Round the string

Question

Some decimal numbers cannot be precisely represented as binary floats due to the internal representation of the binary floats. For example: rounding 14.225 to two decimal digits does not result in 14.23 as one might expect but in 14.22.

Python:

In: round(14.225, 2)
Out: 14.22

Assume, however, that we have a string representation of 14.225 as '14.225', we should be able to achieve our desired rounding '14.23' as a string representation.

This approach can be generalized to arbitrary precision.

Possible Python 2/3 Solution

import sys

def round_string(string, precision):
    assert(int(precision) >= 0)
    float(string)

    decimal_point = string.find('.')
    if decimal_point == -1:
        if precision == 0:
            return string
        return string + '.' + '0' * precision

    all_decimals = string[decimal_point+1:]
    nb_missing_decimals = precision - len(all_decimals)
    if nb_missing_decimals >= 0:
        if precision == 0:
            return string[:decimal_point]
        return string + '0' * nb_missing_decimals

    if int(all_decimals[precision]) < 5:
        if precision == 0:
            return string[:decimal_point]
        return string[:decimal_point+precision+1]

    sign = '-' if string[0] == '-' else '' 
    integer_part = abs(int(string[:decimal_point]))
    if precision == 0:
        return sign + str(integer_part + 1)
    decimals = str(int(all_decimals[:precision]) + 1)
    nb_missing_decimals = precision - len(decimals)
    if nb_missing_decimals >= 0:
        return sign + str(integer_part) + '.' + '0' * nb_missing_decimals + decimals
    return sign + str(integer_part + 1) + '.' + '0' * precision

Try it online!

Usage:

     # No IEEE 754 format rounding
In:  round_string('14.225',2)
Out: '14.23'

     # Trailing zeros
In:  round_string('123.4',5)
Out: '123.40000'

In: round_string('99.9',0)
Out: '100'

    # Negative values
In: round_string('-99.9',0)
Out: '-100'

In: round_string('1',0)
Out: '1'

    # No unnecessary decimal point
In: round_string('1.',0)
Out: '1'

    # No unnecessary decimal point
In: round_string('1.0',0)
Out: '1'

In:  for i in range(8): 
         print(round_string('123456789.987654321',i))
Out: 123456790
     123456790.0
     123456789.99
     123456789.988
     123456789.9877
     123456789.98765
     123456789.987654
     123456789.9876543

Task

Input argument 1: a string containing

• at least one digit (0, 1, 2, 3, 4, 5,6, 7, 8, 9),
• at most one decimal point (.) which must be preceded by at least one digit,
• an optional minus (-) as first character.

Input argument 2: a non-negative integer

Output: the correctly rounded (base 10) string

rounding = Round half away from zero

This is a code-golf. The lowest number of bytes wins!

Answer 1

APL (Dyalog), 4 bytes

Dyalog APL uses enough internal precision.

⎕⍕⍎⍞

Try it online!

⍎⍞ execute string input

⎕⍕ get numeric input and use that as precision for formatting

Answer 2

Perl, 22 20 bytes

printf"%.*f",pop,pop

Using:

perl -e 'printf"%.*f",pop,pop' 123456789.987654321 3

It is Dada’s version of code. Previous:

printf"%*2\$.*f",@ARGV

Answer 3

PHP, 33 31 bytes

PHP rounds correctly too (at least on 64 bit):

printf("%.$argv[2]f",$argv[1]);

takes input from command line arguments. Run with -r.

PHP, no built-ins, 133 bytes

[,$n,$r]=$argv;if($p=strpos(_.$n,46))for($d=$n[$p+=$r],$n=substr($n,0,$p-!$r);$d>4;$n[$p]=(5+$d=$n[$p]-4)%10)$p-=$n[--$p]<"/";echo$n;

Run with -nr or test it online.

breakdown

[,$n,$r]=$argv;             // import arguments
if($p=strpos(_.$n,46))      // if number contains dot
    for($d=$n[$p+=$r],          // 1. $d= ($r+1)-th decimal 
        $n=substr($n,0,$p-!$r); // 2. cut everything behind $r-th decimal
        $d>4;                   // 3. loop while previous decimal needs increment
        $n[$p]=(5+$d=$n[$p]-4)%10   // B. $d=current digit-4, increment current digit
    )
        $p-=$n[--$p]<"/";           // A. move cursor left, skip dot
echo$n;

A null byte doesn´t work; so I have to use substr.

Answer 4

Python, 114 105 103 96 91 89 bytes

Saved 5 bytes thanks to Kevin Cruijssen
Saved 2 bytes thanks to Krazor

from decimal import*
d=Decimal
lambda x,y:d(x).quantize(d('0.'[y>0]+'1'*y),ROUND_HALF_UP)

Try it online!

Answer 5

REXX, 24 bytes

arg n p
say format(n,,p)

Since REXX always uses text representation of numbers, correct rounding of numbers is free.

Try it online!

Answer 6

Ruby 2.3, 12 + 45 = 57

Uses the BigDecimal built-in, but it needs to be required before use, which is cheaper to do as a flag.

the flag: -rbigdecimal

the function:

->(s,i){BigDecimal.new(s).round(i).to_s('f')}

Ruby 2.3 by default uses ROUND_HALF_UP

Answer 7

Javascript (ES6), 44 bytes

n=>p=>(Math.round(n*10**p)/10**p).toFixed(p)

Try it online:

const f = n=>p=>(Math.round(n*10**p)/10**p).toFixed(p)

console.log(f('14.225')(2));

[...Array(8).keys()].map(i=>console.log(f('123456789.987654321')(i)))

console.log(f('123.4')(5))

Answer 8

TI-Basic, 53 16 bytes

TI-Basic does not use IEEE and the below method works for 0-9 (inclusive) decimal positions.

Prompt Str1,N
toString(round(expr(Str1),N

Thanks to @JulianLachniet for showing that CE calcs have the toString( command which I was not aware of (Color Edition calcs OS 5.2 or higher are required).

P.S. I did have a second line with sub(Str1,1,N+inString(Str1,". but then I realized it was useless.

Answer 9

Java 7, 77 72 71 bytes

<T>T c(T n,int d){return(T)"".format("%."+d+"f",new Double(n+""));}

-1 byte thanks to @cliffroot

72-byte answer:

String c(String n,int d){return n.format("%."+d+"f",new Double(n));}

Unlike Python, Java already rounds correctly and already returns a String when you use String.format("%.2f", aDouble) with the 2 replaced with the amount of decimals you want.

EDIT/NOTE: Yes, I'm aware new Float(n) is 1 byte shorter than new Double(n), but apparently it fails for the test cases with 123456789.987654321. See this test code regarding Double vs Float.

Explanation:

<T> T c(T n, int d){               // Method with generic-T & integer parameters and generic-T return-type (generic-T will be String in this case)
  return (T)"".format("%."+d+"f",  //  Return the correctly rounded output as String
    new Double(n+""));             //  After we've converted the input String to a decimal
}                                  // End of method

Test code:

Try it here.

class M{
  static <T>T c(T n,int d){return(T)"".format("%."+d+"f",new Double(n+""));}

  public static void main(String[] a){
    System.out.println(c("14.225", 2));
    System.out.println(c("123.4", 5));
    System.out.println(c("99.9", 0));
    System.out.println(c("-99.9", 0));
    System.out.println(c("1", 0));
    System.out.println(c("1.", 0));
    System.out.println(c("1.0", 0));
    for(int i = 0; i < 8; i++){
      System.out.println(c("123456789.987654321", i));
    }
  }
}

Output:

14.23
123.40000
100
-100
1
1
1
123456790
123456790.0
123456789.99
123456789.988
123456789.9877
123456789.98765
123456789.987654
123456789.9876543

Answer 10

BASH, 26 23 21 bytes

bc<<<"scale=$2;$1/1"

usage

save to round_string.sh, chmod +x round_string.sh

./round_string.sh 23456789.987654321 3

edit:no need to load library

Answer 11

AHK, 25 bytes

a=%1%
Send % Round(a,%2%)

Again I am foiled by the inability of AHK to use passed in parameters directly in functions that accept either a variable name or a number. If I replace a with 1 in the Round function, it uses the value 1. If I try %1%, it tries to use the first argument's contents as a variable name, which doesn't work. Having to set it as another variable first cost me 6 bytes.

Answer 12

Batch, 390 bytes

@echo off
set s=%1
set m=
if %s:~,1%==- set m=-&set s=%s:~1%
set d=%s:*.=%
if %d%==%s% (set d=)else call set s=%%s:.%d%=%%
for /l %%i in (0,1,%2)do call set d=%%d%%0
call set/ac=%%d:~%2,1%%/5
call set d=00%s%%%d:~,%2%%
set z=
:l
set/ac+=%d:~-1%
set d=%d:~,-1%
if %c%==10 set c=1&set z=%z%0&goto l
set d=%m%%d:~2%%c%%z%
if %2==0 (echo %d%)else call echo %%d:~,-%2%%.%%d:~-%2%%

Explanation. Starts by extracting the sign, if applicable. Then, splits the number into integer and fraction digits. The fraction is padded with n+1 zeros to ensure it has more than n digits. The nth (zero-indexed) digit is divided by 5, and this is the initial carry. The integer and n fraction digits are concatenated, and the carry added character by character. (The extra zeros guard against carry ripple.) After the carry stops rippling the number is reconstructed and any decimal point inserted.

Answer 13

Python (2/3), 394 bytes

def rnd(s,p):
    m=s[0]=='-'and'-'or''
    if m:s=s[1:]
    d=s.find('.')
    l=len(s)
    if d<0:
        if p>0:d=l;l+=1;s+='.'
        else:return m+s
    e=(d+p+1)-l
    if e>0:return m+s+'0'*e
    o=''
    c=0
    for i in range(l-1,-1,-1):
        x=s[i]
        if i<=d+p:
            if i!=d:
                n=int(x)+c
                if n>9:n=0;c=1 
                else:c=0
                o+=str(n)
            else:
                if p>0:o+=x
        if i==d+p+1:c=int(x)>4
    if c:o+='1'
    return m+''.join(reversed(o))

Works for arbitrary precision numbers.

Answer 14

JavaScript (ES6), 155 bytes

(s,n)=>s.replace(/(-?\d+).?(.*)/,(m,i,d)=>i+'.'+(d+'0'.repeat(++n)).slice(0,n)).replace(/([0-8]?)([.9]*?)\.?(.)$/,(m,n,c,r)=>r>4?-~n+c.replace(/9/g,0):n+c)

Explanation: The string is first normalised to contain a . and n+1 decimal digits. The trailing digit, any preceding 9s or .s, and any preceding digit, are then considered. If the last digit is less than 5 then it and any immediately preceding . are simply removed but if it is greater than 5 then the 9s are changed to 0s and the previous digit incremented (or 1 prefixed if there was no previous digit).

Answer 15

Python 3 + SymPy, 54 bytes

from sympy import*
lambda s,p:'%.*f'%(p,S(s).round(p))

Try it online!

Answer 16

J, 22 17 bytes

((10 j.[)]@:":".)

NB.    2    ((10 j.[)]@:":".)   '12.45678'
NB.    12.46 

Thanks to @Conor O'Brien for correcting my understanding of the rules.

t=:4 :'(10 j.x)":".y'

    NB.    Examples
    NB.    4 t'12.45678'
    NB.    12.4568
    NB.    4 t'12.456780'
    NB.    12.4568
    NB.    4 t'12.4567801'
    NB.    12.4568
    NB.    2 t'12.45678'
    NB.      12.46
    NB.    2 t'12.4567801'
    NB.      12.46
    NB.    2 (10 j.[)":". '_12.4567801'
    NB.     _12.46

format    
    x t y
where x is a digit number of decimal places required and y
is the character string containing the value to be rounded.

Answer 17

Scala, 44 bytes

(s:String,p:Int)=>s"%.${p}f"format s.toFloat

Test:

scala> var x = (s:String,p:Int)=>s"%.${p}f"format s.toFloat
x: (String, Int) => String = <function2>

scala> x("14.225",2)
res13: String = 14.23

Answer 18

Wonder, 10 bytes

@@fix#1E#0

Usage:

@@fix#1E#0

Set decimal precision and add trailing zeros if necessary.