确定化学式中的原子总数

Question

我有数千个化学式的列表，其中可能包含任何元素的符号。我想确定每个公式中任何元素的原子总数。示例包括：

• CH ₃ NO ₃
• 硒化硒₂
• C ₂氯₂
• C ₂ Cl ₂ O ₂
• C ₂ Cl ₃ F
• C ₂ H ₂ BrF ₃
• C ₂ H ₂ Br ₂
• C ₂ H ₃ Cl ₃ Si

我想要单个式中的原子总数，因此对于第一个示例 (CH ₃ NO ₃ )，答案为 8（1 个碳 + 3 个氢 + 1 个氮 + 3 个氧）。

我找到了 PEH（从化学式中提取数字）的代码，它使用正则表达式来提取化学式中特定元素的实例数。

这可以用来给出总原子吗？

Public Function ChemRegex(ChemFormula As String, Element As String) As Long
    Dim regEx As New RegExp
    With regEx
        .Global = True
        .MultiLine = True
        .IgnoreCase = False
    End With
    
    'first pattern matches every element once
    regEx.Pattern = "([A][cglmrstu]|[B][aehikr]?|[C][adeflmnorsu]?|[D][bsy]|[E][rsu]|[F][elmr]?|[G][ade]|[H][efgos]?|[I][nr]?|[K][r]?|[L][airuv]|[M][cdgnot]|[N][abdehiop]?|[O][gs]?|[P][abdmortu]?|[R][abefghnu]|[S][bcegimnr]?|[T][abcehilms]|[U]|[V]|[W]|[X][e]|[Y][b]?|[Z][nr])([0-9]*)"
    
    Dim Matches As MatchCollection
    Set Matches = regEx.Execute(ChemFormula)
    
    Dim m As Match
    For Each m In Matches
        If m.SubMatches(0) = Element Then
            ChemRegex = ChemRegex + IIf(Not m.SubMatches(1) = vbNullString, m.SubMatches(1), 1)
        End If
    Next m
    
    'second patternd finds parenthesis and multiplies elements within
    regEx.Pattern = "(\((.+?)\)([0-9])+)+?"
    Set Matches = regEx.Execute(ChemFormula)
    For Each m In Matches
        ChemRegex = ChemRegex + ChemRegex(m.SubMatches(1), Element) * (m.SubMatches(2) - 1) '-1 because all elements were already counted once in the first pattern
    Next m
End Function

Answer 1

您可以通过循环遍历所有字符来做到这一点。计算所有大写字符的数量，并将所有数字减去 1。这就是元素的总数。

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Option Explicit\n\nPublic Function ChemCountTotalElements(ByVal ChemFormula As String) As Long\n    Dim RetVal As Long\n\n    Dim c As Long\n    For c = 1 To Len(ChemFormula)\n        Dim Char As String\n        Char = Mid$(ChemFormula, c, 1)\n        \n        If IsNumeric(Char) Then\n            RetVal = RetVal + CLng(Char) - 1\n        ElseIf Char = UCase(Char) Then\n            RetVal = RetVal + 1\n        End If\n        \n    Next c\n    \n    ChemCountTotalElements = RetVal\nEnd Function\n

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请注意，这不处理括号！并且它不会检查该元素是否确实存在。所以XYZ2会被算作4.

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另外，这只能处理以下数字10。如果您有 和10以上的数字，请使用下面的正则表达式解决方案（可以处理该问题）。

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还可以识别带前缀的化学式，例如 Ca(OH)\xe2\x82\x82

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如果您需要更精确的方法（检查元素是否存在）并识别括号，您需要再次使用 RegEx 来完成。

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由于 VBA 不支持开箱即用的正则表达式，因此我们需要首先引用 Windows 库。

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1. 在“工具”下添加对正则表达式的引用，然后在“引用”下
   添加\n

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2. 并选择Microsoft VBScript 正则表达式 5.5
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3. 将此功能添加到模块中

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   Public Function ChemRegexCountTotalElements(ByVal ChemFormula As String) As Long\n    Dim RetVal As Long\n\n    Dim regEx As New RegExp\n    With regEx\n        .Global = True\n        .MultiLine = True\n        .IgnoreCase = False\n    End With\n\n    \'first pattern matches every element once\n    regEx.Pattern = "([A][cglmrstu]|[B][aehikr]?|[C][adeflmnorsu]?|[D][bsy]|[E][rsu]|[F][elmr]?|[G][ade]|[H][efgos]?|[I][nr]?|[K][r]?|[L][airuv]|[M][cdgnot]|[N][abdehiop]?|[O][gs]?|[P][abdmortu]?|[R][abefghnu]|[S][bcegimnr]?|[T][abcehilms]|[U]|[V]|[W]|[X][e]|[Y][b]?|[Z][nr])([0-9]*)"\n\n    Dim Matches As MatchCollection\n    Set Matches = regEx.Execute(ChemFormula)\n\n    Dim m As Match\n    For Each m In Matches\n        RetVal = RetVal + IIf(Not m.SubMatches(1) = vbNullString, m.SubMatches(1), 1)\n    Next m\n\n    \'second patternd finds parenthesis and multiplies elements within\n    regEx.Pattern = "(\\((.+?)\\)([0-9]+)+)+?"\n    Set Matches = regEx.Execute(ChemFormula)\n    For Each m In Matches\n        RetVal = RetVal + ChemRegexCountTotalElements(m.SubMatches(1)) * (m.SubMatches(2) - 1) \'-1 because all elements were already counted once in the first pattern\n    Next m\n\n    ChemRegexCountTotalElements = RetVal\nEnd Function\n

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虽然此代码也可以识别括号，但请注意，它不能识别嵌套括号。

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