整理数据：重命名列，获取非NA列名称，然后收集

Question

I've got a rather ugly bit of data to tidy up and need help! What my data look like now:

countries <- c("Austria", "Belgium", "Croatia")

df <- tibble("age" = c(28,42,19, 67),
         "1_recreate_1"=c(NA,15,NA,NA), 
         "1_recreate_2"=c(NA,10,NA,NA), 
         "1_recreate_3"=c(NA,8,NA,NA),
         "1_recreate_4"=c(NA,4,NA,NA),
         "1_fairness" = c(NA, 7, NA, NA),
         "1_confidence" = c(NA, 5, NA, NA),
         "2_recreate_1"=c(29,NA,NA,30),
         "2_recreate_2"=c(20,NA,NA,24),
         "2_recreate_3"=c(15,NA,NA,15),
         "2_recreate_4"=c(11,NA,NA,9),
         "2_fairness" = c(4, NA, NA, 1),
         "2_confidence" = c(5, NA, NA, 4),
         "3_recreate_1"=c(NA,NA,50,NA), 
         "3_recreate_2"=c(NA,NA,40,NA), 
         "3_recreate_3"=c(NA,NA,30,NA),
         "3_recreate_4"=c(NA,NA,20,NA),
         "3_fairness" = c(NA,  NA, 2, NA),
         "3_confidence" = c(NA, NA, 2, NA),
         "overall" = c(3,3,2,5))    

What I need them to look like at the end (hard-coding it):

df <- tibble(age = rep(c(28,42,19,67), each=4),
         country = rep(c("Belgium", "Austria", "Croatia", "Belgium"), each=4),
         recreate = rep(1:4, times=4),
         fairness = rep(c(4,7,2,1), each=4),
         confidence = rep(c(5,5,2,4), each=4),     
         allocation = c(29, 20, 15, 11,
                        15, 10, 8, 4,
                        50, 40, 30, 20, 
                        30, 24, 15, 9),
         overall = rep(c(3,3,2,5), each=4))

Steps to get there (I think!):

1. Replace the starting numbers for those columns using my list of countries.
The number that starts the string is the index in countries . In other words, 16_recreate_1 would correspond with the 16th country in the vector countries . I think the following code works (though am not sure it's exactly right):

for(i in length(countries):1){
    colnames(df) <- str_replace(colnames(df), paste0(i,"_"), paste0(countries[i],"_"))
}  

2. Create a new variable called "country" by getting the name of the column(s) that is NOT NA for each row.

I tried a BUNCH of experimentation with which.max and names , but couldn't get it fully functional.

3. Create new variables ( recreate_1 ... recreate_4 ) that grab the [country_name]_recreate_1 ... [country_name]_recreate_4 value for each row, whatever country is non-NA for that person.

Maybe rowSums is the way to do this?

4. Make the data long instead of wide I think this is going to require gather , but I'm not sure how to gather from only the variables country and recreate_1 ... recreate_4 .

I'm so sorry this is so complex. Tidyverse solutions are preferred but any help is greatly appreciated!

Answer 1

library(dplyr)
library(tidyr)
df %>% mutate(rid=row_number()) %>% 
       gather(key,val,-c(age,overall,rid, matches('recreate'))) %>% mutate(country=sub('(^\\d)_.*','\\1',key),country=countries[as.numeric(country)]) %>% 
       filter(!is.na(val)) %>% mutate(key=sub('(^\\d\\_)(.*)','\\2',key)) %>%
       spread(key,val) %>% gather(key = recreate,value = allocation,-c(rid,age,overall,Country,confidence,fairness)) %>% 
       filter(!is.na(allocation)) %>% mutate(recreate=sub('.*_(\\d$)','\\1',recreate))

Here (^\\\\d)_.* means get the first digit while .*_(\\\\d$) means get the last digit.

Answer 2

A somehow different tidyverse possibility could be:

df %>%
 gather(variable, allocation, na.rm = TRUE) %>%
 separate(variable, c("ID", "variable", "recreate"), convert = TRUE) %>%
 left_join(data.frame(countries) %>%
            mutate(country = countries,
                   ID = seq_along(countries)) %>%
            select(-countries), by = c("ID" = "ID")) %>%
 select(-variable, -ID) 

   recreate allocation country
      <int>      <dbl> <fct>  
 1        1         15 Austria
 2        2         10 Austria
 3        3          8 Austria
 4        4          4 Austria
 5        1         29 Belgium
 6        1         30 Belgium
 7        2         20 Belgium
 8        2         24 Belgium
 9        3         15 Belgium
10        3         15 Belgium
11        4         11 Belgium
12        4          9 Belgium
13        1         50 Croatia
14        2         40 Croatia
15        3         30 Croatia
16        4         20 Croatia

Here it, first, transforms the data from wide to long format, removing the rows with NA. Second, it separates the variable names into three columns. Third, it transforms the vector of countries into a df and assigns each country a unique ID. Finally, it joins the two and removes the redundant variables.

A solution to the edited question:

df %>%
 select(matches("(recreate)")) %>%
 rowid_to_column() %>%
 gather(var, allocation, -rowid, na.rm = TRUE) %>%
 separate(var, c("ID", "var", "recreate"), convert = TRUE) %>%
 select(-var) %>%
 left_join(data.frame(countries) %>%
            mutate(country = countries,
                   ID = seq_along(countries)) %>%
            select(-countries), by = c("ID" = "ID")) %>% 
 left_join(df %>%
            select(-matches("(recreate)")) %>%
            rowid_to_column() %>%
            gather(var, val, -rowid, na.rm = TRUE) %>%
            mutate(var = gsub("[^[:alpha:]]", "", var)) %>%
            spread(var, val), by = c("rowid" = "rowid")) %>%
 select(-rowid, -ID)

   recreate allocation country   age confidence fairness overall
      <int>      <dbl> <fct>   <dbl>      <dbl>    <dbl>   <dbl>
 1        1         15 Austria    42          5        7       3
 2        2         10 Austria    42          5        7       3
 3        3          8 Austria    42          5        7       3
 4        4          4 Austria    42          5        7       3
 5        1         29 Belgium    28          5        4       3
 6        1         30 Belgium    67          4        1       5
 7        2         20 Belgium    28          5        4       3
 8        2         24 Belgium    67          4        1       5
 9        3         15 Belgium    28          5        4       3
10        3         15 Belgium    67          4        1       5
11        4         11 Belgium    28          5        4       3
12        4          9 Belgium    67          4        1       5
13        1         50 Croatia    19          2        2       2
14        2         40 Croatia    19          2        2       2
15        3         30 Croatia    19          2        2       2
16        4         20 Croatia    19          2        2       2

Here it, first, selects the columns that contain recreate and adds a columns with row ID. Second, it follows the steps from the original solution. Third, it selects the columns that do not contain recreate , performs a wide-to-long data transformation, removes the number from column names and transforms the data back to the original wide format. Finally, it joins the two on row ID and removes the redundant variables.