Convert leading or trailing white space and tab characters to nothing.

This eliminates any characters matched by the regular expression `\s` if they appear at the beginning of a string or at its end. It does not alter spaces in the interior of a string. This was created when I was not aware of R's trimws and the purpose is the same. On our TODO list, we intend to eliminate this function and replace its use with trimws

zapspace(x)

Arguments

x

A character vector

Value

If x is a character vector, return is a character vector with leading and trailing white space values removed. If x is not a character vector, an unaltered x is returned.

Author

Paul Johnson <pauljohn@ku.edu>

Examples

x <-  c("", " ", "   ", "\t", "\t some", "some\t", " space first")
zapspace(x)
#> [1] ""            ""            ""            ""            "some"       
#> [6] "some"        "space first"