Finding strings that are a certain length and contain specific characters

Question

Sample data

a<-c("hour","four","ruoh", "six", "high", "our")

I want to find all strings that contain o & u & h & are 4 characters but the order does not matter.

I want to return "hour","four","ruoh" this is my attempt

grepl("o+u+r", a) nchar(a)==4

Answer 1

To match strings of length 4 containing the characters h , o , and u use:

grepl("(?=^.{4}$)(?=.*h)(?=.*o)(?=.*u)",
      c("hour","four","ruoh", "six", "high", "our"),
      perl = TRUE)
[1]  TRUE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE

• (?=^.{4}$) : string has length 4.
• (?=.*x) : x occurs at any position in string.

Answer 2

You could use strsplit and setdiff , I added an additional edge case to your sample data :

a<-c("hour","four","ruoh", "six", "high", "our","oouh")
a[nchar(a) == 4 &
  lengths(lapply(strsplit(a,""),function(x) setdiff(x, c("o","u","h")))) == 1]
# [1] "hour" "ruoh"

or grepl :

a[nchar(a) == 4 & !rowSums(sapply(c("o","u","h"), Negate(grepl), a))]
# [1] "hour" "ruoh" "oouh"

sapply(c("o","u","h"), Negate(grepl), a) gives you a matrix of which word doesn't contain each letter, then the rowSums acts like any applied by row, as it will be coerced to logical.

Answer 3

Using grepl with your edited method (r instead of h):

a<-c("hour","four","ruoh", "six", "high", "our")

a[grepl(pattern="o", x=a) & grepl(pattern="u", x=a) & grepl(pattern="r", x=a) & nchar(a)==4]

Returns:

[1] "hour" "four" "ruoh"