psyco et matplotlib incompatibles

Bonjour

Savez vous si pylab de matplotlib ou bien numpy ont des incompatibilités avec psyco.

J'ai fait un test avec mon appli en cours c'est bien la présence de psyco qui perturbe l'exécution du code dans pylab...

existe t-il une solution ?
voici la série du problème

Code:

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Traceback (most recent call last):
  File "c:\python26\lib\lib-tk\Tkinter.py", line 1410, in __call__
    return self.func(*args)
  File "c:\python26\lib\site-packages\matplotlib\backends\backend_tkagg.py", line 245, in resize
    self.show()
  File "c:\python26\lib\site-packages\matplotlib\backends\backend_tkagg.py", line 248, in draw
    FigureCanvasAgg.draw(self)
  File "c:\python26\lib\site-packages\matplotlib\backends\backend_agg.py", line 394, in draw
    self.figure.draw(self.renderer)
  File "c:\python26\lib\site-packages\matplotlib\artist.py", line 55, in draw_wrapper
    draw(artist, renderer, *args, **kwargs)
  File "c:\python26\lib\site-packages\matplotlib\figure.py", line 798, in draw
    func(*args)
  File "c:\python26\lib\site-packages\matplotlib\artist.py", line 55, in draw_wrapper
    draw(artist, renderer, *args, **kwargs)
  File "c:\python26\lib\site-packages\matplotlib\axes.py", line 1946, in draw
    a.draw(renderer)
  File "c:\python26\lib\site-packages\matplotlib\artist.py", line 55, in draw_wrapper
    draw(artist, renderer, *args, **kwargs)
  File "c:\python26\lib\site-packages\matplotlib\axis.py", line 1030, in draw
    self._update_label_position(ticklabelBoxes, ticklabelBoxes2)
  File "c:\python26\lib\site-packages\matplotlib\axis.py", line 1587, in _update_label_position
    bottom = self.axes.bbox.ymin
  File "c:\python26\lib\site-packages\matplotlib\transforms.py", line 297, in _get_ymin
    return min(self.get_points()[:, 1])
  File "c:\python26\lib\site-packages\psyco\builtin.py", line 75, in min
    return _min(*args)
  File "c:\python26\lib\site-packages\psyco\builtin.py", line 34, in _min
    if not iterable:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()