How to use stringr functions to remove all empty words?-CodePudding

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?

CodePudding user response：

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"

CodePudding user response：

Two base R alternatives:

b[nzchar(trimws(b))]
# [1] "books"   "animals" "frogs"  
b[grepl("\\S",b)]
# [1] "books"   "animals" "frogs"

CodePudding user response：

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"