How to convert a Variant array to a Range?

Question

I have a 2D array of type Variant. The size and values that populate the array are generated based on data within a worksheet. Further processing is required on this array, the primary being the interpolation of several values. I am using this interpolation function (I know about excel equivalent functions but a design choice was made not to use them) . The problem I am having is the that the Interpolation function requires a Range object.

I have already tried modifying the function to use a Variant (r as Variant) argument. The following line nR = r.Rows.Count can be replaced with nR = Ubound(r). While this works, I would also like to use this function normally within any worksheet and not change the function in any way.

Sub DTOP()
    Dim term_ref() As Variant
    ' snip '
    ReDim term_ref(1 To zeroRange.count, 1 To 2)

    ' values added to term_ref '

    ' need to interpolate x1 for calculated y1 '
    x1 = Common.Linterp(term_ref, y1) 
End Sub

Interpolation Function

Function Linterp(r As Range, x As Double) As Double
    Dim lR As Long, l1 As Long, l2 As Long
    Dim nR As Long

    nR = r.Rows.Count
    ' snipped for brevity ' 
End Function

How do I convert my in-memory variant array to a Range so that it can be used for the interpolate function? (without outputting to a WorkSheet)

Answer

In short, the answer is you can't. A Range object must reference a worksheet.

The changed interpolation function checks the TypeName of the argument and sets the value of nR accordingly. Not the prettiest solution.

As a note, the VarType function proved useless in this situation since both VarType(Variant()) and VarType(Range) returned the same value (i.e vbArray) and could not be used to disambiguate an array from a range

Function Linterp(r As Variant, x As Variant) As Double
    Dim lR As Long, l1 As Long, l2 As Long
    Dim nR As Long

    Dim inputType As String
    inputType = TypeName(r)

    ' Update based on comment from jtolle      
    If TypeOf r Is Range Then
        nR = r.Rows.Count
    Else
        nR = UBound(r) - LBound(r) 'r.Rows.Count
    End If
    ' ....
 End Function 

Answer 1

AFAIK，您无法创建不以某种方式引用工作簿的工作表位置的 Range 对象。它可以是动态的，例如 Named =OFFSET() 函数，但它必须与某处的工作表相关联。

为什么不改变插值函数呢？保持 Linterp 签名不变，但将其放入在数组上插值的函数的包装器中。

像这样的东西：

Function Linterp(rng As Range, x As Double) As Double
' R is a two-column range containing known x, known y
' This is now just a wrapper function, extracting the range values into a variant
    Linterp = ArrayInterp(rng.Value, x)

End Function

Function ArrayInterp(r As Variant, x As Double) As Double

Dim lR As Long
Dim l1 As Long, l2 As Long
Dim nR As Long

    nR = UBound(r) ' assumes arrays are all 1-based

    If nR = 1 Then
        ' code as given would return 0, better would be to either return
        ' the only y-value we have (assuming it applies for all x values)
        ' or perhaps to raise an error.
        ArrayInterp = r(1, 2)
        Exit Function
    End If

    If x < r(1, 1) Then ' x < xmin, extrapolate'
        l1 = 1
        l2 = 2
    ElseIf x > r(nR, 2) Then ' x > xmax, extrapolate'
        l2 = nR
        l1 = l2 - 1
    Else
        ' a binary search might be better here if the arrays are large'
        For lR = 1 To nR
            If r(lR, 1) = x Then ' no need to interpolate if x is a point in the array'
                ArrayInterp = r(lR, 2)
                Exit Function
            ElseIf r(lR, 2) > x Then ' x is between tabulated values, interpolate'
                l2 = lR
                l1 = lR - 1
                Exit For
            End If
        Next
    End If

    ArrayInterp = r(l1, 2) _
           + (r(l2, 2) - r(l1, 2)) _
           * (x - r(l1, 1)) _
           / (r(l2, 1) - r(l1, 1))

End Function