VBA or PostgreSQL: remove unneeded parentheses from a mathematical equation string - regex

Question

I'm looking to remove mathematically unneeded parentheses from a mathematical equation string. I need to do this either, and preferably, in PostgreSQL 6 or VBA.
For example, I have the following string value in a PostgreSQL database:
PercentileRank((([bp47244]+([bp47229][ttm]))/(AvgAeTe([bp48918]))))
And I need it to look like this (edited/corrected):
PercentileRank(([bp47244]+[bp47229][ttm])/AvgAeTe([bp48918]))
I'd prefer a function or query in PostgreSQL, but a VBA solution could work.
Note PercentileRank() and AvgAeTe() are functions. This [bp47244] and [bp47229][ttm] each represent single numbers/variables, but they could be expressed in any way, like [abc123] and [xyz321][ttm]. I see a lot of examples out there, but I don't see one using PostgreSQL or VBA that works for me, so I thought it would be a good question.
Of course I am looking for a general solution that can be applied to any equation.
I'm working on this now, so if I find an answer before one is posted here, I'll share; however, I am not good at regex (not that the solution has to use regex).
Thanks!
UPDATE:
I'm working off this logic:
Let L be operator immediately left of the left parenthesis, or nil
Let R be operator immediately right of the right parenthesis, or nil
If L is nil and R is nil:
Redundant
Else:
Scan the unparenthesized operators between the parentheses
Let X be the lowest priority operator
If X has lower priority than L or R:
Not redundant
Else:
Redundant
from this link:
Remove redundant parentheses from an arithmetic expression
I'll code something up in VBA that follows this logic and post an answer.

Answer 1

This seems to work for my situation:
Function RemoveParens(s As String) As String
'remove unecessary parentheses
'exponents not implemented
'mathematical brackets are not implmented (it is assumed that only parentheses are used to create mathematical order)
'brakets are assumed to identify a variable or calculation on a variable
'[bp47229][ttm] -> one value/variable; [xyz123] -> one value/variable
'logic based on Antti Huima's answer:
'https://stackoverflow.com/questions/44203517/vba-or-postgresql-remove-unneeded-parentheses-from-a-mathematical-equation-stri
's = "PercentileRank((([bp47244]+([bp47229][ttm]))/(AvgAeTe([bp48918]))))"
's = "PercentileRank(2*(1+3)(5*4))"
If InStr(1, s, "^") > 0 Then
msgbox "Exponents are not implemented in RemoveParens"
End If
ReDim arS(1 To Len(s)) As String
Dim i As Integer
For i = 1 To Len(s)
arS(i) = Mid(s, i, 1)
Next i
Dim iCnt As Integer
iCnt = 0
Dim iLen As Integer
iLen = Len(s)
Dim sTmp As String
Dim bRemove As Boolean
bRemove = False
Dim sLfOpr As String
Dim sRtOpr As String
Dim iCntBtwn As Integer
Dim sLast As String
'loop through chars
Do
iCnt = iCnt + 1
sTmp = Mid(s, iCnt, 1)
If sTmp = "(" Then
if iCnt - 1 <= 0 then
sLfOpr = ""
else
sLfOpr = Mid(s, iCnt - 1, 1)
end if
'in case we have "5(...) or (...)(...)
If IsNumeric(sLfOpr) Or sLfOpr = ")" Then
sLfOpr = "*"
End If
'if it isn't an oper then clear it
If sLfOpr <> "+" _
And sLfOpr <> "-" _
And sLfOpr <> "/" _
And ((Not IsAlpha(sLfOpr) = True) Or (Not Mid(s, iCnt, 1) = "(")) _
And sLfOpr <> "*" _
Then
sLfOpr = ""
End If
'find the matching paren to the right of LfOpr
Dim iCntR As Integer
iCntR = iCnt
Dim iCntParen As Integer
iCntParen = 1
Dim sTmpR As String
sTmpR = ""
Do
iCntR = iCntR + 1
sTmpR = Mid(s, iCntR, 1)
If sTmpR = "(" Then
iCntParen = iCntParen + 1
ElseIf sTmpR = ")" Then
iCntParen = iCntParen - 1
End If
'we found the close paren that matches the open paren
If iCntParen = 0 Then
sRtOpr = Mid(s, iCntR + 1, 1)
'in case we have "(...)5 or (...)(...)
If IsNumeric(sRtOpr) Or sRtOpr = "(" Then
sRtOpr = "*"
End If
If sRtOpr <> "+" _
And sRtOpr <> "-" _
And sRtOpr <> "/" _
And ((Not IsAlpha(sRtOpr) = True) Or (Not Mid(s, iCntR, 1) = "(")) _
And sRtOpr <> "*" _
Then
sRtOpr = ""
End If
If sRtOpr = "" And sLfOpr = "" Then
arS(iCnt) = ""
arS(iCntR) = ""
'go to the next overall open paren
Exit Do
Else
' ------------ search btwn parens -------------------
Dim iCntParenOp As Integer
Dim iCntParenCl As Integer
iCntParenOp = 0
iCntParenCl = 0
Dim sTmpB As String
sTmpB = ""
Dim sLowOpr As String
sLowOpr = ""
Dim iCntRLw As Integer
iCntRLw = iCnt
Dim bInSub As Boolean
bInSub = False
Dim bNoOpr As Boolean
bNoOpr = True
'loop through chars between the two parens
For i = iCnt + 1 To iCntR
iCntRLw = iCntRLw + 1
sTmpR = Mid(s, iCntRLw, 1)
If sTmpR = "(" Then
iCntParenOp = iCntParenOp + 1
bInSub = True
ElseIf sTmpR = ")" Then
iCntParenCl = iCntParenCl + 1
If bInSub = True And iCntParenCl = iCntParenOp Then
bInSub = False
End If
End If
'we found the close paren that matches the open paren
'and we are not in a nested/sub paren
If bInSub = False Then
'in case we have "(...)5 or (...)(...)
If (IsNumeric(sTmpR) And Mid(s, iCntRLw + 1, 1) = "(") Or (sTmpR = "(" And Mid(s, iCntRLw + 1, 1) = "(") Then
sTmp = "*"
End If
'it is an operator
If sTmpR = "+" _
Or sTmpR = "-" _
Or sTmpR = "/" _
Or ((IsAlpha(sTmpR) = True) And (Mid(s, iCntRLw + 1, 1) = "(")) _
Or sTmpR = "*" _
Or bNoOpr = True _
Then
'see if sLowROpr operater has lower priority than sLfOpr and sRtOpr
If Not IsLowerPri(sTmpR, sRtOpr, sLfOpr) Then
arS(iCnt) = ""
arS(iCntR) = ""
Exit For
End If
bNoOpr = False
End If
End If
Next i
End If
Exit Do 'always stop loop if iCntParen = 0
End If
Loop While iCntR <> iLen
End If
Loop While iCnt <> iLen
Dim sOut As String
For i = LBound(arS) To UBound(arS)
sOut = sOut & arS(i)
Next i
'Debug.Print s
RemoveParens = sOut
End Function
Function IsLowerPri(sTestOpr As String, sRtOpr As String, sLfOpr As String) As Boolean
'exponents not implemented yet
Dim iTestOpr As Integer
Dim iRtOpr As Integer
Dim iLfOpr As Integer
iTestOpr = 1
If sTestOpr = "+" Or sTestOpr = "-" Then
iTestOpr = 1
ElseIf sTestOpr = "*" Or sTestOpr = "/" Then
iTestOpr = 2
ElseIf IsAlpha(sTestOpr) And sTestOpr <> "" Then
iTestOpr = 3
End If
If sRtOpr = "+" Or sRtOpr = "-" Then
iRtOpr = 1
ElseIf sRtOpr = "*" Or sRtOpr = "/" Then
iRtOpr = 2
ElseIf IsAlpha(sRtOpr) And sRtOpr <> "" Then
iRtOpr = 3
End If
If sLfOpr = "+" Or sLfOpr = "-" Then
iLfOpr = 1
ElseIf sLfOpr = "*" Or sLfOpr = "/" Then
iLfOpr = 2
ElseIf IsAlpha(sLfOpr) And sLfOpr <> "" Then
iLfOpr = 3
End If
If iTestOpr < iRtOpr Or iTestOpr < iLfOpr Then
IsLowerPri = True
Else
IsLowerPri = False
End If
End Function
It needs a lot of clean-up and probably some testing. I will give answer credit to whomever posts the best improvement or a different solution all together that is better.
UPDATE:
Forgot this function:
Public Function IsAlpha(strValue As String) As Boolean
IsAlpha = strValue Like WorksheetFunction.Rept("[a-zA-Z]", Len(strValue))
End Function