This function works, but seems unnecessarily complicated. I would like to simplify but performance is vital as it is called many times.
The conditions are:
1) If either value is negative, then result is -1
2) If both values are zero, then result is zero
3) If NeedMax is true, result is largest value
4) If NeedMax is false, result is smallest non-zero value
CODE:
Public Function SelValue(ValOne As Long, ValTwo As Long, _
Optional NeedMax As Boolean = True) As Long
If ValOne < 0 Or ValTwo < 0 Then
SelValue = -1
ElseIf ValOne = 0 And ValTwo = 0 Then
SelValue = 0
ElseIf NeedMax Then
If ValOne > ValTwo _
Then SelValue = ValOne _
Else SelValue = ValTwo
Else
If ValOne = 0 Then
SelValue = ValTwo
ElseIf ValOne > ValTwo And ValTwo <> 0 Then
SelValue = ValTwo
Else
SelValue = ValOne
End If
End If
End Function
Can anyone suggest a more direct, and hopefully faster, approach?
5 Answers 5
Try to use meaningful names. What's names (SelValue
, ValOne
,...) stands for? It don't give info on their purpose or usage.
Furthermore, you can simplify a lot your conditional branchments, using only this:
If ValOne < 0 Or ValTwo < 0 Then
SelValue = -1
' remove this following condition to discard case 4
Else If ValOne = 0 Or ValTwo = 0 Then
SelValue = ValOne + ValTwo
ElseIf ValOne > ValTwo <> NeedMax Then
SelValue = ValTwo
Else
SelValue = ValOne
End If
not fully tested, but should do the trick
Edit : fixed case (4), where NeedMax
is false and one of values is 0.
-
\$\begingroup\$ Except when one value is negative this always returns the sum of the two values. Doesn't meet my conditions. \$\endgroup\$Rey Juna– Rey Juna2018年12月06日 00:52:32 +00:00Commented Dec 6, 2018 at 0:52
-
\$\begingroup\$ @ReyJunaRey did you tested? If one value (Or both) is negative, it return -1... \$\endgroup\$Calak– Calak2018年12月06日 01:00:45 +00:00Commented Dec 6, 2018 at 1:00
-
\$\begingroup\$ Correct, if one (or both) is negative it will return
-1
. Otherwise it always returns the sum of the two values whereas I would like to either return the maximum or minimum non-zero. \$\endgroup\$Rey Juna– Rey Juna2018年12月06日 01:02:32 +00:00Commented Dec 6, 2018 at 1:02 -
\$\begingroup\$ No, it return the sum only if at least one of value is zero. Test it please ;) \$\endgroup\$Calak– Calak2018年12月06日 01:04:03 +00:00Commented Dec 6, 2018 at 1:04
-
1\$\begingroup\$ Fixed, it should work now. this post was just "phonecrafted", and especially the line after the comment wasn't tested. \$\endgroup\$Calak– Calak2018年12月06日 01:24:42 +00:00Commented Dec 6, 2018 at 1:24
This is a really funky way of formatting an If
statement and is totally inconsistent with the rest of the procedure (and the way most people would format it):
If ValOne > ValTwo _
Then SelValue = ValOne _
Else SelValue = ValTwo
I'd rearrange that as
If ValOne > ValTwo Then
SelValue = ValOne
Else
SelValue = ValTwo
End If
-
\$\begingroup\$
SelValue = If(ValOne > ValTwo, ValOne, ValTwo)
. Using the VBAIf()
function in this context is safe because all components are values, not objects and not likely to throw error in themselves. \$\endgroup\$AJD– AJD2018年12月05日 20:15:34 +00:00Commented Dec 5, 2018 at 20:15 -
\$\begingroup\$ @AJD, I like that approach. You should post as an answer. \$\endgroup\$Rey Juna– Rey Juna2018年12月05日 20:18:14 +00:00Commented Dec 5, 2018 at 20:18
-
\$\begingroup\$ @AJD that's
IIf
;-) \$\endgroup\$Mathieu Guindon– Mathieu Guindon2018年12月05日 20:18:36 +00:00Commented Dec 5, 2018 at 20:18 -
1\$\begingroup\$ @AJD
IIf
is a function, so it involves stack manipulation - i.e. copying memory, pushing the stack pointer, popping back, etc. That's a horrible idea if you're looking for raw performance. \$\endgroup\$Comintern– Comintern2018年12月05日 20:19:51 +00:00Commented Dec 5, 2018 at 20:19 -
\$\begingroup\$ @MathieuGuindon: thanks - yes
IIf()
.If()
works in .Net as well and is safer. \$\endgroup\$AJD– AJD2018年12月05日 20:26:03 +00:00Commented Dec 5, 2018 at 20:26
First, the default value seems like it's overkill here. The function should be a Min
function or a Max
function, not both. It's not going to kill you to have two functions with clear arguments that describe what they are doing.
The naming is suspect too. SelValue
doesn't sound like the name of a function - it sounds like a procedure or property. It invokes the idea of doing something, not returning a value. I'd rename it. Note that I can't even suggest a good name because of the first thing, because FindHigherOrLowerOfTwoNumbers()
seems a little... weird.
As far as performance, that would be the last thing on my mind. This is only performing comparison operations between numbers, and that should be blazingly fast regardless of how many extra comparisons you've managed to sneak in there:
- You don't have to check explicitly to see if they're both zero. If they're equal and both happen to be zero, you'll get zero back anyway.
- If both of them are the same, that's the only check you need to do. Just pick one and return it.
VBA's If
statements don't short circuit, so I'd structure this as a Select Case
and filter away remaining cases on my way down.
Select Case True
Case ValOne < 0 Or ValTwo < 0
SelValue = -1
Case ValOne = ValTwo
SelValue = ValOne
Case NeedMax
If ValOne > ValTwo Then
SelValue = ValOne
Else
SelValue = ValTwo
End If
Case ValOne < ValTwo
SelValue = ValOne
Case Else
SelValue = ValTwo
End Select
-
\$\begingroup\$ Oh well, there goes my draft! :) \$\endgroup\$Mathieu Guindon– Mathieu Guindon2018年12月05日 20:20:47 +00:00Commented Dec 5, 2018 at 20:20
-
\$\begingroup\$ @MathieuGuindon - Reading fail. Edited (although I'll stand by the rest of that bullet). \$\endgroup\$Comintern– Comintern2018年12月05日 20:23:02 +00:00Commented Dec 5, 2018 at 20:23
-
1\$\begingroup\$ Based on the comment conversation though, it appears
-1
is actually an error flag; hence I'd move that first case last, and throw error 5 - eliminating a highly redundant if-check at the call site. \$\endgroup\$Mathieu Guindon– Mathieu Guindon2018年12月05日 20:24:01 +00:00Commented Dec 5, 2018 at 20:24 -
1\$\begingroup\$ @MathieuGuindon That would make a good answer. ;-) \$\endgroup\$Comintern– Comintern2018年12月05日 20:24:37 +00:00Commented Dec 5, 2018 at 20:24
-
\$\begingroup\$ Yeah, I like the
Select Case
approach - I was going to add it to my answer. \$\endgroup\$AJD– AJD2018年12月05日 20:37:50 +00:00Commented Dec 5, 2018 at 20:37
I'm not sure the functional solution below may contribute to better performance, but it's a way of dealing with the three conditional expressions all at once.
The maximum of two numbers A and B can expressed functionally as:
(Abs(A + B) + Abs(A - B)) /2
and the minimum as:
(Abs(A + B) - Abs(A - B)) /2
Noting that the boolean value True is evaluated to -1 and False to 0, thus the expression:
- (1 + 2 * NeedMax)
is evaluated to 1 if NeedMax is True, and -1 otherwise. Combining all the three preceding expressions yields the desired solution:
Public Function SelValue(ValOne As Long, ValTwo As Long, _
Optional NeedMax As Boolean = True) As Long
SelValue = (Abs(ValOne + ValTwo) - (1 + 2 * NeedMax) * Abs(ValOne - ValTwo)) / 2
End Function
This returns the maximum of the two values if NeedMax is True and and the minimum otherwise. Checking wether both values are zero is redundant; the max and min would have the same value; namely Zero. Checking wether both values are negative is left (as in your comment) to the calling procedure.
-
\$\begingroup\$ This doesn't seem to work right when
NeedMax
isFalse
and only one of the values is0
. It returns0
when it should return the non-zero number (Condition 4). \$\endgroup\$Rey Juna– Rey Juna2018年12月07日 17:57:59 +00:00Commented Dec 7, 2018 at 17:57 -
\$\begingroup\$ I see. I've skipped that condition. Anyway, implementing that condtion functionnaly although feasible, is not practical regarding performance. For sake of brieviety denote A and B as the numbers ValOne and ValTwo. Denote P as NeedMax and Q as the Boolean test (A>0 AND B>0) then
SelValue =B + (A>B) * (B-A) + Q*(1+P)*((B-A) - 2*(A>B)*(B-A))
\$\endgroup\$user186038– user1860382018年12月08日 01:10:46 +00:00Commented Dec 8, 2018 at 1:10 -
\$\begingroup\$ Hmm, seems more complex to understand than the answer and, though I'm not sure that I entered it correctly, when I try it with
A=3
,B=1
, andP=False
I get a result of9
. \$\endgroup\$Rey Juna– Rey Juna2018年12月08日 01:33:59 +00:00Commented Dec 8, 2018 at 1:33 -
\$\begingroup\$ You're right as there's a typo (- operater instead of +).
SelValue =B + (A>B) * (B-A) + Q*(1+P)*((B-A) + 2*(A>B)*(B-A))
. As you may see, this implicit boolean casting may not suit your sake after performance optimization. \$\endgroup\$user186038– user1860382018年12月08日 12:05:11 +00:00Commented Dec 8, 2018 at 12:05
This is Excel, so some Excel built-in perks can be used.
Return a Variant, not a defined type because you can then return an error quite easily. Yes this does apply to general VBA as well.
Return early from the function if you can to save further checks - although in this simple example it is not really necessary.
Every time I see an ElseIf
I think that there is something in here that can be improved.
If you are looking to expand this later, then write it for such extension. The first example below just cleans up the code.
Public Function SelValue(ValOne As Long, ValTwo As Long, _
Optional NeedMax As Boolean = True) As Variant
If ValOne < 0 Or ValTwo < 0 Then
SelValue = CVErr(xlErrValue)
Exit Function
End If
If ValOne = 0 And ValTwo = 0 Then
SelValue = CLng(0)
Exit Function
End If
If NeedMax Then
SelValue = IIf(ValOne > ValTwo, CLng(ValOne), CLng(ValTwo)) ' But note comments about a minor performance hit
Exit Function
End If
If ValOne = 0 Then
SelValue = CLng(ValTwo)
Exit Function
End If
' Now that we have cleaned out the special cases above. NeedMax is False
If ValOne > ValTwo And ValTwo <> 0 Then
SelValue = CLng(ValTwo)
Else
SelValue = CLng(ValOne)
End If
End Function
The above code is not the final way I would leave it - personally I would refactor it again because I prefer only one exit/return from a function. But the above iteration highlights something important. It isolates the logic. And you don't handle the case where ValTwo = 0
Addendum: SelValue = CVErr(xlErrValue)
is only useful if the function is a UDF. Raising error #5 (Err.Raise 5
) here and still returning a Long is a valid approach for a non-UDF. Thanks to Mathieu Guindon.
-
\$\begingroup\$ I did it again!
If()
instead ofIIf()
\$\endgroup\$AJD– AJD2018年12月05日 20:39:18 +00:00Commented Dec 5, 2018 at 20:39 -
2\$\begingroup\$
SelValue = CVErr(xlErrValue)
is only useful if the function is a UDF. I would absolutelyErr.Raise 5
here and still return aLong
. \$\endgroup\$Mathieu Guindon– Mathieu Guindon2018年12月05日 20:39:57 +00:00Commented Dec 5, 2018 at 20:39 -
\$\begingroup\$ Returning Variant to be able to return an Error instead of Raising it is bad advice. That's even more true when seen through the lens of performance, where returning Variant incurs overhead for type-coercion and IDispatch/IUnknown. \$\endgroup\$Vogel612– Vogel6122018年12月05日 20:47:54 +00:00Commented Dec 5, 2018 at 20:47
-
\$\begingroup\$ @Vogel612: depends on the scope of the lens. Yes, in the scope of the single function there are performance and storage/space hits, but at the wider scale, performance and functionality is improved due to the improved logic flow ("is that -1 really -1 or is it some sort of coded error result?") An optimal system requires that some or all of the sub-systems are sub-optimal. \$\endgroup\$AJD– AJD2018年12月05日 23:28:02 +00:00Commented Dec 5, 2018 at 23:28
SelValue
is returned as-1
I can check for that. It is called to find an index number for a find result. \$\endgroup\$