我想知道为什么以下语法在 VBA 中不能按照我想象的方式工作,以及我应该做什么来确保它能正常工作;
For a = 1 To 10
For b = 1 To 10
For c = 1 To 10
If a <> b <> c Then
MsgBox (a & " " & b & " " & c)
End If
Next c
Next b
Next a
这是一个简化的示例,仍然可以通过以下方式手动获取:
if a<>b and b<>c and c<>a then
但是我的实际预期代码多次出现 10 个这样的变量,这使得它在 55 个不等条件下不可行,或者我可能会犯错。我认为有更有效的方法,但我还没有找到。
诗。我的目标是仅在所有变量都是唯一的情况下才弹出消息框。
我已经实现了我的目标,尽管它可能比以下方法更有效:
For a = 1 To 10
check(a) = True
For b = 1 To 10
If check(b) = False Then
check(b) = True
For c = 1 To 10
If check(c) = False Then
check(c) = True
For d = 1 To 10
If check(d) = False Then
check(d) = True
For e = 1 To 10
If check(e) = False Then
check(e) = True
MsgBox (a & " " & b & " " & c & " " & d & " " & e)
End If
check(e) = False
check(a) = True
check(b) = True
check(c) = True
check(d) = True
Next e
End If
check(d) = False
check(a) = True
check(b) = True
check(c) = True
Next d
End If
check(c) = False
check(a) = True
check(b) = True
Next c
End If
check(b) = False
check(a) = True
Next b
Next a
最佳答案
这是 Johnson-Trotter algorithm 的实现用于枚举排列。这是我在研究旅行商问题的强力解决方案时编写的一个小修改。请注意,它返回一个二维数组,这可能会消耗大量内存。可以对其进行重构,使其成为消耗而不是存储排列的子组件。只需将靠近底部的代码部分(其中当前排列 perm
存储在数组 perms
中)替换为使用该排列的代码即可。
Function Permutations(n As Long) As Variant
'implements Johnson-Trotter algorithm for
'listing permutations. Returns results as a variant array
'Thus not feasible for n > 10 or so
Dim perm As Variant, perms As Variant
Dim i As Long, j As Long, k As Long, r As Long, D As Long, m As Long
Dim p_i As Long, p_j As Long
Dim state As Variant
m = Application.WorksheetFunction.Fact(n)
ReDim perm(1 To n)
ReDim perms(1 To m, 1 To n) As Integer
ReDim state(1 To n, 1 To 2) 'state(i,1) = where item i is currently in perm
'state(i,2) = direction of i
k = 1 'will point to current permutation
For i = 1 To n
perm(i) = i
perms(k, i) = i
state(i, 1) = i
state(i, 2) = -1
Next i
state(1, 2) = 0
i = n 'from here on out, i will denote the largest moving
'will be 0 at the end
Do While i > 0
D = state(i, 2)
'swap
p_i = state(i, 1)
p_j = p_i + D
j = perm(p_j)
perm(p_i) = j
state(i, 1) = p_j
perm(p_j) = i
state(j, 1) = p_i
p_i = p_j
If p_i = 1 Or p_i = n Then
state(i, 2) = 0
Else
p_j = p_i + D
If perm(p_j) > i Then state(i, 2) = 0
End If
For j = i + 1 To n
If state(j, 1) < p_i Then
state(j, 2) = 1
Else
state(j, 2) = -1
End If
Next j
'now find i for next pass through loop
If i < n Then
i = n
Else
i = 0
For j = 1 To n
If state(j, 2) <> 0 And j > i Then i = j
Next j
End If
'record perm in perms:
k = k + 1
For r = 1 To n
perms(k, r) = perm(r)
Next r
Loop
Permutations = perms
End Function
测试如下:
Sub test()
Range("A1:G5040").Value = Permutations(7)
Dim A As Variant, i As Long, s As String
A = Permutations(10)
For i = 1 To 10
s = s & " " & A(3628800, i)
Next i
Debug.Print s
End Sub
前 20 行输出如下所示:
此外,2 1 3 4 5 6 7 8 9 10
会打印在立即窗口中。我的第一个版本使用了普通变体,并导致 n = 10
出现内存不足错误。我对其进行了调整,以便重新调整 perms
的大小以包含整数(与变体相比,它消耗的内存更少),并且现在能够处理 10
。在我的机器上运行测试代码大约需要10秒。
关于vba - 为什么多个连续不等条件在vba中不起作用?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40162359/