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Update to score_family in font_manager.py #4689

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Merged
merged 3 commits into from
Aug 21, 2015
Merged

Update to score_family in font_manager.py #4689

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78073d3
Update to score_family in font_manager.py
chadawagner Jul 14, 2015
d82c1c6
added parens for clarity
chadawagner Jul 31, 2015
673feb0
Update to score_family in font_manager.py
chadawagner Aug 6, 2015
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12 changes: 7 additions & 5 deletions lib/matplotlib/font_manager.py
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Original file line number Diff line number Diff line change
Expand Up @@ -1099,15 +1099,18 @@ def score_family(self, families, family2):
Returns a match score between the list of font families in
*families* and the font family name *family2*.

An exact match anywhere in the list returns 0.0.
An exact match at the head of the list returns 0.0.

A match by generic font name will return 0.1.
A match further down the list will return between 0 and 1.

No match will return 1.0.
"""
if not isinstance(families, (list, tuple)):
families = [families]
elif len(families) == 0:
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Now checking for empty families list here. The unchecked division by len(families) used to be inside the loop, where it was not reachable if len(families) == 0. Thanks for catching that!

return 1.0
family2 = family2.lower()
step = 1 / len(families)
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I think we need a special case here for len(families) == 0. Unlikely, but technically possible.

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In which case, it should just return 1 immediately.

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Agreed, although it never checked before. Not a bad idea tho to be safe.

for i, family1 in enumerate(families):
family1 = family1.lower()
if family1 in font_family_aliases:
Expand All @@ -1117,12 +1120,11 @@ def score_family(self, families, family2):
options = [x.lower() for x in options]
if family2 in options:
idx = options.index(family2)
return ((0.1 * (idx / len(options))) *
((i + 1) / len(families)))
return (i + (idx / len(options))) * step
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A match for an alias should never be higher than an exact match. Yet here, if i == 0, the return value is 0.

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This is where I get confused, and either don't understand or don't agree with the scoring algorithm. If it matches an alias, and that alias was provided first in the font families list, and the specific font being scored appears first in the alias list, wouldn't that score a 0? How could there be a better match?

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Also, part of the current bug this PR attempts to fix is that an alias match will return 0 if the font is first in the alias list, regardless of where that alias appears in the font families list.

elif family1 == family2:
# The score should be weighted by where in the
# list the font was found.
return i / len(families)
return i * step
return 1.0

def score_style(self, style1, style2):
Expand Down