rpa*_*s21 9 vba ms-word word-vba
我得到的印象是,这是不可能的,但我想如果你正在寻找任何3-4个单词,在一个很长的论文中的任何地方出现相同的序列,我可以找到相同短语的重复.
我从过去的论文中复制并粘贴了大量文档,并希望找到一种简单的方法来查找这个40多页文档中的重复信息有很多不同的格式,但我愿意暂时摆脱格式化顺序找到重复的信息.
Sid*_*out 16
要突出显示所有重复的句子,您也可以使用ActiveDocument.Sentences(i).这是一个例子
逻辑
1)从数组中获取word文档中的所有句子
2)对数组进行排序
3)提取重复项
4)突出显示重复项
码
Option Explicit
Sub Sample()
Dim MyArray() As String
Dim n As Long, i As Long
Dim Col As New Collection
Dim itm
n = 0
'~~> Get all the sentences from the word document in an array
For i = 1 To ActiveDocument.Sentences.Count
n = n + 1
ReDim Preserve MyArray(n)
MyArray(n) = Trim(ActiveDocument.Sentences(i).Text)
Next
'~~> Sort the array
SortArray MyArray, 0, UBound(MyArray)
'~~> Extract Duplicates
For i = 1 To UBound(MyArray)
If i = UBound(MyArray) Then Exit For
If InStr(1, MyArray(i + 1), MyArray(i), vbTextCompare) Then
On Error Resume Next
Col.Add MyArray(i), """" & MyArray(i) & """"
On Error GoTo 0
End If
Next i
'~~> Highlight duplicates
For Each itm In Col
Selection.Find.ClearFormatting
Selection.HomeKey wdStory, wdMove
Selection.Find.Execute itm
Do Until Selection.Find.Found = False
Selection.Range.HighlightColorIndex = wdPink
Selection.Find.Execute
Loop
Next
End Sub
'~~> Sort the array
Public Sub SortArray(vArray As Variant, i As Long, j As Long)
Dim tmp As Variant, tmpSwap As Variant
Dim ii As Long, jj As Long
ii = i: jj = j: tmp = vArray((i + j) \ 2)
While (ii <= jj)
While (vArray(ii) < tmp And ii < j)
ii = ii + 1
Wend
While (tmp < vArray(jj) And jj > i)
jj = jj - 1
Wend
If (ii <= jj) Then
tmpSwap = vArray(ii)
vArray(ii) = vArray(jj): vArray(jj) = tmpSwap
ii = ii + 1: jj = jj - 1
End If
Wend
If (i < jj) Then SortArray vArray, i, jj
If (ii < j) Then SortArray vArray, ii, j
End Sub
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快照
之前

后

我没有使用我自己的 DAWG 建议,我仍然有兴趣看看其他人是否有办法做到这一点,但我能够想出这个:
Option Explicit
Sub test()
Dim ABC As Scripting.Dictionary
Dim v As Range
Dim n As Integer
n = 5
Set ABC = FindRepeatingWordChains(n, ActiveDocument)
' This is a dictionary of word ranges (not the same as an Excel range) that contains the listing of each word chain/phrase of length n (5 from the above example).
' Loop through this collection to make your selections/highlights/whatever you want to do.
If Not ABC Is Nothing Then
For Each v In ABC
v.Font.Color = wdColorRed
Next v
End If
End Sub
' This is where the real code begins.
Function FindRepeatingWordChains(ChainLenth As Integer, DocToCheck As Document) As Scripting.Dictionary
Dim DictWords As New Scripting.Dictionary, DictMatches As New Scripting.Dictionary
Dim sChain As String
Dim CurWord As Range
Dim MatchCount As Integer
Dim i As Integer
MatchCount = 0
For Each CurWord In DocToCheck.Words
' Make sure there are enough remaining words in our document to handle a chain of the length specified.
If Not CurWord.Next(wdWord, ChainLenth - 1) Is Nothing Then
' Check for non-printing characters in the first/last word of the chain.
' This code will read a vbCr, etc. as a word, which is probably not desired.
' However, this check does not exclude these 'words' inside the chain, but it can be modified.
If CurWord <> vbCr And CurWord <> vbNewLine And CurWord <> vbCrLf And CurWord <> vbLf And CurWord <> vbTab And _
CurWord.Next(wdWord, ChainLenth - 1) <> vbCr And CurWord.Next(wdWord, ChainLenth - 1) <> vbNewLine And _
CurWord.Next(wdWord, ChainLenth - 1) <> vbCrLf And CurWord.Next(wdWord, ChainLenth - 1) <> vbLf And _
CurWord.Next(wdWord, ChainLenth - 1) <> vbTab Then
sChain = CurWord
For i = 1 To ChainLenth - 1
' Add each word from the current word through the next ChainLength # of words to a temporary string.
sChain = sChain & " " & CurWord.Next(wdWord, i)
Next i
' If we already have our temporary string stored in the dictionary, then we have a match, assign the word range to the returned dictionary.
' If not, then add it to the dictionary and increment our index.
If DictWords.Exists(sChain) Then
MatchCount = MatchCount + 1
DictMatches.Add DocToCheck.Range(CurWord.Start, CurWord.Next(wdWord, ChainLenth - 1).End), MatchCount
Else
DictWords.Add sChain, sChain
End If
End If
End If
Next CurWord
' If we found any matching results, then return that list, otherwise return nothing (to be caught by the calling function).
If DictMatches.Count > 0 Then
Set FindRepeatingWordChains = DictMatches
Else
Set FindRepeatingWordChains = Nothing
End If
End Function
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我已经在来自该源的258 页文档 ( TheStory.txt)上对此进行了测试,并且只用了几分钟就运行了。
请参阅test()子部分以了解用法。
您需要引用 Microsoft Scripting Runtime 才能使用这些Scripting.Dictionary对象。如果这是不可取的,可以进行一些小的修改Collections来代替使用,但我更喜欢它,Dictionary因为它有有用的.Exists()方法。