Excel 能给我解释一下这个代码吗？

我有下面的工作VBA代码。它在一个新列中列出了序列中缺少的数字，但我现在不知道代码是如何工作的。我不理解For循环和scripting.dictionary

Const rng As String = "$a$2:$a$218"
Dim d As Object, a, c()
Dim i As Long, mx As Long, mn As Long

Set d = CreateObject("scripting.dictionary")
a = Range(rng)

mx = Application.Max(a): mn = Application.Min(a)

ReDim c(1 To mx - mn + 1, 1 To 1)


For i = 1 To UBound(a): d(a(i, 1)) = 1: Next i

For i = mn To mx
    If d(i) <> 1 Then k = k + 1: c(k, 1) = i
Next i


Range("d2").Resize(k) = c

Const rng As String=“$a$2:$a$218”
尺寸d作为对象，a，c（）
Dim i为长，mx为长，mn为长
Set d=CreateObject（“scripting.dictionary”）
a=范围（rng）
mx=Application.Max（a）：mn=Application.Min（a）
雷迪姆c（1对mx-mn+1，1对1）
对于i=1到UBound（a）：d（a（i，1））=1：下一个i
对于i=mn到mx
如果d（i）1，那么k=k+1:c（k，1）=i
接下来我
范围（“d2”）。调整大小（k）=c

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有人能帮我吗？

这个外观将告诉您有关

字典的内容

对象：

您还可以在此处查看文档：

你的代码正在做什么

创建dictionary对象并分配给变量

d

Set d = CreateObject("scripting.dictionary")

a = Range(rng)

mx = Application.Max(a): mn = Application.Min(a)

ReDim c(1 To mx - mn + 1, 1 To 1)

For i = 1 To UBound(a): d(a(i, 1)) = 1: Next i

将指定范围常数中的值分配给范围数组，

a

Set d = CreateObject("scripting.dictionary")

a = Range(rng)

mx = Application.Max(a): mn = Application.Min(a)

ReDim c(1 To mx - mn + 1, 1 To 1)

For i = 1 To UBound(a): d(a(i, 1)) = 1: Next i

确定范围数组中的最小值和最大值，

a

Set d = CreateObject("scripting.dictionary")

a = Range(rng)

mx = Application.Max(a): mn = Application.Min(a)

ReDim c(1 To mx - mn + 1, 1 To 1)

For i = 1 To UBound(a): d(a(i, 1)) = 1: Next i

重新确定数组

c

的尺寸，以便它将在

mn

和

mx

之间保留适当数量的值

Set d = CreateObject("scripting.dictionary")

a = Range(rng)

mx = Application.Max(a): mn = Application.Min(a)

ReDim c(1 To mx - mn + 1, 1 To 1)

For i = 1 To UBound(a): d(a(i, 1)) = 1: Next i

将范围数组

a

中的每个项放入字典对象

d

Set d = CreateObject("scripting.dictionary")

a = Range(rng)

mx = Application.Max(a): mn = Application.Min(a)

ReDim c(1 To mx - mn + 1, 1 To 1)

For i = 1 To UBound(a): d(a(i, 1)) = 1: Next i

从

mn

循环到

mx

，基本上检查字典

d

中是否存在该值，如果不存在，则将该值放入数组

c

For i = mn To mx
    If d(i) <> 1 Then k = k + 1: c(k, 1) = i
Next i

更新：可能修改为：

Const rng As String = "$a$2:$a$218"
Sub DoThis()
Dim d As Object, a, c()
Dim i As Long, mx As Long, mn As Long

Set d = CreateObject("scripting.dictionary")
a = Range(rng)

mx = Application.Max(a): mn = Application.Min(a)

For i = mn To mx: d(i) = True: Next i

For i = 1 To UBound(a, 1)
    '# You should not actually NEED to use the If d.Exists, 
    ' you can probably just do:
    ' d.Remove(a(i, 1))
    ' but just in case, I test to see if it exists:
    If d.Exists(a(i, 1)) Then d.Remove(a(i, 1))
Next i

Range("d2").Resize(UBound(d.Keys()) + 1) = Application.Transpose(d.Keys())
End Sub

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我我并不感到奇怪，因为编写它的人使用了一些非常隐晦的命名约定（例如，

rng

是“传统上”的

范围对象，而不是字符串），
a
不是很有描述性（可以改为“initialValuesArr”）。不管怎样，David Zemens的回答很好地解决了这个问题谢谢你的详细解释，这正是我想要的，现在我可以在理解代码的同时使用它了^ ^没问题！我发布的修订版应该做同样的事情，但它绕过了其中两个循环，通过直接将字典的
Keys（）
（已经是一个数组！）打印到工作表中，我们避免了使用变量
c
。