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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
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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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added parens for clarity
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@chadawagner
chadawagner committed Jul 31, 2015
commit d82c1c69d2f0e22cb58a64e22fb120e414f1adef
2 changes: 1 addition & 1 deletion lib/matplotlib/font_manager.py
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Expand Up @@ -1118,7 +1118,7 @@ def score_family(self, families, family2):
options = [x.lower() for x in options]
if family2 in options:
idx = options.index(family2)
return (i + idx / len(options)) * step
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.
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