How many significant figures?

Question

Given a number as input, determine how many significant figures it has. This number will should be taken as a string because you have to do some special formatting. You'll see what I mean soon (I think).

A digit is a sig-fig if at least one of the following apply:

• Non-zero digits are always significant.
• Any zeros between two significant digits are significant.
• final zero or trailing zeros in the decimal portion only are significant.
• all digits are significant if nothing follows the decimal place.
• when there are only zeroes, all but the last zero are considered leading zeroes

Input

A string or string array of the number. It might have a decimal point at the end without a digit after it. It might not have a decimal point at all.

Output

How many sig-figs there are.

Examples

1.240 -> 4
0. -> 1
83900 -> 3
83900.0 -> 6
0.025 -> 2
0.0250 -> 3
2.5 -> 2
970. -> 3
0.00 -> 1

Answer 1

05AB1E, 11 10 bytes

D.ïi0Ü}þïg

Try it online! or as a Test suite

Explanation

D            # duplicate
 .ïi  }      # if it is an integer
    0Ü       # remove trailing zeroes
       þ     # remove the "." if present
        ï    # convert to int
         g   # length

Answer 2

Retina, 29 27 bytes

Saved 2 bytes thanks to @MartinEnder

^0\.0*.|^\d(\d*?)0+$
1$1
\d

Try it online! | Test suite

Answer 3

Batch, 204 202 bytes

@set/ps=
:t
@if %s:.=%%s:~-1%==%s%0 set s=%s:~,-1%&goto t
@set s=%s:.=%
:l
@if not %s%==0 if %s:~,1%==0 set s=%s:~1%&goto l
@set n=0
:n
@if not "%s%"=="" set/an+=1&set s=%s:~1%&goto n
@echo %n%

Takes input on STDIN. Works by removing trailing zeros if the number does not contain a ., then removing the . and leading zeros, unless there are only zeros, in which case it leaves one zero. Finally it takes the length of the remaining string.

Answer 4

Scala, 90 bytes

& =>(if(&contains 46)&filter(46!=)else&.reverse dropWhile(48==)reverse)dropWhile(48==)size

Explanation:

& =>               //define an anonymous function with a parameter called &
  (
  if(&contains 46) //if & contains a '.'
    &filter(46!=)    //remove all periods
  else             //else
    &.reverse        //reverse the string
    dropWhile(48==)  //remove leading zeros
    reverse          //and reverse again
  )
  dropWhile(48==) //remove leading zeros
  size            //return the length

Answer 5

C#6, 163 bytes

using System.Linq;
int a(string s)=>System.Math.Max(s.Contains('.')?(s[0]<'1'?s.SkipWhile(x=>x<'1').Count():s.Length-1):s.Reverse().SkipWhile(x=>x<'1').Count(),1);

Ungolfed

int a(string s)=>                                  
    System.Math.Max(
        s.Contains('.')                                // Has decimal place?
            ?(                                         // Has decimal place!
                s[0]<'1'                               // Start with '0' or '.'?
                    ?s.SkipWhile(x=>x<'1').Count()     // Yes: Skip leading '0' and '.', then count rest... But gives 0 for "0." and "0.00"
                    :s.Length-1)                       // No: Count length without decimal place
            :s.Reverse().SkipWhile(x=>x<'1').Count()   // No decimal place: Skip trailing '0' and count rest
    ,1);                                               // For "0." and "0.00"