Fixed and decimal places of a positive number - r

Question

I want to write a function for counting the number of fixed digits and the number of decimal digits of each positive number. For whole numbers: for example, for number 459 I would like to see
fixed=3 and decimal=0
as the output of the function.
For decimal numbers: for example, for number 12.657 I would like to see
fixed=2 and decimal=3
(because 12 has two digits and 657 has three digits). For numbers less than 1 I would like to see
0
as the fixed, for example for number 0.4056 I would like to see
fixed=0 and decimal=4
or for number 0.13 I would like to see
fixed=0 and decimal=2
I have a general idea as follows:
digits<-function(x){
length(strsplit(as.character(x),split="")[[1]])
}
I want to extend my code as a new one to work as I explained above.

Answer 1

It seems you kind of stuck hoping that this can be done. Okay here is a crude way:
fun <- function(x){
stopifnot(is.numeric(as.numeric(x)))
s = nchar(unlist(strsplit(as.character(x),".",fixed = TRUE)))
if(as.numeric(x) < 1) s[1] <- s[1]-1
setNames(as.list(s),c("fixed","decimal"))
}
CORRECT:
fun(10.234)
$fixed
[1] 2
$decimal
[1] 3
fun(-10.234)
$fixed
[1] 2
$decimal
[1] 3
fun(0.2346)
$fixed
[1] 0
$decimal
[1] 4
> fun(-0.2346)
$fixed
[1] 0
$decimal
[1] 4
INCORRECT: Note that fixed + decimal <=15!!!
fun(-10000567.2346678901876)
$fixed
[1] 8
$decimal
[1] 7 ## This is incorrect
The correct value is:
fun("-10000567.2346678901876") # Note that the input x is a string
$fixed
[1] 8
$decimal
[1] 13

Answer 2

I do not think this can be done. We cannot assume that a simple numeric value is accurately represented in the computer. Most floating-point values certainly can't.
Enter 0.3 at the R console:
> 0.3
[1] 0.3
Looks alright, doesn't it? But now, let us do this:
> print(0.3, digits=22)
[1] 0.29999999999999999
In essence, if you convert a floating-point number to a string you define how precise you want it. The computer cannot give you that precision because it stores all numbers in bits and, thus, never gives you absolute precision. Even if you see the number as 0.3 and you think it has 0 fixed digits and 1 decimal because of that, this is because R chose to print it that way, not because that is the number represented in the computer's memory.
The other answers prove that a function can handle simple cases. I must admit that I learned that R does an incredible job interpreting the numbers. But we have to use caution! As soon as we start transforming numbers, such function cannot guarantuee meaningful results.

Answer 3

EDITED (based on comments):
The below function will work for numbers with 15 or less significant figures:
digits<-function(x){
## make sure x is a number, if not stop
tryCatch(
{
x/1
},
error=function(cond) {
stop('x must be a number')
}
)
## use nchar() and strsplit() to determine number of digits before decimal point
fixed<-nchar(strsplit(as.character(x),"\\.")[[1]][1])
## check if negative
if(substr(strsplit(as.character(x),"\\.")[[1]][1],1,1)=="-"){fixed<-fixed-1}
## check if -1<x<1
if(as.numeric(strsplit(as.character(x),"\\.")[[1]][1])==0){fixed<-fixed-1}
## use nchar() and strsplit() to determine number of digits after decimal point
decimal<-nchar(strsplit(as.character(x),"\\.")[[1]][2])
## for integers, replace NA decimal result with 0
if(is.na(nchar(strsplit(as.character(x),"\\.")[[1]][2]))){decimal<-0}
## return results
print(paste0("fixed: ",fixed," and decimal: ", decimal))
}
If you want to count negative signs (-) as a digit you'll need to remove:
## check if negative
if(substr(strsplit(as.character(x),"\\.")[[1]][1],1,1)=="-"){fixed<-fixed-1}
Note: The 15 or less significant figures restriction is based on 8 byte floating point representation.
The only way I've been able to overcome this is by using the below code.
digits<-function(x){
tryCatch(
{
as.numeric(x)/1
},
error=function(cond) {
stop('x must be a number')
}
)
j <- 0
num <- x
fixed <- 0
decimal <- 0
for(i in 1:nchar(x)){
if(substr(num, i, i) == "."){
j <- 1
} else if(j==0){
fixed <- fixed + 1
} else{
decimal <- decimal + 1
}
}
if(substr(x,1,1)=="-" & substr(as.numeric(x),2,2)==0){
fixed<-fixed-2
}else if(substr(x,1,1)=="-"){
fixed<-fixed-1
}else if(substr(as.numeric(x),1,1)==0){
fixed<-fixed-1
}else{}
print(paste0("fixed: ",fixed," and decimal: ", decimal))
}
However, this code requires the number to be passed into the function as a character string in order to compute passed 15 significant figures, such as is shown below:
x<-"111111111111111111111111111111111111.22222222222222222222222222222222222222222222222"
digits(x)
Which returns:
[1] "fixed: 36 and decimal: 47"
This then limits this function's applications as it cannot be used in a dplyr pipe, and you will not receive accurate results for numbers that have been previously rounded. For example, using the same number as above, but storing it as a number instead of character string yields the below results:
x<-111111111111111111111111111111111111.22222222222222222222222222222222222222222222222
digits(x)
[1] "fixed: 1 and decimal: 18"
I hope this at least somewhat helps!