Create new lagged data.frame column

Question

I'm not understanding how to create a new 'lagged' column in a data.frame. My current data is collected at the end of the data. One program I need to send this to assumes it's collected first thing in the morning, so I need to lag column 2 by 1 row. The code I wrote just returns the same data.

How can I do this properly?

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 this:

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

It would be a bit easier if you used a time series class. In that case you could use lag :

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

In the other 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 code to lag 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 wish 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