On Tue, 14 Aug 2012 21:40:10 +0200, Virgil Stokes wrote:
> You might find the following useful:
>
> def testFunc(startingList):
> xOnlyList = []; j = -1
> for xl in startingList:
> if (xl[0] == 'x'):
That's going to fail in the starting list contains an empty string. Use
xl.startswith('x') instead.
> xOnlyList.append(xl)
> else:
> j += 1
> startingList[j] = xl
Very cunning, but I have to say that your algorithm fails the "is this
obviously correct without needing to study it?" test. Sometimes that is
unavoidable, but for something like this, there are simpler ways to solve
the same problem.
> if j == -1:
> startingList = []
> else:
> del startingList[j:-1]
> return(xOnlyList)
> And here is another version using list comprehension that I prefer
> def testFunc2(startingList):
> return([x for x in startingList if x[0] == 'x'], [x for x in
> startingList if x[0] != 'x'])
This walks over the starting list twice, doing essentially the same thing
both times. It also fails to meet the stated requirement that
startingList is modified in place, by returning a new list instead.
Here's an example of what I mean:
py> mylist = mylist2 = ['a', 'x', 'b', 'xx', 'cx'] # two names for one
list
py> result, mylist = testFunc2(mylist)
py> mylist
['a', 'b', 'cx']
py> mylist2 # should be same as mylist
['a', 'x', 'b', 'xx', 'cx']
Here is the obvious algorithm for extracting and removing words starting
with 'x'. It walks the starting list only once, and modifies it in place.
The only trick needed is list slice assignment at the end.
def extract_x_words(words):
words_with_x = []
words_without_x = []
for word in words:
if word.startswith('x'):
words_with_x.append(word)
else:
words_without_x.append(word)
words[:] = words_without_x # slice assignment
return words_with_x
The only downside of this is that if the list of words is so enormous
that you can fit it in memory *once* but not *twice*, this may fail. But
the same applies to the list comprehension solution.
--
Steven
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