Excel中银行家的四舍五入公式

Question

It is a well known problem that by default, .NET uses Banker's rounding whereby X.5<\/code> is rounded to nearest even number. Excel, on the other hand uses arithmetic rounding where X.5<\/code> is always rounded up, just like most of us have been taught at school.

Answer 1

Use this formula:

=IF(AND(ISEVEN(A1*10^0),MOD(A1*10^0,1)<=0.5),ROUNDDOWN(A1,0),ROUND(A1,0))

Replace all the 0 , there are 4 of them, with the significance of the desired rounding.

Or you can create a user defined function to use VBA Round, which is Banker's Rounding:

Function BankerRound(rng As Double, sig As Integer) As Double

    BankerRound = Round(rng, sig)
End Function

Then it would simply be:

=BankerRound(A1,0)

Answer 2

=IF(AND(ISEVEN(A1*10^0),MOD(A1*10^0,1)<=0.5),ROUNDDOWN(A1,0),ROUND(A1,0))

I used this formula and it worked great, but in some cases, it was not working because the 0.5 actually was 0.5000000000000000004550000000000000000.

The fix is to ROUND the MOD statement to two(2) decimal places so that the computation reads:

=IF(AND(ISEVEN(A1*10^0),ROUND(MOD(A1*10^0,1),2)<=0.5),ROUNDDOWN(A1,0),ROUND(A1,0))

Answer 3

If for any reason the other codes don't work. This is a brute force variant of the ISEVEN types of formulas.

    =IF(AND(RIGHT(B29,1)="5", OR((RIGHT(LEFT(B29,LEN(B29)-1),1))="0", (RIGHT(LEFT(B29,LEN(B29)-1),1))="2", (RIGHT(LEFT(B29,LEN(B29)-1),1))="4", (RIGHT(LEFT(B29,LEN(B29)-1),1))="6", (RIGHT(LEFT(B29,LEN(B29)-1),1))="8")), ROUNDDOWN(B29,2), IF(AND(RIGHT(B29,1)="5", OR((RIGHT(LEFT(B29,LEN(B29)-1),1))="1", (RIGHT(LEFT(B29,LEN(B29)-1),1))="3", (RIGHT(LEFT(B29,LEN(B29)-1),1))="5", (RIGHT(LEFT(B29,LEN(B29)-1),1))="7", (RIGHT(LEFT(B29,LEN(B29)-1),1))="9")),ROUNDUP(B29,2), ROUND(B29,2)))