b <- c("books", " ", "animals", "frogs")
#My code:
b[!grepl("^\\s+$", b)]
[1] "books" "animals" "frogs"
#Now, I am working to figure out this solution with stringr package.
str_remove_all(b, "^\\s+$")
[1] "books" "" "animals" "frogs"
The output shows "" where my new code fails. Any solution to get the result like my first code?
We may use str_subset in stringr
library(stringr)
str_subset(b, "^\\s+$", negate = TRUE)
[1] "books" "animals" "frogs"
The function that corresponds to grepl is str_detect
b[str_detect(b, "^\\s+$", negate = TRUE)]
[1] "books" "animals" "frogs"
In base R, we may use grep with invert = TRUE
grep("^\\s+$", b, invert = TRUE, value = TRUE)
[1] "books" "animals" "frogs"
Or without regex with trimws (to remove the spaces - leading/lagging) and use nzhcar to create logical vector for subsetting
b[nzchar(trimws(b))]
[1] "books" "animals" "frogs"
Two base R alternatives:
b[nzchar(trimws(b))]
# [1] "books" "animals" "frogs"
b[grepl("\\S",b)]
# [1] "books" "animals" "frogs"
In base R we can do:
b[b !=" "]
Output:
[1] "books" "animals" "frogs"
OR with stringr:
library(stringr)
str_subset(str_squish(b), "")
[1] "books" "animals" "frogs"
b[nzchar(trimws(b))]