Re: Efficiently building ordered dict
OrderedDict is a class in collection module in python 2.7a3+. Perhaps you
can use it from there.
dir(collections)
['Callable', 'Container', 'Counter', 'Hashable', 'ItemsView', 'Iterable',
'Iterator', 'KeysView', 'Mapping', 'MappingView', 'MutableMapping',
'MutableSequence', 'MutableSet', 'OrderedDict', 'Sequence', 'Set', 'Sized',
'ValuesView', '_Link', '__all__', '__builtins__', '__doc__', '__file__',
'__name__', '__package__', '_abcoll', '_chain', '_eq', '_heapq', '_ifilter',
'_imap', '_iskeyword', '_itemgetter', '_proxy', '_repeat', '_starmap',
'_sys', 'defaultdict', 'deque', 'namedtuple']
~l0nwlf
On Mon, Feb 22, 2010 at 10:02 PM, Bryan bryanv...@gmail.com wrote:
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])
orderedDict = OrderedDict()
for k in sorted(unorderedDict.keys()):
orderedDict[k] unorderedDict.pop(k)
--
http://mail.python.org/mailman/listinfo/python-list
--
http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
On Mon, Feb 22, 2010 at 10:32 AM, Bryan bryanv...@gmail.com wrote:
unorderedDict = {}
for thing in unorderedList:
if thing.id in unorderedDict:
UpdateExistingValue(unorderedDict[thing.id])
else:
CreateNewValue(unorderedDict[thing.id])
orderedDict = OrderedDict()
for k in sorted(unorderedDict.keys()):
orderedDict[k] unorderedDict.pop(k)
It's not entirely clear what UpdateExistingValue and CreateNewValue do.
However, it looks like you are trying to create a dictionary where the keys
are sorted. Is that right?
If so, you could use the sorteddict type from my blist package, which is
similar to an OrderDict except it efficiently keeps the keys in sorted order
instead of insertion order. For example:
from blist import sorteddict
my_dict = sorteddict({1: 'a', 6: 'b', -5: 'c'})
my_dict.keys()
[-5, 1, 6]
my_dict[2] = 'd'
my_dict.keys()
[-5, 1, 2, 6]
It's available here:
http://pypi.python.org/pypi/blist/
--
Daniel Stutzbach, Ph.D.
President, Stutzbach Enterprises, LLC http://stutzbachenterprises.com
--
http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
Bryan wrote: 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? Why are you building a dict from a list and then an ordered dict from that? Just build the ordered dict from the list, because it's behaves like a dict, except for remembering the order in which the keys were added. -- http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
On Feb 22, 9:19 am, MRAB pyt...@mrabarnett.plus.com wrote: Bryan wrote: 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? Why are you building a dict from a list and then an ordered dict from that? Just build the ordered dict from the list, because it's behaves like a dict, except for remembering the order in which the keys were added. Could you write some pseudo-code for that? I'm not sure how I would add the items to the OrderedDict while looping through the list. Wouldn't the list need to be sorted first (which in this case isn't practical)? -- http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
Bryan wrote: On Feb 22, 9:19 am, MRAB pyt...@mrabarnett.plus.com wrote: Bryan wrote: 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? Why are you building a dict from a list and then an ordered dict from that? Just build the ordered dict from the list, because it's behaves like a dict, except for remembering the order in which the keys were added. Could you write some pseudo-code for that? I'm not sure how I would add the items to the OrderedDict while looping through the list. Wouldn't the list need to be sorted first (which in this case isn't practical)? ordered != sorted. If you want the ordered dict to be sorted by key then build a dict first and then create the ordered dict from the sorted dict. I think the quickest way to build the sorted dict is: orderedDict = OrderedDict(sorted(unorderedDict.items())) although I haven't tried 'sorteddict' (see Daniel Stutzbach's post). -- http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
Bryan bryanv...@gmail.com writes:
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])
orderedDict = OrderedDict()
for k in sorted(unorderedDict.keys()):
orderedDict[k] unorderedDict.pop(k)
Why don't you sort your unorderedList first then simply create your
orderedDict?
To me your problem is stated in too vague terms anyway and the code you
post could not really work as unorderedDict is bound to remain empty
whatever the values of all the other names.
At any rate, why do you need an ordered dictionary? It seems suspect to
me as you sort the keys before inserting the items.
--
Arnaud
--
http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
Bryan wrote: On Feb 22, 9:19 am, MRAB pyt...@mrabarnett.plus.com wrote: Bryan wrote: 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? Why are you building a dict from a list and then an ordered dict from that? Just build the ordered dict from the list, because it's behaves like a dict, except for remembering the order in which the keys were added. Could you write some pseudo-code for that? I'm not sure how I would add the items to the OrderedDict while looping through the list. Wouldn't the list need to be sorted first (which in this case isn't practical)? Could you please clarify what you are trying to achieve as you appear to be confusing ordering with sorting. Mark Lawrence. -- http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
* 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
--
http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
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.
self.accTotals = {}
for row in self.rows:
if row.acc.code in self.accTotals:
self.accTotals[row.acc.code].addRow(row)
else:
accSummary = Total()
accSummary.addRow(row)
self.accTotals[row.acc.code] = accSummary
self.accTotals = self._getOrderedDict(self.accTotals)
--
http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
On 22 Feb, 21:29, Bryan bryanv...@gmail.com wrote:
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.
Why does the dict need to be sorted?
Why is it impractical to sort the list, but practical to sort the
dict?
Whithout knowing this, it is difficult to get an idea of your problem
an a potential solution.
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.
self.accTotals = {}
for row in self.rows:
if row.acc.code in self.accTotals:
self.accTotals[row.acc.code].addRow(row)
else:
accSummary = Total()
accSummary.addRow(row)
self.accTotals[row.acc.code] = accSummary
self.accTotals = self._getOrderedDict(self.accTotals)
This code is a typical example where defaultdict, which was added in
Python 2.5 [1], would be of use:
accTotals = defaultdict(Total)
for row in self.rows:
accTotals[row.acc.code].addRow(row)
self.accTotals = self._getOrderedDict(accTotals)
However, as you don't explain what self._getOrderedDict(...) does, it
is quite difficult to guess how to improve it!
[1] http://docs.python.org/library/collections.html#collections.defaultdict
--
Arnaud
--
http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
Am 22.02.10 22:29, schrieb Bryan:
On Feb 22, 10:57 am, Alf P. Steinbachal...@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
--
http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
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. Steinbachal...@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.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
Am 22.02.10 23:48, schrieb Bryan:
On Feb 22, 2:16 pm, Diez B. Roggischde...@nospam.web.de wrote:
Am 22.02.10 22:29, schrieb Bryan:
On Feb 22, 10:57 am, Alf P. Steinbachal...@start.nowrote:
* 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.
Cheershth.,
- 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
--
http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
Bryan wrote:
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. Steinbachal...@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()))
[snip]
--
http://mail.python.org/mailman/listinfo/python-list
Re: Efficiently building ordered dict
On Feb 22, 3:00 pm, Diez B. Roggisch de...@nospam.web.de wrote:
Am 22.02.10 23:48, schrieb Bryan:
On Feb 22, 2:16 pm, Diez B. Roggischde...@nospam.web.de wrote:
Am 22.02.10 22:29, schrieb Bryan:
On Feb 22, 10:57 am, Alf P. Steinbachal...@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
Lets say my list has 1 million items. If I sort the list before
summarizing it, I will absolutley have to sort 1 million items.
However, if I summarize first, then just sort the keys in the summary
dict, I could wind up only sorting 4 or 5 items, if within that 1
million item list, there were only 4 or 5 job numbers for example.
The list stays as is, and my summary dictionary is sorted so I can
iterate through it and make little reports that humans will like.
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Re: Efficiently building ordered dict
On 2/22/2010 4:29 PM, Bryan wrote:
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.
Have you looked at this:
http://pypi.python.org/pypi/sorteddict/1.2.1
data = zip('afcedbijhg', range(10))
old = dict(data)
for key in old:
... print key, old[key]
...
a 0
c 2
b 5
e 3
d 4
g 9
f 1
i 6
h 8
j 7
new = sorteddict.sorteddict(data)
for key in new:
... print key, new[key]
...
a 0
b 5
c 2
d 4
e 3
f 1
g 9
h 8
i 6
j 7
-John
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