R：基于字符串替换列值的有效方法（可能使用 case_when 或某种其他形式的 mutate）？

Question

Is there a more efficient way to replace values in columns based on searching for strings within them?

prac <- data.frame(`External Placement` = c(NA, NA, "Spanish words We place outside of our company", NA, NA, "Spanish words We place outside of our company", "Spanish words We place outside of our company", "Spanish words We place outside of our company", NA, "Spanish words We place outside of our company"),
             `Internal Placement` = c(NA, NA, "Spanish words we place inside of our organisation", NA, "Spanish words we place inside of our organisation", "Spanish words we place inside of our organisation", NA, NA, NA, NA),
             `None of the above` = c("Ninguno none", "Ninguno none", NA, NA, NA, NA, NA, NA, NA, NA))

This is the sample data. I have used a series of if/else to get the values I want, but I tried for a long time to figure out a way to do so with mutate, case_when, and so forth to do less repetition. Is there a better way? This is what I did so far:

prac$`Vocational Training` <- ifelse(grepl("vocational training$", prac$`Vocational Training`),
                                 "Vocational training", NA)
prac$`External Placement` <- ifelse(grepl("outside of our company$", prac$`External Placement`),
                                    "External placement", NA)
prac$`Internal Placement` <- ifelse(grepl("inside our organisation$", prac$`Internal Placement`),
                                    "Internal placement", NA)  
prac$`None of the Above` <- ifelse(grepl("^Ninguno", prac$`None of the Above`), "None or other", NA)

Answer 1

Maybe you can try the code like below

v <- c("outside of our company$","inside our organisation$","^Ninguno")
r <- c("External placement","Internal placement","None or other")

prac[]<- mapply(function(x,y) ifelse(x,y,NA),
                data.frame(mapply(grepl, v,prac)),
                r)

such that

> prac
   External.Placement Internal.Placement None.of.the.above
1                <NA>               <NA>     None or other
2                <NA>               <NA>     None or other
3  External placement               <NA>              <NA>
4                <NA>               <NA>              <NA>
5                <NA>               <NA>              <NA>
6  External placement               <NA>              <NA>
7  External placement               <NA>              <NA>
8  External placement               <NA>              <NA>
9                <NA>               <NA>              <NA>
10 External placement               <NA>              <NA>

DATA

prac <- structure(list(External.Placement = structure(c(NA, NA, 1L, NA, 
NA, 1L, 1L, 1L, NA, 1L), .Label = "Spanish words We place outside of our company", class = "factor"), 
    Internal.Placement = structure(c(NA, NA, 1L, NA, 1L, 1L, 
    NA, NA, NA, NA), .Label = "Spanish words we place inside of our organisation", class = "factor"), 
    None.of.the.above = structure(c(1L, 1L, NA, NA, NA, NA, NA, 
    NA, NA, NA), .Label = "Ninguno none", class = "factor")), class = "data.frame", row.names = c(NA, 
-10L))

Answer 2

It depends if the efficiency you want is computing efficiency (time to compute) or programming efficiency (number of lines used)

In terms of programming efficiency, it will be very hard to beat a data.table solution. I can propose you this one:

prac <- data.frame(`External Placement` = c(NA, NA, "Spanish words We place outside of our company", NA, NA, "Spanish words We place outside of our company", "Spanish words We place outside of our company", "Spanish words We place outside of our company", NA, "Spanish words We place outside of our company"),
                   `Internal Placement` = c(NA, NA, "Spanish words we place inside of our organisation", NA, "Spanish words we place inside of our organisation", "Spanish words we place inside of our organisation", NA, NA, NA, NA),
                   `None of the Above` = c("Ninguno none", "Ninguno none", NA, NA, NA, NA, NA, NA, NA, NA))

library(data.table)

list_conditions <- c(#"Vocational.Training" = "vocational training$",
  "External.Placement" = "outside of our company$",
  "Internal.Placement" = "inside our organisation$",
  "None.of.the.Above" = "^Ninguno")


dt <- data.table(prac)

My idea is to use a vector with conditions whose names are the variable for which it is supposed to apply. Then a lapply that updates your dataframe by reference (here you lose some programming efficiency but you will have very good computing efficiency)

dt <- lapply(seq_len(length(list_conditions)), function(i) {

  var <- names(list_conditions)[i]
  cond <- list_conditions[i]
  val <- gsub("\\.","", var)
  dt[, 'tempcol' := NA_character_]
  dt[grepl(as.character(cond), get(var)), tempcol := as.character(val)]
  dt[,c(var) := tempcol]

})
dt <- dt[[length(dt)]]
dt[,'tempcol' := NULL]

The line dt <- dt[[length(dt)]] is here because R returns a list but we are only interested in its last element (last update for the dataframe). You can generalize this program if you prefer creating new columns rather than rewriting existing ones.

The output is:

dt
    External.Placement Internal.Placement None.of.the.Above
 1:               <NA>               <NA> None of the Above
 2:               <NA>               <NA> None of the Above
 3: External Placement               <NA>              <NA>
 4:               <NA>               <NA>              <NA>
 5:               <NA>               <NA>              <NA>
 6: External Placement               <NA>              <NA>
 7: External Placement               <NA>              <NA>
 8: External Placement               <NA>              <NA>
 9:               <NA>               <NA>              <NA>
10: External Placement               <NA>              <NA>