Lag in dataframe

I have a dataframe like

  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

I want the final data frame to have three additional column like

  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

where

Lag_1 is lagged by 1 Month
Lag_2 is lagged by 2 Months
Lag_3 is lagged by 3 Months.

I have used the following code to atleast get Lag_1

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

But this does not give me the desired output for Lag_1.

I have also tried looking at the solutions in Lag in R dataframe

And how can this be done if I have a date object instead of an int column 'Month' as in the current example?

Any help on this will be appreciated.

Solution

From data.table v1.9.6 you can use shift():

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