Basic lag in R vector/dataframe

问题

Will most likely expose that I am new to R, but in SPSS, running lags is very easy. Obviously this is user error, but what I am missing?

x <- sample(c(1:9), 10, replace = T)
y <- lag(x, 1)
ds <- cbind(x, y)
ds

Results in:

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

I figured I would see:

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

Any guidance will be much appreciated.

回答1:

Another way to deal with this is using the zoo package, which has a lag method that will pad the result with NA:

require(zoo)
> set.seed(123)
> x <- zoo(sample(c(1:9), 10, replace = T))
> y <- lag(x, -1, na.pad = TRUE)
> cbind(x, y)
   x  y
1  3 NA
2  8  3
3  4  8
4  8  4
5  9  8
6  1  9
7  5  1
8  9  5
9  5  9
10 5  5

The result is a multivariate zoo object (which is an enhanced matrix), but easily converted to a data.frame via

> data.frame(cbind(x, y))

回答2:

I had the same problem, but I didn't want to use zoo or xts, so I wrote a simple lag function for data frames:

lagpad <- function(x, k) {
  if (k>0) {
    return (c(rep(NA, k), x)[1 : length(x)] );
  }
  else {
    return (c(x[(-k+1) : length(x)], rep(NA, -k)));
  }
}

This can lag forward or backwards:

x<-1:3;
(cbind(x, lagpad(x, 1), lagpad(x,-1)))
     x      
[1,] 1 NA  2
[2,] 2  1  3
[3,] 3  2 NA

回答3:

lag does not shift the data, it only shifts the "time-base". x has no "time base", so cbind does not work as you expected. Try cbind(as.ts(x),lag(x)) and notice that a "lag" of 1 shifts the periods forward.

I would suggesting using zoo / xts for time series. The zoo vignettes are particularly helpful.

回答4:

lag() works with time series, whereas you are trying to use bare matrices. This old question suggests using embed instead, like so:

lagmatrix <- function(x,max.lag) embed(c(rep(NA,max.lag), x), max.lag+1)

for instance

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

回答5:

Using just standard R functions this can be achieved in a much simpler way:

x <- sample(c(1:9), 10, replace = T)
y <- c(NA, head(x, -1))
ds <- cbind(x, y)
ds

回答6:

The easiest way to me now appears to be the following:

require(dplyr)
df <- data.frame(x = sample(c(1:9), 10, replace = T))
df <- df %>% mutate(y = lag(x))

回答7:

tmp<-rnorm(10)
tmp2<-c(NA,tmp[1:length(tmp)-1])
tmp
tmp2

回答8:

This should accommodate vectors or matrices as well as negative lags:

lagpad <- function(x, k=1) {
  i<-is.vector(x)
  if(is.vector(x)) x<-matrix(x) else x<-matrix(x,nrow(x))
  if(k>0) {
      x <- rbind(matrix(rep(NA, k*ncol(x)),ncol=ncol(x)), matrix(x[1:(nrow(x)-k),], ncol=ncol(x)))
  }
  else {
      x <- rbind(matrix(x[(-k+1):(nrow(x)),], ncol=ncol(x)),matrix(rep(NA, -k*ncol(x)),ncol=ncol(x)))
  }
  if(i) x[1:length(x)] else x
}

回答9:

a simple way to do the same may be copying the data to a new data frame and changing the index number. Make sure the original table is indexed sequentially with no gaps

e.g.

tempData <- originalData
rownames(tempData) <- 2:(nrow(tempData)+1)

if you want it in the same data frame as the original use a cbind function

回答10:

Two options, in base R and with data.table:

baseShiftBy1 <- function(x) c(NA, x[-length(x)])
baseShiftBy1(x)
[1] NA  3  8  4  8  9  1  5  9  5

data.table::shift(x)
[1] NA  3  8  4  8  9  1  5  9  5

Data:

set.seed(123)
(x <- sample(c(1:9), 10, replace = T))
[1] 3 8 4 8 9 1 5 9 5 5

回答11:

Just get rid of lag. Change your line for y to:

y <- c(NA, x[-1])

来源：https://stackoverflow.com/questions/3558988/basic-lag-in-r-vector-dataframe

标签

time-series

xts

zoo