Convert decimals to terniary (base3)-CodePudding

I was wondering if there was a way of converting decimal to terniary, since there are functions to binary/octal as intToBin(x).

So i need to convert a chacarcer string as

S0 <- c("Hello Stac")

To base3, so I thought to first convert It to ASCII with:

S01<- utf8ToInt(S0)
> S01
 [1]  72 101 108 108 111  32  83 116  97  99

And then convert It to base3 and obtain something like this:

> S1
[1] 2200 10202 11000 11010  11022 1012 10002 11022 10121 10200

CodePudding user response：

For practice, I guess you can try to write your own converter function like below

f <- function(x, base = 3) {
  q <- c()
  while (x) {
    q <- c(x %% base, q)
    x <- x %/% base
  }
  # as.numeric(paste0(q, collapse = ""))
  sum(q * 10^(rev(seq_along(q) - 1)))
}

or with recursion

f <- function(x, base = 3) {
  ifelse(x < base, x, f(x %/% base) * 10   x %% base)
}

then you can run

> sapply(utf8ToInt(S0),f)
 [1]  2200 10202 11000 11000 11010  1012 10002 11022 10121 10200

CodePudding user response：

You can use cwhmisc::int2B:

library(cwhmisc)
int2B(utf8ToInt(S0), 3)[[1]] |> as.numeric()
# [1]  2200 10202 11000 11000 11010  1012 10002 11022 10121 10200

CodePudding user response：

Nice programming exercise. I have vectorized @ThomasIsCoding's answer and generalized to support any number of strings and any base from 2 to 10.

The function below takes as arguments a character vector x, a base b, and a logical flag double. It returns a list res such that res[[i]] is an nchar(x[i])-length vector giving the base-b representation of x[i]. The list elements are double vectors or character vectors depending on double.

Some caveats:

For efficiency, the function concatenates the strings in x rather than looping over them. It throws an error if the concatenation would exceed 2^31-1 bytes, which is the maximum string size allowed by R.
The largest Unicode code point is "\U10FFFF". The binary representation of this number exceeds 2^53 when interpreted as decimal, so it cannot be stored in a double vector without loss of precision:
```
x <- sub("^0 ", "", paste(rev(as.integer(intToBits(0x10FFFF))), collapse = ""))
x
## [1] "100001111111111111111"
sprintf("%.0f", as.double(x))
## [1] "100001111111111114752"
```
As a defensive measure, the function warns if 2^53 is exceeded when base = 2 and double = TRUE.
The base = 2 and base = 10 cases are already handled by base R's utf8ToInt and intToBits.

utf8ToBase <- function(x, b = 10, double = TRUE) {
    ## Do some basic checks
    stopifnot(is.character(x), !anyNA(x), 
              is.numeric(b), length(b) == 1L, 
              b %% 1 == 0, b >= 2, b <= 10)
    if (length(x) == 0L) {
        return(list())
    }

    ## Require UTF-8 encoding
    x <- enc2utf8(x)
    
    ## Operate on concatenation to avoid loop over strings
    xx <- paste(x, collapse = "")
    ixx <- utf8ToInt(xx)
     
    ## Use common field width determined from greatest integer
    width <- as.integer(ceiling(log(max(ixx), base = b)))
    res <- rep.int(strrep("0", width), length(ixx))
    
    ## Loop over digits
    pos <- 1L
    pow <- b^(width - 1L)
    while (pos <= width) {
        quo <- ixx %/% pow
        substr(res, pos, pos) <- as.character(quo)
        ixx <- ixx - pow * quo
        pos <- pos   1L
        pow <- pow %/% b
    }
    
    ## Discard leading zeros
    if (double) {
        res <- as.double(res)
        if (b == 2 && any(res > 2^53)) {
            warning("binary result not guaranteed due to loss of precision")
        }
    } else {
        res <- sub("^0 ", "", res)
    }
    
    ## Return list
    res <- split(res, rep.int(gl(length(x), 1L), nchar(x)))
    names(res) <- names(x)
    res
}

x <- c(foo = "Hello Stack Overflow!", bar = "Hello world!")
utf8ToBase(x, 2)

$foo
 [1] 1001000 1100101 1101100 1101100 1101111  100000
 [7] 1010011 1110100 1100001 1100011 1101011  100000
[13] 1001111 1110110 1100101 1110010 1100110 1101100
[19] 1101111 1110111  100001

$bar
 [1] 1001000 1100101 1101100 1101100 1101111  100000
 [7] 1110111 1101111 1110010 1101100 1100100  100001

utf8ToBase(x, 3)

$foo
 [1]  2200 10202 11000 11000 11010  1012 10002 11022 10121 10200
[11] 10222  1012  2221 11101 10202 11020 10210 11000 11010 11102
[21]  1020

$bar
 [1]  2200 10202 11000 11000 11010  1012 11102 11010 11020 11000
[11] 10201  1020

utf8ToBase(x, 10)

$foo
 [1]  72 101 108 108 111  32  83 116  97  99 107  32  79 118 101
[16] 114 102 108 111 119  33

$bar
 [1]  72 101 108 108 111  32 119 111 114 108 100  33

utf8ToBase("\U10FFFF", 2)

[[1]]
[1] 1.000011e 20

Warning message:
In str_to_num("\U{10ffff}", 2) :
  binary result not guaranteed due to loss of precision

utf8ToBase("\U10FFFF", 2, double = FALSE)

[[1]]
[1] "100001111111111111111"