Hello,

Look he
http://www.bettersolutions.com/excel...N616805002.htm

HTH,
Bernd

i have tried MEDIAN(IF(SHEET1!A2:A100=SHEET2!A2;SHEET1!B2:B100) )
but it returns either a zero or a #value. even though there is a match the
formula returns false.

entering the matching industry codes in the formula e.g. "100" does not work
either.
do i need to adjust cell format (now"'general")?

thanx.




" wrote:

Hello,

Look he
http://www.bettersolutions.com/excel...N616805002.htm

HTH,
Bernd

Hello Myra,

Did you really enter that formula as array formula (not only ENTER but
CTRL + SHIFT + ENTER)?

Regards,
Bernd

Thank you. took awhile to get the hang of it.
i've been trying the custom function method (since i have to create 300
medians)

but why does the MEDIANIF() function differ from the hands on median formula?
any ideas how i can use the MEDIANIF function to match on the first 3 digits
only and then take a median value?

ciao






" wrote:

Hello Myra,

Did you really enter that formula as array formula (not only ENTER but
CTRL + SHIFT + ENTER)?

Regards,
Bernd

Hello,

If given the same data these functions should not differ.

Could you give a short example how your industry codes and your data
look like?

Either post it here (anonymous data only) or send me an email.

Regards,
Bernd

example:
sheet 1: per firm the industry code in A and data in B (i omitted firmcode)
DNUM MULT1A
1000 1,361632912
1000 1,014911371
1000 1,844271002
1000 0,092151887
1000 0,533133521
1000 0,151594133
1000 0,712074691
1000 0,483009013
1040
1080
1080
........
........

sheet 2 is for the median value per industry
for code=1000 (cell A2)
using =MEDIANIF(sheet1!A2:A9;A2;sheet1!B2:B9)=
0,533133507

=MEDIAN(sheet1!B2:B9)= 0,622604106
the median(if()) function gives this answer also.

Since not all industry codes have enough data( i need at least 5) i want to
take medians of 100* or 10** as well. is this possible with a macro or
formula?
or shall i continue by hand ;)

thnx in advance.

myra







" wrote:

Hello,

If given the same data these functions should not differ.

Could you give a short example how your industry codes and your data
look like?

Either post it here (anonymous data only) or send me an email.

Regards,
Bernd

Hello Myra,

The correct code for the UDF is IMHO:
Function MEDIANIF(ByVal rgeCriteria As Range, _
ByVal sCriteria As String, _
ByVal rgeMaxRange As Range) As Single

Dim iconditioncolno As Integer
Dim inumberscolno As Integer
Dim lrowno As Long
Dim lmatch As Long
Dim arsngvalues() As Single
Dim sngmedian As Single
Dim bsorted As Boolean

iconditioncolno = rgeCriteria.Column
inumberscolno = rgeMaxRange.Column
ReDim arsngvalues(rgeCriteria.Rows.Count)

For lrowno = 1 To rgeCriteria.Rows.Count
If rgeCriteria.Parent.Cells(rgeCriteria.Row + lrowno - 1,
iconditioncolno).Value = sCriteria Then
lmatch = lmatch + 1
arsngvalues(lmatch) = rgeCriteria.Parent.Cells(rgeCriteria.Row
+ lrowno - 1, inumberscolno).Value
End If
Next lrowno
ReDim Preserve arsngvalues(lmatch)
Do
bsorted = True
For lrowno = 2 To lmatch
If arsngvalues(lrowno - 1) arsngvalues(lrowno) Then
sngmedian = arsngvalues(lrowno - 1)
arsngvalues(lrowno - 1) = arsngvalues(lrowno)
arsngvalues(lrowno) = sngmedian
bsorted = False
End If
Next lrowno
Loop Until bsorted = True

If lmatch Mod 2 = 1 Then MEDIANIF = arsngvalues((lmatch + 1) / 2)
If lmatch Mod 2 = 0 Then MEDIANIF = (arsngvalues(lmatch / 2) +
arsngvalues(1 + lmatch / 2)) / 2
End Function

If you enter in sheet2:
A1:
1000
B1 (as array formula!):
=MEDIAN(IF(Sheet1!$A$3:$A$520=Sheet2!A1,Sheet1!$B$ 3:$B$520,""))
C1:
=MEDIAN(Sheet1!B3:B10)
D1:
=medianif(Sheet1!$A$3:$A$520,Sheet2!A1,Sheet1!$B$3 :$B$520)
E1 (as array formula!):
=MEDIAN(IF(LEFT(Sheet1!$A$3:$A$520,3)="100",Sheet1 !$B$3:$B$520,""))

Then cells B1:E1 should all show the correct result 0.622604106. E1
gives you an example how to calculate a median of 100*. For 10* you can
use LEFT(...,2)="10", for example.

Have fun,
Bernd

Looks like this link is no longer valid...

" wrote:

Hello,

Look he
http://www.bettersolutions.com/excel...N616805002.htm

HTH,
Bernd