How can I trim leading and trailing white space?

Question

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.fra

Answer 1

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.

Answer 2

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.

Answer 3

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

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

trim(x)

Answer 4

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.

Answer 5

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"   

Answer 6

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"