I'm new at this. I want to create an array of 16 elements. Let's say that my array is ReDim arr(15) as Integer and in that array I want to put the numbers from 1 to 16 but scrambled, for example arr(0) = 3, arr(5) = 8 and so on.
2 Answers
Give this a try:
Sub MAIN()
Dim ary(1 To 16) As Variant
Dim i As Long, msg As String
For i = 1 To 16
ary(i) = i
Next i
Call Shuffle(ary)
msg = ""
For i = 1 To 16
msg = msg & vbCrLf & ary(i)
Next i
MsgBox msg
End Sub
Sub Shuffle(InOut() As Variant)
Dim HowMany As Long, i As Long, J As Long
Dim tempF As Double, temp As Variant
Hi = UBound(InOut)
Low = LBound(InOut)
ReDim Helper(Low To Hi) As Double
Randomize
For i = Low To Hi
Helper(i) = Rnd
Next i
J = (Hi - Low + 1) \ 2
Do While J > 0
For i = Low To Hi - J
If Helper(i) > Helper(i + J) Then
tempF = Helper(i)
Helper(i) = Helper(i + J)
Helper(i + J) = tempF
temp = InOut(i)
InOut(i) = InOut(i + J)
InOut(i + J) = temp
End If
Next i
For i = Hi - J To Low Step -1
If Helper(i) > Helper(i + J) Then
tempF = Helper(i)
Helper(i) = Helper(i + J)
Helper(i + J) = tempF
temp = InOut(i)
InOut(i) = InOut(i + J)
InOut(i + J) = temp
End If
Next i
J = J \ 2
Loop
End Sub
Comments
Here is some very lazy code:
Dim arr(15) As Integer
Dim i As Integer, j As Integer
i = 1
Do
j = Int(16 * Rnd)
If arr(j) = 0 Then
arr(j) = i
i = i + 1
End If
Loop Until i = 17
Rnd generates a single from 0 to 1, multiply that by 16 and strip the decimal portion with Int and it will give you a random number from 0 to 15.
It isn't efficient, I wouldn't use this in production, but it'll do the job.
Hope this helps!
3 Comments
Andy Wynn
No it wont, I thought of that in this code. If the rnd function returns a value that has already been used, it will just restart the loop.
Scott Craner
You are correct, I apologize and will delete my comment. I missed the if statement. This works great for the requested small subset. I would be interested in knowing the average number loops to get them all filled. By the odds there is a chance of an infinite loop if the subset got to large. That is not what was asked, and this small of a subset would not run that risk.
Andy Wynn
Yeah, as I said it isn't great code but it'll do the job here of you are only going to use it for small sets. I ran it 500 times and it took an average of 35 loops to complete, but all 500 took less than a second, so I'm not too worried. If this was code for a serious production I would write it properly.
arr(15)only gives you 15 elements.arr(15)(akaarr(0 to 15)) gives 16 elements.