ALTO: Extract namespace == ALTO version

master
Gerber, Mike 3 years ago
parent 8285bdb423
commit 3d2e53f739

@ -155,6 +155,7 @@ def process(alto_files: List[str], output_file: str, output_csv: str, output_xls
d = flatten(alto_to_dict(alto, raise_errors=True))
# "meta"
d['alto_file'] = alto_file
d['alto_xmlns'] = ET.QName(alto).namespace
alto_info.append(d)

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