dplyr mutate grouped data without using exact column name - r

Question

I'm trying to wirte a function to process multiple similar dataset, here I want to subtract scores obtained by subject in the second interview by scores obtained by the same subject in the previous interview. In all dataset I want to process, interested score will be stored in the second column. Writing for each specific dataset is simple, simply use the exact column name, everything will go fine.
d <- a %>%
arrange(by_group=interview_date) %>%
dplyr::group_by(subjectkey) %>%
dplyr::mutate(score_change = colname_2nd-lag(colname_2nd))
But since I need a generic function that can be used to process multiple dataset, I can not use exact column name. So I tried 3 approaches, both of them only altered the last line
Approach#1:
dplyr::mutate(score_change = dplyr::vars(2)-lag(dplyr::vars(2)))
Approach#2：
Second column name of interested dataset contains a same string ,so I tried
dplyr::mutate(score_change = dplyr::vars(matches('string'))-lag(dplyr::vars(matches('string'))))
Error messages of the above 2 approaches will be
Error in dplyr::vars(2) - lag(dplyr::vars(2)) :
non-numeric argument to binary operator
Approach#3:
dplyr::mutate(score_change = .[[2]]-lag(.[[2]]))
Error message:
Error: Column `score_change` must be length 2 (the group size) or one, not 10880
10880 is the row number of my sample dataset, so it look like group_by does not work in this approach
Does anyone know how to make the function perform in the desired way?

Answer 1

If you want to use position of the column names use cur_data()[[2]] to refer the 2nd column of the dataframe.
library(dplyr)
d <- a %>%
arrange(interview_date) %>%
dplyr::group_by(subjectkey) %>%
dplyr::mutate(score_change = cur_data()[[2]]-lag(cur_data()[[2]]))
Also note that cur_data() doesn't count the grouped column so if subjectkey is first column in your data and colname_2nd is the second one you may need to use cur_data()[[1]] instead when you group_by.