On 2:59 PM, [email protected] wrote:
Hi,
I have code that works fine when I run it on Python 2.6.5, but I get an "invalid
syntax" error in Python 2.4.3. I'm hoping you can help me fix it.
The line in question splits a chunk of semi-colon separated words into separate
elements.
rgenre = re.split(r';', rf.info["genre"] if "genre" in rf.info else []
I get a syntax error at "if" in 2.4.3.
I tried doing the following but it doesn't work (I don't understand why).
if "genre" in rf.info:
rgenre = re.split(r';', rf.info["genre"])
else:
[]
close. The else clause should read:
rgenre = []
And I tried this, too:
if "genre" in rf.info:
rgenre = re.split(r';', rf.info["genre"])
if "genre" not in rf.info:
[]
Both of these cause problems later on in the program, specifically
"UnboundLocalError: local variable 'rgenre' referenced before assignment",
about this line in the code (where each of these is one of the 30 possible genres):
rg1, rg2, rg3, rg4, rg5, rg6, rg7, rg8, rg9, rg10, rg11, rg12, rg13, rg14, rg15, rg16,
rg17, rg18, rg19, rg20, rg21, rg22, rg23, rg24, rg25, rg26, rg27, rg28, rg29, rg30=
rgenre + ["NA"]*(30-len(rgenre[:30]))
Thanks for any help--I gave a little extra information, but I'm hoping there's
a simple rewrite of the first line I gave that'll work in Python 2.4.3.
Thanks,
Tyler
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