String manipulation in R to create dataframe column

Question

I have an R dataframe(df) which includes a factor column, Team

Team
Baltimore Orioles
Kansas City Chiefs
...

I just want to create a new column, nickname, which just refers to the last name

Nickname
Orioles
Chiefs

As a first stage, I have tried splitting the factor like this

df$Nickname <- strsplit(as.character(df$Team), " ")

which produces a list of character fields which I can reference thus

>df$Nickname[1]

[[1]]
[1] "Baltimore" "Orioles"

and

>str(df$Nickname[1])

List of 1
 $ : chr [1:2] "Baltimore" "Orioles"

but then I do not know how to proceed. Trying to get the length

length(df$Nickname[1])

gives 1 - which flummoxes me

Also, you probably wanted length(df$Nickname[[1]]) or sapply(df$Nickname, length). — Aaron - mostly inactive
– Aaron - mostly inactive, Commented Sep 26, 2011 at 15:43

Andrie · Accepted Answer · 2011-09-26 14:56:30Z

7

Use a regular expression:

text <- c("Baltimore Orioles","Kansas City Chiefs")

gsub("^.*\\s", "", text)
[1] "Orioles" "Chiefs"

The regex searches for:

^ means the start of the string
.* means any character, repeated
\\s means a single white space

gsub finds this pattern and replaces it with an empty string, leaving you with the last word of each string.

answered Sep 26, 2011 at 14:56

Andrie

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Comments

tim riffe · Accepted Answer · 2011-09-26 14:27:09Z

4

you just need to unlist the split strings and take the last one

    full <- c("Baltimore Orioles","Kansas City Chiefs")
    getlast <- function(x){
    parts <- unlist(strsplit(x, split = " "))
    parts[length(parts)]
    }
    sapply(full,getlast)
    > Baltimore Orioles Kansas City Chiefs 
    > "Orioles"           "Chiefs"

answered Sep 26, 2011 at 14:27

tim riffe

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3 Comments

Aaron - mostly inactive Over a year ago

Nice, Tim. A slightly simpler version might be sapply(strsplit(full, split=" "), tail, 1).

pssguy Over a year ago

Thanks Guys This forum is the only one where I can post a question, go brush my teeth, come back and find an answer.make a coffee and have an enhancement

John Over a year ago

Aaron - beautiful, helped me a lot.

David · Accepted Answer · 2014-05-03 00:02:22Z

0

How about this?

require(plyr)
ldply(df$Nickname)

answered May 3, 2014 at 0:02

David

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