如何在每个组中创建滞后变量？ [英] How to create a lag variable within each group?

问题描述

我有一个data.table：

  set.seed（1）
 data<  -  data。表（时间= c（1:3,1:4），
组= c（rep（c（b，a），c（3,4））） = rnorm（7））
 
 data 
＃组时间值
＃1：b 1 -0.6264538 
＃2：b 2 0.1836433 
＃3 ：b 3 -0.8356286 
＃4：a 1 1.5952808 
＃5：a 2 0.3295078 
＃6：a 3 -0.8204684 
＃7：a 4 0.4874291 
  

我想计算每个级别下的value列的滞后版本groups。

结果应类似于

 ＃groups时间价值滞后价值
＃1 a 1 1.5952808 NA 
＃2 a 2 0.3295078 1.5952808 
＃3 a 3 -0.8204684 0.3295078 
＃4 a 4 0.4874291 -0.8204684 
＃5 b 1 -0.6264538 NA 
＃6 b 2 0.1836433 -0.6264538 
＃7 b 3 -0.8356286 0.1836433 
  

我试图直接使用 lag ：

  data $ lag.value<  -  lag（data $ value）
  

...这显然不会工作。

我也尝试过：

  value，data $ groups，lag））
 a1 a2 a3 a4 b1 b2 b3 
 NA -0.1162932 0.4420753 2.1505440 NA 0.5894583 -0.2890288 
  

这几乎是我想要的。但是，生成的向量的排序与数据表中的排序有所不同，这是有问题的。

在base R，plyr，dplyr中执行此操作的最有效的方法是什么，and data.table？

解决方案

您可以在 data.table

  library（data.table）
 data [，lag.value：= c（NA，value [ -.N]），by = groups] 
 data 
＃时间组值lag.value 
＃1：1 a 0.02779005 NA 
＃2：2 a 0.88029938 0.02779005 
＃3：3 a -1.69514201 0.88029938 
＃4：1 b -1.27560288 NA 
＃5：2 b -0.65976434 -1.27560288 
＃6：3 b -1.37804943 -0.65976434 
＃7：4 b 0.12041778 -1.37804943 
  

对于多个列：

  nm1 < -  grep（^ value，colnames（data），value = TRUE）
 nm2 < ，nm1，sep =。）
 data [，（nm2）：= lapply（.SD，function（x）c（NA，x [ nm1] 
 data 
＃时间组值value1 value2 lag.value lag.value1 
＃1：1 b -0.6264538 0.7383247 1.12493092 NA NA 
＃2：2 b 0.1836433 0.5757814  - 0.04493361 -0.6264538 0.7383247 
＃3：3 b -0.8356286 -0.3053884 -0.01619026 0.1836433 0.5757814 
＃4：1 a 1.5952808 1.5117812 0.94383621 NA NA 
＃5：2 a 0.3295078 0.3898432 0.82122120 1.5952808 1.5117812 
＃6：3 a -0.8204684 -0.6212406 0.59390132 0.3295078 0.3898432 
＃7：4 a 0.4874291 -2.2146999 0.91897737 -0.8204684 -0.6212406 
＃lag.value2 
＃1：NA 
＃2：1.12493092 
＃3：-0.04493361 
＃4：NA 
＃5：0.94383621 
＃6：0.82122120 
＃7：0.59390132 
  

更新

从 .table versions> = v1.9.5 ，我们可以使用 shift $ c> type 为 lag 或 lead 。默认情况下，类型为 lag 。

  data [，（nm2）：= shift（.SD）by by groups，.SDcols = nm1] 
＃时间组值value1 value2 lag.value lag.value1 
＃1：1 b -0.6264538 0.7383247 1.12493092 NA NA 
＃2：2 b 0.1836433 0.5757814 -0.04493361 -0.6264538 0.7383247 
＃3 ：3 b -0.8356286 -0.3053884 -0.01619026 0.1836433 0.5757814 
＃4：1 a 1.5952808 1.5117812 0.94383621 NA NA 
＃5：2 a 0.3295078 0.3898432 0.82122120 1.5952808 1.5117812 
＃6：3 a -0.8204684 -0.6212406 0.59390132 0.3295078 0.3898432 
＃7：4 a 0.4874291 -2.2146999 0.91897737 -0.8204684 -0.6212406 
＃lag.value2 
＃1：NA 
＃2：1.12493092 
 ＃3：-0.04493361 
＃4：NA 
＃5：0.94383621 
＃6：0.82122120 
＃7：0.59390132 
  pre> 
 
 
如果需要反向，使用 type = lead  
 
 $ b b 
  nm3 < -  paste（lead，nm1，sep =。）
  
使用原始数据集
  data [，（nm3）：= shift ，type ='lead'），by = groups，.SDcols = nm1] 
＃时间组值value1 value2 lead.value lead.value1 
＃1：1 b -0.6264538 0.7383247 1.12493092 0.1836433 0.5757814 
＃2：2 b 0.1836433 0.5757814 -0.04493361 -0.8356286 -0.3053884 
＃3：3 b -0.8356286 -0.3053884 -0.01619026 NA NA 
＃4：1 a 1.5952808 1.5117812 0.94383621 0.3295078 0.3898432 
 ＃5：2 a 0.3295078 0.3898432 0.82122120 -0.8204684 -0.6212406 
＃6：3 a -0.8204684 -0.6212406 0.59390132 0.4874291 -2.2146999 
＃7：4 a 0.4874291 -2.2146999 0.91897737 NA NA 
＃lead .value2 
＃1：-0.04493361 
＃2：-0.01619026 
＃3：NA 
＃4：0.82122120 
＃5：0.59390132 
＃6 ：0.91897737 
＃7：NA 
  
 
 
 
 data 
 
 
 
  set.seed（1）
 data < -  data.table（time = c（1：3,1：4），groups = c （7），value1 = rnorm（7），value2 = rnorm（7））
（7）  
 
I have a data.table:
set.seed(1)
data <- data.table(time = c(1:3, 1:4),
                   groups = c(rep(c("b", "a"), c(3, 4))),
                   value = rnorm(7))

data
#    groups time      value
# 1:      b    1 -0.6264538
# 2:      b    2  0.1836433
# 3:      b    3 -0.8356286
# 4:      a    1  1.5952808
# 5:      a    2  0.3295078
# 6:      a    3 -0.8204684
# 7:      a    4  0.4874291
I want to compute a lagged version of the "value" column, within each level of "groups".

The result should look like
#   groups time      value  lag.value
# 1      a    1  1.5952808         NA
# 2      a    2  0.3295078  1.5952808
# 3      a    3 -0.8204684  0.3295078
# 4      a    4  0.4874291 -0.8204684
# 5      b    1 -0.6264538         NA
# 6      b    2  0.1836433 -0.6264538
# 7      b    3 -0.8356286  0.1836433
I have tried to use lag directly: 
data$lag.value <- lag(data$value) 
...which clearly wouldn't work. 

I have also tried:
unlist(tapply(data$value, data$groups, lag))
 a1         a2         a3         a4         b1         b2         b3 
 NA -0.1162932  0.4420753  2.1505440         NA  0.5894583 -0.2890288 
Which is almost what I want. However the vector generated is ordered differently from the ordering in the data.table which is problematic.

What is the most efficient way to do this in base R, plyr, dplyr, and data.table?
解决方案
You could do this within data.table
 library(data.table)
 data[, lag.value:=c(NA, value[-.N]), by=groups]
  data
 #   time groups       value   lag.value
 #1:    1      a  0.02779005          NA
 #2:    2      a  0.88029938  0.02779005
 #3:    3      a -1.69514201  0.88029938
 #4:    1      b -1.27560288          NA
 #5:    2      b -0.65976434 -1.27560288
 #6:    3      b -1.37804943 -0.65976434
 #7:    4      b  0.12041778 -1.37804943
For multiple columns:
nm1 <- grep("^value", colnames(data), value=TRUE)
nm2 <- paste("lag", nm1, sep=".")
data[, (nm2):=lapply(.SD, function(x) c(NA, x[-.N])), by=groups, .SDcols=nm1]
 data
#    time groups      value     value1      value2  lag.value lag.value1
#1:    1      b -0.6264538  0.7383247  1.12493092         NA         NA
#2:    2      b  0.1836433  0.5757814 -0.04493361 -0.6264538  0.7383247
#3:    3      b -0.8356286 -0.3053884 -0.01619026  0.1836433  0.5757814
#4:    1      a  1.5952808  1.5117812  0.94383621         NA         NA
#5:    2      a  0.3295078  0.3898432  0.82122120  1.5952808  1.5117812
#6:    3      a -0.8204684 -0.6212406  0.59390132  0.3295078  0.3898432
#7:    4      a  0.4874291 -2.2146999  0.91897737 -0.8204684 -0.6212406
#    lag.value2
#1:          NA
#2:  1.12493092
#3: -0.04493361
#4:          NA
#5:  0.94383621
#6:  0.82122120
#7:  0.59390132


Update

From data.table versions >= v1.9.5, we can use shift with type as lag or lead.  By default, the type is lag.  
data[, (nm2) :=  shift(.SD), by=groups, .SDcols=nm1]
#   time groups      value     value1      value2  lag.value lag.value1
#1:    1      b -0.6264538  0.7383247  1.12493092         NA         NA
#2:    2      b  0.1836433  0.5757814 -0.04493361 -0.6264538  0.7383247
#3:    3      b -0.8356286 -0.3053884 -0.01619026  0.1836433  0.5757814
#4:    1      a  1.5952808  1.5117812  0.94383621         NA         NA
#5:    2      a  0.3295078  0.3898432  0.82122120  1.5952808  1.5117812
#6:    3      a -0.8204684 -0.6212406  0.59390132  0.3295078  0.3898432
#7:    4      a  0.4874291 -2.2146999  0.91897737 -0.8204684 -0.6212406
#    lag.value2
#1:          NA
#2:  1.12493092
#3: -0.04493361
#4:          NA
#5:  0.94383621
#6:  0.82122120
#7:  0.59390132
If you need the reverse, use type=lead
nm3 <- paste("lead", nm1, sep=".")
Using the original dataset
  data[, (nm3) := shift(.SD, type='lead'), by = groups, .SDcols=nm1]
  #  time groups      value     value1      value2 lead.value lead.value1
  #1:    1      b -0.6264538  0.7383247  1.12493092  0.1836433   0.5757814
  #2:    2      b  0.1836433  0.5757814 -0.04493361 -0.8356286  -0.3053884
  #3:    3      b -0.8356286 -0.3053884 -0.01619026         NA          NA
  #4:    1      a  1.5952808  1.5117812  0.94383621  0.3295078   0.3898432
  #5:    2      a  0.3295078  0.3898432  0.82122120 -0.8204684  -0.6212406
  #6:    3      a -0.8204684 -0.6212406  0.59390132  0.4874291  -2.2146999
  #7:    4      a  0.4874291 -2.2146999  0.91897737         NA          NA
 #   lead.value2
 #1: -0.04493361
 #2: -0.01619026
 #3:          NA
 #4:  0.82122120
 #5:  0.59390132
 #6:  0.91897737
 #7:          NA


data

 set.seed(1)
 data <- data.table(time =c(1:3,1:4),groups = c(rep(c("b","a"),c(3,4))),
             value = rnorm(7), value1=rnorm(7), value2=rnorm(7))


                        
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