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bug in font_manager.FontManager.score_family() #4607
Description
If a font to be scored is found first in the font alias list, it currently results in a "perfect" score of 0.
For example:
font_manager.FontManager().score_family(['myfont','sans-serif'], 'bitstream vera sans')
will return 0, and 'myfont' will never be found if it happens to appear later in the list of known fonts.
Commit 314a0cf already attempted to address this issue, but didn't solve it in the case I'm running into.
Activity
chadawagner commentedon Jul 8, 2015
The code for the returned score is currently
and the issue I'm having occurs when idx is 0. The previous commit addressed the right multiplicand, changing
itoi+1so that it would never be 0, but the overall result is still 0 when the left side is 0.mdboom commentedon Jul 9, 2015
Yeah -- I think changing to (idx + 1) makes sense here.
tacaswell commentedon Jul 10, 2015
@chadawagner Would you like to put in a PR to fix this?
chadawagner commentedon Jul 10, 2015
I could make a PR, but I'm still not sure the right way to fix this scoring algorithm. I think it's broken in more ways than just changing to
(idx + 1)would fix.For example, assume I have a font family list of
['font1', 'font2', 'sans-serif'], then any font that is part of the sans-serif alias will score at most 0.1, which is lower (better) than font2.chadawagner commentedon Jul 11, 2015
How about
That way the returned score is always in the half-open interval
[ i/len(families), (i+1)/len(families) )mdboom commentedon Mar 21, 2016
Fixed by #4689