Question

是否使用<条码>VBA或一种公式在<条码>kx+m string中找到“k”和“m”变量?

有若干种假设情景,说明加x+m的扼杀可如何看待,例如:

312*x+12
12+x*2
-4-x

等等。我确信,我可以通过在Excel撰写非常复杂的公式来解决这一问题,但我想也许有人已经解决了这一和类似的问题。我是迄今最直射的,但它没有处理所有尚未发生的情况(例如,在X+m体内有两点:

=TRIM(IF(NOT(ISERROR(SEARCH("~+";F5))); IF(SEARCH("~+";F5)>SEARCH("~*";F5);RIGHT(F5;LEN(F5)-SEARCH("~+";F5));LEFT(F5;SEARCH("~+";F5)-1)); IF(NOT(ISERROR(SEARCH("~-";F5))); IF(SEARCH("~-";F5)>SEARCH("~*";F5);RIGHT(F5;LEN(F5)-SEARCH("~-";F5)+1);LEFT(F5;SEARCH("~*";F5)-1));"")))

Answer 1

我确信这将有助于你:

将这一职能纳入一个单元:

Function FindKXPlusM(ByVal str As String) As String
    Dim K As String, M As String
    Dim regex As Object, matches As Object, sm As Object

       remove unwanted spaces from input string (if any)
    str = Replace(str, " ", "")

       create an instance of RegEx object.
       I m using late binding here, but you can use early binding too.
    Set regex = CreateObject("VBScript.RegExp")
    regex.IgnoreCase = True
    regex.Global = True

       test for kx+m or xk+m types
    regex.Pattern = "^(-?d*)*?x([+-]?d+)?$|^x*(-?d+)([+-]?d+)?$"
    Set matches = regex.Execute(str)
    If matches.Count >= 1 Then
        Set sm = matches(0).SubMatches
        K = sm(0)
        M = sm(1)
        If K = "" Then K = sm(2)
        If M = "" Then M = sm(3)
        If K = "-" Or K = "+" Or K = "" Then K = K & "1"
        If M = "" Then M = "0"
    Else
           test for m+kx or m+xk types
        regex.Pattern = "^(-?d+)[+-]x*([+-]?d+)$|^(-?d+)([+-]d*)*?x$"
        Set matches = regex.Execute(str)
        If matches.Count >= 1 Then
            Set sm = matches(0).SubMatches
            M = sm(0)
            K = sm(1)
            If M = "" Then M = sm(2)
            If K = "" Then K = sm(3)
            If K = "-" Or K = "+" Or K = "" Then K = K & "1"
            If M = "" Then M = "0"
        End If
    End If
    K = Replace(K, "+", "")
    M = Replace(M, "+", "")

       the values found are in K & M.
       I output here in this format only for showing sample.
    FindKXPlusM = " K = " & K & "         M = " & M
End Function

Then you can either call it from a Macro e.g. like this:

Sub Test()
    Debug.Print FindKXPlusM("x*312+12")
End Sub

Or use it like a formula. e.g. by putting this in a cell:

=FindKXPlusM(B1)

我喜欢第二种方式(无工作:P)

我用各种价值观对它进行了测试,在此放映了我获得的以下信息:

“Screenshot

Hope this helps :)

Answer 2

而没有双管齐下,则在VBA中操作一个简单的<代码>LINEST。

Replace StrFunc as needed

Sub Extract()
Dim strFunc As String
Dim X(1 To 2) As Variant
Dim Y(1 To 2) As Variant
Dim C As Variant

X(1) = 0
X(2) = 100

strFunc = "312*x+12"
 strFunc = "12+x*2 "
 strFunc = "-4-X"

Y(1) = Evaluate(Replace(LCase$(strFunc), "x", X(1)))
Y(2) = Evaluate(Replace(LCase$(strFunc), "x", X(2)))
C = Application.WorksheetFunction.LinEst(Y, X)

MsgBox "K is " & C(1) & vbNewLine & "M is " & C(2)

End Sub

Answer 3

它比看起来更为复杂。如果我不错的话,每小时加一米可以有Max7的操作员和1名操作员。在这种情形下,获得“K”和“M”价值确实变得复杂。 —— Siddharth Rout 33mins before

Building on my comment in duffymo s post

This snapshot shows the different combinations that “kx + m” can have

“entergraph

如前所述,实现你想要的东西非常复杂。这里是我的feebletries,目前仅提取“K”。 <><>>> 这部法律绝无任何区别。此外,我没有用不同的假设情景测试该守则,因此可能与其他人失败。然而,这给你带来了如何解决这一问题的公平想法。你们将不得不夸张,以取得你想要的准确结果。

http://www.ohchr.org。 (我正在测试该法典中的7个可能的组合。) 它为这7项工作而努力,但可能会/会因他人而失败。

Option Explicit

Sub Sample()
    Dim StrCheck As String
    Dim posStar As Long, posBrk As Long, pos As Long, i As Long
    Dim strK As String, strM As String
    Dim MyArray(6) As String

    MyArray(0) = "-k*(-x)+(-m)*(-2)"
    MyArray(1) = "-k*x+(-m)*(-2)"
    MyArray(2) = "-k(x)+(-m)*(-2)"
    MyArray(3) = "-k(x)+(-m)(-2)"
    MyArray(4) = "-kx+m"
    MyArray(5) = "kx+m"
    MyArray(6) = "k(x)+m"

    For i = 0 To 6
        StrCheck = MyArray(i)
        Select Case Left(Trim(StrCheck), 1)

        Case "+", "-"
            posBrk = InStr(2, StrCheck, "(")
            posStar = InStr(2, StrCheck, "*")

            If posBrk > posStar Then             <~~ "-k*(-x)+(-m)*(-2)"
                pos = InStr(2, StrCheck, "*")
                If pos <> 0 Then
                    strK = Mid(StrCheck, 1, pos - 1)
                Else
                    strK = Mid(StrCheck, 1, posBrk - 1)
                End If
            ElseIf posBrk < posStar Then         <~~ "-k(-x)+(-m)*(-2)"
                pos = InStr(2, StrCheck, "(")
                strK = Mid(StrCheck, 1, pos - 1)
            Else                                 <~~ "-kx+m"
                 ~~> In such a case I am assuming that you will never use
                 ~~> a >=2 letter variable
                strK = Mid(StrCheck, 1, 2)
            End If
        Case Else
            posBrk = InStr(1, StrCheck, "(")
            posStar = InStr(1, StrCheck, "*")

            If posBrk > posStar Then             <~~ "k*(-x)+(-m)*(-2)"
                pos = InStr(1, StrCheck, "*")
                If pos <> 0 Then
                    strK = Mid(StrCheck, 1, pos - 2)
                Else
                    strK = Mid(StrCheck, 1, posBrk - 1)
                End If
            ElseIf posBrk < posStar Then         <~~ "k(-x)+(-m)*(-2)"
                pos = InStr(1, StrCheck, "(")
                strK = Mid(StrCheck, 1, pos - 2)
            Else                                 <~~ "kx+m"
                 ~~> In such a case I am assuming that you will never use
                 ~~> a >=2 letter variable
                strK = Mid(StrCheck, 1, 1)
            End If
        End Select

        Debug.Print "Found " & strK & " in " & MyArray(i)
    Next i
End Sub

<>SNAPSHOT

“enter

这不是一件好事,但我希望这能让你们走上正确的道路。

Answer 4

我在“x”之后,在“+”之后,用一个或数位数进行搜索。

你的榜样显示了愤怒的价值观。斜坡和拦截是点号? 签字价值? 这比你的例子更为复杂。

我建议,最普遍的解决办法是,为你写一个简单图表的弹性/选择文件。我不知道VB或VB。 NET向您提供。南极海生委将在贾瓦兰找到一个解决办法;有《南极海生委》。 NET。

我不相信这一努力会给你带来什么。你们将做些什么? 我认为,用户在数字类型电池中填充千米和百分数并计算<代码>y = m*x + k时,从这些单位中填满,而不是插入插图和提取。

If your goal is simply to evaluate a String, maybe eval() is your answer:

How to turn a string formula into a "real" formula

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