Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
706 views
in Technique[技术] by (71.8m points)

excel - VBA Removing duplicates values in an array including the same value

There is a way to remove all duplicates in array with VBA, also the first value. Just keeping the not Duplicated values

Example:

Array_1 ['pedro','maria','jose','jesus','pepe','pepe','jose']

Result:

Array_1 ['pedro','maria','jesus']
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Try this code:

Sub Remove_All_Duplicated()
Dim Array_1
    Array_1 = Array("pedro", "maria", "jose", "jesus", "pepe", "pepe", "jose")
Dim Array_2()

Dim eleArr_1, x
x = 0
For Each eleArr_1 In Array_1
    If UBound(Filter(Array_1, eleArr_1)) = 0 Then
        ReDim Preserve Array_2(x)
        Array_2(x) = eleArr_1
        x = x + 1
    End If
Next

End Sub

Additional solution as Filter function doesn't care about 'exact match'. This new one requires reference to Microsoft Scripting Runtime in VBA project.

Sub alternative()
Dim Array_1
    Array_1 = Array("pedro", "pedro maria", "maria", "jose", "jesus", "pepe", "pepe", "jose")
Dim Array_2()
Dim Array_toRemove()

Dim dic As New Scripting.Dictionary
Dim arrItem, x As Long
For Each arrItem In Array_1
    If Not dic.Exists(arrItem) Then
        dic.Add arrItem, arrItem
    Else
        ReDim Preserve Array_toRemove(x)
        Array_toRemove(x) = dic.Item(arrItem)
        x = x + 1
    End If
Next
For Each arrItem In Array_toRemove
    dic.Remove (arrItem)
Next arrItem
Array_2 = dic.Keys

'quic tests to remove when unnecessary
Debug.Print UBound(Array_2), UBound(Array_toRemove)
Debug.Print Join(Array_2, "/")

End Sub

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...