Efficiently building ordered dict

MRAB python at mrabarnett.plus.com
Tue Feb 23 00:08:17 CET 2010

Bryan wrote:
> On Feb 22, 2:16 pm, "Diez B. Roggisch" <de... at nospam.web.de> wrote:
>> Am 22.02.10 22:29, schrieb Bryan:
>>> On Feb 22, 10:57 am, "Alf P. Steinbach"<al... at start.no>  wrote:
>>>> * Bryan:
>>>>> I am looping through a list and creating a regular dictionary.  From
>>>>> that dict, I create an ordered dict.  I can't think of a way to build
>>>>> the ordered dict while going through the original loop.  Is there a
>>>>> way I can avoid creating the first unordered dict just to get the
>>>>> ordered dict?  Also, I am using pop(k) to retrieve the values from the
>>>>> unordered dict while building the ordered one because I figure that as
>>>>> the values are removed from the unordered dict, the lookups will
>>>>> become faster.  Is there a better idiom that the code below to create
>>>>> an ordered dict from an unordered list?
>>>>> unorderedDict = {}
>>>>> for thing in unorderedList:
>>>>>     if thing.id in unorderedDict:
>>>>>             UpdateExistingValue(unorderedDict[thing.id])
>>>>>     else:
>>>>>             CreateNewValue(unorderedDict[thing.id])
>>>> If this were real code the last statement would generate an exception.
>>>>> orderedDict = OrderedDict()
>>>>> for k in sorted(unorderedDict.keys()):
>>>>>     orderedDict[k]  unorderedDict.pop(k)
>>>> This is not even valid syntax.
>>>> Please
>>>>     (1) explain the problem that you're trying to solve, not how you
>>>>         imagine the solution, and
>>>>     (2) if you have any code, please post real code (copy and paste).
>>>> The above code is not real.
>>>> Cheers&  hth.,
>>>> - Alf
>>> Sorry about the sorted != ordered mix up.  I want to end up with a
>>> *sorted* dict from an unordered list.  *Sorting the list is not
>>> practical in this case.*  I am using python 2.5, with an ActiveState
>>> recipe for an OrderedDict.
>>> Given these requirements/limitations, how would you do it?
>>> My solution is to create a regular dict from the list.  Then sort the
>>> keys, and add the keys+values to an OrderedDict.  Since the keys are
>>> being added to the OrderedDict in the correctly sorted order, at the
>>> end I end up with a OrderedDict that is in the correctly *sorted*
>>> order.
>> If that works for you, I don't understand your assertion that you can't
>> sort the list. If you have the space & time to sort the intermediate
>> dict, then it's as easy to create the list & sort & then the ordered
>> dict from it. It should be faster, because you sort the keys anyway.
>> Diez
> Here is how I am converting a regular dict to an ordered dict that is
> sorted by keys.
> def _getOrderedDict(theDict):
> 	ordered = OrderedDict()
> 	for k in sorted(theDict.keys()):
> 		ordered[k] = theDict.pop(k)
> 	return ordered
As I mentioned in an earlier post, you could do:

     def _getOrderedDict(theDict):
         return OrderedDict(sorted(theDict.items()))


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