In R, compute time series difference of lagged values-CodePudding

Hello my dataframe looks like this for a very large number of years and products:

product<-c(1,2,3)
yr1<-c(109,213,30)
yr2<-c(613,488,125)
yr3<-c(729,494,127)
df<-data.frame(product,yr1,yr2,yr3)

I need to transform it so that the values for the years after the first are the difference of the current value and the previous (lag) value, so that it looks like this:

yr2<-c(504,275,95)
yr3<-c(116,6,2)
df<-data.frame(product,yr1,yr2,yr3)

With 613-109=504, 729-613=116, etc.

How can I do this?

CodePudding user response：

dplyr option with cur_data which makes a copy of the data to substract every column with its previous column:

library(dplyr)
df %>% 
  mutate(cur_data()[,c(-2)] - cur_data()[-ncol(.)]) %>%
  mutate(product = df$product)

Output:

  product yr1 yr2 yr3
1       1 109 504 116
2       2 213 275   6
3       3  30  95   2

CodePudding user response：

Try

df[,tail(grep("yr",colnames(df)),-1)]-df[,head(grep("yr",colnames(df)),-1)]

  yr2 yr3
1 504 116
2 275   6
3  95   2

CodePudding user response：

Another dplyr approach:

library(dplyr)

df %>% 
 mutate(yr1 = NULL, across(-product, ~ .x-df[which(cur_column() == names(df))-1]))

#>   product yr1 yr2
#> 1       1 504 116
#> 2       2 275   6
#> 3       3  95   2