CODEWITHSUNDEEP
rgrepl
Diego Gonzalez Avalos
Diego Gonzalez Avalos

Reputation: 138

How to use GREPL with string from a column like pattern in R

I'm working with R and i'm trying to clean my data using. I have the next data:

example<- data.frame(x=c("hungry","fly","day","dog"),
                        y=c("i'm hungry","i believe i can fly","a hard day's night","cat"))

I'm trying to identify if the Y column contains the characters from the column X. I try with grepl() but that function doesn't work with vectors and i try with str_detect() but i don't know why doesn't work. I finally try to get the next table:

      x                   y        Flag
1 hungry          i'm hungry          1
2    fly i believe i can fly          1
3    day  a hard day's night          1
4    dog                 cat          0

I wonder if someone can give me some option or other view to do it.

Thank!

Upvotes: 2

Views: 1148

Answers (3)

ThomasIsCoding
ThomasIsCoding

Reputation: 101538

You can try Vectorize to make grepl vercorized, e.g.,

example <- within(example,Flag <- +Vectorize(grepl)(x,y))

such that

> example
       x                   y Flag
1 hungry          i'm hungry    1
2    fly i believe i can fly    1
3    day  a hard day's night    1
4    dog                 cat    0

Upvotes: 2

Chris Ruehlemann
Chris Ruehlemann

Reputation: 21400

You can use grepl and ifelse in this way:

example$Flag <- ifelse(grepl(paste0(example$x, collapse = "|"), example$y), 1, 0)

Using paste0, this collapses example$x into a single pattern with alternatives separated by | and has grepl check whether the pattern complex matches the values in example$y: if the match is found, the ifelse statement assigns 1; if it doesn't, 0.

Alternatively, you can use str_detectfrom package stringr: Note that the order in which you assign the two variables to the function matters--you need to put the larger strings (i.e. those in example$y) first AND you need to convert both variables to character. On the upside though there's no need for the paste0 transformation:

example$Flag <- ifelse(str_detect(as.character(example$y), as.character(example$x)), 1, 0)

Result:

example
       x                   y  Flag
1 hungry          i'm hungry     1
2    fly i believe i can fly     1
3    day  a hard day's night     1
4    dog                 cat     0

Upvotes: 1

hello_friend
hello_friend

Reputation: 5788

No where near as succinct as @jogo's answer but:

sapply(split(example, rownames(example)), 
       function(z){grepl(as.character(z$x), as.character(z$y))})

Upvotes: 1

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