Question

我已经花了将近一个小时的时间来解决这个问题，但似乎无济于事。

基本上我想尝试一个更好的随机数生成器，并找到了我想尝试在程序中实现的代码

 Function RandNorm(Optional mean As Double = 0, _
              Optional Dev As Double = 1, _
              Optional Corr As Double = 0, _
              Optional bVolatile As Boolean = False) As Double()


Randomize
Dim z(0 To 1)   As Double
Dim U           As Double
Dim V           As Double
Dim S           As Double

If bVolatile Then Application.Volatile

Do
    U = 2 * [rand()] - 1
    V = 2 * [rand()] - 1
    S = U * U + V * V
Loop Until S < 1

S = Sqr(-2 * Log(S) / S)
z(0) = Dev * U * S + mean
z(1) = Dev * V * S + mean

If Corr <> 0 Then z(1) = Corr * z(0) + Sqr(1 - Corr ^ 2) * z(1)
RandNorm = z
End Function

如您所见，该函数返回一个数字。在=RandNorm()之类的Excel中运行时，一切都很好。但是当以这样的简单代码运行时：

 Sub test()

    Dim x() As Double, i As Long

    ReDim x(1 To 10, 1 To 1)
    For i = 1 To 10
          x(i, 1) = RandNorm
    Next i

    Range("A1:A10") = x
End Sub

尽管定义为As Double，但仍显示“类型不匹配”错误。如果我在RandNorm As Double中定义Sub，那么它的值为0。但是，如果我将x(i, 1)替换为Cells(i, 1)，它将起作用。

该功能也位于其自己的模块和Sub中。

我想念什么？任何帮助将非常感激。预先感谢！

Answer 1

Double与Double()不同：RandNorm返回一个 array 。当在单个单元格中用作UDF时，RandNorm返回该数组的 first 元素。例如，考虑以下功能。

Function foo() As Double()
    Dim z(0 To 1) As Double
    z(0) = 1
    z(1) = 2
    foo = z
End Function

在单个单元格中用作UDF时，=foo()将始终返回1。您需要使用 Ctrl + Shift + Enter 在2个单元格中将其作为数组公式输入，以显示第二个元素{{1} }。

简单的解决方法可能是将2更改为Double()，将Double更改为RandNorm = z。

RandNorm = z(0)

很难说这就是您要寻找的-因为您实际上忽略了Function RandNorm(Optional mean As Double = 0, _ Optional Dev As Double = 1, _ Optional Corr As Double = 0, _ Optional bVolatile As Boolean = False) As Double .... If Corr <> 0 Then z(1) = Corr * z(0) + Sqr(1 - Corr ^ 2) * z(1) RandNorm = z(0) End Function。

UDF随机数生成器类型不匹配

1 个答案: