Count number of palindromes within a string - r

Question

I have written the below code to count the number of palindromic strings in a given string:
countPalindromes <- function(str){
len <- nchar(str)
count <- 0
for(i in 1:len){
for(j in i:len){
subs <- substr(str, i, j)
rev <- paste(rev(substring(subs, 1:nchar(subs), 1:nchar(subs))), collapse = "")
if(subs == rev){
count <- count + 1
}
}
}
count
}
This is actually working fine but the code needs to be optimized in such a way so that it executes at a faster rate.
Please suggest some ways to optimize this piece of code.

Answer 1

Here's a solution that uses the wonderful stringi package - just as Andre suggested - together with a wee bit of vectorization.
cp <- function(s) {
lenstr <- stri_length(s) # Get the length
res <- sapply(1:lenstr, function(i) {
# Get all substrings
sub_string <- stringi::stri_sub(s, i, i:lenstr)
# Count matches
sum((sub_string == stringi::stri_reverse(sub_string)))
})
sum(res)
}
This should give the same result as your function
> cp("enafdemderredmedfane")
[1] 30
> countPalindromes("enafdemderredmedfane")
[1] 30
There is not much speedup for short strings, but for longer strings you can really see a benefit:
> microbenchmark::microbenchmark(countPalindromes("howdoyoudo"), cp("howdoyoudo"))
Unit: microseconds
expr min lq mean median uq max neval cld
countPalindromes("howdoyoudo") 480.979 489.6180 508.9044 494.9005 511.201 662.605 100 b
cp("howdoyoudo") 156.117 163.1555 175.4785 169.5640 179.993 324.145 100 a
Compared to
> microbenchmark::microbenchmark(countPalindromes("enafdemderredmedfane"), cp("enafdemderredmedfane"))
Unit: microseconds
expr min lq mean median uq max neval cld
countPalindromes("enafdemderredmedfane") 2031.565 2115.0305 2475.5974 2222.354 2384.151 6696.484 100 b
cp("enafdemderredmedfane") 324.991 357.6055 430.8334 387.242 478.183 1298.390 100 a

Answer 2

Working with a vector the process is faster, I am thinking of eliminating the double for, but I can not find an efficient way.
countPalindromes_new <- function(str){
len <- nchar(str)
strsp <- strsplit(str, "")[[1]]
count <- 0
for(i in 1:len){
for(j in i:len){
if(all(strsp[i:j] == strsp[j:i])){
count <- count + 1
}
}
}
count
}
> microbenchmark::microbenchmark(countPalindromes("howdoyoudo"), cp("howdoyoudo"), countPalindromes_new("howdoyoudo"))
Unit: microseconds
expr min lq mean median uq max neval
countPalindromes("howdoyoudo") 869.121 933.1215 1069.68001 963.201 1022.081 6712.751 100
cp("howdoyoudo") 192.000 202.8805 243.11972 219.308 258.987 477.441 100
countPalindromes_new("howdoyoudo") 49.068 53.3340 62.32815 57.387 63.574 116.481 100
> microbenchmark::microbenchmark(countPalindromes("enafdemderredmedfane"), cp("enafdemderredmedfane"), countPalindromes_new("enafdemderredmedfane"))
Unit: microseconds
expr min lq mean median uq max neval
countPalindromes("enafdemderredmedfane") 3578.029 3800.9620 4170.0888 3987.416 4173.6550 10205.445 100
cp("enafdemderredmedfane") 391.254 438.4010 609.8782 481.708 534.6135 6116.270 100
countPalindromes_new("enafdemderredmedfane") 200.534 214.1875 235.3501 223.148 245.5475 448.854 100
UPDATE (NEW VERSION WIHTOUT LEN 1 COMPARASION):
countPalindromes_new2 <- function(str){
len <- nchar(str)
strsp <- strsplit(str, "")[[1]]
count <- len
for(i in 1:(len-1)){
for(j in (i + 1):len){
if(all(strsp[i:j] == strsp[j:i])){
count <- count + 1
}
}
}
count
}

Answer 3

Simply: normally I'm against using new libraries everywhere. But stringi is THE library for working with strings in R.
string_vec <- c("anna","nothing","abccba")
string_rev <- stringi::stri_reverse(string_vec)
sum(string_vec == string_rev)
#evals 2