Re: sort order for strings of digits
On Thu, 01 Nov 2012 11:53:06 +1100, Chris Angelico wrote: > On Thu, Nov 1, 2012 at 10:44 AM, Steven D'Aprano > wrote: >> On the contrary. If you are using cmp with sort, your sorts are slow, >> and you should upgrade to using a key function as soon as possible. >> >> > But cmp_to_key doesn't actually improve anything. So I'm not sure how > Py3 has achieved anything; Py2 supported key-based sorting already. Yes, but there is a lot of old code pre-dating key-based sorts. There's also some examples of code where it isn't obvious how to write it as a key-based sort, but a comparison function is simple. And people coming from other languages that only support comparison-based sorts (C?) will probably continue with what they know. Even though key-based sorting is better, there's a lot of comparison sorting that falls under "if it ain't broke, don't fix it". So even though key-based sorts are better, there are still comparison- based sorts in the wild. Python 2 has to support them. Python 3, which is allowed to break backwards compatibility, does not. So when porting to 3, you have to change the sorts. Most of the time it is simple to convert a comparison-based sort to a key- based sort. For the cases where you either can't come up with a good key function yourself, or were you want to do so mechanically, Python provides cmp_to_key. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: sort order for strings of digits
Le mercredi 31 octobre 2012 16:17:19 UTC+1, djc a écrit :
> I learn lots of useful things from the list, some not always welcome. No
>
> sooner had I found a solution to a minor inconvenience in my code, than
>
> a recent thread here drew my attention to the fact that it will not work
>
> for python 3. So suggestions please:
>
>
>
> TODO 2012-10-22: sort order numbers first then alphanumeric
>
> >>> n
>
> ('1', '10', '101', '3', '40', '31', '13', '2', '2000')
>
> >>> s
>
> ('a', 'ab', 'acd', 'bcd', '1a', 'a1', '222 bb', 'b a 4')
>
>
>
> >>> sorted(n)
>
> ['1', '10', '101', '13', '2', '2000', '3', '31', '40']
>
> >>> sorted(s)
>
> ['1a', '222 bb', 'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']
>
> >>> sorted(n+s)
>
> ['1', '10', '101', '13', '1a', '2', '2000', '222 bb', '3', '31', '40',
>
> 'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']
>
>
>
>
>
>
>
> Possibly there is a better way but for Python 2.7 this gives the
>
> required result
>
>
>
> Python 2.7.3 (default, Sep 26 2012, 21:51:14)
>
>
>
> >>> sorted(int(x) if x.isdigit() else x for x in n+s)
>
> [1, 2, 3, 10, 13, 31, 40, 101, 2000, '1a', '222 bb', 'a', 'a1', 'ab',
>
> 'acd', 'b a 4', 'bcd']
>
>
>
>
>
> [str(x) for x in sorted(int(x) if x.isdigit() else x for x in n+s)]
>
> ['1', '2', '3', '10', '13', '31', '40', '101', '2000', '1a', '222 bb',
>
> 'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']
>
>
>
>
>
> But not for Python 3
>
> Python 3.2.3 (default, Oct 19 2012, 19:53:16)
>
>
>
> >>> sorted(n+s)
>
> ['1', '10', '101', '13', '1a', '2', '2000', '222 bb', '3', '31', '40',
>
> 'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']
>
>
>
> >>> sorted(int(x) if x.isdigit() else x for x in n+s)
>
> Traceback (most recent call last):
>
>File "", line 1, in
>
> TypeError: unorderable types: str() < int()
>
> >>>
>
>
>
> The best I can think of is to split the input sequence into two lists,
>
> sort each and then join them.
>
>
>
>
>
> --
>
> djc
>>> # Py 3.2.3
>>> z = ['1', '10', '101', '13', '1a', '2', '2000',
... '222 bb', '3', '31', '40', 'a', 'a1', 'ab',
... 'acd', 'b a 4', 'bcd'
... ]
>>> n, s = [], []
>>> for e in z:
... if e.isdigit():
... n.append(int(e))
... else:
... s.append(e)
...
>>> n.sort()
>>> s.sort()
>>> ns = [str(e) for e in n]
>>> ns.extend(s)
>>> ns
['1', '2', '3', '10', '13', '31', '40', '101', '2000', '1a',
'222 bb', 'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']
jmf
--
http://mail.python.org/mailman/listinfo/python-list
Re: sort order for strings of digits
On 31 October 2012 23:09, Steven D'Aprano wrote: > The trick is to take each string and split it into a leading number and a > trailing alphanumeric string. Either part may be "empty". Here's a pure > Python solution: > > from sys import maxsize # use maxint in Python 2 > def split(s): > for i, c in enumerate(s): > if not c.isdigit(): > break > else: # aligned with the FOR, not the IF > return (int(s), '') > return (int(s[:i] or maxsize), s[i:]) > > Now sort using this as a key function: > > py> L = ['9', '1000', 'abc2', '55', '1', 'abc', '55a', '1a'] > py> sorted(L, key=split) > ['1', '1a', '9', '55', '55a', '1000', 'abc', 'abc2'] You don't actually need to split the string, it's enough to return a pair consisting of the number of leading digits followed by the string as the key. Here's an implementation using takewhile: >>> from itertools import takewhile >>> def prefix(s): ... return sum(1 for c in takewhile(str.isdigit, s)) or 1000, s ... >>> L = ['9', '1000', 'abc2', '55', '1', 'abc', '55a', '1a'] >>> sorted(L, key=prefix) ['1', '1a', '9', '55', '55a', '1000', 'abc', 'abc2'] Here's why it works: >>> map(prefix, L) [(1, '9'), (4, '1000'), (1000, 'abc2'), (2, '55'), (1, '1'), (1000, 'abc'), (2, '55a'), (1, '1a')] -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
Re: sort order for strings of digits
On Thu, Nov 1, 2012 at 10:44 AM, Steven D'Aprano wrote: > On the contrary. If you are using cmp with sort, your sorts are slow, and > you should upgrade to using a key function as soon as possible. > But cmp_to_key doesn't actually improve anything. So I'm not sure how Py3 has achieved anything; Py2 supported key-based sorting already. ChrisA -- http://mail.python.org/mailman/listinfo/python-list
Re: sort order for strings of digits
On 31/10/2012 22:24, Ian Kelly wrote: On Wed, Oct 31, 2012 at 3:33 PM, Mark Lawrence wrote: Nope. I'm busy porting my own code from 2.7 to 3.3 and cmp seems to be very dead. This doesn't help either. c:\Users\Mark\Cash\Python>**2to3.py Traceback (most recent call last): File "C:\Python33\Tools\Scripts\**2to3.py", line 3, in from lib2to3.main import main ImportError: No module named main Perhaps you have a sys.path conflict? Correct, now fixed, thanks. Use functools.cmp_to_key for porting cmp functions. "sort(x, my_cmp)" becomes "sort(x, key=cmp_to_key(my_cmp))" The cmp builtin is also gone. If you need it, the suggested replacement for "cmp(a, b)" is "(b < a) - (a < b)". As it's my own small code base I've blown away all references to cmp, it's rich comparisons all the way. Cheers, Ian -- Cheers. Mark Lawrence. -- http://mail.python.org/mailman/listinfo/python-list
Re: sort order for strings of digits
On 31/10/12 23:09, Steven D'Aprano wrote: On Wed, 31 Oct 2012 15:17:14 +, djc wrote: The best I can think of is to split the input sequence into two lists, sort each and then join them. According to your example code, you don't have to split the input because you already have two lists, one filled with numbers and one filled with strings. Sorry for the confusion, the pair of strings was just a way of testing variations on the input. So a sequence with any combination of strings that can be read as numbers and strings of chars that don't look like numbers (even if that string includes digits) is the expected input But I think that what you actually have is a single list of strings, and you are supposed to sort the strings such that they come in numeric order first, then alphanumerical. E.g.: ['9', '1000', 'abc2', '55', '1', 'abc', '55a', '1a'] => ['1', '1a', '9', '55', '55a', '1000', 'abc', 'abc2'] Not quite, what I want is to ensure that if the strings look like numbers they are placed in numerical order. ie 1 2 3 10 100 not 1 10 100 2 3. Cases where a string has some leading digits can be treated as strings like any other. At least that is what I would expect as the useful thing to do when sorting. Well it depends on the use case. In my case the strings are column and row labels for a report. I want them to be presented in a convenient to read sequence. Which the lexical sorting of the strings that look like numbers is not. I want a reasonable do-what-i-mean default sort order that can handle whatever strings are used. The trick is to take each string and split it into a leading number and a trailing alphanumeric string. Either part may be "empty". Here's a pure Python solution: from sys import maxsize # use maxint in Python 2 def split(s): for i, c in enumerate(s): if not c.isdigit(): break else: # aligned with the FOR, not the IF return (int(s), '') return (int(s[:i] or maxsize), s[i:]) Now sort using this as a key function: py> L = ['9', '1000', 'abc2', '55', '1', 'abc', '55a', '1a'] py> sorted(L, key=split) ['1', '1a', '9', '55', '55a', '1000', 'abc', 'abc2'] The above solution is not quite general: * it doesn't handle negative numbers or numbers with a decimal point; * it doesn't handle the empty string in any meaningful way; * in practice, you may or may not want to ignore leading whitespace, or trailing whitespace after the number part; * there's a subtle bug if a string contains a very large numeric prefix, finding and fixing that is left as an exercise. That looks more than general enough for my purposes! I will experiment along those lines, thank you. -- http://mail.python.org/mailman/listinfo/python-list
Re: sort order for strings of digits
On Wed, 31 Oct 2012 19:05:17 -0400, Dennis Lee Bieber wrote: >> The cmp builtin is also gone. If you need it, the suggested >> replacement for "cmp(a, b)" is "(b < a) - (a < b)". >> > OUCH... Just another reason for my to hang onto the 2.x series as > long as possible On the contrary. If you are using cmp with sort, your sorts are slow, and you should upgrade to using a key function as soon as possible. For small lists, you may not notice, but for large lists using a comparison function is a BAD IDEA. Here's an example: sorting a list of numbers by absolute value. py> L = [5, -6, 1, -2, 9, -8, 4, 3, -7, 2, -3] py> sorted(L, key=abs) [1, -2, 2, 3, -3, 4, 5, -6, -7, -8, 9] py> sorted(L, lambda a, b: cmp(abs(a), abs(b))) [1, -2, 2, 3, -3, 4, 5, -6, -7, -8, 9] But the amount of work done is radically different. Let's temporarily shadow the built-ins with patched versions: py> _abs = abs py> _abs, _cmp = abs, cmp py> c1 = c2 = 0 py> def abs(x): ... global c1 ... c1 += 1 ... return _abs(x) ... py> def cmp(a, b): ... global c2 ... c2 += 1 ... return _cmp(a, b) ... Now we can see just how much work is done under the hood using a key function vs a comparison function: py> sorted(L, key=abs) [1, -2, 2, 3, -3, 4, 5, -6, -7, -8, 9] py> c1 11 So the key function is called once for each item in the list. But: py> c1 = 0 # reset the count py> sorted(L, lambda a, b: cmp(abs(a), abs(b))) [1, -2, 2, 3, -3, 4, 5, -6, -7, -8, 9] py> c1, c2 (54, 27) The comparison function is called 27 times for a list of nine items (a average of 2.5 calls to cmp per item), and abs is called twice for each call to cmp. (Well, duh.) If the list is bigger, it gets worse: py> c2 = 0 py> x = sorted(L*10, lambda a, b: cmp(abs(a), abs(b))) py> c2 592 That's an average of 5.4 calls to cmp per item. And it gets even worse as the list gets bigger. As your lists get bigger, the amount of work done calling the comparison function gets ever bigger still. Sorting large lists with a comparison function is SLOOOW. py> del abs, cmp # remove the monkey-patched versions py> L = L*100 py> with Timer(): ... x = sorted(L, key=abs) ... time taken: 9.165448 seconds py> with Timer(): ... x = sorted(L, lambda a, b: cmp(abs(a), abs(b))) ... time taken: 63.579679 seconds The Timer() context manager used can be found here: http://code.activestate.com/recipes/577896 -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: sort order for strings of digits
On Wed, 31 Oct 2012 15:17:14 +, djc wrote: > The best I can think of is to split the input sequence into two lists, > sort each and then join them. According to your example code, you don't have to split the input because you already have two lists, one filled with numbers and one filled with strings. But I think that what you actually have is a single list of strings, and you are supposed to sort the strings such that they come in numeric order first, then alphanumerical. E.g.: ['9', '1000', 'abc2', '55', '1', 'abc', '55a', '1a'] => ['1', '1a', '9', '55', '55a', '1000', 'abc', 'abc2'] At least that is what I would expect as the useful thing to do when sorting. The trick is to take each string and split it into a leading number and a trailing alphanumeric string. Either part may be "empty". Here's a pure Python solution: from sys import maxsize # use maxint in Python 2 def split(s): for i, c in enumerate(s): if not c.isdigit(): break else: # aligned with the FOR, not the IF return (int(s), '') return (int(s[:i] or maxsize), s[i:]) Now sort using this as a key function: py> L = ['9', '1000', 'abc2', '55', '1', 'abc', '55a', '1a'] py> sorted(L, key=split) ['1', '1a', '9', '55', '55a', '1000', 'abc', 'abc2'] The above solution is not quite general: * it doesn't handle negative numbers or numbers with a decimal point; * it doesn't handle the empty string in any meaningful way; * in practice, you may or may not want to ignore leading whitespace, or trailing whitespace after the number part; * there's a subtle bug if a string contains a very large numeric prefix, finding and fixing that is left as an exercise. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: sort order for strings of digits
On Wed, Oct 31, 2012 at 3:33 PM, Mark Lawrence wrote: > Nope. I'm busy porting my own code from 2.7 to 3.3 and cmp seems to be > very dead. > > This doesn't help either. > > c:\Users\Mark\Cash\Python>**2to3.py > > Traceback (most recent call last): > File "C:\Python33\Tools\Scripts\**2to3.py", line 3, in > from lib2to3.main import main > ImportError: No module named main > Perhaps you have a sys.path conflict? Use functools.cmp_to_key for porting cmp functions. "sort(x, my_cmp)" becomes "sort(x, key=cmp_to_key(my_cmp))" The cmp builtin is also gone. If you need it, the suggested replacement for "cmp(a, b)" is "(b < a) - (a < b)". Cheers, Ian -- http://mail.python.org/mailman/listinfo/python-list
Re: sort order for strings of digits
On 31/10/2012 18:17, Dennis Lee Bieber wrote: Why -- I doubt Python 3.x .sort() and sorted() have removed the optional key and cmp keywords. Nope. I'm busy porting my own code from 2.7 to 3.3 and cmp seems to be very dead. This doesn't help either. c:\Users\Mark\Cash\Python>2to3.py Traceback (most recent call last): File "C:\Python33\Tools\Scripts\2to3.py", line 3, in from lib2to3.main import main ImportError: No module named main -- Cheers. Mark Lawrence. -- http://mail.python.org/mailman/listinfo/python-list
Re: sort order for strings of digits
On Wed, Oct 31, 2012 at 9:17 AM, djc wrote:
> The best I can think of is to split the input sequence into two lists, sort
> each and then join them.
In the example you have given they already seem to be split, so you
could just do:
sorted(n, key=int) + sorted(s)
If that's not really the case, then you could construct (str, int)
tuples as sort keys:
sorted(n+s, key=lambda x: ('', int(x)) if x.isdigit() else (x, -1))
Note that the empty string sorts before all numbers here, which may or
may not be desirable.
--
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Re: sort order for strings of digits
On 31/10/12 16:17:14, djc wrote: > Python 3.2.3 (default, Oct 19 2012, 19:53:16) > sorted(n+s) > ['1', '10', '101', '13', '1a', '2', '2000', '222 bb', '3', '31', '40', > 'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd'] > sorted(int(x) if x.isdigit() else x for x in n+s) > Traceback (most recent call last): > File "", line 1, in > TypeError: unorderable types: str() < int() >>> sorted(n+s, key=lambda x:(x.__class__.__name__, x)) ['1', '10', '101', '13', '1a', '2', '2000', '222 bb', '3', '31', '40', 'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd'] >>> > The best I can think of is to split the input sequence into two lists, > sort each and then join them. That might well be the most readable solution. Hope this helps, -- HansM -- http://mail.python.org/mailman/listinfo/python-list
sort order for strings of digits
I learn lots of useful things from the list, some not always welcome. No
sooner had I found a solution to a minor inconvenience in my code, than
a recent thread here drew my attention to the fact that it will not work
for python 3. So suggestions please:
TODO 2012-10-22: sort order numbers first then alphanumeric
>>> n
('1', '10', '101', '3', '40', '31', '13', '2', '2000')
>>> s
('a', 'ab', 'acd', 'bcd', '1a', 'a1', '222 bb', 'b a 4')
>>> sorted(n)
['1', '10', '101', '13', '2', '2000', '3', '31', '40']
>>> sorted(s)
['1a', '222 bb', 'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']
>>> sorted(n+s)
['1', '10', '101', '13', '1a', '2', '2000', '222 bb', '3', '31', '40',
'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']
Possibly there is a better way but for Python 2.7 this gives the
required result
Python 2.7.3 (default, Sep 26 2012, 21:51:14)
>>> sorted(int(x) if x.isdigit() else x for x in n+s)
[1, 2, 3, 10, 13, 31, 40, 101, 2000, '1a', '222 bb', 'a', 'a1', 'ab',
'acd', 'b a 4', 'bcd']
[str(x) for x in sorted(int(x) if x.isdigit() else x for x in n+s)]
['1', '2', '3', '10', '13', '31', '40', '101', '2000', '1a', '222 bb',
'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']
But not for Python 3
Python 3.2.3 (default, Oct 19 2012, 19:53:16)
>>> sorted(n+s)
['1', '10', '101', '13', '1a', '2', '2000', '222 bb', '3', '31', '40',
'a', 'a1', 'ab', 'acd', 'b a 4', 'bcd']
>>> sorted(int(x) if x.isdigit() else x for x in n+s)
Traceback (most recent call last):
File "", line 1, in
TypeError: unorderable types: str() < int()
>>>
The best I can think of is to split the input sequence into two lists,
sort each and then join them.
--
djc
--
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