14425, 14426, 14427, 14428, 14431, 14432, 14433, 14434, 14435), class = "Date"), PL8 = c(0, 0, 0, 0, 76, 0, -334, -974, -104, 356, 378, -1102, -434, 266, -434, 444, 464, 0, 486, 406, -224, -214, 0, -4, 0, -188, 356, 322, -484, 436)), .Names = c("Date", "PL8"), row.names = c(NA, 30L), class = "data.frame") D8 D8[,3] = lag(D8[,2],k=-1) D8 

Answer 1

Try the following:

 transform(D8, PL8.lag = c(PL8[-1], NA)) 

It would be a little easier if you used the time series class. In this case, you can use lag :

 library(zoo) z <- read.zoo(D8) lag(z, 0:1) 

In another direction, we would have:

 transform(D8, PL8.lag = c(NA, head(PL8, -1))) 

and

 lag(z, 0:-1) 

Answer 2

Here is an alternative solution:

 D8$my.PL8.lag <- c(D8$PL8[-1], NA) 

Here is the code to delay a column by group using tapply :

 my.df = read.table(text = " REFNO MONTH DAY YEAR STATE 1 3 5 2012 1 1 3 7 2012 2 1 3 10 2012 3 1 3 14 2012 NA 2 3 1 2012 20 2 3 10 2012 40 2 3 14 2012 60 2 3 17 2012 80 3 4 3 2012 -4 3 4 24 2012 -8 3 4 28 2012 -12 ", header = TRUE, stringsAsFactors = FALSE) desired.result = read.table(text = " REFNO MONTH DAY YEAR STATE STATE.lag 1 3 5 2012 1 NA 1 3 7 2012 2 1 1 3 10 2012 3 2 1 3 14 2012 NA 3 2 3 1 2012 20 NA 2 3 10 2012 40 20 2 3 14 2012 60 40 2 3 17 2012 80 60 3 4 3 2012 -4 NA 3 4 24 2012 -8 -4 3 4 28 2012 -12 -8 ", header = TRUE, stringsAsFactors = FALSE) my.df$STATE.lag <- unlist(tapply(my.df$STATE, my.df$REFNO, function(x) { c(NA, x[-length(x)]) })) all.equal(my.df, desired.result) # [1] TRUE 

If the column you want to lag is in Date format, you can use:

 my.df$MY.DATE <- do.call(paste, list(my.df$MONTH, my.df$DAY, my.df$YEAR)) my.df$MY.DATE <- as.Date(my.df$MY.DATE, format=c("%m %d %Y")) my.df$MY.DATE.lag <- as.Date(unlist(tapply(as.character(my.df$MY.DATE), my.df$REFNO, function(x) { c(NA, x[-length(x)]) } ))) REFNO MONTH DAY YEAR STATE MY.DATE MY.DATE.lag 1 1 3 5 2012 1 2012-03-05 <NA> 2 1 3 7 2012 2 2012-03-07 2012-03-05 3 1 3 10 2012 3 2012-03-10 2012-03-07 4 1 3 14 2012 NA 2012-03-14 2012-03-10 5 2 3 1 2012 20 2012-03-01 <NA> 6 2 3 10 2012 40 2012-03-10 2012-03-01 7 2 3 14 2012 60 2012-03-14 2012-03-10 8 2 3 17 2012 80 2012-03-17 2012-03-14 9 3 4 3 2012 -4 2012-04-03 <NA> 10 3 4 24 2012 -8 2012-04-24 2012-04-03 11 3 4 28 2012 -12 2012-04-28 2012-04-24