Replace set of words with another set of words in R

Question

It is a simple question. I have a list of country names. However, I wanted to change few names with correct names. So, I have two more vectors; one with names to be changed, and second with correct names. See the example:

#country names (names are repetitive in the list)
cn <- c("I", "A", "B", "C", "A", "C", "D", "P")

change <- c("A", "B")
tochange <- c("X", "Y")

Expected Output

cn <- c("I", "X", "Y", "C", "X", "C", "D", "P")

Thanks

Grained

Answer 1

Uisng stringi::stri_replace_all_fixed.

> stringi::stri_replace_all_fixed(cn, change, tochange, vectorize_all=FALSE)
[1] "I" "X" "Y" "C" "X" "C" "D" "P"

Gardening

I have one question: does it work if given the mapping like "A" --> "B" and "B" --> "A"? Seems it cannot give the desired mapping – Fifteen

@Fifteen We probably need vectorize_all=TRUE (default) in this case, x <- rep_len(LETTERS[1:2], 10);stringi::stri_replace_all_fixed(x, c('A', 'B'), c('B', 'A')). – Gardening

yes, that did work, thanks! What is the benefit of vectorize_all=FALSE in your solution (regardless of the edge case here)? Speed reason or something else? – Fifteen

@Fifteen It's the recycling fallacy. We need x <- c("A", "B", "C");stringi::stri_replace_all_fixed(x, c('A', 'B', 'C'), c('B', 'A', 'C'), vectorize_all=TRUE). Without doing C=C it fails. To get the same but using vectorize_all=FALSE we'd need doing x <- c("A", "B", "C"); x <- stringi::stri_replace_all_fixed(x, "A", "TMP", vectorize_all=FALSE);x <- stringi::stri_replace_all_fixed(x, "B", "A", vectorize_all=FALSE);stringi::stri_replace_all_fixed(x, "TMP", "B", vectorize_all=FALSE). – Gardening

Answer 2

You can try replace + match like below

> d <- tochange[match(cn, change)]

> replace(cn, !is.na(d), na.omit(d))
[1] "I" "X" "Y" "C" "X" "C" "D" "P"

Fifteen

Answer 3

Here are some alternatives

1) gsubfn gsubfn is a generalization of gsub in which the second argument can not only be a character string but alternately a named list which we use here (or a function or proto object).

library(gsubfn)
gsubfn("^.*$", setNames(as.list(change), tochange), cn)
## [1] "I" "A" "B" "C" "A" "C" "D" "P"

2) Reduce A base R solution is to use Reduce

dict <- setNames(change, tochange)
Reduce(\(x, y) replace(x, names(y), y), init = cn, dict)
## [1] "I" "A" "B" "C" "A" "C" "D" "P"

3) chartr If the names in the strings are single characters, as in the question, then base R's chartr can be used

chartr(paste0(tochange, collapse = ""), paste0(change, collapse = ""), cn)
## [1] "I" "A" "B" "C" "A" "C" "D" "P"

or hard coding the names

chartr("XY", "AB", cn)
## [1] "I" "A" "B" "C" "A" "C" "D" "P"

Circularity

Although it seems unlikely that the problem here would exhibit circularity such as in where A -> B -> A we can test for it if you think it is possible.

library(igraph)

cnt <- cbind(change, tochange) |>
  graph_from_edgelist() |>
  count_components()

if (cnt != length(change)) stop("circularity found")

Note

Inputs used

cn <- c("I", "A", "B", "C", "A", "C", "D", "P")

change <- c("A", "B")
tochange <- c("X", "Y")

Tolley

Answer 4

You can use the ifelse function in R

cn <- c("I", "A", "B", "C", "A", "C", "D", "P")
cn <- ifelse(cn == "A", "X", ifelse(cn == "B", "Y", cn))

print(cn)

OR

Alternatively, you can use the dplyr package for a more readable solution

library(dplyr)

cn <- c("I", "A", "B", "C", "A", "C", "D", "P")

cn <- cn %>% recode("A" = "X", "B" = "Y")

print(cn)

OUTPUT:

[1] "I" "X" "Y" "C" "X" "C" "D" "P"

Maw

I can not do like this. I have more than 30 words that needs to be replaced. Thanks – Grained

the ifelse solution won't work if you have replacement mapping like "A" --> "X" and "X" --> "A" – Fifteen

Answer 5

As a basic for loop:

cn.new <- cn

for (i in seq_along(change)) {
    cn.new[cn.new == change[i]] <- tochange[i]
}

cn
# [1] "I" "A" "B" "C" "A" "C" "D" "P"
cn.new
# [1] "I" "X" "Y" "C" "X" "C" "D" "P"

Dystrophy

it won't work if you have replacement mapping like "A" --> "X" and "X" --> "A" – Fifteen

@ThomasIsCoding: Depends on what you mean with "won't work", the results might be desired, but in any case the OP doesn't have mappings like these. – Dystrophy

in OP's example your method did work well. My concern is just about the robustness how it deals with some edge cases, as I provided in the comment. – Fifteen

I think that circularity is unlikely to be a problem with this question but in my answer we show how to check for it if you are really worried abou tit. – Tolley