Does VBA have Dictionary Structure?

Question

Yes.

Set a reference to MS Scripting runtime ('Microsoft Scripting Runtime'). As per @regjo's comment, go to Tools->References and tick the box for 'Microsoft Scripting Runtime'.

Create a dictionary instance using the code below:

Set dict = CreateObject("Scripting.Dictionary")

or

Dim dict As New Scripting.Dictionary 

Example of use:

If Not dict.Exists(key) Then 
    dict.Add key, value
End If 

Don't forget to set the dictionary to Nothing when you have finished using it.

Set dict = Nothing 

VBA has the collection object:

    Dim c As Collection
    Set c = New Collection
    c.Add "Data1", "Key1"
    c.Add "Data2", "Key2"
    c.Add "Data3", "Key3"
    'Insert data via key into cell A1
    Range("A1").Value = c.Item("Key2")

The Collection object performs key-based lookups using a hash so it's quick.

---

You can use a Contains() function to check whether a particular collection contains a key:

Public Function Contains(col As Collection, key As Variant) As Boolean
    On Error Resume Next
    col(key) ' Just try it. If it fails, Err.Number will be nonzero.
    Contains = (Err.Number = 0)
    Err.Clear
End Function

Edit 24 June 2015: Shorter Contains() thanks to @TWiStErRob.

Edit 25 September 2015: Added Err.Clear() thanks to @scipilot.

VBA does not have an internal implementation of a dictionary, but from VBA you can still use the dictionary object from MS Scripting Runtime Library.

Dim d
Set d = CreateObject("Scripting.Dictionary")
d.Add "a", "aaa"
d.Add "b", "bbb"
d.Add "c", "ccc"

If d.Exists("c") Then
    MsgBox d("c")
End If

An additional dictionary example that is useful for containing frequency of occurence.

Outside of loop:

Dim dict As New Scripting.dictionary
Dim MyVar as String

Within a loop:

'dictionary
If dict.Exists(MyVar) Then
    dict.Item(MyVar) = dict.Item(MyVar) + 1 'increment
Else
    dict.Item(MyVar) = 1 'set as 1st occurence
End If

To check on frequency:

Dim i As Integer
For i = 0 To dict.Count - 1 ' lower index 0 (instead of 1)
    Debug.Print dict.Items(i) & " " & dict.Keys(i)
Next i