Reverse digits in R

Question

How can you reverse a string of numbers in R?

for instance, I have a vector of about 1000 six digit numbers, and I would like to know if they are palindromes. I would like to create a second set which is the exact reverse, so I could do a matchup.

Answer 1

It is actually the decimial representation of the number that you are testing to be a palindrome, not the number itself (255 is a palendrome in hex and binary, but not decimal).

You can do this fairly simply using pattern matching:

> tmp <- c(100001, 123321, 123456)
> grepl( '^([0-9])([0-9])([0-9])\\3\\2\\1$', tmp )
[1]  TRUE  TRUE FALSE
> 

you could convert the numbers to character, split into individual characters (strsplit), reverse each number (sapply and rev), then paste the values back together (paste) and covert back to numbers (as.numeric). But I think the above is better if you are just interested in 6 digit palendromes.

Answer 2

I don't think rev quite does it. It reverses the elements of the vector, while the question is how to reverse the elements in the vector.

> nums <- sapply(1:10,function(i)as.numeric(paste(sample(1:9,6,TRUE),collapse="")))
> nums
 [1] 912516 568934 693275 835117 155656 378192 343266 685182 298574 666354
> sapply(strsplit(as.character(nums),""), function(i) paste(rev(i),collapse=""))
 [1] "615219" "439865" "572396" "711538" "656551" "291873" "662343" "281586" "475892" "453666"

Answer 3

Edit: I misread the question. Here's my answer for posterity.

---

You can use the rev function:

> 1:10
 [1]  1  2  3  4  5  6  7  8  9 10
> rev(1:10)
 [1] 10  9  8  7  6  5  4  3  2  1

Answer 4

There is function in stringi package for that - stri_reverse

require(stringi)
stri_reverse("123456")
## [1] "654321"

Now palindrome function might as simple as that:

palindrome <- function(x) stri_reverse(x)==x
palindrome(c("651156","1234321"))
## [1] TRUE  TRUE

Answer 5

This should work in the general case, with any choice of base:

is.palindromic <- function(x, base=10)
{
    p <- 0
    m <- floor(log(x,base))
    sig <- -1
    for (i in m:0)
        {
        tp <- floor(x/base^i)
        a <- i+1
        b <- m+1-i
        if(a==b){c<-0}else{c<-a*b;sig<-sig*-1}
        p <- p + tp*c*sig
        x <- x - tp*base^i
        }
    return(!as.logical(p))
}

Answer 6

If you are interested in the reversals for their own sake, you can use sub with a longer version of Greg's regexp:

> x
[1] 123321 343324 563660
> sub( '^([0-9])([0-9])([0-9])([0-9])([0-9])([0-9])','\\6\\5\\4\\3\\2\\1', x)
[1] "123321" "423343" "066365"

Although is this quicker than split/rev/paste?

Answer 7

Using intToUtf8 to split, then reverse:

tmp <- c(100001, 123321, 123456)

res <- sapply(tmp, function(i) intToUtf8(rev(utf8ToInt(as.character(i)))))

res
# [1] "100001" "123321" "654321"

To check if it is a palindrome:

tmp == res
# [1]  TRUE  TRUE FALSE

Benchmark

# bigger vector
tmpBIG <- rep(c(100001, 123321, 123456), 4000)

bench::mark(
  GregSnow = { grepl( '^([0-9])([0-9])([0-9])\\3\\2\\1$', tmpBIG) },
  bartektartanus = { tmpBIG == stringi::stri_reverse(tmpBIG) },
  Spacedman = { tmpBIG == sub( '^([0-9])([0-9])([0-9])([0-9])([0-9])([0-9])','\\6\\5\\4\\3\\2\\1', tmpBIG) },
  Joshua = { tmpBIG == sapply(strsplit(as.character(tmpBIG),""), function(i) paste(rev(i),collapse="")) },
  zx8754 = { tmpBIG == sapply(tmpBIG, function(i) intToUtf8(rev(utf8ToInt(as.character(i))))) },
  relative = TRUE)[, 1:9]

# expression         min median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time
# <bch:expr>       <dbl>  <dbl>     <dbl>     <dbl>    <dbl> <int> <dbl>   <bch:tm>
# 1 GregSnow        1      1         6.82      1         NaN    23     0      517ms
# 2 bartektartanus  1.38   1.34      5.02      2.33      NaN    17     0      520ms
# 3 Spacedman       1.58   1.55      4.52      2.33      NaN    15     0      509ms
# 4 Joshua          5.82   5.56      1.24      5.29      Inf     5     6      617ms
# 5 zx8754          6.06   6.17      1         3.98      Inf     4     4      614ms