滞后面板数据与数据表 [英] lagging panel data with data.table
问题描述
我目前使用 data.table
以下列方式滞后面板数据:
require(data.table)
x < - data.table(id = 1:10,t = rep(1:10,each = 10),v = 1:100)
setkey(x,id,t)#以使事情按增加的顺序
x [,lag_v:= c(NA,v [1:(length(v)-1)]),by = id]
我想知道是否有更好的方法来做到这一点?我在网上找到了关于cross-join的东西,这是有道理的。但是,交叉连接会为大型数据集生成相当大的 data.table
,所以我不太愿意使用它。
我不知道这是否与你的方法有很大的不同,但你可以使用事实 x
由 id
x [J(1:10),lag_v := c(NA,head(v,-1))]
$ b
的事实, c> t
(不要使用函数作为变量名称!)是时间id x < - data.table(id = 1:10,t = rep(1:10,each = 10),v = 1:100)
pre>
setkey(x,t)
J(setdiff(x [,unique(t)],1))
x [replacement,lag_v:= x [replaced,v] [,v]]
但是,使用双连接似乎效率低下
I currently lag panel data using
data.table
in the following manner:require(data.table) x <- data.table(id=1:10, t=rep(1:10, each=10), v=1:100) setkey(x, id, t) #so that things are in increasing order x[,lag_v:=c(NA, v[1:(length(v)-1)]),by=id]
I am wondering if there is a better way to do this? I had found something online about cross-join, which makes sense. However, a cross-join would generate a fairly large
data.table
for a large dataset so I am hesitant to use it.解决方案I'm not sure this is that much different from your approach, but you can use the fact that
x
is keyed byid
x[J(1:10), lag_v := c(NA,head(v, -1)) ]
I have not tested whether this is faster than
by
, especially if it is already keyed.Or, using the fact that
t
(don't use functions as variable names!) is the time idx <- data.table(id=1:10, t=rep(1:10, each=10), v=1:100) setkey(x, t) replacing <- J(setdiff(x[, unique(t)],1)) x[replacing, lag_v := x[replacing, v][,v]]
but again, using a double join here seems inefficient
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