On Feb 22, 2:16 pm, "Diez B. Roggisch" <de...@nospam.web.de> wrote:
> Am 22.02.10 22:29, schrieb Bryan:
>
>
>
> > On Feb 22, 10:57 am, "Alf P. Steinbach"<al...@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
The list is a bunch of objects that represent hours worked by
employees on particular jobs, and accounts, and client purchase orders
etc. From this list, I am generating these sorted dicts that contain
summarizing information about the list. So one of the sorted dicts
will give a summary of hours worked by job number. Another one will
give summary information by client PO number etc. So instead of
sorting the list a bunch of different ways, I keep the list as is,
generate the summaries I need into dictionaries, and then sort those
dictionaries as appropriate.
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