Use strsplit to get last character in r - regex

Question

I have a file of baby names that I am reading in and then trying to get the last character in the baby name. For example, the file looks like..
Name Sex
Anna F
Michael M
David M
Sarah F
I read this in using
sourcenames = read.csv("babynames.txt", header=F, sep=",")
I ultimately want to end up with my result looking like..
Name Last Initial Sex
Michael l M
Sarah h F
I've managed to split the name into separate characters..
sourceout = strsplit(as.character(sourcenames$Name),'')
But now where I'm stuck is how to get the last letter, so in the case of Michael, how to get 'l'. I thought tail() might work but its returning the last few records, not the last character in each Name element.
Any help or advice is greatly appreciated.
Thanks :)

Answer 1

For your strsplit method to work, you can use tail with sapply
df$LastInit <- sapply(strsplit(as.character(df$Name), ""), tail, 1)
df
# Name Sex LastInit
# 1 Anna F a
# 2 Michael M l
# 3 David M d
# 4 Sarah F h
Alternatively, you can use substring
with(df, substring(Name, nchar(Name)))
# [1] "a" "l" "d" "h"

Answer 2

Try this function from stringi package:
require(stringi)
x <- c("Ala", "Sarah","Meg")
stri_sub(x, from = -1, to = -1)
This function extracts substrings between from and to index. If indexes are negative, then it counts characters from the end of a string. So if from=-1 and to=-1 it means that we want substring from last to last character :)
Why use stringi? Just look at this benchmarks :)
require(microbenchmark)
x <- sample(x,1000,T)
microbenchmark(stri_sub(x,-1), str_extract(x, "[a-z]{1}$"), gsub(".*(.)$", "\\1", x),
sapply(strsplit(as.character(x), ""), tail, 1), substring(x, nchar(x)))
Unit: microseconds
expr min lq median uq max neval
stri_sub(x, -1) 56.378 63.4295 80.6325 85.4170 139.158 100
str_extract(x, "[a-z]{1}$") 718.579 764.4660 821.6320 863.5485 1128.715 100
gsub(".*(.)$", "\\\\1", x) 478.676 493.4250 509.9275 533.8135 673.233 100
sapply(strsplit(as.character(x), ""), tail, 1) 12165.470 13188.6430 14215.1970 14771.4800 21723.832 100
substring(x, nchar(x)) 133.857 135.9355 141.2770 147.1830 283.153 100

Answer 3

Here is another option using data.table (for relatively clean syntax) and stringr (easier grammar).
library(data.table); library(stringr)
df = read.table(text="Name Sex
Anna F
Michael M
David M
Sarah F", header=T)
setDT(df) # convert to data.table
df[, "Last Initial" := str_extract(Name, "[a-z]{1}$") ][]
Name Sex Last Initial
1: Anna F a
2: Michael M l
3: David M d
4: Sarah F h

Answer 4

One liner:
x <- c("abc","123","Male")
regmatches(x,regexpr(".$", x))
## [1] "c" "3" "e"

Answer 5

You can do it with a Regular Expression and gsub:
sourcenames$last.letter = gsub(".*(.)$", "\\1", sourcenames$Name)
sourcenames
Name Sex last.letter
1 Anna F a
2 Michael M l
3 David M d
4 Sarah F h

Answer 6

you can try this one... str_sub() function in stringr package would help you.
library(dplyr)
library(stringr)
library(babynames)
babynames %>%
select(name,sex) %>%
mutate(last_letter = str_sub(name,-1,-1)) %>%
head()

Answer 7

dplyr approach:
sourcenames %>% rowwise() %>% mutate("Last Initial" = strsplit(as.character(Name),'') %>% unlist() %>% .[length(.)])