How can I trim leading and trailing white space?


I am having some troubles with leading and trailing white space in a data.frame.

For example, I like to take a look at a specific row in a data.frame based on a certain condition:

> myDummy[myDummy$country == c("Austria"),c(1,2,3:7,19)] 

[1] codeHelper     country        dummyLI    dummyLMI       dummyUMI       

[6] dummyHInonOECD dummyHIOECD    dummyOECD      

<0 rows> (or 0-length row.names)

I was wondering why I didn't get the expected output since the country Austria obviously existed in my data.frame. After looking through my code history and trying to figure out what went wrong I tried:

> myDummy[myDummy$country == c("Austria "),c(1,2,3:7,19)]
   codeHelper  country dummyLI dummyLMI dummyUMI dummyHInonOECD dummyHIOECD
18        AUT Austria        0        0        0              0           1
18         1

All I have changed in the command is an additional white space after Austria.

Further annoying problems obviously arise. For example, when I like to merge two frames based on the country column. One data.frame uses "Austria " while the other frame has "Austria". The matching doesn't work.

  1. Is there a nice way to 'show' the white space on my screen so that I am aware of the problem?
  2. And can I remove the leading and trailing white space in R?

So far I used to write a simple Perl script which removes the whites pace, but it would be nice if I can somehow do it inside R.

8/23/2020 11:22:42 PM

Accepted Answer

Probably the best way is to handle the trailing white spaces when you read your data file. If you use read.csv or read.table you can set the parameterstrip.white=TRUE.

If you want to clean strings afterwards you could use one of these functions:

# Returns string without leading white space
trim.leading <- function (x)  sub("^\\s+", "", x)

# Returns string without trailing white space
trim.trailing <- function (x) sub("\\s+$", "", x)

# Returns string without leading or trailing white space
trim <- function (x) gsub("^\\s+|\\s+$", "", x)

To use one of these functions on myDummy$country:

 myDummy$country <- trim(myDummy$country)

To 'show' the white space you could use:


which will show you the strings surrounded by quotation marks (") making white spaces easier to spot.

8/23/2020 11:29:43 PM

To manipulate the white space, use str_trim() in the stringr package. The package has manual dated Feb 15, 2013 and is in CRAN. The function can also handle string vectors.

install.packages("stringr", dependencies=TRUE)

(Credit goes to commenter: R. Cotton)


A simple function to remove leading and trailing whitespace:

trim <- function( x ) {
  gsub("(^[[:space:]]+|[[:space:]]+$)", "", x)


> text = "   foo bar  baz 3 "
> trim(text)
[1] "foo bar  baz 3"

Ad 1) To see white spaces you could directly call with modified arguments:

print(head(iris), quote=TRUE)
#   Sepal.Length Sepal.Width Petal.Length Petal.Width  Species
# 1        "5.1"       "3.5"        "1.4"       "0.2" "setosa"
# 2        "4.9"       "3.0"        "1.4"       "0.2" "setosa"
# 3        "4.7"       "3.2"        "1.3"       "0.2" "setosa"
# 4        "4.6"       "3.1"        "1.5"       "0.2" "setosa"
# 5        "5.0"       "3.6"        "1.4"       "0.2" "setosa"
# 6        "5.4"       "3.9"        "1.7"       "0.4" "setosa"

See also ? for other options.


Use grep or grepl to find observations with white spaces and sub to get rid of them.

names<-c("Ganga Din\t", "Shyam Lal", "Bulbul ")
grep("[[:space:]]+$", names)
[1] 1 3
grepl("[[:space:]]+$", names)
sub("[[:space:]]+$", "", names)
[1] "Ganga Din" "Shyam Lal" "Bulbul"

Removing leading and trailing blanks might be achieved through the trim() function from the gdata package as well:


Usage example:

> trim("   Remove leading and trailing blanks    ")
[1] "Remove leading and trailing blanks"

I'd prefer to add the answer as comment to user56's, but I am yet unable so writing as an independent answer.


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