CODEWITHSUNDEEP
rdataframerow
Annemarie
Annemarie

Reputation: 609

Using grep in R to delete rows from a data.frame

I have a dataframe such as this one:

d <- data.frame(x=1,
                y=1:10,
                z=c("apple","pear","banana","A","B","C","D","E","F","G"),
                stringsAsFactors = FALSE)

I'd like to delete some rows from this dataframe, depending on the content of column z:

new_d <- d[-grep("D",d$z),]

This works fine; row 7 is now deleted:

    new_d
     x  y      z
  1  1  1  apple
  2  1  2   pear
  3  1  3 banana
  4  1  4      A
  5  1  5      B
  6  1  6      C
  8  1  8      E
  9  1  9      F
  10 1 10      G

However, when I use grep to search for content that is not present in column z, it seems to delete all content of the dataframe:

new_d <- d[-grep("K",d$z),]
new_d
# [1] x y z
# <0 rows> (or 0-length row.names)

I would like to search and delete rows in this or another way, even if the character string I am searching for is not present. How to go about this?

Upvotes: 14

Views: 24599

Answers (5)

andschar
andschar

Reputation: 3973

If you don't want to use ! or invert = TRUE you can also use a REGEX only:

d[ grep('^((?!K).)*$', d$z, perl = TRUE), ]

Taken from here.

Upvotes: 0

Hugh
Hugh

Reputation: 16089

For completeness, since R 3.3.0, grep and friends come with an invert argument:

new_d <- d[grep("K", d$z, invert = TRUE), ]

Upvotes: 4

Ari B. Friedman
Ari B. Friedman

Reputation: 72741

Here's your problem:

> grep("K",c("apple","pear","banana","A","B","C","D","E","F","G"))
integer(0)

Try grepl() instead:

d[!grepl("K",d$z),]

This works because the negated logical vector has an entry for every row:

> grepl("K",d$z)
 [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
> !grepl("K",d$z)
 [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE

Upvotes: 7

GSee
GSee

Reputation: 49810

You can use TRUE/FALSE subsetting instead of numeric.

grepl is like grep, but it returns a logical vector. Negation works with it.

 d[!grepl("K",d$z),]
   x  y      z
1  1  1  apple
2  1  2   pear
3  1  3 banana
4  1  4      A
5  1  5      B
6  1  6      C
7  1  7      D
8  1  8      E
9  1  9      F
10 1 10      G

Upvotes: 26

Matthew Plourde
Matthew Plourde

Reputation: 44614

You want to use grepl in this case, e.g., new_d <- d[! grepl("K",d$z),].

Upvotes: 1

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