how to remove last text in parenthesis (elegantly and simply) - f#

Question

So I have some working code.
I want to remove the last parenthesized (?) text from a string...as long as its "the last" (i.e. isnt followed by whitespace or a full stop).
so
let x1 = remove "hello how are you. "
let x2 = remove "hello how are you. (remove me) "
let x3 = remove "hello how are you(don't remove me!)., "
goes to
val x1 : string = "hello how are you. "
val x2 : string = "hello how are you. "
val x3 : string = "hello how are you(don't remove me!)., "
I need to do this in language that doesnt support regex and is (sort of) functional, and is pretty basic.
(its XSLT 1.0)
I don't really like hacking in XSLT, so I've hacked together a simple routine in f#.
let remove s =
match (s : string).LastIndexOf '(' with
| -1 ->
s
| lastOpen ->
let rest = s.Substring lastOpen
match rest.LastIndexOf ')' with
| -1 ->
s
| lastClose ->
let after = rest.Substring (lastClose + 1)
if after.Replace(".","") |> String.IsNullOrWhiteSpace then
s.Substring (0,lastOpen) + after
else
s
I can't use Regex, (or write some sort of parser), its just a basic as simple as possible algorithm.
The above broadly works (I can see an edge case it doesnt...but don't get too hung up).
Anything simpler (but less ugly)?
EDIT #1 post accepting reply.
as a sobering thought this is the xslt version of a 4 line f# program!
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<xsl:variable name="foo">
<xsl:call-template name="removeParenthesisIf">
<xsl:with-param name="s" select="'foo bar dfsfsfffs)'"/>
<xsl:with-param name="maxCount" select="20"/>
</xsl:call-template>
</xsl:variable>
</xsl:template>
<xsl:template name="removeParenthesisIf">
<xsl:param name="s"/>
<xsl:param name="maxCount"/>
<xsl:choose>
<xsl:when test="substring($s,string-length($s)) = ')'">
<xsl:variable name="lastOpen">
<xsl:call-template name="lastCharIndex">
<xsl:with-param name="pText" select="$s"/>
<xsl:with-param name="pChar" select="'('"/>
</xsl:call-template>
</xsl:variable>
<xsl:choose>
<xsl:when test="($maxCount > string-length($s) - $lastOpen + 1) and not(string-length($s) = $lastOpen )">
<xsl:value-of select="substring($s,1,$lastOpen - 1)"/>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$s"/>
</xsl:otherwise>
</xsl:choose>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="$s"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
<xsl:template name="lastCharIndex">
<xsl:param name="pText"/>
<xsl:param name="pChar" select="' '"/>
<xsl:variable name="vRev">
<xsl:call-template name="reverse">
<xsl:with-param name="pStr" select="$pText"/>
</xsl:call-template>
</xsl:variable>
<xsl:value-of select="string-length($pText) - string-length(substring-before($vRev, $pChar))"/>
</xsl:template>
<xsl:template name="reverse">
<xsl:param name="pStr"/>
<xsl:variable name="vLength" select="string-length($pStr)"/>
<xsl:choose>
<xsl:when test="$vLength = 1">
<xsl:value-of select="$pStr"/>
</xsl:when>
<xsl:otherwise>
<xsl:variable name="vHalfLength" select="floor($vLength div 2)"/>
<xsl:variable name="vrevHalf1">
<xsl:call-template name="reverse">
<xsl:with-param name="pStr"
select="substring($pStr, 1, $vHalfLength)"/>
</xsl:call-template>
</xsl:variable>
<xsl:variable name="vrevHalf2">
<xsl:call-template name="reverse">
<xsl:with-param name="pStr"
select="substring($pStr, $vHalfLength+1)"/>
</xsl:call-template>
</xsl:variable>
<xsl:value-of select="concat($vrevHalf2, $vrevHalf1)"/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>

Answer 1

The simplest way of doing something along those lines that I can think of, using mainly the built-in string functions, is the following:
let remove (s:string) =
if s.TrimEnd(' ').EndsWith(")") then
let last = s.LastIndexOf('(')
s.Remove(last, s.LastIndexOf(')') - last + 1)
else s
There are two tricks:
First, we check if the string ends with ) if we remove all spaces from the end. This is detecting the case where you have something to remove.
If that's the case, you can remove a substring from the middle using Remove. This assumes that the string is well-formed and does actually have an opening parenthesis - so that's an extra check you may need to add.