Replace Nth occurrence of a word (substring) in a string in R, N is the value of an integer column

user_19000340 0
Viewed 61

I want to find the Nth occurence of a word in an utterance and put [brackets] around it. I tried with various things but I think the closest I'm getting is with gsub but I can't have {copy-1} for the number of times in my regex. Any ideas? Can we put a variable in there? Thanks!

#creating my df
utterance <- c("we are not who we think we are", "they know who we are")
df <- data.frame(utterance)
df$occurences = str_count(df$utterance, "we")
df <- df %>% mutate(ID = row_number())
df <- df %>% uncount(occurences) %>% group_by(ID) %>% mutate(copy = row_number()) 

#this is my gsub
gsub("((?:we){copy-1}.*)we", "\\[we\\]", df$utterance)

This would be my result

    utterance                         ID  copy
    <chr>                          <int> <int>
1 [we] are not who we think we are     1     1
2 we are not who [we] think we are     1     2
3 we are not who we think [we] are     1     3
4 they know who [we] are               2     1
r
3 Answers

How about just this:

library(tidyverse)

f <- function(s,c,target) {
 g = gregexpr(target,s)[[1]][c]
 if(is.na(g) | g<0) return(s)
 paste0(str_sub(s,1,g-1),"[",target,"]",str_sub(s,1+g+length(target)))
}

df %>% rowwise() %>% mutate(utterance = f(utterance,copy, "we"))

Output:

  utterance                           ID  copy
  <chr>                            <int> <int>
1 [we] are not who we think we are     1     1
2 we are not who [we] think we are     1     2
3 we are not who we think [we] are     1     3
4 they know who [we] are               2     1

Note that this will also find targets that are not whole words. For example the second of occurrence of "we" in "We went where we went yesterday" is the first two letters of "went", not the second occurrence of the word "we". If you want to restrict to whole words, you can update the gregexpr() call to this:

 g = gregexpr(paste0("\\b",target, "\\b"),s)[[1]][c]
user_44475400

Here is a string splitting approach. We can split the input string on we, and then piece together, using [we] as the nth connector.

repn <- function(x, find, repl, n) {
    parts <- strsplit(x, paste0("\\b", find, "\\b"))[[1]]
    output <- paste0(
        paste0(parts[1:n], collapse=find),
        repl,
        paste0(parts[(n+1):length(parts)], collapse="we")
    )

    return(output)
}

x <- "we are not who we think we are"
repn(x, "we", "[we]", 1)
repn(x, "we", "[we]", 2)
repn(x, "we", "[we]", 3)

[1] "[we] are not who we think we are"
[1] "we are not who [we] think we are"
[1] "we are not who we think [we] are"
user_18632290

Here's a mixed approach using a number of additional packages:

library(data.table)
library(tibble)
library(dplyr)
library(tidyr)
df %>%
  rowid_to_column() %>%
  separate_rows(utterance, sep = " ") %>%
  group_by(rowid) %>%
  mutate(wordcount = ifelse(utterance == "we", rleid(rowid), NA), # simpler: wordcount = ifelse(utterance == "we", 1, NA)
         wordcount = cumsum(!is.na(wordcount))) %>% 
  mutate(utterance = ifelse(utterance == "we" & wordcount == copy, paste0("[", utterance, "]"), utterance)) %>% 
  summarise(utterance = paste0(utterance, collapse = " ")) %>%
  bind_cols(.,df[,2:3])
# A tibble: 4 × 4
  rowid utterance                           ID  copy
  <int> <chr>                            <int> <int>
1     1 [we] are not who we think we are     1     1
2     2 we are not who [we] think we are     1     2
3     3 we are not who we think [we] are     1     3
4     4 they know who [we] are               2     1
user_80399780
Related