Question

我有一个像

这样的数据框

  ID_CASE   Month   
CS00000026A 201301  
CS00000026A 201302  
CS00000026A 201303  
CS00000026A 201304  
CS00000026A 201305  
CS00000026A 201306  
CS00000026A 201307  
CS00000026A 201308  
CS00000026A 201309  
CS00000026A 201310  
CS00000191C 201302  
CS00000191C 201303  
CS00000191C 201304  
CS00000191C 201305  
CS00000191C 201306  
CS00000191C 201307  
CS00000191C 201308  
CS00000191C 201309  
CS00000191C 201310

我希望最终数据框有三个额外的列，如

  ID_CASE   Month   Lag_1   Lag_2   Lag_3
CS00000026A 201301  NA      NA      NA
CS00000026A 201302  201301  NA      NA
CS00000026A 201303  201202  201201  NA
CS00000026A 201304  201203  201202  201201
CS00000026A 201305  201204  201203  201202
CS00000026A 201306  201305  201304  201303
CS00000026A 201307  201306  201305  201304
CS00000026A 201308  201307  201306  201305
CS00000026A 201309  201308  201307  201306
CS00000026A 201310  201309  201308  201307
CS00000191C 201302  NA       NA     NA
CS00000191C 201303  201302   NA     NA
CS00000191C 201304  201303  201302      NA
CS00000191C 201305  201304  201303  201302
CS00000191C 201306  201305  201304  201303
CS00000191C 201307  201306  201305  201304
CS00000191C 201308  201307  201306  201305
CS00000191C 201309  201308  201307  201306
CS00000191C 201310  201309  201308  201307

其中

Lag_1滞后1个月
Lag_2滞后2个月
Lag_3滞后3个月。

我已使用以下代码至少获取Lag_1

df <- ddply(df,.(ID_CASE),transform,
                  Lag_1 <- c(NA,Month[-nrow(df)]))

但这并不能为我提供Lag_1所需的输出。

我也尝试过查看解决方案 Lag in R dataframe

如果我有一个日期对象而不是 int 列＆＃39;月＆＃39;那怎么办呢？如在当前的例子中？

对此有任何帮助将不胜感激。

Answer 1

尝试data.table

library(data.table)
setDT(df)[, `:=` (Lag_1 = c(NA, Month[-.N]),
                  Lag_2 = c(rep(NA, 2), Month[-.N]),
                  Lag_3 = c(rep(NA, 3), Month[-.N])), by = ID_CASE]
df
#         ID_CASE  Month  Lag_1  Lag_2  Lag_3
#  1: CS00000026A 201301     NA     NA     NA
#  2: CS00000026A 201302 201301     NA     NA
#  3: CS00000026A 201303 201302 201301     NA
#  4: CS00000026A 201304 201303 201302 201301
#  5: CS00000026A 201305 201304 201303 201302
#  6: CS00000026A 201306 201305 201304 201303
#  7: CS00000026A 201307 201306 201305 201304
#  8: CS00000026A 201308 201307 201306 201305
#  9: CS00000026A 201309 201308 201307 201306
# 10: CS00000026A 201310 201309 201308 201307
# 11: CS00000191C 201302     NA     NA     NA
# 12: CS00000191C 201303 201302     NA     NA
# 13: CS00000191C 201304 201303 201302     NA
# 14: CS00000191C 201305 201304 201303 201302
# 15: CS00000191C 201306 201305 201304 201303
# 16: CS00000191C 201307 201306 201305 201304
# 17: CS00000191C 201308 201307 201306 201305
# 18: CS00000191C 201309 201308 201307 201306
# 19: CS00000191C 201310 201309 201308 201307

Answer 2

您可以使用lag.zoo，其中k可以是滞后矢量。

library(plyr)
library(zoo)

ddply(df, .(ID_CASE), function(x){
  z <- zoo(x$Month)
  lag(z, k = 0:-3)
})

#        ID_CASE   lag0  lag-1  lag-2  lag-3
# 1  CS00000026A 201301     NA     NA     NA
# 2  CS00000026A 201302 201301     NA     NA
# 3  CS00000026A 201303 201302 201301     NA
# 4  CS00000026A 201304 201303 201302 201301
# 5  CS00000026A 201305 201304 201303 201302
# 6  CS00000026A 201306 201305 201304 201303
# 7  CS00000026A 201307 201306 201305 201304
# 8  CS00000026A 201308 201307 201306 201305
# 9  CS00000026A 201309 201308 201307 201306
# 10 CS00000026A 201310 201309 201308 201307
# 11 CS00000191C 201302     NA     NA     NA
# 12 CS00000191C 201303 201302     NA     NA
# 13 CS00000191C 201304 201303 201302     NA
# 14 CS00000191C 201305 201304 201303 201302
# 15 CS00000191C 201306 201305 201304 201303
# 16 CS00000191C 201307 201306 201305 201304
# 17 CS00000191C 201308 201307 201306 201305
# 18 CS00000191C 201309 201308 201307 201306
# 19 CS00000191C 201310 201309 201308 201307

评论后

修改。

如果只有一个日期的组，则上面的代码将生成错误。一个小例子：

df <- data.frame(ID_CASE = c(1, 1, 1, 2), Month = 1:4)
df
#   ID_CASE Month
# 1       1     1
# 2       1     2
# 3       1     3
# 4       2     4

ddply(df, .(ID_CASE), function(x){
  z <- zoo(x$Month)
  lag(z, k = 0:-3)
})

# Error in list_to_dataframe(res, attr(.data, "split_labels"), .id, id_as_factor) : 
#   Results do not have equal lengths

这是由于“一个仅限注册的团体”被强制转换为单变量时间序列。要避免此类强制，请使用[子集和drop = FALSE

ddply(df, .(ID_CASE), function(x){
  z <- zoo(x[ , "Month", drop = FALSE])
  lag(z, k = 0:-3)
})

#   ID_CASE Month.lag0 Month.lag-1 Month.lag-2 Month.lag-3
# 1       1          1          NA          NA          NA
# 2       1          2           1          NA          NA
# 3       1          3           2           1          NA
# 4       2          4          NA          NA          NA

Answer 3

从data.table v1.9.6，您可以使用shift()：

require(data.table)
setDT(df)[, paste("lag", 1:3, sep="_") := shift(Month, 1:3), by=ID_CASE]

Answer 4

使用dplyr：

library(dplyr)

 df %.%
  group_by(ID_CASE) %.%
  mutate(lag_1 = lag(Month, 1),
         lag_2 = lag(Month, 2),
         lag_3 = lag(Month, 3))

数据帧滞后

4 个答案: