我有一個數組prLst,它是一個整數列表。整數沒有排序,因爲它們在數組中的位置表示電子表格上的特定列。我想知道如何在數組中找到一個特定的整數,並返回它的索引。數組Excel中的元素的返回索引VBA
似乎沒有任何資源向我展示如何將數組轉換爲工作表上的範圍。這似乎有點複雜。這對於VBA來說是不可能的嗎?
我有一個數組prLst,它是一個整數列表。整數沒有排序,因爲它們在數組中的位置表示電子表格上的特定列。我想知道如何在數組中找到一個特定的整數,並返回它的索引。數組Excel中的元素的返回索引VBA
似乎沒有任何資源向我展示如何將數組轉換爲工作表上的範圍。這似乎有點複雜。這對於VBA來說是不可能的嗎?
Dim pos, arr, val
arr=Array(1,2,4,5)
val = 4
pos=Application.Match(val, arr, False)
if not iserror(pos) then
Msgbox val & " is at position " & pos
else
Msgbox val & " not found!"
end if
更新使用匹配(帶的.index),以找到一個二維陣列的尺寸值顯示:
Dim arr(1 To 10, 1 To 2)
Dim x
For x = 1 To 10
arr(x, 1) = x
arr(x, 2) = 11 - x
Next x
Debug.Print Application.Match(3, Application.Index(arr, 0, 1), 0)
Debug.Print Application.Match(3, Application.Index(arr, 0, 2), 0)
編輯:這是值得說明這裏什麼@ARich在評論中指出 - 如果你在一個循環中使用Index()
來切片數組會有可怕的表現。
在測試中(下面的代碼)Index()方法比使用嵌套循環慢將近2000倍。
Sub PerfTest()
Const VAL_TO_FIND As String = "R1800:C8"
Dim a(1 To 2000, 1 To 10)
Dim r As Long, c As Long, t
For r = 1 To 2000
For c = 1 To 10
a(r, c) = "R" & r & ":C" & c
Next c
Next r
t = Timer
Debug.Print FindLoop(a, VAL_TO_FIND), Timer - t
' >> 0.00781 sec
t = Timer
Debug.Print FindIndex(a, VAL_TO_FIND), Timer - t
' >> 14.18 sec
End Sub
Function FindLoop(arr, val) As Boolean
Dim r As Long, c As Long
For r = 1 To UBound(arr, 1)
For c = 1 To UBound(arr, 2)
If arr(r, c) = val Then
FindLoop = True
Exit Function
End If
Next c
Next r
End Function
Function FindIndex(arr, val)
Dim r As Long
For r = 1 To UBound(arr, 1)
If Not IsError(Application.Match(val, Application.Index(arr, r, 0), 0)) Then
FindIndex = True
Exit Function
End If
Next r
End Function
這是你在找什麼?
public function GetIndex(byref iaList() as integer, byval iInteger as integer) as integer
dim i as integer
for i=lbound(ialist) to ubound(ialist)
if iInteger=ialist(i) then
GetIndex=i
exit for
end if
next i
end function
這裏的另一種方式:
Option Explicit
' Just a little test stub.
Sub Tester()
Dim pList(500) As Integer
Dim i As Integer
For i = 0 To UBound(pList)
pList(i) = 500 - i
Next i
MsgBox "Value 18 is at array position " & FindInArray(pList, 18) & "."
MsgBox "Value 217 is at array position " & FindInArray(pList, 217) & "."
MsgBox "Value 1001 is at array position " & FindInArray(pList, 1001) & "."
End Sub
Function FindInArray(pList() As Integer, value As Integer)
Dim i As Integer
Dim FoundValueLocation As Integer
FoundValueLocation = -1
For i = 0 To UBound(pList)
If pList(i) = value Then
FoundValueLocation = i
Exit For
End If
Next i
FindInArray = FoundValueLocation
End Function
循環找到一個值? – egidiocs
陣列的變體:
Public Function GetIndex(ByRef iaList() As Variant, ByVal value As Variant) As Long
Dim i As Long
For i = LBound(iaList) To UBound(iaList)
If value = iaList(i) Then
GetIndex = i
Exit For
End If
Next i
End Function
一個最快的版本爲整數(如PREF下面測試)
Public Function GetIndex(ByRef iaList() As Integer, ByVal value As Integer) As Integer
Dim i As Integer
For i = LBound(iaList) To UBound(iaList)
If iaList(i) = value Then: GetIndex = i: Exit For:
Next i
End Function
' a snippet, replace myList and myValue to your varible names: (also have not tested)
一個片段,讓我們測試一下這個假設,即通過引用傳遞參數意味着什麼。 (答案是否定的),用它代替myList中和myvalue的到你的變量名:
Dim found As Integer, foundi As Integer ' put only once
found = -1
For foundi = LBound(myList) To UBound(myList):
If myList(foundi) = myValue Then
found = foundi: Exit For
End If
Next
result = found
爲了證明這一點我已經取得了一些基準
這裏的結果:
---------------------------
Milliseconds
---------------------------
result0: 5 ' just empty loop
result1: 2702 ' function variant array
result2: 1498 ' function integer array
result3: 2511 ' snippet variant array
result4: 1508 ' snippet integer array
result5: 58493 ' excel function Application.Match on variant array
result6: 136128 ' excel function Application.Match on integer array
---------------------------
OK
---------------------------
模塊:
Public Declare Function GetTickCount Lib "kernel32.dll"() As Long
#If VBA7 Then
Public Declare PtrSafe Sub Sleep Lib "kernel32" (ByVal dwMilliseconds As LongPtr) 'For 64 Bit Systems
#Else
Public Declare Sub Sleep Lib "kernel32" (ByVal dwMilliseconds As Long) 'For 32 Bit Systems
#End If
Public Function GetIndex1(ByRef iaList() As Variant, ByVal value As Variant) As Long
Dim i As Long
For i = LBound(iaList) To UBound(iaList)
If value = iaList(i) Then
GetIndex = i
Exit For
End If
Next i
End Function
'maybe a faster variant for integers
Public Function GetIndex2(ByRef iaList() As Integer, ByVal value As Integer) As Integer
Dim i As Integer
For i = LBound(iaList) To UBound(iaList)
If iaList(i) = value Then: GetIndex = i: Exit For:
Next i
End Function
' a snippet, replace myList and myValue to your varible names: (also have not tested)
Public Sub test1()
Dim i As Integer
For i = LBound(iaList) To UBound(iaList)
If iaList(i) = value Then: GetIndex = i: Exit For:
Next i
End Sub
Sub testTimer()
Dim myList(500) As Variant, myValue As Variant
Dim myList2(500) As Integer, myValue2 As Integer
Dim n
For n = 1 To 500
myList(n) = n
Next
For n = 1 To 500
myList2(n) = n
Next
myValue = 100
myValue2 = 100
Dim oPM
Set oPM = New PerformanceMonitor
Dim result0 As Long
Dim result1 As Long
Dim result2 As Long
Dim result3 As Long
Dim result4 As Long
Dim result5 As Long
Dim result6 As Long
Dim t As Long
Dim a As Long
a = 0
Dim i
't = GetTickCount
oPM.StartCounter
For i = 1 To 1000000
Next
result0 = oPM.TimeElapsed() ' GetTickCount - t
a = 0
't = GetTickCount
oPM.StartCounter
For i = 1 To 1000000
a = GetIndex1(myList, myValue)
Next
result1 = oPM.TimeElapsed()
'result1 = GetTickCount - t
a = 0
't = GetTickCount
oPM.StartCounter
For i = 1 To 1000000
a = GetIndex2(myList2, myValue2)
Next
result2 = oPM.TimeElapsed()
'result2 = GetTickCount - t
a = 0
't = GetTickCount
oPM.StartCounter
Dim found As Integer, foundi As Integer ' put only once
For i = 1 To 1000000
found = -1
For foundi = LBound(myList) To UBound(myList):
If myList(foundi) = myValue Then
found = foundi: Exit For
End If
Next
a = found
Next
result3 = oPM.TimeElapsed()
'result3 = GetTickCount - t
a = 0
't = GetTickCount
oPM.StartCounter
For i = 1 To 1000000
found = -1
For foundi = LBound(myList2) To UBound(myList2):
If myList2(foundi) = myValue2 Then
found = foundi: Exit For
End If
Next
a = found
Next
result4 = oPM.TimeElapsed()
'result4 = GetTickCount - t
a = 0
't = GetTickCount
oPM.StartCounter
For i = 1 To 1000000
a = pos = Application.Match(myValue, myList, False)
Next
result5 = oPM.TimeElapsed()
'result5 = GetTickCount - t
a = 0
't = GetTickCount
oPM.StartCounter
For i = 1 To 1000000
a = pos = Application.Match(myValue2, myList2, False)
Next
result6 = oPM.TimeElapsed()
'result6 = GetTickCount - t
MsgBox "result0: " & result0 & vbCrLf & "result1: " & result1 & vbCrLf & "result2: " & result2 & vbCrLf & "result3: " & result3 & vbCrLf & "result4: " & result4 & vbCrLf & "result5: " & result5 & vbCrLf & "result6: " & result6, , "Milliseconds"
End Sub
名爲PerformanceMonitor
類Option Explicit
Private Type LARGE_INTEGER
lowpart As Long
highpart As Long
End Type
Private Declare Function QueryPerformanceCounter Lib "kernel32" (lpPerformanceCount As LARGE_INTEGER) As Long
Private Declare Function QueryPerformanceFrequency Lib "kernel32" (lpFrequency As LARGE_INTEGER) As Long
Private m_CounterStart As LARGE_INTEGER
Private m_CounterEnd As LARGE_INTEGER
Private m_crFrequency As Double
Private Const TWO_32 = 4294967296# ' = 256# * 256# * 256# * 256#
Private Function LI2Double(LI As LARGE_INTEGER) As Double
Dim Low As Double
Low = LI.lowpart
If Low < 0 Then
Low = Low + TWO_32
End If
LI2Double = LI.highpart * TWO_32 + Low
End Function
Private Sub Class_Initialize()
Dim PerfFrequency As LARGE_INTEGER
QueryPerformanceFrequency PerfFrequency
m_crFrequency = LI2Double(PerfFrequency)
End Sub
Public Sub StartCounter()
QueryPerformanceCounter m_CounterStart
End Sub
Property Get TimeElapsed() As Double
Dim crStart As Double
Dim crStop As Double
QueryPerformanceCounter m_CounterEnd
crStart = LI2Double(m_CounterStart)
crStop = LI2Double(m_CounterEnd)
TimeElapsed = 1000# * (crStop - crStart)/m_crFrequency
End Property
使用循環性能不佳... – Holene
我認爲糟糕的表現是使用variant作爲參數。由於預取效果,因此可能會出現這種情況。即如果可以提前讀取所有的存儲器。像所有變量都是相同的,並且讀取的順序很好。如果它在內存位置跳轉使用引用可能會工作得更慢。每次它跳過一個參考時,它會降低o(1)的性能。對於許多參考文獻而言,它就像(o(1)+ o(1)+ o(1)+ o(1))* nloop。 –
variant是一種封裝格式。像bstr和安全數組這樣的ole對象通常是系統內存中進程外的引用。並動態分配。在記憶中的不同位置。而一個安全的數組可以很容易地成爲一個引用數組。變體也可能是對參考的參考。所以它的定義應該很慢。 我想excel函數是破解系統,並針對這種類型的問題進行了優化,並以某種方式更快地忽略幾個引用和檢查時,這是可能的 –
注意數組是否從零開始。 此外,當函數返回位置0或1時,請確保函數返回的True或False不會混淆。
Function array_return_index(arr As Variant, val As Variant, Optional array_start_at_zero As Boolean = True) As Variant
Dim pos
pos = Application.Match(val, arr, False)
If Not IsError(pos) Then
If array_start_at_zero = True Then
pos = pos - 1
'initializing array at 0
End If
array_return_index = pos
Else
array_return_index = False
End If
End Function
Sub array_return_index_test()
Dim pos, arr, val
arr = Array(1, 2, 4, 5)
val = 1
'When array starts at zero
pos = array_return_index(arr, val)
If IsNumeric(pos) Then
MsgBox "Array starting at 0; Value found at : " & pos
Else
MsgBox "Not found"
End If
'When array starts at one
pos = array_return_index(arr, val, False)
If IsNumeric(pos) Then
MsgBox "Array starting at 1; Value found at : " & pos
Else
MsgBox "Not found"
End If
End Sub
它的工作原理! +1我真的不知道可以在VBA陣列上使用匹配匹配方法! –
許多Excel工作表函數都具有通過Application.WorksheetFunction提供的類似表單。[FunctionName]請注意,如果刪除WorksheetFunction部分(如在我的示例中那樣),則可以使用IsError()測試函數的返回值。如果你包含了WorksheetFunction部分,那麼(例如)在Match()沒有找到匹配的地方,它會拋出一個錯誤,你需要使用錯誤處理器來捕獲它。 –
整潔!匹配是否也適用於多維數組? – aevanko