vb.net 生成2位小数的随机数
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Generate random number to 2 decimal places
提问by finisterre
blah = CInt(Int((7 * Rnd()) + 0))
Generates a random integer between 0 and 6.
生成 0 到 6 之间的随机整数。
How can I modify this to give me a random number with 2 decimal places, still between 0 and 6?
如何修改它以给我一个带有 2 个小数位的随机数，仍然在 0 到 6 之间？
As suggested below, I'm now using this code, and it seems to work:
正如下面所建议的，我现在正在使用此代码，它似乎有效：
Dim prng As New Random
Private Function aRand() As Double
Return Math.Round(prng.Next(0, 601) / 100, 2)
End Function
currentApp.statements(Pick, 7) = aRand()
currentApp.statements(Pick, 8) = aRand()
Thanks for all the suggestions.
感谢所有的建议。
采纳答案by dbasnett
Like this
像这样
Dim prng As New Random
Private Function aRand() As Double
Return prng.Next(0, 601) / 100
End Function
note that the location of the random.
注意位置随机。
Your code would look like
你的代码看起来像
currentApp.statements(Pick, 7) = aRand()
currentApp.statements(Pick, 8) = aRand()
回答by Steve
The OP says between 0.00 and 6.00, so I believe that the advice from @HansPassant is the best to try but enlarging the upper limit to 601 ( if he means the limits are inclusive of course)
OP 说between 0.00 and 6.00，所以我相信@HansPassant 的建议是最好的尝试，但将上限扩大到 601（如果他的意思是限制当然包括在内）
Dim rnd As New Random
Dim x As Integer = rnd.Next(0, 601)
Dim d = x / 100
Console.WriteLine(d)
回答by Neolisk
Based on @Steve's answer, here is a generic implementation:
根据@Steve 的回答，这是一个通用实现：
Function RandomDouble(maxValue As Integer, precision As Integer) As Double
Static rnd As New Random
Dim scale As Double = 10 ^ precision
Dim x As Integer = rnd.Next(0, maxValue * scale + 1)
Return x / scale
End Function
And the usage (tested in an empty console app, sub main):
和用法（在一个空的控制台应用程序中测试，sub main）：
Dim dbl As Double = RandomDouble(6, 2)
Debug.WriteLine(dbl)
So you can reuse it like this:
所以你可以像这样重用它：
currentApp.statements(Pick, 7) = RandomDouble(6, 2)
currentApp.statements(Pick, 8) = RandomDouble(6, 2)
Or with DRY principle (=don't repeat yourself):
或者使用DRY 原则（=不要重复自己）：
For i As Integer = 7 To 8
currentApp.statements(Pick, i) = RandomDouble(6, 2)
Next
回答by ARidder101
I thought I would throw my hat in the ring on this question because I just expanded on @dbasnett answer to make a nice generic generator.
我想我会在这个问题上大放异彩，因为我只是扩展了@dbasnett 的答案以制作一个不错的通用生成器。
Note: I used Decimal but Double can be substituted as the output with no seen issues.
注意：我使用了 Decimal，但 Double 可以被替换为输出，没有任何问题。
''' <summary>
''' Random Decimal generator with variable precision"
''' </summary>
''' <param name="L">Minimum Value</param>
''' <param name="U">Maximum Value</param>
''' <param name="P">Precision</param>
''' <returns>Decimal</returns>
''' <remarks></remarks>
Private Function DRandom(L As Integer, U As Integer, P As Integer) As Decimal
Dim Rand As New Random
Dim Upper As String = U.ToString
Dim Precision As String = "1"
For I = 0 To P
If I > 0 Then
If I = P Then
Upper = Upper + "1"
Else
Upper = Upper + "0"
End If
Precision = Precision + "0"
End If
Next
Return Rand.Next(L, Upper.toInteger) / Precision.toInteger
End Function
With my generic toInteger extension:
使用我的通用 toInteger 扩展名：
''' <summary>
''' Handles conversion of variable to Integer.
''' </summary>
''' <param name="X"></param>
''' <param name="I">Returned if conversion fails.</param>
''' <returns>Signed 32bit Integer</returns>
''' <remarks></remarks>
<Runtime.CompilerServices.Extension()> _
Public Function toInteger(Of T)(ByRef X As T, Optional I As Integer = 0) As Integer
Dim S As String = X.ToString
Try
If S = String.Empty Then
Return I
Else
Return Integer.Parse(S)
End If
Catch
Dim result As String = String.Empty
Dim ReturnInt As Integer
Dim Parsed As Byte
For Each Character In S.ToCharArray
If Character = "-" Then
If S.Substring(0, 1).ToString <> "-" Then
result = Character + result
End If
End If
If Character = "." Then
Exit For
End If
If Byte.TryParse(Character, Parsed) Then
result = result + Parsed.ToString
End If
Next
If result <> String.Empty Then
If Integer.TryParse(result, ReturnInt) Then
Return Integer.Parse(ReturnInt)
Else
If Double.Parse(result) > Double.Parse(Integer.MaxValue.ToString) Then
Return Integer.MaxValue
ElseIf Double.Parse(result) < Double.Parse(Integer.MinValue.ToString) Then
Return Integer.MinValue
Else
Return I
End If
End If
Else
Return I
End If
End Try
End Function
This generator takes a variable precision as well as letting you choose in the code what the upper and lower bounds are allowing for (in my opinion) maximum re-usability for a random number generator.
该生成器采用可变精度，并让您在代码中选择上限和下限允许（在我看来）随机数生成器的最大可重用性。
Hope this helps people.
希望这可以帮助人们。
