Question

编辑：我修复它，ReDim和所有从0开始而不是1，所以我有一个空的单元格，不应该在那里！它现在有效，谢谢你的帮助！

我正在尝试取一个矩阵并将其反转，但出于某种原因我得到了这个错误：

无法获取WorksheetFunction类的MInverse属性。

我的（相关）代码如下：

Dim covar() As Variant
ReDim covar(UBound(assetNames), UBound(assetNames))
Dim corr() As Double
ReDim corr(UBound(assetNames), UBound(assetNames))

Dim covarTmp As Double
For i = 0 To UBound(assetNames) - 1
    For j = 0 To UBound(assetNames) - 1
        covarTmp = 0
        For t = 1 To wantedT
            covarTmp = covarTmp + (Log((prices(histAmount + 1 - t, i + 1)) / (prices(histAmount - t, i + 1))) - mu(i) * dt) * (Log((prices(histAmount + 1 - t, j + 1)) / (prices(histAmount - t, j + 1))) - mu(j) * dt)
        Next t
        covar(i, j) = covarTmp * (1 / ((wantedT - 1) * dt))
        corr(i, j) = covar(i, j) / (sigma(i) * sigma(j))
    Next j
Next i


Dim covarInv() As Variant
ReDim covarInv(UBound(assetNames), UBound(assetNames))
'ReDim covar(1 To UBound(assetNames), 1 To UBound(assetNames))

covarInv = Application.WorksheetFunction.MInverse(covar)

最后一行是发生错误的地方。

我尝试过很多东西，covar和covarInv dim为double，variant等。在covar和covarInv上有不同的ReDims。

Answer 1

您没有说出您正在使用的Excel版本，但是使用Excel 2010时，似乎有一个Minverse最大限制为200 * 200（对于Excel 2003，它可能大约为80 * 80）：您有多少资产名称？有？

我的MInverse功能在VBA中不起作用

1 个答案: