Using grep in R to delete rows from a data.frame
Solution 1
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$z),]
x y z
1 1 1 apple
2 1 2 pear
3 1 3 banana
4 1 4 A
5 1 5 B
6 1 6 C
7 1 7 D
8 1 8 E
9 1 9 F
10 1 10 G
Solution 2
Here's your problem:
> grep("K",c("apple","pear","banana","A","B","C","D","E","F","G"))
integer(0)
Try grepl() instead:
d[!grepl("K",d$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
Solution 3
For completeness, since R 3.3.0, grep
and friends come with an invert
argument:
new_d <- d[grep("K", d$z, invert = TRUE)]
Solution 4
You want to use grepl
in this case, e.g., new_d <- d[! grepl("K",d$z),]
.
Annemarie
Updated on August 15, 2020Comments
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Annemarie almost 4 years
I have a dataframe such as this one:
d <- data.frame(cbind(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?