Sort the values of a dict
mattia
pairs. I need to sort the values of the dict and insert everything in a
tuple. The additional problem is that I need to sort the values looking
at the i-th element of the list. I'm not that good at python (v3.1), but
this is my solution:
>>> d = {1:('a', 1, 12), 5:('r', 21, 10), 2:('u', 9, 8)}
>>> t = [x for x in d.values()]
>>> def third(mls):
... return mls[2]
...
>>> s = sorted(t, key=third)
>>> pres = []
>>> for x in s:
... for k in d.keys():
... if d[k] == x:
... pres.append(k)
... break
...
>>> res = []
>>> for x in pres:
... res.append((x, d[x]))
...
>>> res
[(2, ('u', 9, 8)), (5, ('r', 21, 10)), (1, ('a', 1, 12))]
>>>
Can you provide me a much pythonic solution (with comments if possible,
so I can actually learn something)?
Thanks, Mattia
mattia
>>> import copy
>>> d = {1:('a', 1, 12), 5:('r', 21, 10), 2:('u', 9, 8), 3:('u', 9, 8) }
>>> dc = copy.deepcopy(d)
>>> t = [x for x in d.values()]
>>> def third(mls):
... return mls[2]
...
>>> s = sorted(t, key=third)
>>> pres = []
>>> for x in s:
... for k in d.keys():
... if d[k] == x:
... pres.append(k)
... del d[k] # speedup and use duplicate values
... break
...
>>> res = []
>>> for x in pres:
... res.append((x, dc[x]))
...
>>> res
[(2, ('u', 9, 8)), (3, ('u', 9, 8)), (5, ('r', 21, 10)), (1, ('a', 1,
12))]
>>>
David Robinow
> http://mail.python.org/mailman/listinfo/python-list
>
I won't engage in any arguments about pythonicity but it seems simpler
if you convert to a list of tuples right away.
d = {1:('a', 1, 12), 5:('r',21,10), 2:('u',9,8)}
l = [(x, d[x]) for x in d.keys()]
def third(q):
return q[1][2]
s = sorted(l, key=third)
print s
mattia
Thanks, I'm not yet aware of all the wonderful conversions python can do,
amazing.
Rory Campbell-Lange
How about
>>> mylist = [z for z in zip(list(d), list(d.values()))]
and then sort on your sort criteria, either i[0], i[0][1], i[0][2] or
i[0][3].
Rory
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MRAB
>
>>>> mylist = [z for z in zip(list(d), list(d.values()))]
>
>>> mylist = list(d.items())
Lie Ryan
> Can you provide me a much pythonic solution (with comments if possible,
> so I can actually learn something)?
If you only need to get i'th element sometimes, sorting the dict is
fine. Otherwise, you might want to use collections.OrderedDict.
mattia
Well, in the python doc OrderedDict is described as a dict that remembers
the order that keys were first inserted and I don't need this. The fact
is that I use a structure composed by a date and a list of possible
solutions found, like
(2009/12/21, (('e', 12, 33), ('r', 4, 11), ('r', 1, 33))) then every
solution is inserted concurrently in a dictionary. I want to sort the
solution found to provide, e.g., the first 10 dates found and the best
result of every date based on the i-th element of the date's list.
Rory Campbell-Lange
> The Pythonic way for the above line is:
> >>> mylist = list(d.items())
Quite right. Thanks for pointing that out.
I liked Chris Kaynor's solution:
s = sorted(d.items(), key=lambda i: i[1][2])
I'm a bit rusty, and have just picked up the new "Learning Python" book
but I'm finding that it isn't providing a useful refresher to the
language or a particularly good introduction to version 3. Have you any
suggestions.
Regards
Rory
--
Rory Campbell-Lange
John Bokma
> I'm a bit rusty, and have just picked up the new "Learning Python" book
> but I'm finding that it isn't providing a useful refresher to the
> language or a particularly good introduction to version 3. Have you any
> suggestions.
While you aren't addressing me, I do have 2 suggestions:
* Programming in Python 3 by Mark Summerfield (make sure you get the
2nd edition)
* Dive into Python 3 by Mark Pilgrim
I am halfway in the first book (but the 1st edition) and like it a
lot. I know Dive into Python from an earlier version (2.x) and have no
problem at all recommending the Python 3 edition. You can read/download it
here: http://diveintopython3.org/
--
John Bokma
Read my blog: http://johnbokma.com/
Hire me (Perl/Python): http://castleamber.com/
Rhodri James
wrote:
> I won't engage in any arguments about pythonicity but it seems simpler
> if you convert to a list of tuples right away.
>
> d = {1:('a', 1, 12), 5:('r',21,10), 2:('u',9,8)}
> l = [(x, d[x]) for x in d.keys()]
which is a long way of writing "l = d.items()", or "l = list(d.items())"
if you're using Python 3. :-)
> def third(q):
> return q[1][2]
>
> s = sorted(l, key=third)
> print s
--
Rhodri James *-* Wildebeeste Herder to the Masses