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Avoid recomputing findReferencedElements in removeUnusedDefs

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The removeUnusedDefs function does not actually remove anything (that
is left for its callers to do).  This implies that
findReferencedElements will return the same value before, during and
after a call to removeUnusedDefs.  Therefore, we can reuse the value
from findReferencedElements when recursing into child nodes.

Signed-off-by: Niels Thykier <niels@thykier.net>
This commit is contained in:
Niels Thykier 2018-02-17 10:14:49 +00:00 • committed by Eduard Braun
parent 633b381d87
commit b916a189e9
1 changed files with 6 additions and 3 deletions
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9
scour/scour.py
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@ -641,11 +641,14 @@ def findReferencingProperty(node, prop, val, ids):
ids[id] = [1, [node]]
def removeUnusedDefs(doc, defElem, elemsToRemove=None):
def removeUnusedDefs(doc, defElem, elemsToRemove=None, referencedIDs=None):
if elemsToRemove is None:
elemsToRemove = []
referencedIDs = findReferencedElements(doc.documentElement)
# removeUnusedDefs do not change the XML itself; therefore there is no point in
# recomputing findReferencedElements when we recurse into child nodes.
if referencedIDs is None:
referencedIDs = findReferencedElements(doc.documentElement)
keepTags = ['font', 'style', 'metadata', 'script', 'title', 'desc']
for elem in defElem.childNodes:
@ -655,7 +658,7 @@ def removeUnusedDefs(doc, defElem, elemsToRemove=None):
# we only inspect the children of a group in a defs if the group
# is not referenced anywhere else
if elem.nodeName == 'g' and elem.namespaceURI == NS['SVG']:
elemsToRemove = removeUnusedDefs(doc, elem, elemsToRemove)
elemsToRemove = removeUnusedDefs(doc, elem, elemsToRemove, referencedIDs=referencedIDs)
# we only remove if it is not one of our tags we always keep (see above)
elif elem.nodeName not in keepTags:
elemsToRemove.append(elem)