How to make a conditional Median function?

G

Guest

for my thesis i need to create median industry multipliers. i have a list of
companies with their industry codes and multiples, but now i need to create a
list with medians per industry. is there a function similar to SUMIF for
medians?
 
G

Guest

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.
 
B

bplumhoff

Hello Myra,

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

Regards,
Bernd
 
G

Guest

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
 
B

bplumhoff

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
 
G

Guest

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
 
B

bplumhoff

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
 
G

Guest

hi Bernd,

sorry for being a complete dummy. but now it only spits out #value
with a message of a syntax error for:

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

i made the module exactly as you posted it and at first it seems to be
working.
so where did I go wrong?
the fact that i have a crappy old win98 compu with excel 2000 probably isn't
helping either.
guess is should have paid more attention during IT at uni.

regards.
myra
 
B

bplumhoff

Hello Myra,

If lines are broken after being copied into a module, either combine
them again or insert a blank and an underscore (" _") at the end of a
broken line.

Citation of MS Visual Basic Help:
MyVar = "This is an " _
& "example" _
& " of how to continue code."

HTH,
Bernd
 
G

Guest

Using the Array Formula does work but you must use a comma separator between
your function arguments and not a semi-colon

Please use the following formula
MEDIAN(IF(SHEET1!A2:A100=SHEET2!A2,SHEET1!B2:B100))

and enter it using (Ctrl + Shift + Enter).

regards
BetterSolutions.com
 
G

Guest

Many thanks to Bernd for finding a mistake in the VBA MEDIANIF function.

The function was returning an incorrect value when the total number of
matches was an even number:

The incorrect line was:
If lmatch Mod 2 = 0 Then MEDIANIF = arsngvalues((lmatch + 1) / 2)

The correct line should read:
If lmatch Mod 2 = 0 Then MEDIANIF = (arsngvalues(lmatch / 2) +arsngvalues(1
+ (lmatch / 2))) / 2

This has been corrected on the website.
 
G

Guest

Dear BetterSolutions:

I successfully used the array formula for a conditional median. I would
like to create a conditional median formula with more than one condition
(about 4 "and" conditions). I've tried what seemed reasonable but no luck.
I'd appreciate any thoughts.
 
T

T. Valko

Try something like this (array entered):

=MEDIAN(IF((A1:A10="a")*(B1:B10="b")*(C1:C10="c")*(D1:D10="d"),E1:E10))

Biff
 
G

Guest

Thank you very much. It worked fine. I had gotten obsessed with using the
AND function and missed this tidy approach.
 
T

T. Valko

You're welcome. Thanks for the feedback!

Biff

Ed Ford said:
Thank you very much. It worked fine. I had gotten obsessed with using
the
AND function and missed this tidy approach.
 

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