Is there an easy way to convert a number to a letter?
For example,
3 => "C" and 23 => "W"?
-
two options - create an array of all the letters and get it's index+1 from an array. Second - use ascii table and subtract needed number.Mike B– Mike B2016-04-22 22:15:40 +00:00Commented Apr 22, 2016 at 22:15
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4 Answers
For simplicity range check is omitted from below solutions.
They all can be tried on the Go Playground.
Number -> rune
Simply add the number to the const 'A' - 1 so adding 1 to this you get 'A', adding 2 you get 'B' etc.:
func toChar(i int) rune {
return rune('A' - 1 + i)
}
Testing it:
for _, i := range []int{1, 2, 23, 26} {
fmt.Printf("%d %q\n", i, toChar(i))
}
Output:
1 'A'
2 'B'
23 'W'
26 'Z'
Number -> string
Or if you want it as a string:
func toCharStr(i int) string {
return string('A' - 1 + i)
}
Output:
1 "A"
2 "B"
23 "W"
26 "Z"
This last one (converting a number to string) is documented in the Spec: Conversions to and from a string type:
Converting a signed or unsigned integer value to a string type yields a string containing the UTF-8 representation of the integer.
Number -> string (cached)
If you need to do this a lot of times, it is profitable to store the strings in an array for example, and just return the string from that:
var arr = [...]string{"A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M",
"N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"}
func toCharStrArr(i int) string {
return arr[i-1]
}
Note: a slice (instead of the array) would also be fine.
Note #2: you may improve this if you add a dummy first character so you don't have to subtract 1 from i:
var arr = [...]string{".", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M",
"N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"}
func toCharStrArr(i int) string { return arr[i] }
Number -> string (slicing a string constant)
Also another interesting solution:
const abc = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
func toCharStrConst(i int) string {
return abc[i-1 : i]
}
Slicing a string is efficient: the new string will share the backing array (it can be done because strings are immutable).
3 Comments
afreeland
Might be a dumb question but how did you know that you could add a number to a constant character ( "A" ) to navigate the characters? Is this something in most languages, go specific, or rune specific? Never messed with rune() before...
icza
@afreeland
rune in Go is an alias for int32. If you have a constant like 'A', that is equivalent to a value 65. The types rune and int32 are one and the same, completely interchangable. So when you write 'A' + 1, it's the same as if you'd write 65 + 1.afreeland
Awesome, thanks for breaking it down a little further for me...I feel much better now =)
If you need not a rune, but a string and also more than one character for e.g. excel column
package main
import (
"fmt"
)
func IntToLetters(number int32) (letters string){
number--
if firstLetter := number/26; firstLetter >0{
letters += IntToLetters(firstLetter)
letters += string('A' + number%26)
} else {
letters += string('A' + number)
}
return
}
func main() {
fmt.Println(IntToLetters(1))// print A
fmt.Println(IntToLetters(26))// print Z
fmt.Println(IntToLetters(27))// print AA
fmt.Println(IntToLetters(1999))// print BXW
}
preview here: https://play.golang.org/p/GAWebM_QCKi
I made also package with this: https://github.com/arturwwl/gointtoletters
Comments
The simplest solution would be
func stringValueOf(i int) string {
var foo = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
return string(foo[i-1])
}
Hope this will help you to solve your problem. Happy Coding!!
answered Mar 18, 2018 at 15:35
Aravinda Meewalaarachchi
2,6691 gold badge31 silver badges27 bronze badges
Comments
If you want to represent a number bigger than 26(Z), this is the simplest solution. Give a try.
func convertToAlphabetic(n int) string {
result := ""
for n > 0 {
mod := (n - 1) % 26
result = string('A'+mod) + result
n = (n - mod) / 26
}
return result
}
Example results:
- 3: C
- 32: AF
- 876: AGR
