mjhjmtu

I have the following strings:

strings <- c("ABBSDGNHNGA", "AABSDGDRY", "AGNAFG", "GGGDSRTYHG") 

I want to cut off the string, as soon as the number of occurances of A, G and N reach a certain value, say 3. In that case, the result should be:

some_function(strings)

c("ABBSDGN", "AABSDG", "AGN", "GGG") 

I tried to use the stringi, stringr and regex expressions but I can't figure it out.

You can accomplish your task with a simple call to str_extract from the stringr package:

library(stringr)

strings <- c("ABBSDGNHNGA", "AABSDGDRY", "AGNAFG", "GGGDSRTYHG")

str_extract(strings, '([^AGN]*[AGN])3')
# [1] "ABBSDGN" "AABSDG" "AGN" "GGG"

The [^AGN]*[AGN] portion of the regex pattern says to look for zero or more consecutive characters that are not A, G, or N, followed by one instance of A, G, or N. The additional wrapping with parenthesis and braces, like this ([^AGN]*[AGN])3, means look for that pattern three times consecutively. You can change the number of occurrences of A, G, N, that you are looking for by changing the integer in the curly braces:

str_extract(strings, '([^AGN]*[AGN])4')
# [1] "ABBSDGNHN" NA "AGNA" "GGGDSRTYHG"

There are a couple ways to accomplish your task using base R functions. One is to use regexpr followed by regmatches:

m <- regexpr('([^AGN]*[AGN])3', strings)
regmatches(strings, m)
# [1] "ABBSDGN" "AABSDG" "AGN" "GGG"

Alternatively, you can use sub:

sub('(([^AGN]*[AGN])3).*', '\1', strings)
# [1] "ABBSDGN" "AABSDG" "AGN" "GGG"

Here is a base R option using strsplit

sapply(strsplit(strings, ""), function(x)
 paste(x[1:which.max(cumsum(x %in% c("A", "G", "N")) == 3)], collapse = ""))
#[1] "ABBSDGN" "AABSDG" "AGN" "GGG"

Or in the tidyverse

library(tidyverse)
map_chr(str_split(strings, ""), 
 ~str_c(.x[1:which.max(cumsum(.x %in% c("A", "G", "N")) == 3)], collapse = ""))

Identify positions of pattern using gregexpr then extract n-th position (3) and substring everything from 1 to this n-th position using subset.

nChars <- 3
pattern <- "A|G|N"
# Using sapply to iterate over strings vector
sapply(strings, function(x) substr(x, 1, gregexpr(pattern, x)[[1]][nChars]))

PS:

If there's a string that doesn't have 3 matches it will generate NA, so you just need to use na.omit on the final result.

This is just a version without strsplit to Maurits Evers neat solution.

sapply(strings,
 function(x) 
 raw <- rawToChar(charToRaw(x), multiple = TRUE)
 idx <- which.max(cumsum(raw %in% c("A", "G", "N")) == 3)
 paste(raw[1:idx], collapse = "")
 )
## ABBSDGNHNGA AABSDGDRY AGNAFG GGGDSRTYHG 
## "ABBSDGN" "AABSDG" "AGN" "GGG"

Or, slightly different, without strsplit and paste:

test <- charToRaw("AGN")
sapply(strings,
 function(x) 
 raw <- charToRaw(x)
 idx <- which.max(cumsum(raw %in% test) == 3)
 rawToChar(raw[1:idx])
 )

Interesting problem. I created a function (see below) that solves your problem. It's assumed that there are just letters and no special characters in any of your strings.

 reduce_strings = function(str, chars, cnt)

 # Replacing chars in str with "!"
 chars = paste0(chars, collapse = "")
 replacement = paste0(rep("!", nchar(chars)), collapse = "")
 str_alias = chartr(chars, replacement, str) 

 # Obtain indices with ! for each string
 idx = stringr::str_locate_all(pattern = '!', str_alias)

 # Reduce each string in str
 reduce = function(i) substr(str[i], start = 1, stop = idx[[i]][cnt, 1])
 result = vapply(seq_along(str), reduce, "character")
 return(result)


# Example call
str = c("ABBSDGNHNGA", "AABSDGDRY", "AGNAFG", "GGGDSRTYHG") 
chars = c("A", "G", "N") # Characters that are counted
cnt = 3 # Count of the characters, at which the strings are cut off
reduce_strings(str, chars, cnt) # "ABBSDGN" "AABSDG" "AGN" "GGG"