"Lowell Kirsh"
> I'm trying to emulate the sorted() method introduced in python 2.4. The
> only difference is that it takes a sequence as one of its arguments
> rather than being a method of the sequence class. Does my method do the
> same as the sorted()?
Almost. This is closer to the mark:
def sorted(iterable, cmp=None, key=None, reverse=False):
"return a sorted copy of its input"
if sys.version_info >= (2,4):
return sorted(iterable, cmp, key, reverse)
seq = list(iterable)
if reverse:
seq.reverse() # preserve stability
if key is not None:
seq = [(key(elem), i, elem) for i, elem in enumerate(seq)]
seq.sort(cmp)
if key is not None:
seq = [elem for (key, i, elem) in seq]
if reverse:
seq.reverse()
return seq
Try it against the tests in Lib/test/test_builtin.py.
The differences from your version:
* >= 2.4 rather than just > 2.4
* renamed the parameter to iterable
* handle the case where both cmp and key are defined
* add an enumerated tie breaker to prevent full key comparisons
* preserve by using reverse twice
The real sorted() does the same thing but is implemented a bit differently. A
custom key wrapper is applied to each object so that only the key value gets
compared (no need for a full tuple with a tie breaker value).
Raymond Hettinger
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