How does Excel successfully round floating point n

问题:

For example, this blog says 0.005 is not exactly 0.005, but rounding that number yields the right result.

I have tried all kinds of rounding in C++ and it fails when rounding numbers to certain decimal places. For example, Round(x,y) rounds x to a multiple of y. So Round(37.785,0.01) should give you 37.79 and not 37.78.

I am reopening this question to ask the community for help. The problem is with the impreciseness of floating point numbers (37,785 is represented as 37.78499999999).

The question is how does Excel get around this problem?

The solution in this round() for float in C++ is incorrect for the above problem.

回答1:

"Round(37.785,0.01) should give you 37.79 and not 37.78."

First off, there is no consensus that 37.79 rather than 37.78 is the "right" answer here? Tie-breakers are always a bit tough. While always rounding up in the case of a tie is a widely-used approach, it certainly is not the only approach.

Secondly, this isn't a tie-breaking situation. The numerical value in the IEEE binary64 floating point format is 37.784999999999997 (approximately). There are lots of ways to get a value of 37.784999999999997 besides a human typing in a value of 37.785 and happen to have that converted to that floating point representation. In most of these cases, the correct answer is 37.78 rather than 37.79.

Addendum
Consider the following Excel formulae:

=ROUND(37785/1000,2)
=ROUND(19810222/2^19+21474836/2^47,2)

Both cells will display the same value, 37.79. There is a legitimate argument over whether 37785/1000 should round to 37.78 or 37.79 with two place accuracy. How to deal with these corner cases is a bit arbitrary, and there is no consensus answer. There isn't even a consensus answer inside Microsoft: "the Round() function is not implemented in a consistent fashion among different Microsoft products for historical reasons." ( http://support.microsoft.com/kb/196652 ) Given an infinite precision machine, Microsoft's VBA would round 37.785 to 37.78 (banker's round) while Excel would yield 37.79 (symmetric arithmetic round).

There is no argument over the rounding of the latter formula. It is strictly less than 37.785, so it should round to 37.78, not 37.79. Yet Excel rounds it up. Why?

The reason has to do with how real numbers are represented in a computer. Microsoft, like many others, uses the IEEE 64 bit floating point format. The number 37785/1000 suffers from precision loss when expressed in this format. This precision loss does not occur with 19810222/2^19+21474836/2^47; it is an "exact number".

I intentionally constructed that exact number to have the same floating point representation as does the inexact 37785/1000. That Excel rounds this exact value up rather than down is the key to determining how Excel's ROUND() function works: It is a variant of symmetric arithmetic rounding. It rounds based on a comparison to the floating point representation of the corner case.

The algorithm in C++:

#include <cmath> // std::floor

// Compute 10 to some positive integral power.
// Dealing with overflow (exponent > 308) is an exercise left to the reader.
double pow10 (unsigned int exponent) { 
   double result = 1.0;
   double base = 10.0;
   while (exponent > 0) {
      if ((exponent & 1) != 0) result *= base;
      exponent >>= 1;
      base *= base;
   }
   return result;
}   

// Round the same way Excel does.
// Dealing with nonsense such as nplaces=400 is an exercise left to the reader.
double excel_round (double x, int nplaces) {
   bool is_neg = false;

   // Excel uses symmetric arithmetic round: Round away from zero.
   // The algorithm will be easier if we only deal with positive numbers.
   if (x < 0.0) {
      is_neg = true;
      x = -x; 
   }

   // Construct the nearest rounded values and the nasty corner case.
   // Note: We really do not want an optimizing compiler to put the corner
   // case in an extended double precision register. Hence the volatile.
   double round_down, round_up;
   volatile double corner_case;
   if (nplaces < 0) {
      double scale = pow10 (-nplaces);
      round_down  = std::floor (x * scale);
      corner_case = (round_down + 0.5) / scale;
      round_up    = (round_down + 1.0) / scale;
      round_down /= scale;
   }
   else {
      double scale = pow10 (nplaces);
      round_down  = std::floor (x / scale);
      corner_case = (round_down + 0.5) * scale;
      round_up    = (round_down + 1.0) * scale;
      round_down *= scale;
   }

   // Round by comparing to the corner case.
   x = (x < corner_case) ? round_down : round_up;

   // Correct the sign if needed.
   if (is_neg) x = -x; 

   return x;
}   

回答2:

For very accurate arbitrary precision and rounding of floating point numbers to a fixed set of decimal places, you should take a look at a math library like GNU MPFR. While it's a C-library, the web-page I posted also links to a couple different C++ bindings if you want to avoid using C.

You may also want to read a paper entitled "What every computer scientist should know about floating point arithmetic" by David Goldberg at the Xerox Palo Alto Research Center. It's an excellent article demonstrating the underlying process that allows floating point numbers to be approximated in a computer that represents everything in binary data, and how rounding errors and other problems can creep up in FPU-based floating point math.

回答3:

I don't know how Excel does it, but printing floating point numbers nicely is a hard problem: http://www.serpentine.com/blog/2011/06/29/here-be-dragons-advances-in-problems-you-didnt-even-know-you-had/

回答4:

So your actual question seems to be, how to get correctly rounded floating point -> string conversions. By googling for those terms you'll get a bunch of articles, but if you're interested in something to use, most platforms provide reasonably competent implementations of sprintf()/snprintf(). So just use those, and if you find bugs, file a report to the vendor.

回答5:

A function that takes a floating point number as argument and returns another floating point number, rounded exactly to a given number of decimal digits cannot be written, because there are many numbers with a finite decimal representation that have an infinite binary representation; one of the simplest examples is 0.1 .

To achieve what you want you must accept to use a different type as a result of your rounding function. If your immediate need is printing the number you can use a string and a formatting function: the problem becomes how to obtain exactly the formatting you expect. Otherwise if you need to store this number in order to perform exact calculations on it, for instance if you are doing accounting, you need a library that's capable of representing decimal numbers exactly. In this case the most common approach is to use a scaled representation: an integer for the value together with the number of decimal digits. Dividing the value by ten raised to the scale gives you the original number.

If any of these approaches is suitable, I'll try and expand my answer with practical suggestions.

回答6:

Excel rounds numbers like this "correctly" by doing WORK. They started in 1985, with a fairly "normal" set of floating-point routines, and added some scaled-integer fake floating point, and they've been tuning those things and adding special cases ever since. The app DID used to have most of the same "obvious" bugs that everybody else did, it's just that it mostly had them a long time ago. I filed a couple myself, back when I was doing tech support for them in the early 90s.

回答7:

As mjfgates says, Excel does hard work to get this "right". The first thing to do when you trying to reimplement this, is define what you mean by "right". Obvious solutions: - implement rational arithmetic Slow but reliable. - implement a bunch of heuristics Fast but tricky to get right (think "years of bug reports").

It really depends on your application.

回答8:

What you NEED is this :

 double f = 22.0/7.0;
    cout.setf(ios::fixed, ios::floatfield);
    cout.precision(6); 
    cout<<f<<endl;  

How it can be implemented (just a overview for rounding last digit) :

long getRoundedPrec(double d,   double precision = 9)
{
    precision = (int)precision;
    stringstream s;
    long l = (d - ((double)((int)d)))* pow(10.0,precision+1);
    int lastDigit = (l-((l/10)*10));
    if( lastDigit >= 5){
        l = l/10 +1;
    }
    return l;
}

回答9:

Just as base-10 numbers must be rounded as they are converted to base-2, it is possible to round a number as it is converted from base-2 to base-10. Once the number has a base-10 representation it can be rounded again in a straightforward manner by looking at the digit to the right of the one you wish to round.

While there's nothing wrong with the above assertion, there's a much more pragmatic solution. The problem is that the binary representation tries to get as close as possible to the decimal number, even if that binary is less than the decimal. The amount of error is within [-0.5,0.5] least significant bits (LSB) of the true value. For rounding purposes you'd rather it be within [0,1] LSB so that the error is always positive, but that's not possible without changing all the rules of floating point math.

The one thing you can do is add 1 LSB to the value, so the error is within [0.5,1.5] LSB of the true value. This is less accurate overall, but only by a very tiny amount; when the value is rounded for representation as a decimal number it is much more likely to be rounded to a proper decimal number because the error is always positive.

To add 1 LSB to the value before rounding it, see the answers to this question. For example in Visual Studio C++ 2010 the procedure would be:

Round(_nextafter(37.785,37.785*1.1),0.01);

回答10:

There are many ways to optimize the result of a floating point using statistical, numerical... algorithms

The easiest one is maybe searching for repetitive 9s or 0s in the range of precision. If there is, maybe those 9s are reduntant, just round them up. But this may not work in many case

2.67899999 → 2.679
12.3499999 → 12.35
1.20000001 → 1.2

Or you can include a number of precision along with the floating point number. After each step, adjust the accuracy depend on the precision of operands. For example

1.113   → 3 decimal digits
6.15634 → 5 decimal digits

Since both number are inside the double's 16-17 digits precision range, their sum will be accurate to the larger of them, which is 5 digits. Similarly, 3+5 < 16, so their product will be precise to 8 decimal numbers

1.113 + 6.15634 = 7.26934    → 5 decimal digits
1.113 * 6.15634 = 6.85200642 → 8 decimal digits

But 4.1341677841 * 2.251457145 will only take double's accuracy because the real result's decimal numbers exceed double's precision

Another efficient algorithm is Grisu but I haven't had an opportunity to try.

In 2010, Florian Loitsch published a wonderful paper in PLDI, "Printing floating-point numbers quickly and accurately with integers", which represents the biggest step in this field in 20 years: he mostly figured out how to use machine integers to perform accurate rendering! Why do I say "mostly"? Because although Loitsch's "Grisu3" algorithm is very fast, it gives up on about 0.5% of numbers, in which case you have to fall back to Dragon4 or a derivative

http://www.serpentine.com/blog/2011/06/29/here-be-dragons-advances-in-problems-you-didnt-even-know-you-had/

In fact I think Excel must combine many different methods to achieve the best result of all

Example When a Value Reaches Zero

In Excel 95 or earlier, enter the following into a new workbook:

A1: =1.333+1.225-1.333-1.225

Right-click cell A1, and then click Format Cells. On the Number tab, click Scientific under Category. Set the Decimal places to 15.

Rather than displaying 0, Excel 95 displays -2.22044604925031E-16.

Excel 97, however, introduced an optimization that attempts to correct for this problem. Should an addition or subtraction operation result in a value at or very close to zero, Excel 97 and later will compensate for any error introduced as a result of converting an operand to and from binary. The example above when performed in Excel 97 and later correctly displays 0 or 0.000000000000000E+00 in scientific notation.

http://support.microsoft.com/kb/78113

回答11:

I believe the following C# code rounds numbers as they are rounded in Excel. To exactly replicate the behavior in C++ you might need to use a special decimal type.

In plain English, the double-precision number is converted to a decimal and then rounded to fifteen significant digits (not to be confused with fifteen decimal places). The result is rounded a second time to the specified number of decimal places.

That might seem weird, but what you have to understand is that Excel always displays numbers that are rounded to 15 significant figures. If the ROUND() function weren't using that display value as a starting point, and used the internal double representation instead, then there would be cases where ROUND(A1,N) did not seem to correspond to the actual value in A1. That would be very confusing to a non-technical user.

The double which is closest to 37.785 has an exact decimal value of 37.784999999999996589394868351519107818603515625. (Any double can be represented precisely by a finite base ten decimal because one quarter, one eighth, one sixteenth, and so forth all have finite decimal expansions.) If that number were rounded directly to two decimal places, there would be no tie to break and the result would be 37.78. If you round to 15 significant figures first you get 37.7850000000000. If this is further rounded to two decimal places, then you get 37.79, so there is no real mystery after all.

    // Convert to a floating decimal point number, round to fifteen 
    // significant digits, and then round to the number of places
    // indicated.
    static decimal SmartRoundDouble(double input, int places)
    {
        int numLeadingDigits = (int)Math.Log10(Math.Abs(input)) + 1;

        decimal inputDec = GetAccurateDecimal(input);

        inputDec = MoveDecimalPointRight(inputDec, -numLeadingDigits);

        decimal round1 = Math.Round(inputDec, 15);

        round1 = MoveDecimalPointRight(round1, numLeadingDigits);

        decimal round2 = Math.Round(round1, places, MidpointRounding.AwayFromZero);

        return round2;
    }

    static decimal MoveDecimalPointRight(decimal d, int n)
    {
        if (n > 0)
            for (int i = 0; i < n; i++)
                d *= 10.0m;
        else
            for (int i = 0; i > n; i--)
                d /= 10.0m;

        return d;
    }

    // The constructor for decimal that accepts a double does
    // some rounding by default. This gets a more exact number.
    static decimal GetAccurateDecimal(double r)
    {
        string accurateStr = r.ToString("G17", CultureInfo.InvariantCulture);
        return Decimal.Parse(accurateStr, CultureInfo.InvariantCulture);
    }

回答12:

Most decimal fractions can't be accurately represented in binary.

double x = 0.0;
for (int i = 1; i <= 10; i++)
{
  x += 0.1;
}
// x should now be 1.0, right?
//
// it isn't. Test it and see.

One solution is to use BCD. It's old. But, it's also tried and true. We have a lot of other old ideas that we use every day (like using a 0 to represent nothing...).

Another technique uses scaling upon input/output. This has the advantage of nearly all math being integer math.