A rounding oddity

[Bernard Liengme]

To help answer a question on this group, I wanted to convert the decimal
part of a number (in the set 1, 1.1, 1.2, ...... 9.7, 9.8, 9.9 ) to a
letter. Thus 2.1 would yield a, 2.2 would give b.

I experimented with
=CHOOSE(MOD(A1,1)*10+1,"z","a","b","c","d","e","f","g","h","i")
but it did not quite work (for example 3.8 gave 'g' and not 'h')
So I tried =CHOOSE(MOD(A1*10,10)+1,"z","a","b","c","d","e","f","g","h","i")
and this worked.

Odd since, at the integer level, =MOD(A1,1)*10+1 and =MOD(A1*10,10)+1 give
the same result.

Using the Evaluate Formula tool (I work with XL 2003) I was unable to see
why my original formula did not work.
=CHOOSE(MOD(3.8,1)*10+1,"z","a","b","c","d","e","f","g","h","i") yields
=CHOOSE(9,"z","a","b","c","d","e","f","g","h","i") but the next result is g
not h
Very odd!!!
Happy New Year to all

 

[Bernie Deitrick]

Bernard,

Starting at 3.1, and incrementing by .1, I get

a b b c e f g h

Anyway, instead of

=CHOOSE(MOD(3.8,1)*10+1,"z","a","b","c","d","e","f","g","h","i")

try

=CHOOSE(ROUND(MOD(3.8,1)*10+1,0),"z","a","b","c","d","e","f","g","h","i")

I'm sure MOD is affected by binary math, and ....

HTH,
Bernie
MS Excel MVP

 

[Peo Sjoblom]

To help answer a question on this group, I wanted to convert the decimal
part of a number (in the set 1, 1.1, 1.2, ...... 9.7, 9.8, 9.9 ) to a
letter. Thus 2.1 would yield a, 2.2 would give b.

I experimented with
=CHOOSE(MOD(A1,1)*10+1,"z","a","b","c","d","e","f","g","h","i")
but it did not quite work (for example 3.8 gave 'g' and not 'h')
So I tried =CHOOSE(MOD(A1*10,10)+1,"z","a","b","c","d","e","f","g","h","i")
and this worked.

Odd since, at the integer level, =MOD(A1,1)*10+1 and =MOD(A1*10,10)+1 give
the same result.

Using the Evaluate Formula tool (I work with XL 2003) I was unable to see
why my original formula did not work.
=CHOOSE(MOD(3.8,1)*10+1,"z","a","b","c","d","e","f","g","h","i") yields
=CHOOSE(9,"z","a","b","c","d","e","f","g","h","i") but the next result is g
not h
Very odd!!!
Happy New Year to all

Hi Bernard,

interesting,


this works fine

=INDEX({"z";"a";"b";"c";"d";"e";"f";"g";"h";"i"},MOD(A1,1)*10+1)

and is non volatile

this works as well

=CHOOSE(INT(MOD(A1,1)*10)+1,"z","a","b","c","d","e","f","g","h","i")

another option might be

=CHAR(MOD(A1,1)*10+96)

but it needs editing depending on what the result should be if
A1 is an integer or is blank



Regards,

Peo Sjoblom

 

[Roger Govier]

Hi Bernard

Yes, I get the same result as you with XL2003.
With the series 2.0,2.1 etc through to 3.0 in A1:A11 I get the letter
series
z,a,b,b,c,e,f,g,g,i,z
whereas with your second formula I get the correct series
z,a,b,c,d,e,f,g,h,i,z

Looking at the =MOD(A1,1)*10+1 part of the formula and the
=MOD(A1*10,10)+1without the Choose function the return is the series of
numbers 1,2,3,4,5,6,7,8,9,10,1 in both cases, with no detectable
difference even when extending decimal places to 30.

It therefore appears as though Choose is giving rise to the error.

Amending the first formula to
=CHOOSE(ROUND(MOD(A1,1)*10+1,0),"z","a","b","c","d","e","f","g","h","i")
gives the correct results.
As you say, very odd indeed.

I would also like to send best wishes to you (and everyone else) for the
New Year.

 

[Guest]

Requesting 30 decimal places is wasted effort. As documented, Excel will
display no more than 15 significant figures. If you ask for more, Excel pads
them with zeros regardless of the actual values.

On the other hand, it takes 17 significant figures to uniquely identify the
binary representation of an IEEE double precision number (used by Excel and
almost all other software). Therefore, to determine what is going on, you
either need to use subtraction, such as
=(MOD(A1*10,10)+1-9)
(note the outer parentheses that prevent Excel from being overly "helpful"
and returning zero when Bernard's formula isn't really 9), or you can use
functions such as those at
http://groups.google.com/group/microsoft.public.excel/msg/b106871cf92f8465

Jerry

 

[Roger Govier]

Hi John

I'm not sure what you mean here.
The Mod(5.1,1)*10+1,0) returns 2, which Choose then correctly selects
"a" from the series as a result.

 

[Roger Govier]

Hi Jerry

Many thanks for the response.
Whilst I had always thought that there were only 15 significant places
of decimal, I was very surprised that formatting allowed me to select
30 - I have never attempted to select that number of decimals before.
I don't understand why it allows you to select above 15, but that's
another matter.

I follow what you are saying, and note your reply earlier in the thread
(which was not available to me at the time of my posting).

Thank you for the generosity of the code as published in the link you
provided, and best wishes for 2007