R: Return a vector using %in% operator

Question

I have a vector with delimiters and I want to generate a vector of the same length with boolean values based on whether or not one of the delimited values contains what I am after. I cannot find a way to do this neatly in vector-based logic. As an example:

x <- c('a', 'a; b', 'ab; c', 'b; c', 'c; a', 'c')

Using some magic asking whether 'a' %in% x, I want to get the vector:

TRUE, TRUE, FALSE, FALSE, TRUE, FALSE

I initially tried the following:

'a' %in% trimws(strsplit(x, ';'))

But this unexpectedly collapses the entire list and returns TRUE, rather than a vector, since one of the elements in x is 'a'. Is there a way to get the vector I am looking for without rewriting the code into a for-loop?

Answer 1

Update: To consider white spaces:

library(stringr)
x <- str_replace_all(string=x, pattern=" ", repl="")
x
[1] "a"    "a;b"  "ab;c" "b;c"  "c;a"  "c" 

str_detect(x, 'a$|a;')

[1]  TRUE  TRUE FALSE FALSE  TRUE FALSE

First answer: If you want to use str_detect we have to account on a + delimiter ;:

library(stringr)
str_detect(x, 'a$|a;')

[1]  TRUE  TRUE FALSE FALSE  TRUE FALSE

Answer 2

Base R:

grepl("a", x)

or (when you want to use explicitly %in%):

sapply(strsplit(x,""), function(x){ "a" %in% x})

When working with strings and letters I always use the great library stringr

library(stringr)
x <- c('a', 'a; b', 'ab; c', 'b; c', 'c; a', 'c')
str_detect(x, "a")

Answer 3

If you would like to use %in%, here is a base R option

> mapply(`%in%`, list("a"), strsplit(x, ";\\s+"))
[1]  TRUE  TRUE FALSE FALSE  TRUE FALSE

---

A more efficient way might be using grepl like below

> grepl("\\ba\\b",x)
[1]  TRUE  TRUE FALSE FALSE  TRUE FALSE

Answer 4

You can read each item separately with scan, trim leading and trailing WS as you attempted, and test each resulting character vector in turn with:

 sapply(x, function(x){"a" %in% trimws( scan( text=x, what="",sep=";", quiet=TRUE))})
    a  a; b ab; c  b; c  c; a     c 
 TRUE  TRUE FALSE FALSE  TRUE FALSE 

The top row of the result is just the names and would not affect a logical test that depended on this result. There is an unname function if needed.