Question

我无法弄清楚Filter功能如何如此快速地工作。我已经对所有类型的数据使用了Filter，无论数据类型如何，Filter都会删除我使用的任何替代方法。我经常使用二进制搜索算法和由Stephen Bullen编写的QuickArraySort算法（在Professional Excel Development中找到）。二进制搜索是快速的（与Filter函数一样快，假设数组已排序），Quick Sort算法是已知的最快排序算法之一。

我在下面编写了一些测试代码，比较了在一个非常大的数组中查找随机元素的速度（大小= 2,000,000）。我故意以无序的方式填充数组（应该注意我已尝试过各种无序的赋值方法，无论分配方法如何，结果都相似）。

Sub SearchTest()

Dim i As Long, strMyArray() As String, lngSize As Long, strTest As String
Dim TimeBinarySearch As Long, TimeFilterSearch As Long
Dim lngResultBinary As Long, lngResultFilter As Long

Dim StartHour As Long, StartMinute As Long, StartSecond As Long
Dim StartMiliSecond As Long, StartTime As Long

Dim EndHour As Long, EndMinute  As Long, EndSecond As Long
Dim EndMiliSecond As Long, EndTime As Long

    lngSize = 2000000

    strTest = CStr(1735674 * 987)

    ReDim strMyArray(lngSize)

    For i = 1 To UBound(strMyArray)
        If i Mod 2 = 0 Then
            strMyArray(i) = CStr((i - 1) * 987)
        Else
            strMyArray(i) = CStr((i + 1) * 987)
        End If
    Next i

''Filter Test
'*******************************************************************
    StartHour = Hour(Now()) * 60 * 60 * 1000
    StartMinute = Minute(Now()) * 60 * 1000
    StartSecond = Second(Now()) * 1000
    StartMiliSecond = Format(Now(), "ms")

    StartTime = StartHour + StartMinute + StartSecond + StartMiliSecond

    lngResultFilter = CLng(Filter(strMyArray, strTest)(0))

    EndHour = Hour(Now()) * 60 * 60 * 1000
    EndMinute = Minute(Now()) * 60 * 1000
    EndSecond = Second(Now()) * 1000
    EndMiliSecond = Format(Now(), "ms")

    EndTime = EndHour + EndMinute + EndSecond + EndMiliSecond

    TimeFilterSearch = EndTime - StartTime
'*******************************************************************

''Binary Test
'*******************************************************************
    StartHour = Hour(Now()) * 60 * 60 * 1000
    StartMinute = Minute(Now()) * 60 * 1000
    StartSecond = Second(Now()) * 1000
    StartMiliSecond = Format(Now(), "ms")

    StartTime = StartHour + StartMinute + StartSecond + StartMiliSecond

    QuickSortString1D strMyArray

    lngResultBinary = strMyArray(CLng(BinarySearchString(strTest, strMyArray)))

    EndHour = Hour(Now()) * 60 * 60 * 1000
    EndMinute = Minute(Now()) * 60 * 1000
    EndSecond = Second(Now()) * 1000
    EndMiliSecond = Format(Now(), "ms")

    EndTime = EndHour + EndMinute + EndSecond + EndMiliSecond

    TimeBinarySearch = EndTime - StartTime
'*******************************************************************

    MsgBox lngResultFilter & vbCr & vbCr & lngResultBinary 

    MsgBox TimeFilterSearch & vbCr & vbCr & TimeBinarySearch

End Sub

两种方法都返回相同的结果，但Filter方法的返回时间为0 ms，QuickSort / BinarySearch方法的返回时间接近20秒。这是一个巨大的差异!!如前所述，如果对数组进行排序，则二进制搜索方法也会返回0 ms（大多数人都知道，二进制搜索需要对数组进行排序才能开始）

那么，Filter函数如何查看2,000,000个未排序的条目并立即找到正确的结果？它不能简单地遍历每个条目并将其与filtervalue进行比较（这是迄今为止最慢的方法），但基于这些初步测试，它也不能使用二进制搜索，因为它必须排序数组首先。即使有一个已编译的令人敬畏的排序算法，我发现很难相信它可以立即对大小超过一百万的数组进行排序。

顺便说一句，下面是QuickSort算法和二进制搜索算法。

    Sub QuickSortString1D(ByRef saArray() As String, _
                Optional ByVal bSortAscending As Boolean = True, _
                Optional ByVal lLow1 As Variant, _
                Optional ByVal lHigh1 As Variant)

    'Dimension variables
    Dim lLow2 As Long
    Dim lHigh2 As Long
    Dim sKey As String
    Dim sSwap As String

        On Error GoTo ErrorExit
        'If not provided, sort the entire array
        If IsMissing(lLow1) Then lLow1 = LBound(saArray)
        If IsMissing(lHigh1) Then lHigh1 = UBound(saArray)

        'Set new extremes to old extremes
        lLow2 = lLow1
        lHigh2 = lHigh1

        'Get value of array item in middle of new extremes
        sKey = saArray((lLow1 + lHigh1) \ 2)

        'Loop for all the items in the array between the extremes
        Do While lLow2 < lHigh2

            If bSortAscending Then
                'Find the first item that is greater than the mid-point item
                Do While saArray(lLow2) < sKey And lLow2 < lHigh1
                    lLow2 = lLow2 + 1
                Loop

                'Find the last item that is less than the mid-point item
                Do While saArray(lHigh2) > sKey And lHigh2 > lLow1
                    lHigh2 = lHigh2 - 1
                Loop
            Else
                'Find the first item that is less than the mid-point item
                Do While saArray(lLow2) > sKey And lLow2 < lHigh1
                    lLow2 = lLow2 + 1
                Loop

                'Find the last item that is greater than the mid-point item
                Do While saArray(lHigh2) < sKey And lHigh2 > lLow1
                    lHigh2 = lHigh2 - 1
                Loop

            End If

            'If the two items are in the wrong order, swap the rows
            If lLow2 < lHigh2 Then
                sSwap = saArray(lLow2)
                saArray(lLow2) = saArray(lHigh2)
                saArray(lHigh2) = sSwap
            End If

            'If the pointers are not together, advance to the next item
            If lLow2 <= lHigh2 Then
                lLow2 = lLow2 + 1
                lHigh2 = lHigh2 - 1
            End If
        Loop

        'Recurse to sort the lower half of the extremes
        If lHigh2 > lLow1 Then
            QuickSortString1D saArray, bSortAscending, lLow1, lHigh2
        End If

        'Recurse to sort the upper half of the extremes
        If lLow2 < lHigh1 Then
            QuickSortString1D saArray, bSortAscending, lLow2, lHigh1
        End If

    ErrorExit:

    End Sub

    '***********************************************************
    ' Comments: Uses a binary search algorithm to quickly locate
    ' a string within a sorted array of strings
    '
    ' Arguments: sLookFor The string to search for in the array
    ' saArray An array of strings, sorted ascending
    ' lMethod Either vbBinaryCompare or vbTextCompare
    ' Defaults to vbTextCompare
    ' lNotFound The value to return if the text isn’t
    ' found. Defaults to -1
    '
    ' Returns: Long The located position in the array,
    ' or lNotFound if not found
    '
    ' Date Developer Action
    ' ———————————————————————————————-
    ' 02 Jun 04 Stephen Bullen Created
    '
    Function BinarySearchString(ByRef sLookFor As String, _
                ByRef saArray() As String, _
                Optional ByVal lMethod As VbCompareMethod = vbTextCompare, _
                Optional ByVal lNotFound As Long = -1) As Long

    Dim lLow As Long
    Dim lMid As Long
    Dim lHigh As Long
    Dim lComp As Long

        On Error GoTo ErrorExit

        'Assume we didn’t find it
        BinarySearchString = lNotFound

        'Get the starting positions
        lLow = LBound(saArray)
        lHigh = UBound(saArray)

        Do
            'Find the midpoint of the array
            lMid = (lLow + lHigh) \ 2

            'Compare the mid-point element to the string being searched for
            lComp = StrComp(saArray(lMid), sLookFor, lMethod)

            If lComp = 0 Then
                'We found it, so return the location and quit
                BinarySearchString = lMid
                Exit Do
            ElseIf lComp = 1 Then
                'The midpoint item is bigger than us - throw away the top half
                lHigh = lMid - 1
            Else
                'The midpoint item is smaller than us - throw away the bottom half
                lLow = lMid + 1
            End If

            'Continue until our pointers cross
        Loop Until lLow > lHigh

    ErrorExit:

    End Function

编辑：看来我应该先做一些“野蛮”的力量测试。通过简单地以线性方式循环遍历数组，如John Coleman建议Filter函数执行的那样，上述相同场景的返回时间为0 ms。见下文：

Sub Test3()

Dim i As Long, strMyArray() As String, lngSize As Long, strTest As String
Dim lngResultBrute As Long, TimeBruteSearch As Long

    lngSize = 2000000
    strTest = CStr(936740 * 97)
    ReDim strMyArray(lngSize)

    For i = 1 To UBound(strMyArray)
        If i Mod 2 = 0 Then
            strMyArray(i) = CStr((i - 1) * 97)
        Else
            strMyArray(i) = CStr((i + 1) * 97)
        End If
    Next i

    StartTime = Timer

    ' Brute force search
    For i = 1 To UBound(strMyArray)
        If strMyArray(i) = strTest Then
            lngResultBrute = CLng(strTest)
            Exit For
        End If
    Next i

    EndTime = Timer

    TimeBruteSearch = EndTime - StartTime
    MsgBox TimeBruteSearch

End Sub

Answer 1

Filter确实使用线性搜索 - 它只是快速执行它，因为它是在高度优化的C / C ++代码中实现的。要查看此内容，请运行以下代码：

Function RandString(n As Long) As String
    'returns a random string in B-Z
    Dim i As Long
    Dim s As String
    For i = 1 To n
        s = s & Chr(66 + Int(25 * Rnd()))
    Next i
    RandString = s
End Function

Sub test()
    Dim times(1 To 20) As Double
    Dim i As Long, n As Long
    Dim A() As String
    Dim start As Double
    Dim s As String
    Randomize
    s = RandString(99)
    ReDim A(1 To 2000000)
    For i = 1 To 2000000
        A(i) = s + RandString(1)
    Next i
    s = s & "A"
    For i = 20 To 1 Step -1
        n = i * 100000
        ReDim Preserve A(1 To n)
        start = Timer
        Debug.Print UBound(Filter(A, s)) 'should be -1
        times(i) = Timer - start
    Next i
    For i = 1 To 20
        Cells(i, 1) = i
        Cells(i, 2) = times(i)
    Next i
End Sub

此代码创建一个2,000,000个长度为100的随机字符串数组，每个字符串与最后一个位置的目标字符串不同。然后它将大小为100,000的倍数的子阵列输入Filter，计算所需的时间。输出如下：

明确的线性趋势并未完全证明，但有力证据表明VBA的Filter正在执行简单的线性搜索。

Answer 2

我相信你在这里比较苹果和橘子。看起来当您测试Filter函数时，将无序数组作为输入，然后使用Filter查找与测试值的匹配项。 Intuition说O（N）= 200万次操作---你测试每个数组元素一次。然后你就完成了。

当您使用自定义VBA函数进行过滤时，首先排序，这是非常昂贵的O（N * Log2（N））= 2900万。一旦对数组进行了排序，你就可以获得搜索有序数组的好处，即O（Log2（N））= 14.即使你加快了搜索速度，排序的惩罚也会导致你死亡。

希望有所帮助。

使用VBA过滤器函数时的性能注意事项

2 个答案: