I have a data frame like the one below:

dat<-data.frame (col1= c(rep("A", 10), rep("B", 10)),
                 test= c(rep("pre", 5), rep("post", 5), rep("pre", 5), rep("post", 5)), 
                 ID= c("ID_A","ID_B","ID_C","ID_D","ID_E","ID_A","ID_B","ID_C", "ID_D","ID_E","ID_F","ID_G","ID_H","ID_I","ID_J","ID_F","ID_G","ID_H","ID_I","ID_J"),
                 answer= c("1_2_3", "4_6_8", "9_4_1", "3_7_2", "5_7_4", "1_2_3", "6_8_4", "9_4_2", "3_9_1", "5_4_7", "6_5_4", "4_6_8", "3_7_1", "6_7_9", "5_7_4", "8_9_1", "6_8_4", "7_1_4", "6_8_3", "5_4_7"))

Each level of ID has two levels in test - which represent responses to a pre and post test. answer represents the respondents answer.

I wish to summarize the following result data frame res:

res<-data.frame(ID= c("ID_A","ID_B","ID_C", "ID_D","ID_E", "ID_F","ID_G","ID_H","ID_I","ID_J"),
               res1=c("yes", "yes","no", "no", "yes", "no", "yes","no", "no", "yes"), #match? yes or no
               res2=c(0,0,1, 2, 0, 3, 0,1, 2, 0 ),#count of the new answer components
               res3= c(NA, NA, 2, "9_1", NA,"8_9_1",NA, 4, "8_3", NA)) #string specifying new answer components

res summarizes matches in answer for each level of ID irrespective of order and returns:

• col1 a binary variable that summarizes the presence or absence of a match between the pre and post test
• col2 a count of the number of new entries in the post test. Values can only be between 0-3 since there is a limit of 3 values for each answer.
• col3 a string that specifies the new answer components present in the post test but not in the pre test.

CodePudding user response：

• We can use strsplit to split the answer string column and match the pre with post responses

library(dplyr)

dat |> group_by(ID) |> select(-col1) |> 
       mutate(res1 = case_when(all(strsplit(nth(answer , 1) , "_")[[1]] %in% strsplit(nth(answer , 2) , "_")[[1]]) ~ "yes" , TRUE ~ "no")) |>
       group_by(ID , res1) |> summarise(res2 = 3 - sum(strsplit(nth(answer , 1) , "_")[[1]] %in% strsplit(nth(answer , 2) , "_")[[1]]) ,
       res3 = paste0(setdiff(strsplit(nth(answer , 2) , "_")[[1]] , strsplit(nth(answer , 1) , "_")[[1]]) , collapse = "_"))

• Output

# A tibble: 10 × 4
# Groups:   ID [10]
   ID    res1   res2 res3   
   <chr> <chr> <dbl> <chr>  
 1 ID_A  yes       0 ""     
 2 ID_B  yes       0 ""     
 3 ID_C  no        1 "2"    
 4 ID_D  no        2 "9_1"  
 5 ID_E  yes       0 ""     
 6 ID_F  no        3 "8_9_1"
 7 ID_G  yes       0 ""     
 8 ID_H  no        1 "4"    
 9 ID_I  no        2 "8_3"  
10 ID_J  yes       0 ""     

CodePudding user response：

You can create a small helper function f

f <- function(ans) {
  p =  strsplit(ans, "_")
  pdiff = setdiff(p[[2]], p[[1]])
  if(length(pdiff)==0) return(list("res1"="yes", "res2"=0L, "res3"=NA_character_))
  return(list("res1"="no", "res2"=length(pdiff), "res3" = paste0(pdiff, collapse="_")))
}

and apply it by ID:

library(tidyverse)
unnest_wider(at %>% group_by(ID) %>% summarize(res=list(f(answer))),res)

Output:

   ID    res1   res2 res3 
   <chr> <chr> <int> <chr>
 1 ID_A  yes       0 NA   
 2 ID_B  yes       0 NA   
 3 ID_C  no        1 2    
 4 ID_D  no        2 9_1  
 5 ID_E  yes       0 NA   
 6 ID_F  no        3 8_9_1
 7 ID_G  yes       0 NA   
 8 ID_H  no        1 4    
 9 ID_I  no        2 8_3  
10 ID_J  yes       0 NA