有没有办法从 R 中的数据字典创建因子？

Question

Am trying to create factors from the data dictionary ? I tried using the Map but all the variables are converted to missing. How best was to approach this approach this? Doing it the purrr way would also be welcome.

library(dplyr)

mydata <- tibble(
  a_1 = c(20,22, 13,14,44),
  a_2 = c(42, 13, 32, 31, 14),
  b = c(1, 2, 1, 1, 2),
  c = c(1, 2, 1, 3, 1)
)



dictionary <- tibble(
  variable = c("a", "b", "c"),
  label = c("Age", "Gender", "Education"),
  type = c("mselect", "select", "select"),
  values = c(NA, "1, 2", "1, 2,3" ),
  valuelabel = c(NA, "Male, Female", "Primary, Secondary, Tertiary")

)

# Expected results 
expectedata <- mydata %>% 
  mutate(
    b = factor(b, levels = c(1, 2), labels = c("Male", "Female")),
    c = factor(c, levels = c(1, 2, 3), 
               labels = c("Primary", "Secondary", "Tertiary"))
  )
expectedata 


# Select the factor variables

factor_vars <- dictionary %>%
  filter(type == "select") %>% pull(variable)


mydata[] <- Map(
  function(x, fctvalues, fctlabels)  factor(x, fctvalues,  fctlabels) ,
                mydata,
                dictionary$values[ match(factor_vars,
                                                 dictionary$variable) ],

                dictionary$valuelabel[ match(factor_vars,
                                             dictionary$variable) ]
)

Answer 1

Via pivot_ , left_join , and a bit of data wrangling:

Data

library(tidyverse)

mydata <- tibble(
    a_1 = c(20,22, 13,14,44),
    a_2 = c(42, 13, 32, 31, 14),
    b = c(1, 2, 1, 1, 2),
    c = c(1, 2, 1, 3, 1)
)



dictionary <- tibble(
    variable = c("a", "b", "c"),
    label = c("Age", "Gender", "Education"),
    type = c("mselect", "select", "select"),
    values = c(NA, "1, 2", "1, 2, 3" ),
    valuelabel = c(NA, "Male, Female", "Primary, Secondary, Tertiary")
    
)

Code

target_dictionary <- dictionary %>%
    # optional: filter(type == "select") %>%
    separate_rows(values, valuelabel) %>% 
    select(variable, values, valuelabel)

target_mydata <- mydata %>%
    # Assuming you have no unique identifier
    rownames_to_column("id") %>%
    pivot_longer(
        cols = c("b", "c"),
        names_to = "var_name",
        values_to = "var_value"
    ) %>%
    # because the data types don't match here
    mutate(
        var_value = as.character(var_value)
    ) %>%
    left_join(
        target_dictionary,
        by = c("var_name" = "variable", "var_value" = "values")
    ) %>%
    pivot_wider(
        names_from = var_name,
        values_from = valuelabel, 
        id_cols = c("id", "a_1", "a_2")
    ) %>%
    select(-id)

Result:

> target_mydata
# A tibble: 5 × 4
    a_1   a_2 b      c        
  <dbl> <dbl> <chr>  <chr>    
1    20    42 Male   Primary  
2    22    13 Female Secondary
3    13    32 Male   Primary  
4    14    31 Male   Tertiary 
5    44    14 Female Primary  

---

Edit: You cpuld also go one step further and rename the factor column names.

Renaming the columns

target_mydata %>%
    rename_with(
        .fn = ~ setNames(dictionary$label, dictionary$variable)[.x], 
        .cols = intersect(names(mydata), setNames(dictionary$variable, dictionary$label))
    )

Result:

# A tibble: 5 × 4
    a_1   a_2 Gender Education
  <dbl> <dbl> <chr>  <chr>    
1    20    42 Male   Primary  
2    22    13 Female Secondary
3    13    32 Male   Primary  
4    14    31 Male   Tertiary 
5    44    14 Female Primary