How to create a new column using the values of the next n rows?-CodePudding

I have a data.frame object and a parameter parm which represents how many lines after it will be used with itself.

data is looking so;

parm <- 3

df <- data.frame(a=LETTERS[1:5])

df

 a    
  <chr>
1 A    
2 B    
3 C    
4 D    
5 E

The number of rows to be used should be reduced if not possible, If I need to explain it with the desired output;

  a     des_column
  <chr> <chr>     
1 A     A,B,C     
2 B     B,C,D     
3 C     C,D,E     
4 D     D,E       
5 E     E

base R functions would be much better.

Thanks in advance.

CodePudding user response：

You could just apply across all the rows, ensuring your desired length doesn't overrun the number of rows and paste together.

n <- nrow(df)

df$des_column <- sapply(
  1:n,
  \(x) paste(df$a[x:min(x parm-1, n)], collapse = ",")
)

df
#>   a des_column
#> 1 A      A,B,C
#> 2 B      B,C,D
#> 3 C      C,D,E
#> 4 D        D,E
#> 5 E          E

^{\(x) is shorthand for function(x) released in R 4.1}

CodePudding user response：

Another possible solution, based on zoo::rollapply:

library(zoo)

parm <- 3
df <- data.frame(a=LETTERS[1:5])

df$des_column <- rollapply(df$a, parm, paste, collapse = ",",
                    partial=T, align="left")

df

#>   a des_column
#> 1 A      A,B,C
#> 2 B      B,C,D
#> 3 C      C,D,E
#> 4 D        D,E
#> 5 E          E

CodePudding user response：

Using the same logic from above, another way would be:

library(tidyverse)

df$des_column <- map_chr(seq_len(nrow(df)),
                         function(i) glue_collapse(df$a[i:min(i parm-1, nrow(df))], sep = ','))

#>   a des_column
#> 1 A      A,B,C
#> 2 B      B,C,D
#> 3 C      C,D,E
#> 4 D        D,E
#> 5 E          E