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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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Update to score_family in font_manager.py 
This PR fixes the score calculation for the case of matching a generic font name given in the `families` list.

The returned score is now constrained by the position `i` of the match in the list. If the font being scored (`family2`) is the first choice for that generic name in rcParams, it will receive the same score as if it were an exact match for that position in the `families` list, namely `i / len(families)`. If the font being scored is in the alias list but not the first choice, the returned score will be greater than `i / len(families)` but less than `(i + 1) / len(families)`, ensuring that the preferred order of fonts listed in both `families` and rcParams font alias list are respected.
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@chadawagner
chadawagner committed Jul 14, 2015
commit 78073d3c5bcccf3755aafde3149e37ca7e559bff
10 changes: 5 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 +1099,16 @@ 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]
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 +1118,11 @@ 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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Can you add some () here to be explicit about the grouping intended?

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):
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