Question

我有一个Variant类型的2D数组。填充数组的大小和值是根据工作表中的数据生成的。此阵列需要进一步处理，主要是几个值的插值。我正在使用这个interpolation function（我知道excel等效函数，但设计选择不使用它们）。我遇到的问题是Interpolation函数需要一个Range对象。

我已经尝试修改函数以使用Variant（r as Variant）参数。以下行nR = r.Rows.Count可以替换为nR = Ubound(r)。虽然这有效，但我还想在任何工作表中正常使用此功能，而不是以任何方式更改功能。

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

插值功能

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

如何将我的内存变量数组转换为Range，以便它可以用于插值函数？（不输出到工作表）

答案

总之，答案是你做不到的。 Range对象必须引用工作表。

更改的插值函数检查参数的TypeName并相应地设置nR的值。不是最漂亮的解决方案。

作为一个注释，VarType函数在这种情况下被证明是无用的，因为VarType(Variant())和VarType(Range)都返回了相同的值（即vbArray），并且无法用于消除数组的歧义。范围

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，您无法创建一个Range对象，该对象不以某种方式引用工作簿的工作表位置。它可以是动态的，例如，喜欢Named = OFFSET（）函数，但它必须绑定到某个地方的工作表。

为什么不更改插值功能？按原样保留Linterp签名，但将其作为在数组上插值的函数的包装器。

这样的事情：

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

Answer 2

这是一个在新工作表中创建范围的功能。您可以通过添加另一个范围参数来修改此功能，以提供用于容纳阵列的单元格范围的起点。

首先输入代码，然后使用调试器遍历Sub Test（），看看它能为你做些什么......

Function Array2Range(MyArray() As Variant) As Range
Dim X As Integer, Y As Integer
Dim Idx As Integer, Jdx As Integer
Dim TmpSht As Worksheet, TmpRng As Range, PrevRng As Range

    X = UBound(MyArray, 1) - LBound(MyArray, 1)
    Y = UBound(MyArray, 2) - LBound(MyArray, 2)

    Set PrevRng = Selection
    Set TmpSht = ActiveWorkbook.Worksheets.Add
    Set TmpRng = TmpSht.[A1]


    For Idx = 0 To X
        For Jdx = 0 To Y
            TmpRng(Idx + 1, Jdx + 1) = MyArray(LBound(MyArray, 1) + Idx, LBound(MyArray, 2) + Jdx)
        Next Jdx
    Next Idx

    Set Array2Range = TmpRng.CurrentRegion
    PrevRng.Worksheet.Activate

End Function

Sub Test()
Dim MyR As Range
Dim MyArr(3, 3) As Variant

MyArr(0, 0) = "'000"
MyArr(0, 1) = "'0-1" ' demo correct row/column
MyArr(1, 0) = "'1-0" ' demo correct row/column
MyArr(1, 1) = 111
MyArr(2, 2) = 222
MyArr(3, 3) = 333

Set MyR = Array2Range(MyArr) ' to range
Range2Array MyR, MyOther     ' and back

End Sub

编辑============= 修改了sub test（）以演示转换回数组并添加了快速＆amp;用于将范围转换为数组的脏代码

Sub Range2Array(MyRange As Range, ByRef MyArr() As Variant)
Dim X As Integer, Y As Integer
Dim Idx As Integer, Jdx As Integer
Dim MyArray() As Variant, PrevRng As Range

    X = MyRange.CurrentRegion.Rows.Count - 1
    Y = MyRange.CurrentRegion.Columns.Count - 1
    ReDim MyArr(X, Y)

    For Idx = 0 To X
        For Jdx = 0 To Y
            MyArr(Idx, Jdx) = MyRange(Idx + 1, Jdx + 1)
        Next Jdx
    Next Idx
    MyRange.Worksheet.Delete

End Sub

如何将Variant数组转换为Range？

2 个答案: