Am 22.02.10 23:48, schrieb Bryan:
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.
Again - why? Unless there is some filtering going on that reduces the
number of total entries before the sorting when building the
intermediate dict, a simple
ordered = OrderedDict(sorted(the_list, key=lambda v: v['some_key']))
won't cost you a dime more.
I think you believe in building the dict so that ou can have the key for
sorting. As shown abov - you don't need to.
It might even benefitial to really re-sort the original list, because
that spares you the intermediate list.
Diez
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