我把Kevin Strikers优秀的解决方案转换为VB。由于我被锁定在。net 3.5，我还必须写IsNullOrWhiteSpace。这通过了他所有的测试

<Extension()>
Public Function IsNullOrWhiteSpace(value As String) As Boolean
    If value Is Nothing Then
        Return True
    End If
    For i As Integer = 0 To value.Length - 1
        If Not Char.IsWhiteSpace(value(i)) Then
            Return False
        End If
    Next
    Return True
End Function

<Extension()>
Public Function UnPascalCase(text As String) As String
    If text.IsNullOrWhiteSpace Then
        Return String.Empty
    End If

    Dim newText = New StringBuilder()
    newText.Append(text(0))
    For i As Integer = 1 To text.Length - 1
        Dim currentUpper = Char.IsUpper(text(i))
        Dim prevUpper = Char.IsUpper(text(i - 1))
        Dim nextUpper = If(text.Length > i + 1, Char.IsUpper(text(i + 1)) Or Char.IsWhiteSpace(text(i + 1)), prevUpper)
        Dim spaceExists = Char.IsWhiteSpace(text(i - 1))
        If (currentUpper And Not spaceExists And (Not nextUpper Or Not prevUpper)) Then
            newText.Append(" ")
        End If
        newText.Append(text(i))
    Next
    Return newText.ToString()
End Function

我开始做一个简单的扩展方法，基于二进制Worrier的代码，它将正确地处理首字母缩略词，并且是可重复的(不会破坏已经间隔的单词)。这是我的结果。

public static string UnPascalCase(this string text)
{
    if (string.IsNullOrWhiteSpace(text))
        return "";
    var newText = new StringBuilder(text.Length * 2);
    newText.Append(text[0]);
    for (int i = 1; i < text.Length; i++)
    {
        var currentUpper = char.IsUpper(text[i]);
        var prevUpper = char.IsUpper(text[i - 1]);
        var nextUpper = (text.Length > i + 1) ? char.IsUpper(text[i + 1]) || char.IsWhiteSpace(text[i + 1]): prevUpper;
        var spaceExists = char.IsWhiteSpace(text[i - 1]);
        if (currentUpper && !spaceExists && (!nextUpper || !prevUpper))
                newText.Append(' ');
        newText.Append(text[i]);
    }
    return newText.ToString();
}

下面是这个函数通过的单元测试用例。我把他建议的大部分案例都加到了这个清单上。其中三个没有通过的(两个只是罗马数字)被注释掉了:

Assert.AreEqual("For You And I", "ForYouAndI".UnPascalCase());
Assert.AreEqual("For You And The FBI", "ForYouAndTheFBI".UnPascalCase());
Assert.AreEqual("A Man A Plan A Canal Panama", "AManAPlanACanalPanama".UnPascalCase());
Assert.AreEqual("DNS Server", "DNSServer".UnPascalCase());
Assert.AreEqual("For You And I", "For You And I".UnPascalCase());
Assert.AreEqual("Mount Mᶜ Kinley National Park", "MountMᶜKinleyNationalPark".UnPascalCase());
Assert.AreEqual("El Álamo Tejano", "ElÁlamoTejano".UnPascalCase());
Assert.AreEqual("The Ævar Arnfjörð Bjarmason", "TheÆvarArnfjörðBjarmason".UnPascalCase());
Assert.AreEqual("Il Caffè Macchiato", "IlCaffèMacchiato".UnPascalCase());
//Assert.AreEqual("Mister Dženan Ljubović", "MisterDženanLjubović".UnPascalCase());
//Assert.AreEqual("Ole King Henry Ⅷ", "OleKingHenryⅧ".UnPascalCase());
//Assert.AreEqual("Carlos Ⅴº El Emperador", "CarlosⅤºElEmperador".UnPascalCase());
Assert.AreEqual("For You And The FBI", "For You And The FBI".UnPascalCase());
Assert.AreEqual("A Man A Plan A Canal Panama", "A Man A Plan A Canal Panama".UnPascalCase());
Assert.AreEqual("DNS Server", "DNS Server".UnPascalCase());
Assert.AreEqual("Mount Mᶜ Kinley National Park", "Mount Mᶜ Kinley National Park".UnPascalCase());

这个问题有点老了，但现在在Nuget上有一个很好的库，它可以做到这一点，以及许多其他转换到人类可读的文本。

在GitHub或Nuget上检查Humanizer。

例子

"PascalCaseInputStringIsTurnedIntoSentence".Humanize() => "Pascal case input string is turned into sentence"
"Underscored_input_string_is_turned_into_sentence".Humanize() => "Underscored input string is turned into sentence"
"Underscored_input_String_is_turned_INTO_sentence".Humanize() => "Underscored input String is turned INTO sentence"

// acronyms are left intact
"HTML".Humanize() => "HTML"

请确保您没有在字符串的开头放置空格，而是将它们放在连续的大写字母之间。这里的一些答案并没有解决其中的一个或两个问题。除了regex，还有其他方法，但如果你更喜欢使用它，试试这个:

Regex.Replace(value, @"\B[A-Z]", " $0")

\B是一个负的\B，所以它代表一个非单词边界。这意味着模式匹配XYzabc中的“Y”，但不匹配Yzabc或XYzabc。作为一个小奖励，你可以在一个有空格的字符串上使用它，它不会使它们加倍。

你的解决方案有一个问题，它在第一个字母T之前放了一个空格，所以你得到

" This String..." instead of "This String..."

要绕开这个问题，请寻找前面的小写字母，然后在中间插入空格:

newValue = Regex.Replace(value, "([a-z])([A-Z])", "$1 $2");

编辑1:

如果你使用@"(\p{Ll})(\p{Lu})"，它也会拾取重音字符。

编辑2:

如果你的字符串可以包含首字母缩略词，你可能想使用这个:

newValue = Regex.Replace(value, @"((?<=\p{Ll})\p{Lu})|((?!\A)\p{Lu}(?>\p{Ll}))", " $0");

所以driveisscsiccompatible变成了DriveIsSCSICompatible

在Ruby中，通过Regexp:

"FooBarBaz".gsub(/(?!^)(?=[A-Z])/, ' ') # => "Foo Bar Baz"