无法使用 vba 函数中的索引结果进行进一步计算

Question

我正在 Excel VBA 中构建自定义函数以接收具有 Webull 分布的数组的第 n 个（此处：第 5 个）元素。这行得通，但是我无法对结果进行进一步的计算，例如简单乘法而不会出错。

Function weib(xr, shape, scaler)
    n = Application.Weibull_Dist(xr, shape, scaler * 100, 0)
    weib = n
End Function

Function good(xr, shape, scaler)
    n = Application.Index(Application.Weibull_Dist(xr, shape, scaler * 100, 0), 5, 0)
    good = n
End Function

Function nogood(xr, shape, scaler)
    n = Application.Index(Application.Weibull_Dist(xr, shape, scaler * 100, 0), 5, 0) * 1
    nogood = n
End Function

为什么“nogood”功能不起作用？它只在末尾添加 * 1 - 我该如何解决？

Answer 1

在测试中，n 是一个 Variant 数组，下限为 1。尝试乘以 1 时会导致 Type Mismatch。因此一个（较差的）解决方案是：

Function good(xr, shape, scaler)
    n = Application.Index(Application.Weibull_Dist(xr, shape, scaler * 100, 0), 5, 0)
    good = n(1)
End Function

Function nogood(xr, shape, scaler)
    n = Application.Index(Application.Weibull_Dist(xr, shape, scaler * 100, 0), 5, 0)
    nogood = n(1) * 1
End Function

更好的解决方案是在 Index 调用中使用 1 而不是 0：

Function good(xr, shape, scaler)
    n = Application.Index(Application.Weibull_Dist(xr, shape, scaler * 100, 0), 5, 1)
    good = n
End Function

Function nogood(xr, shape, scaler)
    n = Application.Index(Application.Weibull_Dist(xr, shape, scaler * 100, 0), 5, 1) * 1
    nogood = n
End Function

您甚至可以完全跳过使用 Index（也许是最佳解决方案）：

Function good(xr, shape, scaler)
    good = Application.Weibull_Dist(xr, shape, scaler * 100, 0)(5, 1)
End Function

Function nogood(xr, shape, scaler)
    nogood = Application.Weibull_Dist(xr, shape, scaler * 100, 0)(5, 1) * 1
End Function

无法使用 vba 函数中的索引结果进行进一步计算

Cannot use result of index in vba function for further computation

indexing

vba

function

excel-formula

weibull