Evan, there are oodles of ways to do many things in R, and mcu of what the tidyverse supplies can often be done as easily, or easier, outside it.
Before presenting a solution, I need to make sure I am answering the same question or problem you intend. Here is the string you have as an example: st <- "abaaabbaaaaabaaab" Is the string a string testing for single characters called "b" with any other characters being either just "a" or at least non-"b" and of any length but at least a few? If so, ONE METHOD is to convert the string to a vector for reasons that will become clear. For oddball reasons, this is a way to do it: > unlist(strsplit(st,"")) [1] "a" "b" "a" "a" "a" "b" "b" "a" "a" "a" "a" "a" "b" "a" "a" "a" "b" The result is a vector you can examine to see if they are equal to "b" or not as a TRUE/FALSE vector: > unlist(strsplit(st,"")) == "b" [1] FALSE TRUE FALSE FALSE FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE [17] TRUE Now you can ask for the indices which are TRUE, meaning at what offset from the beginning are there instances of the letter "b": > which(unlist(strsplit(st,"")) == "b") [1] 2 6 7 13 17 This shows the second the integer offsets for the letter "b" are the second, sixth and so on to seventeenth. Again, if I understood you, you want a measure of how far apart instances of "b" are with adjacent ones being 1 apart. Again, many methods but I chose one where I sort of slid over the above values by sliding in a zero from the front and removing the last entry. So save that in a variable first: indices <- which(unlist(strsplit(st,"")) == "b") indices_shifted <- c(0, head(indices, -1)) The two contain: > indices [1] 2 6 7 13 17 > indices_shifted [1] 0 2 6 7 13 > indices - indices_shifted [1] 2 4 1 6 4 The above is the same as your intended result. If you want to be cautious, handle edge cases like not having any "b" or an empty string. Here is the consolidated code: st <- "abaaabbaaaaabaaab" indices <- which(unlist(strsplit(st,"")) == "b") indices_shifted <- c(0, head(indices, -1)) result <- indices - indices_shifted There are many other ways to do this and of course some are more straightforward and some more complex. Consider a loop using a vector version of the string where each time you see a b", you remember the last index you saw it and put out the number representing the gap. Fairly low tech. -----Original Message----- From: R-help <r-help-boun...@r-project.org> On Behalf Of Evan Cooch Sent: Friday, December 2, 2022 12:19 PM To: r-help@r-project.org Subject: [R] interval between specific characters in a string... Was wondering if there is an 'efficient/elegant' way to do the following (without tidyverse). Take a string abaaabbaaaaabaaab Its easy enough to count the number of times the character 'b' shows up in the string, but...what I'm looking for is outputing the 'intervals' between occurrences of 'b' (starting the counter at the beginning of the string). So, for the preceding example, 'b' shows up in positions 2, 6, 7, 13, 17 So, the interval data would be: 2, 4, 1, 6, 4 My main approach has been to simply output positions (say, something like unlist(gregexpr('b', target_string))), and 'do the math' between successive positions. Can anyone suggest a more elegant approach? Thanks in advance... ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.