我有一个与此类似的数据集
User Date Value
A 2012-01-01 4
A 2012-01-02 5
A 2012-01-03 6
A 2012-01-04 7
B 2012-01-01 2
B 2012-01-02 3
B 2012-01-03 4
B 2012-01-04 5
我想创建一个Value
的滞后,同时尊重User
。
User Date Value Value.lag
A 2012-01-01 4 NA
A 2012-01-02 5 4
A 2012-01-03 6 5
A 2012-01-04 7 6
B 2012-01-01 2 NA
B 2012-01-02 3 2
B 2012-01-03 4 3
B 2012-01-04 5 4
我在循环中效率非常低下
df$value.lag1<-NA
levs<-levels(as.factor(df$User))
levs
for (i in 1:length(levs)) {
temper<- subset(df,User==as.numeric(levs[i]))
temper<- rbind(NA,temper[-nrow(temper),])
df$value.lag1[df$User==as.numeric(as.character(levs[i]))]<- temper
}
但是这很慢。我看过使用by
和tapply
,但没有弄清楚如何使它们工作。
我认为XTS或TS不会因为User元素而起作用。
有什么建议吗?
您可以使用ddply
:它将data.frame切成小块并转换每块。
d <- data.frame(
User = rep( LETTERS[1:3], each=10 ),
Date = seq.Date( Sys.Date(), length=30, by="day" ),
Value = rep(1:10, 3)
)
library(plyr)
d <- ddply(
d, .(User), transform,
# This assumes that the data is sorted
Value = c( NA, Value[-length(Value)] )
)
我认为最简单的方法,尤其是考虑进行进一步分析的方法是,将数据帧从pdata.frame
包转换为plm
类。
从diff()
和lag()
转换后,可用于创建面板差异和滞后。
df<-pdata.frame(df,index=c("id","date"))
df<-transform(df, l_value=lag(value,1))
我偶然发现了类似的问题并编写了一个函数。
#df needs to be a structured balanced paneldata set sorted by id and date
#OBS the function deletes the row where the NA value would have been.
df <- data.frame(id = c(1, 1, 1, 1, 1, 2, 2,2,2,2),
date = c(1992, 1993, 1991, 1990, 1994, 1992, 1991
,1994,1990,1993),
value = c(4.1, 4.5, 3.3, 5.3, 3.0, 3.2, 5.2,5.3,3.4,5.6))
# sort paneldata set
library(dplyr)
df<-arrange(df,id,date)
#Function
# a=df
# b=colname of variable/variables that you want to lag
# q=number of lag years
# t=colname of date/time column
retraso<-function(a,b,q,t){
sto<-max(as.numeric(unique(a[[t]])))
sta<-min(as.numeric(unique(a[[t]])))
yo<-a[which(a[[t]]>=(sta+q)),]
la<-function(a,d,t,sto,sta){
ja<-data.frame(a[[d]],a[[t]])
colnames(ja)<-c(d,t)
ja<-ja[which(ja[[t]]<=(sto-q)),1]
return(ja)
}
for (i in 1:length(b)){
yo[[b[i]]] <-la(a,b[i],t,sto,sta)
}
return(yo)
}
#lag df 1 year
df<-retraso(df,"value",1,"date")
如果时间变量中没有空格,请执行
df %>% group_by(User) %>% mutate(value_lag = lag(value, order_by =Date)
如果时间变量中确实有间隔,请参见此答案https://stackoverflow.com/a/26108191/3662288
对于不遗漏obs的面板,这是一个直观的解决方案:
df <- data.frame(id = c(1, 1, 1, 1, 1, 2, 2),
date = c(1992, 1993, 1991, 1990, 1994, 1992, 1991),
value = c(4.1, 4.5, 3.3, 5.3, 3.0, 3.2, 5.2))
df<-df[with(df, order(id,date)), ] # sort by id and then by date
df$l_value=c(NA,df$value[-length(df$value)]) # create a new var with data displaced by 1 unit
df$l_value[df$id != c(NA, df$id[-length(df$id)])] =NA # NA data with different current and lagged id.
df
id date value l_value
4 1 1990 5.3 NA
3 1 1991 3.3 5.3
1 1 1992 4.1 3.3
2 1 1993 4.5 4.1
5 1 1994 3.0 4.5
7 2 1991 5.2 NA
6 2 1992 3.2 5.2
类似地,您可以使用tapply
# Create Data
user = c(rep('A',4),rep('B',4))
date = rep(seq(as.Date('2012-01-01'),as.Date('2012-01-04'),1),2)
value = c(4:7,2:5)
df = data.frame(user,date,value)
# Get lagged values
df$value.lag = unlist(tapply(df$value, df$user, function(x) c(NA,x[-length(df$value)])))
想法是完全相同的:取值,按用户分割,然后在每个子集上运行一个函数。取消列表将其恢复为矢量格式。
[该表按用户和日期排序,可以用zoo
完成。诀窍是此时不指定索引。
library(zoo)
df <-read.table(text="User Date Value
A 2012-01-01 4
A 2012-01-02 5
A 2012-01-03 6
A 2012-01-04 7
B 2012-01-01 2
B 2012-01-02 3
B 2012-01-03 4
B 2012-01-04 5", header=TRUE, as.is=TRUE,sep = " ")
out <-zoo(df)
Value.lag <-lag(out,-1)[out$User==lag(out$User)]
res <-merge.zoo(out,Value.lag)
res <-res[,-(4:5)] # to remove extra columns
User.out Date.out Value.out Value.Value.lag
1 A 2012-01-01 4 <NA>
2 A 2012-01-02 5 4
3 A 2012-01-03 6 5
4 A 2012-01-04 7 6
5 B 2012-01-01 2 <NA>
6 B 2012-01-02 3 2
7 B 2012-01-03 4 3
8 B 2012-01-04 5 4