[Python] How to pass dynamic or external variable's (path) text to the class

Below code taken from Rolfbly, with yours permission, I requested here for how to pass the dynamic path for the XSLT.

from saxonche import PySaxonProcessor
from pathlib import Path
import argparse

stylesheet_file1=r'D:\Test\Stream.xsl' # instead of static path, need to pass dynamic or global variable text,
#how to pass here, see code-101 in commented text below
#stylesheet_file1=fxsl # this path or variable text needs to assign

class Xslt_proc():
proc = PySaxonProcessor(license = False)
nuproc = proc.new_xslt30_processor()
xform = nuproc.compile_stylesheet(stylesheet_file=stylesheet_file1)

def transform(processor, infile, sfx):
outfname = r'D:\Test\Stream_out.xml'
doc = processor.proc.parse_xml(xml_file_name=infile)
out = processor.xform.transform_to_string(xdm_node=doc)

with open(outfname, 'w') as f:
f.write(out)

def parse_arguments():
parser = argparse.ArgumentParser(description="XML XSLT ResultName")
parser.add_argument('param1', type=str, help="XML file")
parser.add_argument('param2', type=str, help="XSL file")

def main():
try:
args = parse_arguments()
f_xml=args.param1
f_xsl=args.param2
except:
# Define file names
f_xml = r'D:\Test\Stream.xml'
f_xsl = r'D:\Test\Stream.xsl' #code-101, this path needs to add in class 'Xslt_proc'

P = Xslt_proc()
transform(P, f_xml, '_done')

if __name__ == "__main__":
main()


XSLT3

<xsl:stylesheet version="3.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="@*|node()">
<xsl:copy><xsl:apply-templates select="@*|node()"/></xsl:copy>
</xsl:template>

<xsl:template match="title">
<xsl:element name="mainTitle">
<xsl:apply-templates select="@*"/>
<xsl:apply-templates/>
</xsl:element>
</xsl:template>

</xsl:stylesheet>


InPut XML

<root><title>Solar System</title></root>

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