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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def _get_tags_by_num(self):
 		""" 		Return a dictionary mapping revision number to tags for that number. 		"""

		by_revision = operator.attrgetter('revision')
		tags = sorted(self.get_tags(), key=by_revision)
		revision_tags = itertools.groupby(tags, key=by_revision)

		def get_id(rev):
			return rev.split(':', 1)[0]
		return dict(
			(get_id(rev), [tr.tag for tr in tr_list])
			for rev, tr_list in revision_tags
		)