Using grepl to match characters in a dataframe column

[cp1998]

I am trying to use grepl to match a pattern in a column of a dataframe. The data frame column is a list of irish peoples surnames and I want to return the first letter of the surname. However some surnames start with Mc, Mac and O'. in those cases I want to return the prefix and the next letter after that.
I have some code that successfully does this for the names that have the Mc and Mac prefixes. But I can't get it to work for cases where the name begins with O'.

I have the following code:
ifelse(grepl("^Mc", DF$surname), substr(DF$surname, 1, 3),
ifelse(grepl("^Mac", DF$surname), substr(DF$surname, 1, 4),
ifelse(grepl("^O'", DF$surname), substr(DF$surname, 1, 3), substr(DF$surname, 1, 1))))

This code will work if I run it using a vector I created myself such as surnames <- c("O'Connell, "O'Callaghan")
But doesn't work for a dataframe column
What is the difference between a vector and a dataframe column

Any help would be appreciated
Thanks!

[FJCC]

I can't think why a data frame would act differently. Can you post a few rows of data? To post rows 2, 6, and 10 of the two columns surname and givenname, use syntax like this

dput(DF[c(2,6,10), c("surname", "givenname")])

Post the output of that, putting three back ticks on each side of the output, like this
```
output of dput() goes here
```

[cp1998]

structure(list(Op_surname = c("Fulp", "Pyatt", "O’Dwyer"),
Op_forename = c("Pia", "Drusilla", "Mac")), row.names = c(2L,
6L, 51L), class = "data.frame")

Replaced index 10 with index 51 as that corresponds to one of the values where the code won't pick up the fact it starts with O'D

[FJCC]

The problem is that in the data frame, the O' names have a "curly" single quote. Notice the difference between O' and O’. When you tried the vector, you probably typed the entries manually and thus had a straight single quote in O'. We humans don't care about the difference, unless we are very carefully proofreading, but the two quote characters are different to the computer.

DF <- structure(list(Op_surname = c("Fulp", "Pyatt", "O’Dwyer"),
               Op_forename = c("Pia", "Drusilla", "Mac")), row.names = c(2L,
                                                                         6L, 51L), class = "data.frame")
ifelse(grepl("^Mc", DF$Op_surname), substr(DF$Op_surname, 1, 3),
       ifelse(grepl("^Mac", DF$Op_surname), substr(DF$Op_surname, 1, 4),
              ifelse(grepl("^O'", DF$Op_surname), substr(DF$Op_surname, 1, 3), substr(DF$Op_surname, 1, 1))))
#> [1] "F" "P" "O"

ifelse(grepl("^Mc", DF$Op_surname), substr(DF$Op_surname, 1, 3),
       ifelse(grepl("^Mac", DF$Op_surname), substr(DF$Op_surname, 1, 4),
              ifelse(grepl("^O’", DF$Op_surname), substr(DF$Op_surname, 1, 3), substr(DF$Op_surname, 1, 1))))
#> [1] "F"   "P"   "O’D"

^{Created on 2023-10-25 with reprex v2.0.2}

[cp1998]

Yes that did the trick. Thank you for the help. A very subtle difference!