Remind me how to for loop in a few lines (like apply etc)

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I have a for loop using grepl to match an acronym in a column of a dataframe with a dictionary key and creating a column with the dictionary value of those keys to bind to the original dataframe.

I have a working for loop, and code that works / gives me what I need.

#Example dataframe to loop over 
loop <- data.frame(
  acronym = c("cmr", "cmr", "den", "den", "nmw", "nmw"),
  profession = c("chinese medical practitioner", "chinese medical practitioner",
                 "dentist", "dentist", "medical practitioner", "nurse"),
  stringsAsFactors = FALSE
#The dictionary I created to provide values to each key
dic <- list(
  cmr = "chinese medical practitioner",
  den = "dentist",
  med = "medical practitioner",
  nmw = "nurse"

The for loop with a list created to store the dictionary values for each iteration where there is a match between the acronym in the dataframe and the key in the dictionary. I then bind the unlisted list of values to the dataframe loop. This example 'corrects' the profession column in the loop, so kind of a data clean leveraging the dictionary and the acronym column.

column <- list()
for (i in 1:length(loop$acronym)){
  for (j in 1:length(dic)){
    if (grepl(names(dic)[j], loop$acronym[i], loop$code)){
      column[i] <- dic[j]

BUT, I want to solve the problem using apply or something from the tidyverse world. I don't really want a solution from data.table, unless its amazing then I might start considering learning data.table.

Using base R one way is to use to use stack and then merge.

merge(loop, stack(dic), by.x = "acronym", by.y = "ind")

#  acronym                   profession                       values
#1     cmr chinese medical practitioner chinese medical practitioner
#2     cmr chinese medical practitioner chinese medical practitioner
#3     den                      dentist                      dentist
#4     den                      dentist                      dentist
#5     nmw         medical practitioner                        nurse
#6     nmw                        nurse                        nurse

Where stack(dic) turns a named list to a dataframe

#                        values ind
#1 chinese medical practitioner cmr
#2                      dentist den
#3         medical practitioner med
#4                        nurse nmw

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If we replace list() with c() in your code, to make dic a named vector instead of a list, then we can do this in a single line using the vector names as indices:

dic <- c(
  cmr = "chinese medical practitioner",
  den = "dentist",
  med = "medical practitioner",
  nmw = "nurse"

loop$code = dic[loop$acronym]
#   acronym                   profession                         code
# 1     cmr chinese medical practitioner chinese medical practitioner
# 2     cmr chinese medical practitioner chinese medical practitioner
# 3     den                      dentist                      dentist
# 4     den                      dentist                      dentist
# 5     nmw         medical practitioner                        nurse
# 6     nmw                        nurse                        nurse

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I often have better luck with these tasks when I treat objects of the same or similar type together—lists with lists, data frames with data frames, etc.

There's a couple quick tidyverse ways to get the dictionary into a data frame, which will make it easier to join with the loop data. The first just takes the names of the list and a flattened version of the list, and creates columns of both.


dict_df <- tibble(
  acronym = names(dic), 
  profession = flatten_chr(dic)
#> # A tibble: 4 x 2
#>   acronym profession                  
#>   <chr>   <chr>                       
#> 1 cmr     chinese medical practitioner
#> 2 den     dentist                     
#> 3 med     medical practitioner        
#> 4 nmw     nurse

You might also use the newer function tibble::enframe, which creates data frames from single vectors (like you get after unlist) and uses the vector's names as a column as well. The advantage here is that it might fit well within a larger piped workflow—gets the same output as above.

unlist(dic) %>% 
  tibble::enframe(name = "acronym", value = "profession")

Then join the original data with the dictionary. dplyr's *_join functions optionally take suffixes which will be appended to columns that aren't the join-by columns but have the same names. Here that lets you see which profession column comes from the original data and which comes from the correction.

loop %>%
  left_join(dict_df, by = "acronym", 
            suffix = c("_loop", "_dict"))
#>   acronym              profession_loop              profession_dict
#> 1     cmr chinese medical practitioner chinese medical practitioner
#> 2     cmr chinese medical practitioner chinese medical practitioner
#> 3     den                      dentist                      dentist
#> 4     den                      dentist                      dentist
#> 5     nmw         medical practitioner                        nurse
#> 6     nmw                        nurse                        nurse

Created on 2019-04-19 by the reprex package (v0.2.1)

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  • grepl is great for searching inside strings---but you're not doing that here, you just need to find exact matches, "cmr" to "cmr", so you don't need grep, you just need == or %in% or match.
  • Similarly, lists` are great when you need to combine a bunch of objects with different classes or sizes. But when you just have a mapping of names and values, lists are overkill. Using a named vector or a 2-column data frame is simpler.
  • This is the option I would use