How to check if a word is a palindrome?

Question

I would like to write a function that checks if a word is a palindrome. return should tell me how

palindrome "love" is not a palindrome 

the function should contain the word (and not as a vector) as an argument (I do not quite understand what this means)

I tried

 Palindrome <- function(character){ charsplit <- strsplit(as.character(character), "") revchar <- rev(unlist(charsplit)) palinum <- as.numeric(paste(revchar, collapse="")) character==palinum } 

but

 Palindrome ("love") 

returns

 NA 

Thanks for any help!

Answer 1

Without using strsplit :

 is.palindrome <- function(word) { rawWord <- charToRaw(tolower(word)) ## converts to lower case sprintf("%s is %sa palindrome", word, c("not ", "")[identical(rawWord, rev(rawWord)) + 1]) } > is.palindrome("otto") [1] "otto is a palindrome" > is.palindrome("love") [1] "love is not a palindrome" > is.palindrome("Otto") [1] "Otto is a palindrome" 

Answer 2

 is.palindrome <- function (word) { identical(word, paste(rev(strsplit(word, "")[[1]]), collapse="")) } is.palindrome("kayak") [1] TRUE is.palindrome("love") [1] FALSE 

Answer 3

This will work:

 a = "blahalb" identical(strsplit(a, "")[[1]], rev(strsplit(a, "")[[1]])) #[1] TRUE a = "love" identical(strsplit(a, "")[[1]], rev(strsplit(a, "")[[1]])) #[1] FALSE 

Also check out the Kmisc package for supposedly faster results:

 library(Kmisc) a = c("blahalb", "love") str_rev(a) == a #[1] TRUE FALSE 

Answer 4

 is_palindrome <- function(word){ charsplit <- strsplit(word, "")[[1]] revchar <- rev(charsplit) all(charsplit==revchar) } is_palindrome("love") is_palindrome("otto") 

Answer 5

You can also use the substring() and rev() functions:

 is_palindrome <- function(x){ a <- substring(x,seq(1,nchar(x),1) , seq(1,nchar(x),1)) paste(rev(a),sep="",collapse="") == paste(a,sep="",collapse="") } 

Answer 6

Here's the C ++ implementation:

 sourceCpp(' bool isPalindrome(String x) { std::string y(x); int n = y.size(); for(int i = 0; i < n / 2; ++i) { if (y[i] != y[n - i - 1]) return false; } return true; } ') 

It also does not work with non-ASCII strings, but it is about 10 times faster than a pure R solution:

 library(microbenchmark) options(digits = 3) is.palindrome <- function (word) { identical(word, paste(rev(strsplit(word, "")[[1]]), collapse="")) } x <- paste(letters, rev(letters), collapse = "") y <- paste("a", x) microbenchmark( is.palindrome(x), isPalindrome(x), is.palindrome(y), isPalindrome(y) ) # Unit: microseconds # expr min lq median uq max neval # is.palindrome(x) 24.62 25.99 27.14 28.29 36.38 100 # isPalindrome(x) 2.38 2.68 2.82 3.58 4.03 100 # is.palindrome(y) 24.68 26.44 27.78 28.46 80.94 100 # isPalindrome(y) 2.33 2.67 3.41 3.64 33.60 100 

Answer 7

The stringi package has a function - stri_reverse

 require(stringi) stri_reverse("eye") ## [1] "eye" stri_reverse("1234") ## [1] "4321" 

Now the function of the palindrome can be simple:

 palindrome <- function(x) stri_reverse(x)==x palindrome(c("love","1234321","eye","oço","ąłą")) ## [1] FALSE TRUE TRUE TRUE TRUE