Bengt Richter enlightened us with:
> I suspect it's not possible to get '' in the list from
> somestring.split()
Time to adjust your suspicions:
>>> ';abc;'.split(';')
['', 'abc', '']
>> countDict[w] += 1
>> else:
>> countDict[w] = 1
> does that beat the try and get versions? I.e., (untested)
> try: countDict[w] += 1
> except KeyError: countDict[w] = 1
OPs way is faster. A 'try' block is very fast, but trowing & handling
an exception is slow.
> countDict[w] = countDict.get(w, 0) + 1
I think this has the potential of being faster than OPs method,
because it's likely to be fully implemented in C.
Sybren
--
The problem with the world is stupidity. Not saying there should be a
capital punishment for stupidity, but why don't we just take the
safety labels off of everything and let the problem solve itself?
Frank Zappa
--
http://mail.python.org/mailman/listinfo/python-list
