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#1
August 27th 05, 02:43 PM
 Warren Smith Posts: n/a
how to count the number of decimal places in a cell?

The worksheet I am doing needs to look at cell values, and in another return
the number of decimal places used in the cell.

for example
if A1 contained a number of 34.45,
I want B1 to tell me 2.
I need to know this because I want to enter the data in a table against
another set of figures which will have 1 more decimal place than the first
set,
and I can't set the decimal places up before because I don't know how many
there will be.

Thanks in advance for any help
Warren

#2
August 27th 05, 03:03 PM
 Bob Phillips Posts: n/a

=IF(ISNUMBER(FIND(".",A20)),LEN(A20)-FIND(".",A20),0)

--

HTH

RP
(remove nothere from the email address if mailing direct)

"Warren Smith" wrote in message
...
The worksheet I am doing needs to look at cell values, and in another

return
the number of decimal places used in the cell.

for example
if A1 contained a number of 34.45,
I want B1 to tell me 2.
I need to know this because I want to enter the data in a table against
another set of figures which will have 1 more decimal place than the first
set,
and I can't set the decimal places up before because I don't know how many
there will be.

Thanks in advance for any help
Warren

#3
August 27th 05, 05:12 PM
 KL Posts: n/a

....just in case the file might be used in other language environments, the
following formula avoids using the decimal separator explicitly:

=IF(ISNUMBER(A1),LEN(MOD(ABS(A1),1))-1-(MOD(A1,1)0),0)

It also uses IF(ISNUMBER(A1),...,0) construct to check if there is text in
the cell A1. If there can only be numbers or empty cells then a shorter
version can be used:

=LEN(MOD(ABS(A1),1))-1-(MOD(A1,1)0)

Regards,
KL

"Bob Phillips" wrote in message
...
=IF(ISNUMBER(FIND(".",A20)),LEN(A20)-FIND(".",A20),0)

--

HTH

RP
(remove nothere from the email address if mailing direct)

"Warren Smith" wrote in message
...
The worksheet I am doing needs to look at cell values, and in another

return
the number of decimal places used in the cell.

for example
if A1 contained a number of 34.45,
I want B1 to tell me 2.
I need to know this because I want to enter the data in a table against
another set of figures which will have 1 more decimal place than the
first
set,
and I can't set the decimal places up before because I don't know how
many
there will be.

Thanks in advance for any help
Warren

#4
August 27th 05, 05:25 PM
 Warren Smith Posts: n/a

Thank you very much , that worked great!

"KL" wrote in message
...
...just in case the file might be used in other language environments, the
following formula avoids using the decimal separator explicitly:

=IF(ISNUMBER(A1),LEN(MOD(ABS(A1),1))-1-(MOD(A1,1)0),0)

It also uses IF(ISNUMBER(A1),...,0) construct to check if there is text in
the cell A1. If there can only be numbers or empty cells then a shorter
version can be used:

=LEN(MOD(ABS(A1),1))-1-(MOD(A1,1)0)

Regards,
KL

"Bob Phillips" wrote in message
...
=IF(ISNUMBER(FIND(".",A20)),LEN(A20)-FIND(".",A20),0)

--

HTH

RP
(remove nothere from the email address if mailing direct)

"Warren Smith" wrote in message
...
The worksheet I am doing needs to look at cell values, and in another

return
the number of decimal places used in the cell.

for example
if A1 contained a number of 34.45,
I want B1 to tell me 2.
I need to know this because I want to enter the data in a table against
another set of figures which will have 1 more decimal place than the
first
set,
and I can't set the decimal places up before because I don't know how
many
there will be.

Thanks in advance for any help
Warren

#5
August 27th 05, 05:54 PM
 Bob Phillips Posts: n/a

KL,

This doesn't work for me at all, it returns 15 for 34.45.

Looking at it MOD(ABS(A1),1) evaluated to .450000000000003, which I would
take is caused by lack of precision when using MOD.

--

HTH

RP
(remove nothere from the email address if mailing direct)

"KL" wrote in message
...
...just in case the file might be used in other language environments, the
following formula avoids using the decimal separator explicitly:

=IF(ISNUMBER(A1),LEN(MOD(ABS(A1),1))-1-(MOD(A1,1)0),0)

It also uses IF(ISNUMBER(A1),...,0) construct to check if there is text in
the cell A1. If there can only be numbers or empty cells then a shorter
version can be used:

=LEN(MOD(ABS(A1),1))-1-(MOD(A1,1)0)

Regards,
KL

"Bob Phillips" wrote in message
...
=IF(ISNUMBER(FIND(".",A20)),LEN(A20)-FIND(".",A20),0)

--

HTH

RP
(remove nothere from the email address if mailing direct)

"Warren Smith" wrote in message
...
The worksheet I am doing needs to look at cell values, and in another

return
the number of decimal places used in the cell.

for example
if A1 contained a number of 34.45,
I want B1 to tell me 2.
I need to know this because I want to enter the data in a table against
another set of figures which will have 1 more decimal place than the
first
set,
and I can't set the decimal places up before because I don't know how
many
there will be.

Thanks in advance for any help
Warren

#6
August 28th 05, 08:22 AM
 Harlan Grove Posts: n/a

"Bob Phillips" wrote...
This doesn't work for me at all, it returns 15 for 34.45.

Looking at it MOD(ABS(A1),1) evaluated to .450000000000003, which I
would take is caused by lack of precision when using MOD.

....

The motivation may have been sound. The implementation wasn't. It should be
as simple as

=LEN(x)-LEN(INT(x))-1

though that'd work with values stored and used but not displayed.

What should the result be for, say, =32+1/3?

#7
August 28th 05, 01:45 PM
 KL Posts: n/a

Opps! You're right Bob. In my testing I hadn't run into this issue and I
didn't suspect any precision issue with MOD - it definetely returns
..450000000000003, which I believe has to do with the floating-point
limitations ( http://support.microsoft.com/kb/78113/en-us )

Regards,
KL

"Bob Phillips" wrote in message
...
KL,

This doesn't work for me at all, it returns 15 for 34.45.

Looking at it MOD(ABS(A1),1) evaluated to .450000000000003, which I would
take is caused by lack of precision when using MOD.

--

HTH

RP
(remove nothere from the email address if mailing direct)

"KL" wrote in message
...
...just in case the file might be used in other language environments,
the
following formula avoids using the decimal separator explicitly:

=IF(ISNUMBER(A1),LEN(MOD(ABS(A1),1))-1-(MOD(A1,1)0),0)

It also uses IF(ISNUMBER(A1),...,0) construct to check if there is text
in
the cell A1. If there can only be numbers or empty cells then a shorter
version can be used:

=LEN(MOD(ABS(A1),1))-1-(MOD(A1,1)0)

Regards,
KL

"Bob Phillips" wrote in message
...
=IF(ISNUMBER(FIND(".",A20)),LEN(A20)-FIND(".",A20),0)

--

HTH

RP
(remove nothere from the email address if mailing direct)

"Warren Smith" wrote in message
...
The worksheet I am doing needs to look at cell values, and in another
return
the number of decimal places used in the cell.

for example
if A1 contained a number of 34.45,
I want B1 to tell me 2.
I need to know this because I want to enter the data in a table
against
another set of figures which will have 1 more decimal place than the
first
set,
and I can't set the decimal places up before because I don't know how
many
there will be.

Thanks in advance for any help
Warren

#8
August 28th 05, 01:52 PM
 KL Posts: n/a

Harlan,

Yours is clearly a much better (and neater) implementation of the idea. As
to =32+1/3, I guess it is going to be the same issue for all possible
solutions given the IEEE 754 specification, so as long as one is aware of
that, =LEN(x)-LEN(INT(x))-1 is probably the best option.

Thanks and regards,
KL

"Harlan Grove" wrote in message
...
"Bob Phillips" wrote...
This doesn't work for me at all, it returns 15 for 34.45.

Looking at it MOD(ABS(A1),1) evaluated to .450000000000003, which I
would take is caused by lack of precision when using MOD.

...

The motivation may have been sound. The implementation wasn't. It should
be as simple as

=LEN(x)-LEN(INT(x))-1

though that'd work with values stored and used but not displayed.

What should the result be for, say, =32+1/3?

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