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Legend autopositioning with "spiraling" lines. #7625

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Jan 30, 2017
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Legend autopositioning with "spiraling" lines. #7625

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4 changes: 4 additions & 0 deletions doc/api/api_changes/2016-12-14-AL_legend-autoposition.rst
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Original file line number Diff line number Diff line change
@@ -0,0 +1,4 @@
More accurate legend autopositioning
````````````````````````````````````

Automatic positioning of legends now prefers using the area surrounded by a `Line2D` rather than placing the legend over the line itself.
46 changes: 12 additions & 34 deletions lib/matplotlib/legend.py
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Original file line number Diff line number Diff line change
Expand Up @@ -939,47 +939,25 @@ def _find_best_position(self, width, height, renderer, consider=None):
renderer)
for x in range(1, len(self.codes))]

# tx, ty = self.legendPatch.get_x(), self.legendPatch.get_y()

candidates = []
for l, b in consider:
for idx, (l, b) in enumerate(consider):
legendBox = Bbox.from_bounds(l, b, width, height)
badness = 0
# XXX TODO: If markers are present, it would be good to
# take their into account when checking vertex overlaps in
# take them into account when checking vertex overlaps in
# the next line.
badness = legendBox.count_contains(verts)
badness += legendBox.count_contains(offsets)
badness += legendBox.count_overlaps(bboxes)
for line in lines:
# FIXME: the following line is ill-suited for lines
# that 'spiral' around the center, because the bbox
# may intersect with the legend even if the line
# itself doesn't. One solution would be to break up
# the line into its straight-segment components, but
# this may (or may not) result in a significant
# slowdown if lines with many vertices are present.
if line.intersects_bbox(legendBox):
badness += 1

ox, oy = l, b
badness = (legendBox.count_contains(verts)
+ legendBox.count_contains(offsets)
+ legendBox.count_overlaps(bboxes)
+ sum(line.intersects_bbox(legendBox, filled=False)
for line in lines))
if badness == 0:
return ox, oy

candidates.append((badness, (l, b)))

# rather than use min() or list.sort(), do this so that we are assured
# that in the case of two equal badnesses, the one first considered is
# returned.
# NOTE: list.sort() is stable.But leave as it is for now. -JJL
minCandidate = candidates[0]
for candidate in candidates:
if candidate[0] < minCandidate[0]:
minCandidate = candidate

ox, oy = minCandidate[1]
return l, b
# Include the index to favor lower codes in case of a tie.
candidates.append((badness, idx, (l, b)))

return ox, oy
_, _, (l, b) = min(candidates)
return l, b

def contains(self, event):
return self.legendPatch.contains(event)
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