The Art of R Programming

But what if you want to search for a period usinggrep()? Here’s the

naive approach:

> grep(".",c("abc","de","f.g"))

[1]123

The result should have been 3 , not(1,2,3). This call failed because peri-

ods are metacharacters. You need toescapethe metacharacter nature of the

period, which is done via a backslash:

> grep("\\.",c("abc","de","f.g"))

[1] 3

Now, didn’t I sayabackslash? Then why are there two? Well, the sad

truth is that the backslash itself must be escaped, which is accomplished by

its own backslash! This goes to show how arcanely complex regular expres-

sions can become. Indeed, a number of books have been written on the sub-

ject of regular expressions (for various programming languages). As a start

in learning about the topic, refer to R’s online help (type?regex).

11.2.1 Extended Example: Testing a Filename for a Given Suffix...........

Suppose we wish to test for a specified suffix in a filename. We might, for

instance, want to find all HTML files (those with suffix.html,.htm, and so

on). Here is code for that:

1 testsuffix <- function(fn,suff) {
2 parts <- strsplit(fn,".",fixed=TRUE)
3 nparts <- length(parts[[1]])
4 return(parts[[1]][nparts] == suff)
5 }

Let’s test it.

> testsuffix("x.abc","abc")

[1] TRUE

> testsuffix("x.abc","ac")

[1] FALSE

> testsuffix("x.y.abc","ac")

[1] FALSE

> testsuffix("x.y.abc","abc")

[1] TRUE

How does the function work? First note that the call tostrsplit()on

line 2 returns a list consisting of one element (becausefnis a one-element

vector)—a vector of strings. For example, callingtestsuffix("x.y.abc","abc")

will result inpartsbeing a list consisting of a three-element vector with ele-

mentsx,y, andabc. We then pick up the last element and compare it tosuff.

String Manipulation 255