[Python-Dev] Issue? (original) (raw)

Chris Barker chris.barker at noaa.gov
Wed Aug 22 12:55:36 EDT 2018

python used the "timsort" sorting routine:

https://en.wikipedia.org/wiki/Timsort

So you can look at that and confirm that this is correct behaviour (I'm betting it is :-)

But in general, sorting is O(n log(n)) -- there are going to be more than n comparisons.

If comparing is slow, you want to use a key function, to reduce your comparison to a simple and fast one:

sorted(L, key=lambda i: (i.name, i.score))

or something like that.

personally, I advocate adding a "key_fun" attribute to classes you want to make efficiently sortable, so you'd have:

sorted(L, key=Student.key_fun)

There was a discussion on python-ideas about adding a sort_key protocol to python, but there were too many downsides.

-CHB

On Wed, Aug 22, 2018 at 3:40 AM, 楼晓峰 <1520306395 at qq.com> wrote:

Why compare twice? class Student(object): def init(self, name, score): self.name = name self.score = score def str(self): return '(%s: %s)' % (self.name, self.score) repr_ = _str def lt(self, s): #print(self, '--', s) if(self.score<s.score):_ _print(self, '<', s)_ _return True_ _if(self.score>s.score): print(self, '>', s) return False if(self.score==s.score): if(self.name>s.name): print(self, '>', s) return False if(self.name<s.name):_ _print(self, '<', s)_ _return True_ _if(self.name==s.name):_ _print(self, '==', s)_ _return False_ _def _eq_(self, s):_ _return (self.score == s.score) and (self.name == s.name)_ _def _gt_(self, s):_ _return not ((self == s) or (self < s))_ _def _le_(self, s):_ _return ((self == s) or (self < s))_ _def _ge_(self, s):_ _return ((self == s) or (self > s)) def nq(self, s): return not (self == s) L = [Student('Tim', 22), Student('Bob', 33), Student('Kevin', 11), Student('Alice', 11)] print(sorted(L)) Output: (Bob: 33) > (Tim: 22) (Kevin: 11) < (Bob: 33) (Kevin: 11) < (Bob: 33) (Kevin: 11) < (Tim: 22) (Alice: 11) < (Tim: 22) (Alice: 11) < (Kevin: 11) [(Alice: 11), (Kevin: 11), (Tim: 22), (Bob: 33)] Best regards, Xiaofeng

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