gsub: replace regex match with regex replacement string

Question

I have a requirement to replace more than 2 continuous 1's with equal number of zeros. Currently, I can find the match as below but I don't know how to replace with the exact number of zeros as the match is found

ind<-c(1,1,0,0,0,1,1,1,1,0,1,1,0,0,0,1,1,0,1,0,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0)
gsub("([1])\\1\\1+","0",paste0(ind,collapse=""))

gives

"11000001100011010010101000101000"

as it replaces the match with just one 0 but I need

"11000000001100011010010101000000010100000"

You begin with a vector, do you also want to end with a vector or a string? — Colonel Beauvel, Sep 04 '15 at 10:27
@ColonelBeauvel either should be fine as I can convert between any. — Saksham, Sep 04 '15 at 11:09

score 2 · Accepted Answer · edited May 23 '17 at 12:14

2

You can use the following gsub replacement:

ind<-c(1,1,0,0,0,1,1,1,1,0,1,1,0,0,0,1,1,0,1,0,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0)
gsub("1(?=1{2,})|(?!^)\\G1","0",paste(ind,collapse=""), perl=T)

See IDEONE demo, the result is [1] "11000000001100011010010101000000010100000".

The regex is Perl-based since it uses look-aheads and the \G operator.

This regex matches:

1 - a literal 1 if...
(?=1{2,}) - it is followed by 2 or more 1s or...
(?!^)\\G1 - any 1 that is following the previous match.

For more details on the \G operator, see What good is \G in a regular expression? at perldoc.perl.org, and When is \G useful application in a regex? SO post.

edited May 23 '17 at 12:14

Community

1
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answered Sep 04 '15 at 10:19

Wiktor Stribiżew

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what if i want to do reverse. I want to replace sequence of 3 or less 1's with 0? – Saksham Sep 05 '15 at 07:42
and could you tell me what to google for `\\G1`. I am seeing this for the first time – Saksham Sep 05 '15 at 07:44
I added some links to the `\G` operator descriptions. When you need to replace sequences of 1 to 3 `1`s, you can use nested look-ahead with a look-behind to restrict the number of `1`s: [`(? – Wiktor Stribiżew Sep 05 '15 at 20:35

Colonel Beauvel · Answer 2 · 2015-09-04T11:10:37.110

1

A solution not using regex but rle:

x = rle(ind)
x$values[x$lengths>2 & x$values] <- 0
inverse.rle(x)

#[1] 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0

edited Sep 04 '15 at 11:10

answered Sep 04 '15 at 10:23

Colonel Beauvel

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gsub: replace regex match with regex replacement string

2 Answers2