Efficient way to split a vector of a full name in

问题:

I have a vector consisting of full names with the first and last name separated by a comma this is what the first few elements look like:

> head(val.vec)
[1] "Aabye,ֲ Edgar"        "Aaltonen,ֲ Arvo"      "Aaltonen,ֲ Paavo"    
[4] "Aalvik Grimsb,ֲ Kari" "Aamodt,ֲ Kjetil Andr" "Aamodt,ֲ Ragnhild

I am looking for a way to split them in to 2 separate columns of first and last name. My final intention is to have both of them as a part of a bigger data frame.

I tried using strsplit function like this

names<-unlist(strsplit(val.vec,','))

but it gave me one long vector instead of 2 separate sets, I know it is Possible to use a loop and go over all the elements and place the first and last name in 2 separate vectors, but it is a little time consuming considering the fact that there are about 25000 records.

I saw a few similar questions but the discussion was how to do it on C+ and Java

回答1:

We can use read.csv to convert the vector into a data.frame with 2 columns

read.csv(text=val.vec, header=FALSE, stringsAsFactors=FALSE)

---

Or if we are using strsplit, instead of unlisting (which will convert the whole list to a single vector), we can extract the first and second elements in the list separately to create two vectors ('v1' and 'v2').

lst <- strsplit(val.vec,',')
v1 <- lapply(lst, `[`, 1)
v2 <- lapply(lst, `[`, 2)

---

Yet another option would be sub

v1 <- sub(",.*", "", val.vec)
v2 <- sub("[^,]+,", "", val.vec)

data

val.vec <- c("Aabye,ֲ Edgar", "Aaltonen,ֲ Arvo", "Aaltonen,ֲ Paavo", 
        "Aalvik Grimsb,ֲ Kari", "Aamodt,ֲ Kjetil Andr", "Aamodt,ֲ Ragnhild")

回答2:

Another option:

library(stringi)
stri_split_fixed(val.vec, ",", simplify = TRUE)

Which gives:

#     [,1]            [,2]          
#[1,] "Aabye"         "ֲ Edgar"      
#[2,] "Aaltonen"      "ֲ Arvo"       
#[3,] "Aaltonen"      "ֲ Paavo"      
#[4,] "Aalvik Grimsb" "ֲ Kari"       
#[5,] "Aamodt"        "ֲ Kjetil Andr"
#[6,] "Aamodt"        "ֲ Ragnhild"  

Should you want the result in a data.frame, you could wrap it in as.data.frame()

回答3:

Just encase your function call into a sapply call:

val.vec = c("Aabye,ֲ Edgar", "Aaltonen,ֲ Arvo", "Aaltonen,ֲ Paavo", "Aalvik Grimsb,ֲ Kari", "Aamodt,ֲ Kjetil Andr", "Aamodt,ֲ Ragnhild")

names = t(sapply(val.vec, function(x) unlist(strsplit(x,','))))
names

#> names
#                     [,1]            [,2]           
#Aabye,? Edgar        "Aabye"         "? Edgar"      
#Aaltonen,? Arvo      "Aaltonen"      "? Arvo"       
#Aaltonen,? Paavo     "Aaltonen"      "? Paavo"      
#Aalvik Grimsb,? Kari "Aalvik Grimsb" "? Kari"       
#Aamodt,? Kjetil Andr "Aamodt"        "? Kjetil Andr"
#Aamodt,? Ragnhild    "Aamodt"        "? Ragnhild"  

---

Using the solution you tried we can coerce it to two columns.

val.vec = c("Aabye,ֲ Edgar", "Aaltonen,ֲ Arvo", "Aaltonen,ֲ Paavo", "Aalvik Grimsb,ֲ Kari", "Aamodt,ֲ Kjetil Andr", "Aamodt,ֲ Ragnhild")
names = matrix(unlist(strsplit(val.vec,',')), ncol = 2L, byrow = TRUE)
#> names
#     [,1]            [,2]           
#[1,] "Aabye"         "? Edgar"      
#[2,] "Aaltonen"      "? Arvo"       
#[3,] "Aaltonen"      "? Paavo"      
#[4,] "Aalvik Grimsb" "? Kari"       
#[5,] "Aamodt"        "? Kjetil Andr"
#[6,] "Aamodt"        "? Ragnhild"   

Testing it against the (very fast) solution proposed by Richard Scriven we can see yours and his are equivalent:

#> library(microbenchmark)
#> microbenchmark(
#+   names_1 = do.call(rbind, strsplit(val.vec, ",")),
#+   names_2 = matrix(unlist(strsplit(val.vec,',')), ncol = 2L, byrow = TRUE),
#+   times = 10000L
#+ )
#Unit: microseconds
#    expr    min     lq     mean median     uq      max neval cld
# names_1 12.596 13.530 15.08867 13.996 14.463  513.185 10000   b
# names_2 11.663 12.131 14.03413 12.597 13.530 1436.917 10000  a 

回答4:

If you are into the dplyr way of doing things, have a look at separate from the tidyr package:

library(dplyr)
library(tidyr)

dat = data.frame(val = c("Lee, John", "Lee, Spike", "Doe, John", 
        "Longstocking, Pippy", "Bond, James", "Jordan, Michael"))
#                   val
# 1           Lee, John
# 2          Lee, Spike
# 3           Doe, John
# 4 Longstocking, Pippy
# 5         Bond, James
# 6     Jordan, Michael
dat %>% 
  separate(val, c('last_name', 'first_name'), sep = ',') %>% 
  mutate(first_name = trimws(first_name))
#      last_name first_name
# 1          Lee       John
# 2          Lee      Spike
# 3          Doe       John
# 4 Longstocking      Pippy
# 5         Bond      James
# 6       Jordan    Michael

Added in the call to trimws to get rid of the leading whitespace.