offset in dragging a legend

Community matplotlib-users
1

Hi Che, I think you got bit by the "reply to list" non-feature of this list...

In ours, we catch the mpl button down event and after establishing a
hit on the legend do:

I was using the pick event, not the button down event. How do you
"establish a hit on the legend" in the button down event handler?

So, keep in mind that we're using 0.90.1. This was before the nice
hit-test routines. We wrote our own hit-test routines loosely based
around the "object_picker.py" demo that was current at the time.

       bbox = self\.\_legend\.get\_window\_extent\(\)
       self\.\_dragOffset = \(bbox\.xmin\(\) \- mplEvent\.x, bbox\.ymin\(\)

- mplEvent.y)

Then in the mousemove event handler we do:

       self\.\_moveLegend\(mplEvent\.x \+ self\.\_dragOffset\[0\],

mplEvent.y + self._dragOffset[1])

where self._moveLegend is:

def _moveLegend(self, x, y, autoDraw=True):

What is the purpose of autoDraw = True?

This is something we carry around in our application to limit the
number of redraws to reduce flicker.

   height = float\(self\.figure\.bbox\.height\(\)\)
   width = float\(self\.figure\.bbox\.width\(\)\)

Can someone confirm that in the most recent version of mpl,
one can no longer do bbox.height(), but must just do bbox.height ?
(when I do it the first way it says float is not callable).

I'm pretty sure this is true.

   dx = x/width
   dy = y/height

   \# The following two lines are specific to our implementation

and our internal data classes
self._legendLocation = (dx, dy)
self.plot.getLegend().setLocation(self._legendLocation)

So can I exclude them and still get the legend to move?

Yes.

   \# This is the line that "moves" the legend
   self\.\_legend\.\_loc=\(dx,dy\)

   if autoDraw:
       self\.draw\(\)

Now we're using an ancient version of MPL (0.90.1) and many things
have changed in the meantime. YMMV.

Thanks very much,
Che

No worries.

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