| abbreviate {base} | R Documentation |
Abbreviate strings to at least minlength characters,
such that they remain unique (if they were).
abbreviate(names.arg, minlength = 4, use.classes = TRUE,
dot = FALSE)
names.arg |
a character vector of names to be abbreviated, or an
object to be coerced to a character vector by as.character. |
minlength |
the minimum length of the abbreviations. |
use.classes |
logical (currently ignored by R). |
dot |
logical: should a dot (".") be appended? |
The algorithm used is similar to that of S. For a single string it
works as follows.
First all spaces at the beginning of the string are stripped.
Then (if necessary) any other spaces are stripped.
Next, lower case vowels are removed (starting at the right)
followed by lower case consonants.
Finally if the abbreviation is still longer than minlength
upper case letters are stripped.
Characters are always stripped from the end of the word first.
If an element of names.arg contains more than one word (words
are separated by space) then at least one letter from each word will be
retained.
Missing (NA) values are unaltered.
If use.classes is FALSE then the only distinction is to
be between letters and space. This has NOT been implemented.
A character vector containing abbreviations for the strings in its
first argument. Duplicates in the original names.arg will be
given identical abbreviations. If any non-duplicated elements have
the same minlength abbreviations then minlength is
incremented by one and new abbreviations are found for those elements
only. This process is repeated until all unique elements of
names.arg have unique abbreviations.
The character version of names.arg is attached to the returned
value as a names argument: no other attributes are retained.
This is really only suitable for English, and does not work correctly with non-ASCII characters in multibyte locales. It will warn if used with non-ASCII characters.
x <- c("abcd", "efgh", "abce")
abbreviate(x, 2)
(st.abb <- abbreviate(state.name, 2))
table(nchar(st.abb))# out of 50, 3 need 4 letters