To only find the first locations, use lapply()
with min()
:
my_string <- c("test1", "test1test1", "test1test1test1")
unlist(lapply(gregexpr(pattern = '1', my_string), min))
#> [1] 5 5 5
# or the readable tidyverse form
my_string %>%
gregexpr(pattern = '1') %>%
lapply(min) %>%
unlist()
#> [1] 5 5 5
To only find the last locations, use lapply()
with max()
:
unlist(lapply(gregexpr(pattern = '1', my_string), max))
#> [1] 5 10 15
# or the readable tidyverse form
my_string %>%
gregexpr(pattern = '1') %>%
lapply(max) %>%
unlist()
#> [1] 5 10 15
Here's another straightforward alternative.
> which(strsplit(string, "")[[1]]=="2")
[1] 4 24
find the position of the nth occurrence of str2 in str1(same order of parameters as Oracle SQL INSTR), returns 0 if not found
instr <- function(str1,str2,startpos=1,n=1){
aa=unlist(strsplit(substring(str1,startpos),str2))
if(length(aa) < n+1 ) return(0);
return(sum(nchar(aa[1:n])) + startpos+(n-1)*nchar(str2) )
}
instr('xxabcdefabdddfabx','ab')
[1] 3
instr('xxabcdefabdddfabx','ab',1,3)
[1] 15
instr('xxabcdefabdddfabx','xx',2,1)
[1] 0
You can make the output just 4 and 24 using unlist:
unlist(gregexpr(pattern ='2',"the2quickbrownfoxeswere2tired"))
[1] 4 24
You could use grep
as well:
grep('2', strsplit(string, '')[[1]])
#4 24
Source: Stackoverflow.com