John Hunter wrote:
> Anyone know of a simple fix for this
>
>
>>python unicode_demo.py -dSVG
>
>
> File "/usr/lib/python2.4/site-packages/matplotlib/backends/backend_svg.py", line 196, in draw_text
> self._svgwriter.write (svg)
> UnicodeEncodeError: 'ascii' codec can't encode character u'\xe9' in position 136: ordinal not in range(128)
I recently had to fix the same bug for ipython. Here's what my
stumble-in-the-dark approach to unicode bug fixing (no net access at home that
day) produced:
def _str(self,glob,loc):
"""Evaluate to a string in the given globals/locals.
The final output is built by calling str(), but if this fails, the
result is encoded with the instance's codec and error handling policy,
via a call to out.encode(self.codec,self.encoding_errors)"""
result = []
app = result.append
for live, chunk in self.chunks:
if live: app(str(eval(chunk,glob,loc)))
else: app(chunk)
out = ''.join(result)
try:
return str(out)
except UnicodeError:
return out.encode(self.codec,self.encoding_errors)
where the codec stuff is set via:
Optional arguments:
- codec('utf_8'): a string containing the name of a valid Python
codec.
- encoding_errors('backslashreplace'): a string with a valid error
handling
policy. See the codecs module documentation for details.
These are used to encode the format string if a call to str() fails on
the expanded result."""
if not isinstance(format,basestring):
raise TypeError, "needs string initializer"
self.format = format
self.codec = codec
self.encoding_errors = encoding_errors
This is all in ipython's Itpl.py module, in case you want to look closer (the
crash was there b/c that's the machinery used for prompt printing).
cheers,
f