Replace part of strings by the corresponding code R-CodePudding

In a large dataset, I need to replace the territory names by their corresponding code.

Here below a small replicable example:

library(tidyverse)
library(stringr)
library(dplyr)
library(tidyr)
library(purrr)
library(stringi)

adrs_data <- data.frame (adress  = c("6 Frien Street, Paris", "Toulouse, 7 Hospital street", "10 market avenue (Bordeaux)") )

dep_code <- data.frame (code  = c("75", "31", "33"), names  = c("Paris", "Toulouse", "Bordeaux"))

This is what I have tried:

d_search<-c(dep_code$names)
d_search <- paste(paste0(d_search[order(-nchar(d_search))]), collapse = "|")
c_search<-c(dep_code $code)

df<-adrs_data %>% 
  dplyr::mutate(c_adress = case_when(adress %in% d_search ~ 
            str_replace_all(adress,  d_search, c_search), TRUE ~ adress))

But it does not produce the wanted output which is:

df <- data.frame (adress  = c("6 Frien Street, 75", "31, 7 Hospital street", "10 market avenue (33)")

Thank you for your help, Best regards

CodePudding user response：

After merging both data frames, you can use pmap to replace the pattern for each row:

library(dplyr)
library(stringr)
library(purrr)
library(fuzzyjoin)

fuzzy_left_join(adrs_data, dep_code, match_fun = str_detect, 
                by = c("adress" = "names")) %>% 
  mutate(adress = pmap(., ~ str_replace(..1, ..3, ..2)))

#                  adress code    names
# 1    6 Frien Street, 75   75    Paris
# 2 31, 7 Hospital street   31 Toulouse
# 3 10 market avenue (33)   33 Bordeaux

CodePudding user response：

This also works and removes parenthesis:

I added one more entry to adrs_data

adrs_data <- data.frame (adress  = c("6 Frien Street, Paris", "Toulouse, 7 Hospital street", "10 market avenue (Bordeaux)", "9 Test Street, Paris") )

dep_code <- data.frame (code  = c("75", "31", "33"), names  = c("Paris", "Toulouse", "Bordeaux")) 

sapply(seq(nrow(adrs_data)), \(i){ 
  adrs_data[i,] %>% 
    str_replace_all(dep_code$names, dep_code$code) %>% 
    .[.!= adrs_data[i,]] %>% 
    sub(pattern = "\\(", replacement = "", x = .) %>% 
    sub(pattern = "\\)", replacement = "", x = .) 
})

     [,1]                   
[1,] "6 Frien Street, 75"   
[2,] "31, 7 Hospital street"
[3,] "10 market avenue 33"  
[4,] "9 Test Street, 75"

For very short data this even provides decent speed.

Unit: milliseconds
       min        lq      mean    median        uq      max neval
  1.508001  1.671950  2.481109  1.823151  2.534501  11.9309   100 # lapply
 58.665500 67.577851 85.267257 76.948251 89.716950 231.3375   100 # fuzzy

Would be interesting to compare with @Maël's solution for a big data set. I assume the larger the data the slower the lapply solution.