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rybad80
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Quote rybad80 Replybullet Topic: Evaluating the Median of formula-derived values
     Posted: 23 Dec 2011 at 6:25am

I'm looking to find the median value of a list of times (durations).  I am using the formula previous({endtime})-{starttime} to derive the {@duration}.  Then, I was looking to find the median value of the formula field called {@duration}.  Since this field is using the 'previous' condition, you cannot just summarize the field using median({@duration}) . 

I was thinking if I could build an array from the duration values and then evaluated that array, it could work.  I'm just not sure how to do that, nor can I find it after much searching on forums.
 
Any help would be greatly appreciated.
 
thanks!
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rybad80
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Quote rybad80 Replybullet Posted: 23 Dec 2011 at 6:41am
Here is what I have so far
 
WhilePrintingRecords;
Numbervar Array u :=[{@turnovermins}];
numbervar array y := 0;
Numbervar Array Final;
Numbervar m;
NumberVar n;
numbervar p := 0;
NumberVar MedianValue := 0;
for m := 1 to ubound(u) do (
if u[m] <> 0 then (
p := p + 1;
redim preserve y

;
y

:= u[m]
));
//sorting
for p := 1 to ubound(y) do
  (
    redim preserve Final[ubound(y)];
    Final

:= Maximum(y);
    for n := 1 to ubound(y) do
      (
        if y[n] = Final

then
           (
             y[n] := 0;
             Exit For;
           );
      );
   );
MedianValue := Final[round(p/2,0)]; ;

 
 
 
Problem is, the array seems to be resetting on each row, so in essence, each row has an array with only 1 value in it.
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lockwelle
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Quote lockwelle Replybullet Posted: 27 Dec 2011 at 3:47am
redim is for setting the dimension(size of the array) and you are doing that every row.  Preserve just keeps the existing values.
 
with that said:
for m := 1 to ubound(u) do (
if u[m] <> 0 then (
p := p + 1;
redim preserve y;
y := u[m]
));
 I would think would be like:
for m := 1 to ubound(u) do (
if u[m] <> 0 then (
p := p + 1;
redim preserve y

;
y := u[m]
));

 redim preserve Final[ubound(y)];
//sorting
for p := 1 to ubound(y) do
...
 
as you only need to set the size of Final once
 
for sorting, wouldn't you want to arrange the values in Final in order?  I don't see that happening in the code.
 
HTH
HTH
 
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