How can I trim leading and trailing white space?

As of R 3.2.0 a new function was introduced for removing leading/trailing white spaces:

trimws()

See: Remove Leading/Trailing Whitespace

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:

 paste(myDummy$country)

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

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)
require(stringr)
example(str_trim)
d4$clean2<-str_trim(d4$V2)

(Credit goes to commenter: R. Cotton)

A simple function to remove leading and trailing whitespace:

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

Usage:

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

Ad 1) To see white spaces you could directly call print.data.frame 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 ?print.data.frame 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)
[1]  TRUE FALSE  TRUE
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:

require(gdata)
example(trim)

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.

Another option is to use the stri_trim function from the stringi package which defaults to removing leading and trailing whitespace:

> x <- c("  leading space","trailing space   ")
> stri_trim(x)
[1] "leading space"  "trailing space"

For only removing leading whitespace, use stri_trim_left. For only removing trailing whitespace, use stri_trim_right. When you want to remove other leading or trailing characters, you have to specify that with pattern =.

See also ?stri_trim for more info.

Another related problem occurs if you have multiple spaces in between inputs:

> a <- "  a string         with lots   of starting, inter   mediate and trailing   whitespace     "

You can then easily split this string into "real" tokens using a regular expression to the split argument:

> strsplit(a, split=" +")
[[1]]
 [1] ""           "a"          "string"     "with"       "lots"
 [6] "of"         "starting,"  "inter"      "mediate"    "and"
[11] "trailing"   "whitespace"

Note that if there is a match at the beginning of a (non-empty) string, the first element of the output is ‘""’, but if there is a match at the end of the string, the output is the same as with the match removed.

I created a trim.strings () function to trim leading and/or trailing whitespace as:

# Arguments:    x - character vector
#            side - side(s) on which to remove whitespace 
#                   default : "both"
#                   possible values: c("both", "leading", "trailing")

trim.strings <- function(x, side = "both") { 
    if (is.na(match(side, c("both", "leading", "trailing")))) { 
      side <- "both" 
      } 
    if (side == "leading") { 
      sub("^\\s+", "", x)
      } else {
        if (side == "trailing") {
          sub("\\s+$", "", x)
    } else gsub("^\\s+|\\s+$", "", x)
    } 
} 

For illustration,

a <- c("   ABC123 456    ", " ABC123DEF          ")

# returns string without leading and trailing whitespace
trim.strings(a)
# [1] "ABC123 456" "ABC123DEF" 

# returns string without leading whitespace
trim.strings(a, side = "leading")
# [1] "ABC123 456    "      "ABC123DEF          "

# returns string without trailing whitespace
trim.strings(a, side = "trailing")
# [1] "   ABC123 456" " ABC123DEF"   

The best method is trimws().

The following code will apply this function to the entire dataframe.

mydataframe<- data.frame(lapply(mydataframe, trimws),stringsAsFactors = FALSE)

Use dplyr/tidyverse mutate_all with str_trim to trim the entire data frame:

myDummy %>%
  mutate_all(str_trim)

library(tidyverse)
set.seed(335)
df <- mtcars %>%
        rownames_to_column("car") %>%
        mutate(car = ifelse(runif(nrow(mtcars)) > 0.4, car, paste0(car, " "))) %>%
        select(car, mpg)

print(head(df), quote = T)
#>                    car    mpg
#> 1         "Mazda RX4 " "21.0"
#> 2      "Mazda RX4 Wag" "21.0"
#> 3        "Datsun 710 " "22.8"
#> 4    "Hornet 4 Drive " "21.4"
#> 5 "Hornet Sportabout " "18.7"
#> 6           "Valiant " "18.1"

df_trim <- df %>%
  mutate_all(str_trim)

print(head(df_trim), quote = T)  
#>                   car    mpg
#> 1         "Mazda RX4"   "21"
#> 2     "Mazda RX4 Wag"   "21"
#> 3        "Datsun 710" "22.8"
#> 4    "Hornet 4 Drive" "21.4"
#> 5 "Hornet Sportabout" "18.7"
#> 6           "Valiant" "18.1"

^{Created on 2021-05-07 by the reprex package (v0.3.0)}

I tried trim(). It works well with white spaces as well as the '\n'.

x = '\n              Harden, J.\n              '

trim(x)

myDummy[myDummy$country == "Austria "] <- "Austria"

After this, you'll need to force R not to recognize "Austria " as a level. Let's pretend you also have "USA" and "Spain" as levels:

myDummy$country = factor(myDummy$country, levels=c("Austria", "USA", "Spain"))

It is a little less intimidating than the highest voted response, but it should still work.

For whom it may be of interest, here a simple benchmarking. This is of course not capturing all weird cases, but so far we are still lacking the example where str_trim removes whitespace and trimws doesn't (see Richard Telford's comment to this answer). Doesn't seem to matter - the gsub option seems to be fastest :)

x <- c(" lead", "trail ", " both ", " both and middle ", " _special")
gsub_trim <- function (x) gsub("^\\s+|\\s+$", "", x)
res <- microbenchmark::microbenchmark(
  gsub = gsub_trim(x),
  trimws = trimws(x),
  str_trim = stringr::str_trim(x),
  times = 10^5
)
res
#> Unit: microseconds
#>      expr    min     lq      mean median       uq       max neval cld
#>      gsub 20.201 22.788  31.43943 24.654  28.4115  5303.741 1e+05 a  
#>    trimws 38.204 41.980  61.92218 44.420  51.1810 40363.860 1e+05  b 
#>  str_trim 88.672 92.347 116.59186 94.542 105.2800 13618.673 1e+05   c
ggplot2::autoplot(res)

https://i.imgur.com/vqKlsms.png

sessionInfo()
#> R version 4.0.3 (2020-10-10)
#> Platform: x86_64-apple-darwin17.0 (64-bit)
#> Running under: macOS Big Sur 10.16
#> 
#> locale:
#> [1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8
#> 
#> attached base packages:
#> [1] stats     graphics  grDevices utils     datasets  methods   base     
#> 
#> loaded via a namespace (and not attached):
#>  stringr_1.4.0