R преобразовать строку символов в фрейм данных

Вот решение, использующее tidyverse, которое объединяет различные stringr функции для очистки строки, прежде чем readr прочитать ее, в основном как CSV:

library(dplyr, warn.conflicts = FALSE) # for pipes

df <- 
  txt %>% 
  
  # Replace "." sep with newline
  stringr::str_replace_all(
    "\\.[A-Z]", 
    function(x) stringr::str_replace(x, "\\.", "\n")
  ) %>% 
  
  # Replace all commas in (First[,Middle1,Middle2,...]) with space
  stringr::str_replace_all(
    # Match anything inside brackets, but as few times as possible, so we don't
    # match multiple brackets
    "\\(.*?\\)", 
    # Inside the regex that was matched, replace comma with space
    function(x) stringr::str_replace_all(x, ",", " ")
  ) %>% 
  
  # Replace ( with ,
  stringr::str_replace_all("\\(", ",") %>%
  
  # Remove )
  stringr::str_remove_all("\\)") %>%
  
  # Replace @ with ,
  stringr::str_replace_all("@", ",") %>%
  
  # Remove the last "."
  stringr::str_replace_all("\\.$", "\n") %>% 
  
  # Add , after female/male
  stringr::str_replace_all("male", "male,") %>% 
  
  # Read as comma delimited file (works since string contains \n)
  readr::read_delim(
    file = .,
    delim = ",",
    col_names = FALSE,
    show_col_types = FALSE
  )

# Add names (could also be done directly in read_delim with col_names argument)
names(df) <- c(
  "family_name",
  "first_names",
  "gender",
  "birthday",
  "birth_city",
  "birth_country",
  "acc_type",
  "acc_num",
  "district",
  "code"
)

df
#> # A tibble: 4 × 10
#>   family_name first_names      gender birthday birth_city birth_country acc_type
#>   <chr>       <chr>            <chr>  <chr>    <chr>      <chr>         <chr>   
#> 1 EREKSON     Andrew Hélène    female 10/06/2… Geneva     Switzerland   PPF     
#> 2 BOUKAR      Mohamed El-Hadi  male   04/12/1… London     England       PPF     
#> 3 HARIMA      Olak N’nassik G… female 25/06/2… Paris      France        PPF     
#> 4 THOMAS      Hajil Pau Joëli  female 03/03/1… Berlin     Germany       VAT     
#> # … with 3 more variables: acc_num <chr>, district <chr>, code <chr>

^{Created on 2022-03-20 by the reprex package (v2.0.1)}

Обратите внимание, что, вероятно, существуют более эффективные регулярные выражения, которые можно было бы использовать, но я считаю, что это проще и легче изменить позже.

Вот аккуратное решение с функциями tidyrseparate_rows и extract:

library(tidyr)
data.frame(txt) %>%
  # separate `txt` into rows using the dot `.` *if* 
  # preceded by `Dt\\.\\d{3}/\\d{3}` as splitting pattern:
  separate_rows(txt, sep = "(?<=Dt\\.\\d{3}/\\d{3})\\.(?!$)") %>%
  extract(
          # select column from which to extract:
          txt,
          # define column names into which to extract:
          into = c("family_name","first_names","gender",
                   "birthday","birth_city","birth_country",
                   "acc_type","acc_num","district","code"),
          # describe the string exhaustively using capturing groups
          # `(...)` to delimit what's to be extracted:
          regex = "([A-Z]+)\\(([\\w,]+)\\),([a-z]+)([\\d/]+)@(\\w+)\\((\\w+)\\),([A-Z]+),(\\w+),dist.(\\d+),Dt\\.([\\d/]+)")
# A tibble: 4 × 10
  family_name first_names    gender birthday   birth_city birth_country acc_type acc_num  
  <chr>       <chr>          <chr>  <chr>      <chr>      <chr>         <chr>    <chr>    
1 EREKSON     Andrew,Peter   male   10/06/2011 Geneva     Switzerland   PPF      2000X007…
2 OBAMA       Barack,Hussian male   04/12/1956 London     England       PPF      2001X005…
3 CLINTON     Hillary        female 25/06/2013 Paris      France        PPF      2009X005…
4 GATES       Melinda        female 03/03/1980 Berlin     Germany       VAT      2010X006…
# … with 2 more variables: district <chr>, code <chr>