CODEWITHSUNDEEP
r
melange164
melange164

Reputation: 143

In R, compute time series difference of lagged values

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?

Upvotes: 1

Views: 232

Answers (3)

Quinten
Quinten

Reputation: 41225

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

Upvotes: 2

PaulS
PaulS

Reputation: 25313

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

Upvotes: 1

user2974951
user2974951

Reputation: 10375

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

Upvotes: 1

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