Combine data file and label file together to have one single labelled dataframe in R

Question

I have two data frames, one is survey data (data.csv) and another one is label data (label.csv). Here is the sample data (My original data has about 150 variables)

#sample data

df <- tibble::tribble(
  ~id, ~House_member, ~dob, ~age_quota, ~work, ~sex, ~pss,
  1L,            4L,  1983L,  2L,        2L,     1,      1,
  2L,            1L,  1940L,  7L,        2L,     1,      2,
  3L,            2L,  1951L,  5L,        6L,     1,      1,
  4L,            4L,  1965L,  2L,        2L,     1,      4,
  5L,            3L,  1965L,  2L,        3L,     1,      1,
  6L,            1L,  1951L,  3L,        1L,     1,      3,
  7L,            1L,  1955L,  1L,        1L,     1,      3,
  8L,            4L,  1982L,  2L,        2L,     2,      5,
  9L,            2L,  1990L,  2L,        4L,     2,      3,
  10L,            2L,  1953L, 3L,        2L,     2,      4
)


#sample label data
label <- tibble::tribble(
                ~variable, ~value,                           ~label,
           "House_member",     NA, "How many people live with you?",
           "House_member",     1L,                       "1 person",
           "House_member",     2L,                      "2 persons",
           "House_member",     3L,                      "3 persons",
           "House_member",     4L,                      "4 persons",
           "House_member",     5L,                      "5 persons",
           "House_member",     6L,                      "6 persons",
           "House_member",     7L,                      "7 persons",
           "House_member",     8L,                      "8 persons",
           "House_member",     9L,                      "9 persons",
           "House_member",    10L,                     "10 or more",
                    "dob",     NA,                  "date of brith",
              "age_quota",     NA,                      "age_quota",
              "age_quota",     1L,                          "10-14",
              "age_quota",     2L,                          "15-19",
              "age_quota",     3L,                          "20-29",
              "age_quota",     4L,                          "30-39",
              "age_quota",     5L,                          "40-49",
              "age_quota",     6L,                          "50-70",
              "age_quota",     7L,                           "70 +",
                   "work",     NA,        "what is your occupation?",
                   "work",     1L,                      "full time",
                   "work",     2L,                      "part time",
                   "work",     3L,                        "retired",
                   "work",     4L,                        "student",
                   "work",     5L,                      "housewife",
                   "work",     6L,                     "unemployed",
                   "work",     7L,                          "other",
                   "work",     8L,                   "kid under 15",
                    "sex",     NA,                        "gender?",
                    "sex",     1L,                            "Man",
                    "sex",     2L,                          "Woman",
                    "pss",     NA,       "How often do you use PS?",
                    "pss",     1L,                          "Daily",
                    "pss",     2L,         "several times per week",
                    "pss",     3L,                  "once per week",
                    "pss",     4L,         "several time per month",
                    "pss",     5L,                          "Rarly"
           )

I am wondering is there any way that I can combine these file together to have a one labelled dataframe like SPSS’s style format (dbl+lbl format). I know labelled package which can add a value label to a non labelled vector, like this example:

v <- labelled::labelled(c(1,2,2,2,3,9,1,3,2,NA), c(yes = 1, maybe = 2, no = 3))

I was hoping there is a better/faster way than adding label to each variable one by one.

Answer 1

Another imap_dfc solution:

library(tidyverse)

df %>% imap_dfc(~{ 
                  label[label$variable==.y,c('label','value')] %>%
                  deframe() %>% # to named vector
                  haven::labelled(.x,.)
                 })

# A tibble: 10 x 7
          id  House_member       dob age_quota           work       sex                        pss
   <int+lbl>     <int+lbl> <int+lbl> <int+lbl>      <int+lbl> <dbl+lbl>                  <dbl+lbl>
 1         1 4 [4 persons]      1983 2 [15-19] 2 [part time]  1 [Man]   1 [Daily]                 
 2         2 1 [1 person]       1940 7 [70 +]  2 [part time]  1 [Man]   2 [several times per week]
 3         3 2 [2 persons]      1951 5 [40-49] 6 [unemployed] 1 [Man]   1 [Daily]                 
 4         4 4 [4 persons]      1965 2 [15-19] 2 [part time]  1 [Man]   4 [several time per month]
 5         5 3 [3 persons]      1965 2 [15-19] 3 [retired]    1 [Man]   1 [Daily]                 
 6         6 1 [1 person]       1951 3 [20-29] 1 [full time]  1 [Man]   3 [once per week]         
 7         7 1 [1 person]       1955 1 [10-14] 1 [full time]  1 [Man]   3 [once per week]         
 8         8 4 [4 persons]      1982 2 [15-19] 2 [part time]  2 [Woman] 5 [Rarly]                 
 9         9 2 [2 persons]      1990 2 [15-19] 4 [student]    2 [Woman] 3 [once per week]         
10        10 2 [2 persons]      1953 3 [20-29] 2 [part time]  2 [Woman] 4 [several time per month]

Used tibble::deframe and haven::labelled which are included in tidyverse

Speed comparison after replacing filter/select by direct access to label:

Waldi <- function() {
df %>% imap_dfc(~{ 
    label[label$variable==.y,c('label','value')] %>%
      deframe() %>% # to named vector
      haven::labelled(.x,.)})}

Waldi_old <- function() {   
    df %>% imap_dfc(~{ 
      label %>% filter(variable==.y) %>%
        select(label, value) %>%
        deframe() %>% # to named vector
        haven::labelled(.x,.)
    })}

#EDIT : Included TIC33() for-loop solution

microbenchmark::microbenchmark(TIC3(),Waldi(),Anil(),TIC1(),Waldi_old(),Sinh())
Unit: microseconds
        expr     min       lq      mean   median       uq     max neval   cld
      TIC3()   688.0   871.80   982.280   920.95  1005.55  1801.6   100 a    
     Waldi()  1345.5  1543.60  1804.758  1635.45  1893.75  4306.8   100  b   
      Anil()  4006.8  4476.65  5188.519  4862.95  5439.10 10163.6   100   c  
      TIC1()  3898.2  4278.80  5009.927  4774.95  5277.05 12916.2   100   c  
 Waldi_old() 18712.3 20091.75 21756.140 20609.35 22169.75 33359.8   100    d 
      Sinh() 22730.9 24093.45 25931.412 24946.00 26614.00 38735.3   100     e

Answer 2

Here is an approach using purrr::imap_dfc

library(dplyr)
library(purrr)

# custom function for taking the column data and column name and reformat the values using factor
custom_function <- function(value, col_name) { 
  matching_vairable <- label %>%
    filter(variable == col_name & !is.na(value)) %>%
    select(label, value)
  column_data <- tibble(!!sym(col_name) := value)
  if (nrow(matching_vairable) > 0) {
    column_data[[1]] <- labelled::labelled(column_data[[1]],
      tibble::deframe(matching_vairable))
  }
  column_data
}

new_df <- imap_dfc(df, .f = custom_function)

Output

new_df
#> # A tibble: 10 x 7
#>       id  House_member   dob age_quota         work      sex                 pss
#>    <int>     <int+lbl> <int> <int+lbl>    <int+lbl> <dbl+lb>           <dbl+lbl>
#>  1     1 4 [4 persons]  1983 2 [15-19] 2 [part tim… 1 [Man]  1 [Daily]          
#>  2     2 1 [1 person]   1940 7 [70 +]  2 [part tim… 1 [Man]  2 [several times p…
#>  3     3 2 [2 persons]  1951 5 [40-49] 6 [unemploy… 1 [Man]  1 [Daily]          
#>  4     4 4 [4 persons]  1965 2 [15-19] 2 [part tim… 1 [Man]  4 [several time pe…
#>  5     5 3 [3 persons]  1965 2 [15-19] 3 [retired]  1 [Man]  1 [Daily]          
#>  6     6 1 [1 person]   1951 3 [20-29] 1 [full tim… 1 [Man]  3 [once per week]  
#>  7     7 1 [1 person]   1955 1 [10-14] 1 [full tim… 1 [Man]  3 [once per week]  
#>  8     8 4 [4 persons]  1982 2 [15-19] 2 [part tim… 2 [Woma… 5 [Rarly]          
#>  9     9 2 [2 persons]  1990 2 [15-19] 4 [student]  2 [Woma… 3 [once per week]  
#> 10    10 2 [2 persons]  1953 3 [20-29] 2 [part tim… 2 [Woma… 4 [several time pe…

new_df %>% pull(House_member)
#> <labelled<integer>[10]>
#>  [1] 4 1 2 4 3 1 1 4 2 2
#> 
#> Labels:
#>  value      label
#>      1   1 person
#>      2  2 persons
#>      3  3 persons
#>      4  4 persons
#>      5  5 persons
#>      6  6 persons
#>      7  7 persons
#>      8  8 persons
#>      9  9 persons
#>     10 10 or more

^{Created on 2021-05-16 by the reprex package (v2.0.0)}

Answer 3

Though not as fast as other answer proposed by @Waldi, yet this may be considered as an option due to use of minimum external packages.

Alternate without using purrr::imap_* & tibble::deframe, this may be done in dplyr only using mutate(across(.. as shown below-

library(dplyr)
library(labelled)

df %>%
  mutate(across(everything(), ~labelled::labelled(as.double(.), 
                                                  setNames(label$value[label$variable == cur_column()], 
                                                           label$label[label$variable == cur_column()])
                                                  )))


# A tibble: 10 x 7
          id  House_member       dob age_quota           work       sex                        pss
   <dbl+lbl>     <dbl+lbl> <dbl+lbl> <dbl+lbl>      <dbl+lbl> <dbl+lbl>                  <dbl+lbl>
 1         1 4 [4 persons]      1983 2 [15-19] 2 [part time]  1 [Man]   1 [Daily]                 
 2         2 1 [1 person]       1940 7 [70 +]  2 [part time]  1 [Man]   2 [several times per week]
 3         3 2 [2 persons]      1951 5 [40-49] 6 [unemployed] 1 [Man]   1 [Daily]                 
 4         4 4 [4 persons]      1965 2 [15-19] 2 [part time]  1 [Man]   4 [several time per month]
 5         5 3 [3 persons]      1965 2 [15-19] 3 [retired]    1 [Man]   1 [Daily]                 
 6         6 1 [1 person]       1951 3 [20-29] 1 [full time]  1 [Man]   3 [once per week]         
 7         7 1 [1 person]       1955 1 [10-14] 1 [full time]  1 [Man]   3 [once per week]         
 8         8 4 [4 persons]      1982 2 [15-19] 2 [part time]  2 [Woman] 5 [Rarly]                 
 9         9 2 [2 persons]      1990 2 [15-19] 4 [student]    2 [Woman] 3 [once per week]         
10        10 2 [2 persons]      1953 3 [20-29] 2 [part time]  2 [Woman] 4 [several time per month]

As noticed, in comments, you required output columns as dbl + lbl therefore first argument has been used as as.double(.) instead of just . in which the output will be int + lbl when input cols are of integer type.

Answer 4

Here are some variants to the answer by @AnilGoyal. It seems for loop (see TIC3()) provides nice speed.

• Variant1

TIC1 <- function() {
  df %>%
    mutate(
      across(everything(), ~ labelled(
        .,
        with(label, setNames(value, label)[variable == cur_column()])
      ))
    )
}

• Variant 2 (If you don't want NA being labelled)

TIC2 <- function() {
  df %>%
    mutate(
      across(
        with(label, unique(variable[!is.na(value)])),
        ~ labelled(
          .,
          with(label, setNames(value, label)[variable == cur_column()])
        )
      )
    )
}

• Variant 3 (for loop version of TIC1())

TIC3 <- function() {
  nms <- names(df)
  for (k in nms[nms %in% label$variable]) {
    df[[k]] <- labelled(df[[k]], with(label, setNames(value, label)[variable == k]))
  }
  df
}

---

Benchmarking

TIC1 <- function() {
  df %>%
    mutate(
      across(everything(), ~ labelled(
        .,
        with(label, setNames(value, label)[variable == cur_column()])
      ))
    )
}

TIC2 <- function() {
  df %>%
    mutate(
      across(
        with(label, unique(variable[!is.na(value)])),
        ~ labelled(
          .,
          with(label, setNames(value, label)[variable == cur_column()])
        )
      )
    )
}

TIC3 <- function() {
  nms <- names(df)
  for (k in nms[nms %in% label$variable]) {
    df[[k]] <- labelled(df[[k]], with(label, setNames(value, label)[variable == k]))
  }
  df
}


Waldi1 <- function() {
  df %>% imap_dfc(~ {
    label %>%
      filter(variable == .y) %>%
      select(label, value) %>%
      deframe() %>%
      # to named vector
      haven::labelled(.x, .)
  })
}

Waldi2 <- function() {
  df %>% imap_dfc(~ {
    label[label$variable == .y, c("label", "value")] %>%
      deframe() %>% # to named vector
      haven::labelled(.x, .)
  })
}

Anil <- function() {
  df %>%
    mutate(across(everything(), ~ labelled::labelled(
      as.double(.),
      setNames(
        label$value[label$variable == cur_column()],
        label$label[label$variable == cur_column()]
      )
    )))
}

custom_function <- function(value, col_name) {
  matching_vairable <- label %>%
    filter(variable == col_name & !is.na(value)) %>%
    select(label, value)
  column_data <- tibble(!!sym(col_name) := value)
  if (nrow(matching_vairable) > 0) {
    column_data[[1]] <- labelled::labelled(
      column_data[[1]],
      tibble::deframe(matching_vairable)
    )
  }
  column_data
}

Sinh <- function(x) {
  imap_dfc(df, .f = custom_function)
}

microbenchmark(
  Waldi1(),
  Waldi2(),
  Anil(),
  Sinh(),
  TIC1(),
  TIC2(),
  TIC3(),
  unit = "relative"
)

gives

Unit: relative
     expr       min        lq      mean    median        uq        max neval
 Waldi1() 17.540613 17.359550 17.019266 17.238594 18.502584  4.7788575   100
 Waldi2()  1.355634  1.350547  1.338517  1.352509  1.342408  0.7033271   100
   Anil()  3.996836  4.011826  3.902559  4.029819  3.937232  1.2877871   100
   Sinh() 20.756122 20.595253 20.637410 20.452746 21.484992 13.0362139   100
   TIC1()  3.617278  3.617310  3.480283  3.609973  3.526703  1.0682179   100
   TIC2()  3.315545  3.384422  3.282862  3.389645  3.325616  1.0474304   100
   TIC3()  1.000000  1.000000  1.000000  1.000000  1.000000  1.0000000   100