Using grepl in R to search for an asterisk

Question

I am reading a phrase into an R script as an argument. If the phrase contains an asterisk (*), I do not want the script to run.

However, I am having issues recognising the asterisk when using grepl. For example:

> asterisk="*"
> phrase1="hello"
> phrase2="h*llo"
> grepl(asterisk,phrase1)
[1] TRUE
> grepl(asterisk,phrase2)
[1] TRUE

The outcome for grepl(asterisk,phrase1) should be FALSE. Does anyone know how I can get grepl to recognise if there is or isn't an asterisk in the phrase?

note https://en.wikipedia.org/wiki/Asterix vs https://en.wikipedia.org/wiki/Asterisk — Ben Bolker, Apr 06 '16 at 16:22
Some others to watch out for: http://stackoverflow.com/q/399078/ — Frank, Apr 06 '16 at 16:24

score 11 · Accepted Answer · answered Apr 06 '16 at 16:17

11

Try this:

p <- c("Hello", "H*llo")
grepl("\\*", p)

[1] FALSE  TRUE

This works because the * asterisk has special meaning in a regular expresssion. Specifically, * means find zero or more of the previous element.

Thus you have to escape the asterisk using \\*. The double escape is necessary because the \ already has the meaning of escape in R.

answered Apr 06 '16 at 16:17

Andrie

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or `grepl("*",p,fixed=TRUE)` – Ben Bolker Apr 06 '16 at 16:21
I prefer this approach, but an alternative to avoid the use of the escape characters is `grepl("*", p, fixed=TRUE)` – tblznbits Apr 06 '16 at 16:21

Using grepl in R to search for an asterisk

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