From: David M. <dav...@ho...> - 2007-09-17 17:12:54
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Hi, =20 I am new to matplotlib. =20 I am trying to save a sample graph I created as a jpeg image. It seems for= m the documentation that the type of image can be determined by the extensi= on that is used. Commands seem simple enough. =20 ind =3D arange(len(importance_IncNodePurity)) # the x locations for the gro= ups width =3D 0.35 # the width of the bars p1 =3D bar(ind, importance_IncNodePurity, width, color=3D'r') ylabel('Importance') title('Importance Node Purity') xticks(ind+width, importance_row_names) psave=3D"c:\eclipse\dafaf.jpg" savefig(psave) show() =20 =20 However, I seem to get the following errors and I have no idea what they me= an. =20 Traceback (most recent call last): File "C:\Documents and Settings\dmontgomery\workspace\Test\src\create_table= .py", line 259, in <module> savefig(psave) File "C:\Python25\Lib\site-packages\matplotlib\pylab.py", line 796, in save= fig return fig.savefig(*args, **kwargs) File "C:\Python25\Lib\site-packages\matplotlib\figure.py", line 759, in sav= efig self.canvas.print_figure(*args, **kwargs) File "C:\Python25\Lib\site-packages\matplotlib\backends\backend_tkagg.py", = line 188, in print_figure **kwargs) File "C:\Python25\Lib\site-packages\matplotlib\backends\backend_agg.py", li= ne 493, in print_figure raise IOError('Do not know know to handle extension *%s' % ext) IOError: Do not know know to handle extension *.jpg =20 Thanks, =20 David= |