Extract only characters, including spaces, from a column

Question

My data is structured as:

df <- data.frame(Athlete = c('02 Paul Jones', '02 Paul Jones', '02 Paul Jones', '02 Paul Jones',
                             '02 Paul Jones', '02 Paul Jones', '02 Paul Jones', '02 Paul Jones',
                             '01 Joe Smith', '01 Joe Smith', '01 Joe Smith', '01 Joe Smith',
                             '01 Joe Smith', '01 Joe Smith', '01 Joe Smith', '01 Joe Smith'),
                 Period = c('P1', 'P1', 'P1', 'P1',
                            'P2', 'P2', 'P2', 'P2',
                            'P1', 'P1', 'P1', 'P1',
                            'P2', 'P2', 'P2', 'P2'))
# Make `Athlete` column a character
df$Athlete <- as.character(df$Athlete)

How do I extract the first and last names of each athlete whilst keeping the space between first and last name? I do not want the leading space including either. For example, "Paul Jones" not " Paul Jones"

How about removing all the numbers from the column? `gsub('[0-9]+', '', df$Athlete)` — Ronak Shah, Mar 22 '17 at 07:10
[This](http://stackoverflow.com/questions/30912199/extract-letters-from-a-string-in-r) is almost what you want. — Ronak Shah, Mar 22 '17 at 07:24

Sathish · Accepted Answer · 2017-03-22T07:43:12.577

3

remove all except alphabets [:alpha:] and space characters [:space:] using POSIX locale type interpretation of regular expression pattern.

df$Athlete <- as.character(df$Athlete)  # convert factor to character

df$Athlete <- gsub("[^[:alpha:][:space:]]", '', df$Athlete) 
df$Athlete <- gsub("^[[:space:]]+", '', df$Athlete )  # removing leading spaces

head(df)
#       Athlete Period
# 1  Paul Jones     P1
# 2  Paul Jones     P1
# 3  Paul Jones     P1
# 4  Paul Jones     P1
# 5  Paul Jones     P2
# 6  Paul Jones     P2

edited Mar 22 '17 at 07:43

answered Mar 22 '17 at 07:11

Sathish

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If I convert this back to a factor, the levels still show a blank space: `> levels(df$Athlete) [1] " Joe Smith" " Paul Jones"` – user2716568 Mar 22 '17 at 07:26

akrun · Answer 2 · 2017-03-22T07:17:50.737

We can use sub to match one or more numbers ([0-9]+) followed by one or more space (\\s+) from the start (^) of the string and replace it with ""

df$Athlete <- sub("^[0-9]+\\s+", "", df$Athlete)
df
#      Athlete Period
#1  Paul Jones     P1
#2  Paul Jones     P1
#3  Paul Jones     P1
#4  Paul Jones     P1
#5  Paul Jones     P2
#6  Paul Jones     P2
#7  Paul Jones     P2
#8  Paul Jones     P2
#9   Joe Smith     P1
#10  Joe Smith     P1
#11  Joe Smith     P1
#12  Joe Smith     P1
#13  Joe Smith     P2
#14  Joe Smith     P2
#15  Joe Smith     P2
#16  Joe Smith     P2

Extract only characters, including spaces, from a column

2 Answers2