Finding the number of digits of an integer

问题:

What is the best method to find the number of digits of a positive integer?

I have found this 3 basic methods:

• conversion to string

  String s = new Integer(t).toString(); 
  int len = s.length();
  

• for loop

  for(long long int temp = number; temp >= 1;)
  {
      temp/=10;
      decimalPlaces++;
  } 
  

• logaritmic calculation

  digits = floor( log10( number ) ) + 1;
  

where you can calculate log10(x) = ln(x) / ln(10) in most languages.

First I thought the string method is the dirtiest one but the more I think about it the more I think it's the fastest way. Or is it?

回答1:

There's always this method:

n = 1;
if ( i >= 100000000 ) { n += 8; i /= 100000000; }
if ( i >= 10000     ) { n += 4; i /= 10000; }
if ( i >= 100       ) { n += 2; i /= 100; }
if ( i >= 10        ) { n += 1; }

回答2:

I don't know, and the answer may well be different depending on how your individual language is implemented.

So, stress test it! Implement all three solutions. Run them on 1 through 1,000,000 (or some other huge set of numbers that's representative of the numbers the solution will be running against) and time how long each of them takes.

Pit your solutions against one another and let them fight it out. Like intellectual gladiators. Three algorithms enter! One algorithm leaves!

回答3:

Well the correct answer would be to measure it - but you should be able to make a guess about the number of CPU steps involved in converting strings and going through them looking for an end marker

Then think how many FPU operations/s your processor can do and how easy it is to calculate a single log.

edit: wasting some more time on a monday morning :-)

String s = new Integer(t).toString(); 
int len = s.length();

One of the problems with high level languages is guessing how much work the system is doing behind the scenes of an apparently simple statement. Mandatory Joel link

This statement involves allocating memory for a string, and possibly a couple of temporary copies of a string. It must parse the integer and copy the digits of it into a string, possibly having to reallocate and move the existing memory if the number is large. It might have to check a bunch of locale settings to decide if your country uses "," or ".", it might have to do a bunch of unicode conversions.
Then finding the length has to scan the entire string, again considering unicode and any local specific settings such as - are you in a right->left language?.

Alternatively:

digits = floor( log10( number ) ) + 1;

Just because this would be harder for you to do on paper doesn't mean it's hard for a computer! In fact a good rule in high performance computing seems to have been - if something is hard for a human (fluid dynamics, 3d rendering) it's easy for a computer, and if it's easy for a human (face recognition, detecting a voice in a noisy room) it's hard for a computer!

You can generally assume that the builtin maths functions log/sin/cos etc - have been an important part of computer design for 50years. So even if they don't map directly into a hardware function in the FPU you can bet that the alternative implementation is pretty efficient.

回答4:

This algorithm might be good also, assuming that:

• Number is integer and binary encoded (<< operation is cheap)

• We don't known number boundaries

  var num = 123456789L;
  var len = 0;
  var tmp = 1L;
  while(tmp < num)
  {
      len++;
      tmp = (tmp << 3) + (tmp << 1);
  }
  

This algorithm, should have speed comparable to for-loop (2) provided, but a bit faster due to (2 bit-shifts, add and subtract, instead of division).

As for Log10 algorithm, it will give you only approximate answer (that is close to real, but still), since analytic formula for computing Log function have infinite loop and can't be calculated precisely Wiki.

回答5:

Test conditions

• Decimal numeral system
• Positive integers
• Up to 10 digits
• Language: ActionScript 3

Results

digits: [1,10],

no. of runs: 1,000,000

random sample: 8777509,40442298,477894,329950,513,91751410,313,3159,131309,2

result: 7,8,6,6,3,8,3,4,6,1

CONVERSION TO STRING: 724ms

LOGARITMIC CALCULATION: 349ms

DIV 10 ITERATION: 229ms

MANUAL CONDITIONING: 136ms

Note: Author refrains from making any conclusions for numbers with more than 10 digits.

---

Script

package {
    import flash.display.MovieClip;
    import flash.utils.getTimer;
    /**
     * @author Daniel
     */
    public class Digits extends MovieClip {
        private const NUMBERS : uint = 1000000;
        private const DIGITS : uint = 10;

        private var numbers : Array;
        private var digits : Array;

        public function Digits() {
            // ************* NUMBERS *************
            numbers = [];
            for (var i : int = 0; i < NUMBERS; i++) {
                var number : Number = Math.floor(Math.pow(10, Math.random()*DIGITS));
                numbers.push(number);
            }   
            trace('Max digits: ' + DIGITS + ', count of numbers: ' + NUMBERS);
            trace('sample: ' + numbers.slice(0, 10));

            // ************* CONVERSION TO STRING *************
            digits = [];
            var time : Number = getTimer();

            for (var i : int = 0; i < numbers.length; i++) {
                digits.push(String(numbers[i]).length);
            }

            trace('\nCONVERSION TO STRING - time: ' + (getTimer() - time));
            trace('sample: ' + digits.slice(0, 10));

            // ************* LOGARITMIC CALCULATION *************
            digits = [];
            time = getTimer();

            for (var i : int = 0; i < numbers.length; i++) {
                digits.push(Math.floor( Math.log( numbers[i] ) / Math.log(10) ) + 1);
            }

            trace('\nLOGARITMIC CALCULATION - time: ' + (getTimer() - time));
            trace('sample: ' + digits.slice(0, 10));

            // ************* DIV 10 ITERATION *************
            digits = [];
            time = getTimer();

            var digit : uint = 0;
            for (var i : int = 0; i < numbers.length; i++) {
                digit = 0;
                for(var temp : Number = numbers[i]; temp >= 1;)
                {
                    temp/=10;
                    digit++;
                } 
                digits.push(digit);
            }

            trace('\nDIV 10 ITERATION - time: ' + (getTimer() - time));
            trace('sample: ' + digits.slice(0, 10));

            // ************* MANUAL CONDITIONING *************
            digits = [];
            time = getTimer();

            var digit : uint;
            for (var i : int = 0; i < numbers.length; i++) {
                var number : Number = numbers[i];
                if (number < 10) digit = 1;
                else if (number < 100) digit = 2;  
                else if (number < 1000) digit = 3;  
                else if (number < 10000) digit = 4;  
                else if (number < 100000) digit = 5;  
                else if (number < 1000000) digit = 6;  
                else if (number < 10000000) digit = 7;  
                else if (number < 100000000) digit = 8;  
                else if (number < 1000000000) digit = 9;  
                else if (number < 10000000000) digit = 10;  
                digits.push(digit);
            }

            trace('\nMANUAL CONDITIONING: ' + (getTimer() - time));
            trace('sample: ' + digits.slice(0, 10));
        }
    }
}

回答6:

• conversion to string: This will have to iterate through each digit, find the character that maps to the current digit, add a character to a collection of characters. Then get the length of the resulting String object. Will run in O(n) for n=#digits.

• for-loop: will perform 2 mathematical operation: dividing the number by 10 and incrementing a counter. Will run in O(n) for n=#digits.

• logarithmic: Will call log10 and floor, and add 1. Looks like O(1) but I'm not really sure how fast the log10 or floor functions are. My knowledge of this sort of things has atrophied with lack of use so there could be hidden complexity in these functions.

So I guess it comes down to: is looking up digit mappings faster than multiple mathematical operations or whatever is happening in log10? The answer will probably vary. There could be platforms where the character mapping is faster, and others where doing the calculations is faster. Also to keep in mind is that the first method will creats a new String object that only exists for the purpose of getting the length. This will probably use more memory than the other two methods, but it may or may not matter.

回答7:

You can obviously eliminate the method 1 from the competition, because the atoi/toString algorithm it uses would be similar to method 2.

Method 3's speed depends on whether the code is being compiled for a system whose instruction set includes log base 10.

回答8:

Use the simplest solution in whatever programming language you're using. I can't think of a case where counting digits in an integer would be the bottleneck in any (useful) program.

C, C++:

char buffer[32];
int length = sprintf(buffer, "%ld", (long)123456789);

Haskell:

len = (length . show) 123456789

JavaScript:

length = String(123456789).length;

PHP:

$length = strlen(123456789);

Visual Basic (untested):

length = Len(str(123456789)) - 1

回答9:

import math
def numdigits(n):
    return ( int(math.floor(math.log10(n))) + 1 )

回答10:

Keep it simple:

long long int a = 223452355415634664;

int x;
for (x = 1; a >= 10; x++)
{
   a = a / 10;
}

printf("%d", x);

回答11:

For very large integers, the log method is much faster. For instance, with a 2491327 digit number (the 11920928th Fibonacci number, if you care), Python takes several minutes to execute the divide-by-10 algorithm, and milliseconds to execute 1+floor(log(n,10)).

回答12:

You can use a recursive solution instead of a loop, but somehow similar:

@tailrec
def digits (i: Long, carry: Int=1) : Int =  if (i < 10) carry else digits (i/10, carry+1)

digits (8345012978643L)

With longs, the picture might change - measure small and long numbers independently against different algorithms, and pick the appropriate one, depending on your typical input. :)

Of course nothing beats a switch:

switch (x) {
  case 0:  case 1:  case 2:  case 3:  case 4:  case 5:  case 6:  case 7:  case 8:  case 9: return 1;
  case 10: case 11: // ...
  case 99: return 2;
  case 100: // you get the point :) 
  default: return 10; // switch only over int
}

except a plain-o-array:

   int [] size = {1,1,1,1,1,1,1,1,1,2,2,2,2,2,... };
   int x = 234561798;
   return size [x];

Some people will tell you to optimize the code-size, but yaknow, premature optimization ...

回答13:

Regarding the three methods you propose for "determining the number of digits necessary to represent a given number in a given base", I don't like any of them, actually; I prefer the method I give below instead.

Re your method #1 (strings): Anything involving converting back-and-forth between strings and numbers is usually very slow.

Re your method #2 (temp/=10): This is fatally flawed because it assumes that x/10 always means "x divided by 10". But in many programming languages (eg: C, C++), if "x" is an integer type, then "x/10" means "integer division", which isn't the same thing as floating-point division, and it introduces round-off errors at every iteration, and they accumulate in a recursive formula such as your solution #2 uses.

Re your method #3 (logs): it's buggy for large numbers (at least in C, and probably other languages as well), because floating-point data types tend not to be as precise as 64-bit integers.

Hence I dislike all 3 of those methods: #1 works but is slow, #2 is broken, and #3 is buggy for large numbers. Instead, I prefer this:

unsigned NumberOfDigits (uint64_t Number, unsigned Base)
{
   unsigned Digits = 1;
   uint64_t Power  = 1;
   while ( Number / Power >=  Base )
   {
      ++Digits;
      Power *= Base;
   }
   return Digits;
}

回答14:

I was curious after seeing @daniel.sedlacek results so I did some testing using Swift for numbers having more than 10 digits. I ran the following script in the playground.

let base = [Double(100090000000), Double(100050000), Double(100050000), Double(100000200)]
var rar = [Double]()
for i in 1...10 {
    for d in base {
        let v = d*Double(arc4random_uniform(UInt32(1000000000)))
        rar.append(v*Double(arc4random_uniform(UInt32(1000000000))))
        rar.append(Double(1)*pow(1,Double(i)))
    }
}

print(rar)

var timeInterval = NSDate().timeIntervalSince1970

for d in rar {
    floor(log10(d))
}

var newTimeInterval = NSDate().timeIntervalSince1970

print(newTimeInterval-timeInterval)


timeInterval = NSDate().timeIntervalSince1970
for d in rar {
    var c = d
    while c > 10 {
        c = c/10
    }
}

newTimeInterval = NSDate().timeIntervalSince1970

print(newTimeInterval-timeInterval)

Results of 80 elements

0.105069875717163 for floor(log10(x))
0.867973804473877 for div 10 iterations

回答15:

log(x,n)-mod(log(x,n),1)+1

Where x is a the base and n is the number.

回答16:

Here is the measurement in Swift 4.

Algorithms code:

extension Int {
    var numberOfDigits0: Int {
        var currentNumber = self
        var n = 1
        if (currentNumber >= 100000000) {
            n += 8
            currentNumber /= 100000000
        }
        if (currentNumber >= 10000) {
            n += 4
            currentNumber /= 10000
        }
        if (currentNumber >= 100) {
            n += 2
            currentNumber /= 100
        }
        if (currentNumber >= 10) {
            n += 1
        }
        return n
    }

    var numberOfDigits1: Int {
        return String(self).count
    }

    var numberOfDigits2: Int {
        var n = 1
        var currentNumber = self
        while currentNumber > 9 {
            n += 1
            currentNumber /= 10
        }
        return n
    }

}

Measurement code:

var timeInterval0 = Date()
for i in 0...10000 {
    i.numberOfDigits0
}
print("timeInterval0: \(Date().timeIntervalSince(timeInterval0))")

var timeInterval1 = Date()
for i in 0...10000 {
    i.numberOfDigits1
}
print("timeInterval1: \(Date().timeIntervalSince(timeInterval1))")

var timeInterval2 = Date()
for i in 0...10000 {
    i.numberOfDigits2
}
print("timeInterval2: \(Date().timeIntervalSince(timeInterval2))")

Output

timeInterval0: 1.92149806022644

timeInterval1: 0.557608008384705

timeInterval2: 2.83262193202972

On this measurement basis String conversion is the best option for the Swift language.