[英]How to generate a series of randomly distributed numbers in sequence using Excel
我有一个号码,通常想分配到15个垃圾箱或单元格中。 我希望15个数字按顺序排列
例:
分配数量-340输出:6 9 12 16 20 24 27 30 32 32 32 32 30 27 24 20
...是的,我的系列作品发行不完善,但目前我正在这样做,
有一个更好的方法吗? 因为我必须为此生成多个矩阵,但有些不正确...
这是一种命中方法,它搜索独立正态变量的随机向量,这些向量的和在目标和的给定公差内,并且如果是,则对所有数字进行重新缩放以使其精确等于和:
Function RandNorm(mu As Double, sigma As Double) As Double
'assumes that Ranomize has been called
Dim r As Double
r = Rnd()
Do While r = 0
r = Rnd()
Loop
RandNorm = Application.WorksheetFunction.Norm_Inv(r, mu, sigma)
End Function
Function RandSemiNormVect(target As Double, n As Long, mu As Double, sigma As Double, Optional tol As Double = 1) As Variant
Dim sum As Double
Dim rescale As Double
Dim v As Variant
Dim i As Long, j As Long
Randomize
ReDim v(1 To n)
For j = 1 To 10000 'for safety -- can increase if wanted
sum = 0
For i = 1 To n
v(i) = RandNorm(mu, sigma)
sum = sum + v(i)
Next i
If Abs(sum - target) < tol Then
rescale = target / sum
For i = 1 To n
v(i) = rescale * v(i)
Next i
RandSemiNormVect = v
Exit Function
End If
Next j
RandSemiNormVect = CVErr(xlErrValue)
End Function
像这样测试:
Sub test()
On Error Resume Next
Range("A1:A15").Value = Application.WorksheetFunction.Transpose(RandSemiNormVect(340, 15, 20, 3))
If Err.Number > 0 Then MsgBox "No Solution Found"
End Sub
具有这些参数的典型输出:
另一方面,如果将标准偏差更改为1,则只会得到一条消息,即找不到解决方案,因为在目标总和的指定公差范围内获得解决方案的可能性很小。
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