how to find if pattern is in string

Question

I have some strings and I'd like to convert each string in a number, so I'd like to use regular expression. My strings can be one of like:

["star"]
["near-star"]
["shared"]
["near-shared"]
["complete"]
["near-complete"]
["null"]
["near-null"]

my problem is that both these statements are true:

> grepl("star", "[\"near-star\"]")
[1] TRUE
> grepl("near-star", "[\"near-star\"]")
[1] TRUE

and this applies also to the other labels... any advice on how to write the right code to match each label is much appreciated.

best regards, Simone

It seems to me that this question calls for a regex tutorial, not a simple answer. — Jon Kiparsky
– Jon Kiparsky, Commented Feb 5, 2014 at 18:17
What do you mean by "convert each string in a number"? I don't see any numbers... If you want to convert N different strings to the numbers 1 to N then you can go via factors... — Spacedman
– Spacedman, Commented Feb 5, 2014 at 18:19
You can test for absolute string equality with ==. "star" == "near-star" returns FALSE. — Señor O
– Señor O, Commented Feb 5, 2014 at 18:22

Spacedman · Accepted Answer · 2014-02-05 18:29:50Z

3

Trying to answer what I think might be your real problem (convert each string "to" a number)...

Given data:

> strings = c('["star"]', '["near-stat"]', '["shared"]', '["near-shared"]')
> data = sample(strings,20,TRUE)

such that:

> head(data)
[1] "[\"near-stat\"]"   "[\"star\"]"        "[\"near-shared\"]"
[4] "[\"near-shared\"]" "[\"shared\"]"      "[\"star\"]"

Simply do:

> dataf=factor(data)
> as.numeric(dataf)
 [1] 2 4 1 1 3 4 1 2 2 1 2 3 4 4 3 4 4 1 1 4

the mapping being given by:

> levels(dataf)
[1] "[\"near-shared\"]" "[\"near-stat\"]"   "[\"shared\"]"     
[4] "[\"star\"]"

answered Feb 5, 2014 at 18:29

Spacedman

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Greg Snow · Accepted Answer · 2014-02-05 19:09:55Z

2

Others have mentioned just using factors or the fixed argument (either of which will work fine for your stated question). But in general if you want to match a string or pattern, but only if it is not preceded by a given string then you can use negative look behind, an extension in Perl regular expressions:

> test <- c('star','near-star')
> grepl('(?<!near-)star', test, perl=TRUE )
[1]  TRUE FALSE

The regular expression here say to match the string "star", but only if not preceded by the string "near-". The help page ?regexp has details (you need to scroll almost all the way to the bottom).

answered Feb 5, 2014 at 19:09

Greg Snow

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Sven Hohenstein · Accepted Answer · 2014-02-05 18:17:40Z

1

You can include the square brackets and quotes in your pattern. Furthermore, you can use fixed = TRUE for matching the string as is.

> grepl("[\"star\"]", "[\"near-star\"]", fixed = TRUE)
[1] FALSE
> grepl("[\"star\"]", "[\"star\"]", fixed = TRUE)
[1] TRUE

answered Feb 5, 2014 at 18:17

Sven Hohenstein

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