asked on

Returning a (multi-dimensional) Variant Array as a function result

I have a function that processes and returns a variant array. so far the array is one dimensional (I'll call it an array with one column), and everything works fine. bwLPF function always takes in a one-column variant array, and will return a one column variant array. See working code below for one-column arrays. (For one column array: TimeHistory[0] := 0, TimeHistory[1] := 0.25, and so on..)

I want to expand this to array of variable number of columns. Basically, I want the function to process each column by recursion.

I tried "NewButterworthLPF" (see below) without success. Input for the new function was created as

V := VarArrayCreate([0, 2], varVariant);
V[0] := VarArrayCreate([0, VarArrayHighBound(FHistoryOriginal, 1)], varDouble);
V[1] := VarArrayCreate([0, VarArrayHighBound(FHistoryOriginal, 1)], varDouble);
V[2] := VarArrayCreate([0, VarArrayHighBound(FHistoryOriginal, 1)], varDouble);
V[0] := History1;
V[1] := History2;
V[2] := History3;
V := NewButterworthLPF(V, fs, fc, fOrder);

so, TimeHistory will look like:
TimeHistory[0][0] := 0, TimeHistory[0][1] := 0.21 and so on for column 0
TimeHistory[1][0] := 0, TimeHistory[1][1] := 0.05 and so on for column 1, etc

What am I missing ? (also would it be better to use a multi-dimensional Variant array, instead of using a 1-D varVariant array in which each column is an array of varDouble ?). A working sample that uses recursion to solve the problem will get the points.

Thanks.

function ButterworthLPF(TimeHistory:OleVariant; SamplingFrequency, CutOffFrequency:real; FilterOrder:byte=8):OleVariant;
//This works for 1-column input
 
var fs, fc, odr   : OleVariant;
 
begin;
    fs  := SamplingFrequency;
    fc  := CutOffFrequency;
    odr := FilterOrder div 2;
    result := bwLPF(TimeHistory, fs, fc, odr);
end;
 
function NewButterworthLPF(TimeHistory:OleVariant; SamplingFrequency, CutOffFrequency:real; FilterOrder:byte=8):OleVariant;
//This does NOT work for 3-column input
 
var fs, fc, odr   : OleVariant;
    i, j : integer;
    vt : TVarType;
 
begin;
  vt := VarType(TimeHistory);
  if (vt and varVariant = varVariant) then
  for i := 1 to VarArrayHighBound(TimeHistory, 1) do
      result[i] := NewButterworthLPF(TimeHistory[i], SamplingFrequency, CutOffFrequency, FilterOrder)
  else
 
  fs  := SamplingFrequency;
  fc  := CutOffFrequency;
  odr := FilterOrder div 2;
  result := bwLPF(TimeHistory, fs, fc, odr);
end;

Open in new window

SOLUTION

atul_parmar

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

ahalya

ASKER

I'll try your code when I get a chance, but you should give comments on why you suggest, what you suggest as well please.

ahalya

ASKER

Nope, doesn't work.

I don't understand why you have
(TVarData(TimeHistory[VarArrayLowBound(TimeHistory, 1)]).VType and VarVariant = varVariant) ??

My FHistoryOriginal is a Variant array of type varDouble.

atul_parmar

>>My FHistoryOriginal is a Variant array of type varDouble
I guess you are using the array you posted earlier. i.e.
V := VarArrayCreate([0, 2], varVariant);
V[0] := VarArrayCreate([0, VarArrayHighBound(FHistoryOriginal, 1)], varDouble);
V[1] := VarArrayCreate([0, VarArrayHighBound(FHistoryOriginal, 1)], varDouble);
V[2] := VarArrayCreate([0, VarArrayHighBound(FHistoryOriginal, 1)], varDouble);

Here, V[0] is an array of variant. NO matter what data type you specify. The data type of its individual element would be Double. i.e. V[0][0]. And thus V is a variant array of variant array.

>>(TVarData(TimeHistory[VarArrayLowBound(TimeHistory, 1)]).VType and VarVariant = varVariant) ??

In your original function (which doesn't work for multi column) you have
vt := VarType(TimeHistory);
if (vt and varVariant = varVariant) ...

So if you pass V (variant array of variant array) to NewButterworthLPF function, in the first pass, "if (vt and varVariant = varVariant)" will return true and will call itself with its V[i] element. Now if I understand correctly, V[i] should be executed on same code which you use in your ButterworthLPF function. i.e.
fs := SamplingFrequency;
fc := CutOffFrequency;
...

But in your original code "if (vt and varVariant = varVariant)" will again return true for V[i] and will call itself again. Which I thought INCORRECT.

Based on that understanding, I checked V[i][j] element of V using
(TVarData(TimeHistory[VarArrayLowBound(TimeHistory, 1)]).VType and VarVariant = varVariant)

Which will branch the V[i] element to the else part of your function.

If that's what you want; my function should work for variant array as well as variant array of variant array. If not, tell me how you use the function with some DUMMY data which I can test.

Atul

ahalya

ASKER

Atul,

I was using the array I posted earlier. I understand that V is a variant array of variant array (of Double).

I want to be able to pass either of the following:
Case 1 (Recursion) : V -> Variant array of variant array
Case 2: (No recursion) V -> Variant Array.

I will make a short sample and post it to show you what I am after. Thanks.

ahalya

ASKER

Atul,

While exploring you code, found the following.

Say
U := VarArrayCreate([0, 2047], varDouble);
vt := VarType(U); //vt is not varVariant, instead it is varDouble

V := VarArrayCreate([0, 4], varVariant);
for i := 0 to 4 do
begin;
U := VarArrayCreate([0, 2047], varDouble);
for j := 0 to 2047 do U[j] := random;
V[i] := U;
end;

vt := VarType(V); //vt is varVariant
vt := VarType(V[1]); //vt is not varVariant, it is varDouble

This suggests that your suggestion regarding varType is not correct. Can you confirm.

ahalya

ASKER

The check that is needed is the following:

vt1 := VarType(TimeHistory);
vt2 := VarType(TimeHistory[VarArrayLowBound(TimeHistory, 1)]);

if (vt2 and varArray = varArray) then ...

atul_parmar

Correct;
vt2 := VarType(TimeHistory[VarArrayLowBound(TimeHistory, 1)]);
if (vt2 and varArray = varArray)

IS SAME AS

IF (TVarData(TimeHistory[VarArrayLowBound(TimeHistory, 1)]).VType and VarVariant = varVariant)

atul_parmar

My misatake; it should be
IF (TVarData(TimeHistory[VarArrayLowBound(TimeHistory, 1)]).VType and varArray= varArray)

ASKER CERTIFIED SOLUTION

ahalya

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

atul_parmar

>>yes, but you don't need this check after all.
It depends on the variant array you are passing. e.g. try with following array

var
V, V1 : Variant;
begin
V := VarArrayCreate([0, 2], varVariant);
V1 := VarArrayCreate([0, 1], varDouble);
V[0] := VarArrayOf(V1);
V[1] := VarArrayOf(V1);
V[2] := VarArrayOf(V1);
NewButterworthLPF(V, 0, 0); // call to your function
end;

Atul.

atul_parmar

It works well with the code I posted first.

...
if (vt and varVariant = varVariant) and (TVarData(TimeHistory[VarArrayLowBound(TimeHistory, 1)]).VType and varVariant = varVariant) then
...

ahalya

ASKER

I haven't had a chance to try with an array as you have defined above. But, my question had the arrays defined in a specific way, and your solution doesn't work with that.

But, I managed to tweak my code (adding a line to create result as a VarArray) after seeing your solution. Thanks.

atul_parmar

>>The solution I posted in the question required the addition one just one line of >>code to make it work. (I figured it out myself after discussions).

My first post has the line (line#10) which makes this work.

>>The answer posted posted might be correct for a different question, but it's not >>the answer for the posted question.

There are numerous ways to solve a problem. It doesn't matter which one u like or dislike.

If, the posted comment doesn't lead to solve the problem. I would rather suggest to delete this question.

ahalya

ASKER

Atul,

>> My first post has the line (line#10) which makes this work.

Even though you had that line, your solution was wrong, and did not work. I accepted your suggestion as an "Assisted Solution" because I picked up your line 10. I cannot accept your solution as a full solution, because it did not work with my array.

Accepting your solution as an "Assisted Solution" was fair, I thought. Don't you think so ?