Re: building a dict
Andreas Waldenburger wrote: On Sat, 13 Mar 2010 13:42:12 -0800 (PST) vsoler vicente.so...@gmail.com wrote: By the way, I suppose I am the OP. Since I am not an native English speaking person, I do not know what it stands for. Perhaps you can tell me. Perhaps you can find out yourself: http://www.urbandictionary.com/define.php?term=op Possibly so, and that's a useful site, but I hope you aren't suggesting that Vicente shouldn't have asked. It seemed like a perfectly acceptable question to me. regards Steve -- Steve Holden +1 571 484 6266 +1 800 494 3119 See PyCon Talks from Atlanta 2010 http://pycon.blip.tv/ Holden Web LLC http://www.holdenweb.com/ UPCOMING EVENTS:http://holdenweb.eventbrite.com/ -- http://mail.python.org/mailman/listinfo/python-list
Re: building a dict
On Sun, 14 Mar 2010 08:36:55 -0400 Steve Holden st...@holdenweb.com wrote: Andreas Waldenburger wrote: On Sat, 13 Mar 2010 13:42:12 -0800 (PST) vsoler vicente.so...@gmail.com wrote: By the way, I suppose I am the OP. Since I am not an native English speaking person, I do not know what it stands for. Perhaps you can tell me. Perhaps you can find out yourself: http://www.urbandictionary.com/define.php?term=op Possibly so, and that's a useful site, but I hope you aren't suggesting that Vicente shouldn't have asked. It seemed like a perfectly acceptable question to me. I was a bit trigger happy there, I admit. But with vocabulary questions like these, my first reflex is to ask the web rather than people, because it's usually quicker than Usenet (albeit often only by a minute or so). I somehow felt the need to bestow that bit of wisdom on the world. But yeah, the question is fine, of course. /W -- INVALID? DE! -- http://mail.python.org/mailman/listinfo/python-list
building a dict
Say that m is a tuple of 2-tuples
m=(('as',3), ('ab',5), (None, 1), ('as',None), ('as',6))
and I need to build a d dict where each key has an associated list
whose first element is the count, and the second is the sum. If a 2-
tuple contains a None value, it should be discarded.
The expected result is:
d={'as':[2, 9], 'ab': [1,5]}
How should I proceed? So far I have been unsuccessful. I have tried
with a for loop.
Thank you for your help
--
http://mail.python.org/mailman/listinfo/python-list
Re: building a dict
On 13 Mar, 15:05, vsoler vicente.so...@gmail.com wrote:
Say that m is a tuple of 2-tuples
m=(('as',3), ('ab',5), (None, 1), ('as',None), ('as',6))
and I need to build a d dict where each key has an associated list
whose first element is the count, and the second is the sum. If a 2-
tuple contains a None value, it should be discarded.
The expected result is:
d={'as':[2, 9], 'ab': [1,5]}
How should I proceed? So far I have been unsuccessful. I have tried
with a for loop.
Thank you for your help
Something like:
d = defaultdict( lambda: [0,0] )
for key, val in filter(lambda L: not any(i is None for i in L), m):
d[key][0] += 1
d[key][1] += val
hth
Jon
--
http://mail.python.org/mailman/listinfo/python-list
Re: building a dict
On Mar 13, 9:05 am, vsoler vicente.so...@gmail.com wrote:
Say that m is a tuple of 2-tuples
m=(('as',3), ('ab',5), (None, 1), ('as',None), ('as',6))
and I need to build a d dict where each key has an associated list
whose first element is the count, and the second is the sum. If a 2-
tuple contains a None value, it should be discarded.
The expected result is:
d={'as':[2, 9], 'ab': [1,5]}
How should I proceed? So far I have been unsuccessful. I have tried
with a for loop.
Post your first try at a for loop, and people might be willing to
point out problems, but this is such a basic for loop that it is
unlikely that anybody is going to write your ENTIRE homework for you.
Regards,
Pat
--
http://mail.python.org/mailman/listinfo/python-list
Re: building a dict
vsoler wrote:
Say that m is a tuple of 2-tuples
m=(('as',3), ('ab',5), (None, 1), ('as',None), ('as',6))
and I need to build a d dict where each key has an associated list
whose first element is the count, and the second is the sum. If a 2-
tuple contains a None value, it should be discarded.
The expected result is:
d={'as':[2, 9], 'ab': [1,5]}
How should I proceed? So far I have been unsuccessful. I have tried
with a for loop.
Thank you for your help
Here's a fairly simple-minded approach using a defaultdict, which calls
the dflt() function to create a value when the key is absent.
from collections import defaultdict
def dflt():
... return [0, 0]
...
m = (('as',3), ('ab',5), (None, 1), ('as',None), ('as',6))
d = defaultdict(dflt)
for key, n in m:
... if key is not None and n is not None:
... c, t = d[key]
... d[key] = [c+1, t+n]
...
d
defaultdict(function dflt at 0x7f0bcb1b0ed8,
{'as': [2, 9], 'ab': [1, 5]})
regards
Steve
--
Steve Holden +1 571 484 6266 +1 800 494 3119
See PyCon Talks from Atlanta 2010 http://pycon.blip.tv/
Holden Web LLC http://www.holdenweb.com/
UPCOMING EVENTS:http://holdenweb.eventbrite.com/
--
http://mail.python.org/mailman/listinfo/python-list
Re: building a dict
On 13 Mar, 15:28, Patrick Maupin pmau...@gmail.com wrote:
On Mar 13, 9:05 am, vsoler vicente.so...@gmail.com wrote:
Say that m is a tuple of 2-tuples
m=(('as',3), ('ab',5), (None, 1), ('as',None), ('as',6))
and I need to build a d dict where each key has an associated list
whose first element is the count, and the second is the sum. If a 2-
tuple contains a None value, it should be discarded.
The expected result is:
d={'as':[2, 9], 'ab': [1,5]}
How should I proceed? So far I have been unsuccessful. I have tried
with a for loop.
Post your first try at a for loop, and people might be willing to
point out problems, but this is such a basic for loop that it is
unlikely that anybody is going to write your ENTIRE homework for you.
Regards,
Pat
I was thinking it's possibly homework, but looking at previous posts
it's fairly unlikely.
(If it is, then mea culpa, but as Steve has replied, I think I'll
manage to sleep tonight not worrying about the influx of uneducated,
incompetent and otherwise useless developers to the market).
However, they're receiving some 'elegant' solutions which no professor
(unless they're a star pupil - in which case they wouldn't be asking)
would take as having been done by their selves.
(Or at least I hope not)
But yes, I would certainly be interested in the 'unsuccessful
attempt'.
(To the OP, do post your attempts, it does give more validity).
Cheers,
Jon.
--
http://mail.python.org/mailman/listinfo/python-list
Re: building a dict
On Mar 13, 8:28 am, Patrick Maupin pmau...@gmail.com wrote:
On Mar 13, 9:05 am, vsoler vicente.so...@gmail.com wrote:
Say that m is a tuple of 2-tuples
m=(('as',3), ('ab',5), (None, 1), ('as',None), ('as',6))
and I need to build a d dict where each key has an associated list
whose first element is the count, and the second is the sum. If a 2-
tuple contains a None value, it should be discarded.
The expected result is:
d={'as':[2, 9], 'ab': [1,5]}
How should I proceed? So far I have been unsuccessful. I have tried
with a for loop.
Post your first try at a for loop, and people might be willing to
point out problems, but this is such a basic for loop that it is
unlikely that anybody is going to write your ENTIRE homework for you.
This is probably what you (OP) were trying to come up with?
[untested]
d = {}
for item in m:
key = m[0]; value = m[1]
if key is None or value is None: continue
if key not in dict:
d[key] = [value]
else:
d[key].append (value)
You can replace the
for item in m:
key = m[0]; value = m[1]
above with
for key, value in m:
which is a little nicer.
However, as other responses point out, when you want
to accumulate results in a dict, collections.defaultdict
should pop into your mind first.
--
http://mail.python.org/mailman/listinfo/python-list
Re: building a dict
On Mar 13, 9:13 am, ru...@yahoo.com wrote:
On Mar 13, 8:28 am, Patrick Maupin pmau...@gmail.com wrote:
On Mar 13, 9:05 am, vsoler vicente.so...@gmail.com wrote:
Say that m is a tuple of 2-tuples
m=(('as',3), ('ab',5), (None, 1), ('as',None), ('as',6))
and I need to build a d dict where each key has an associated list
whose first element is the count, and the second is the sum. If a 2-
tuple contains a None value, it should be discarded.
The expected result is:
d={'as':[2, 9], 'ab': [1,5]}
How should I proceed? So far I have been unsuccessful. I have tried
with a for loop.
Post your first try at a for loop, and people might be willing to
point out problems, but this is such a basic for loop that it is
unlikely that anybody is going to write your ENTIRE homework for you.
This is probably what you (OP) were trying to come up with?
[untested]
d = {}
for item in m:
key = m[0]; value = m[1]
if key is None or value is None: continue
if key not in dict:
d[key] = [value]
else:
d[key].append (value)
You can replace the
for item in m:
key = m[0]; value = m[1]
above with
for key, value in m:
which is a little nicer.
However, as other responses point out, when you want
to accumulate results in a dict, collections.defaultdict
should pop into your mind first.
Oops, didn't read very carefully, did I?
That should be:
d = {}
for item in m:
key = m[0]; value = m[1]
if key is None or value is None: continue
if key not in dict:
d[key] = [1, value]
else:
d[key][0] += 1
d[key][1] += value
--
http://mail.python.org/mailman/listinfo/python-list
Re: building a dict
On Mar 13, 9:26 am, ru...@yahoo.com wrote:
That should be:
d = {}
for item in m:
key = item[0]; value = item[1]
if key is None or value is None: continue
if key not in dict:
d[key] = [1, value]
else:
d[key][0] += 1
d[key][1] += value
That's it. Any other mistakes, you find 'em.
--
http://mail.python.org/mailman/listinfo/python-list
Re: building a dict
On 13 mar, 18:16, ru...@yahoo.com wrote:
On Mar 13, 9:26 am, ru...@yahoo.com wrote: That should be:
d = {}
for item in m:
key = item[0]; value = item[1]
if key is None or value is None: continue
if key not in dict:
d[key] = [1, value]
else:
d[key][0] += 1
d[key][1] += value
That's it. Any other mistakes, you find 'em.
Thank you all. Your answers are more than valuable to me. I'll study
them carefully, but no doubt, my post has been answered.
By the way, I suppose I am the OP. Since I am not an native English
speaking person, I do not know what it stands for. Perhaps you can
tell me.
From what I see from your posts, you would have preferred that I
included in my original post my for loop, so that the post is not so
abstract. I have taken note and I'll make it better next time.
Thank you for your help.
Vicente Soler
--
http://mail.python.org/mailman/listinfo/python-list
Re: building a dict
On Mar 13, 2:42 pm, vsoler vicente.so...@gmail.com wrote: By the way, I suppose I am the OP. Since I am not an native English speaking person, I do not know what it stands for. Perhaps you can tell me. OP means Original Poster (the person who started the discussion) or sometimes Original Post, depending on context. -- http://mail.python.org/mailman/listinfo/python-list
Re: building a dict
On Sat, 13 Mar 2010 13:42:12 -0800 (PST) vsoler vicente.so...@gmail.com wrote: By the way, I suppose I am the OP. Since I am not an native English speaking person, I do not know what it stands for. Perhaps you can tell me. Perhaps you can find out yourself: http://www.urbandictionary.com/define.php?term=op /W -- INVALID? DE! -- http://mail.python.org/mailman/listinfo/python-list
Efficiently building ordered dict
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
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.
--
http://mail.python.org/mailman/listinfo/python-list
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
--
http://mail.python.org/mailman/listinfo/python-list
Building a dict from a tuple of tuples
Hello everyone!
I have a tuple of tuples, coming from an Excel range, such as this:
((None, u'x', u'y'),
(u'a', 1.0, 7.0),
(u'b', None, 8.0))
I need to build a dictionary that has, as key, the row and column
header.
For example:
d={ (u'a',u'x'):1.0, (u'a',u'y'): 7.0, (u'b',u'y'):8.0 }
As you can see, if the value in the matrix is None, no key has to be
added to the dictionary.
Of course, my tuple of tuples is a lot bigger.
How can I possibly do this?
Thank you
--
http://mail.python.org/mailman/listinfo/python-list
Re: Building a dict from a tuple of tuples
vsoler wrote:
Hello everyone!
I have a tuple of tuples, coming from an Excel range, such as this:
((None, u'x', u'y'),
(u'a', 1.0, 7.0),
(u'b', None, 8.0))
I need to build a dictionary that has, as key, the row and column
header.
For example:
d={ (u'a',u'x'):1.0, (u'a',u'y'): 7.0, (u'b',u'y'):8.0 }
As you can see, if the value in the matrix is None, no key has to be
added to the dictionary.
Of course, my tuple of tuples is a lot bigger.
How can I possibly do this?
Thank you
Does this help?
matrix = ((None, u'x', u'y'),
(u'a', 1.0, 7.0),
(u'b', None, 8.0))
for row in matrix[1 : ]:
for col, val in zip(matrix[0][1 : ], row[1 : ]):
print row[0], col, val
--
http://mail.python.org/mailman/listinfo/python-list
Re: Building a dict from a tuple of tuples
On Feb 20, 7:00 pm, MRAB pyt...@mrabarnett.plus.com wrote:
vsoler wrote:
Hello everyone!
I have a tuple of tuples, coming from an Excel range, such as this:
((None, u'x', u'y'),
(u'a', 1.0, 7.0),
(u'b', None, 8.0))
I need to build a dictionary that has, as key, the row and column
header.
For example:
d={ (u'a',u'x'):1.0, (u'a',u'y'): 7.0, (u'b',u'y'):8.0 }
As you can see, if the value in the matrix is None, no key has to be
added to the dictionary.
Of course, my tuple of tuples is a lot bigger.
How can I possibly do this?
Thank you
Does this help?
matrix = ((None, u'x', u'y'),
(u'a', 1.0, 7.0),
(u'b', None, 8.0))
for row in matrix[1 : ]:
for col, val in zip(matrix[0][1 : ], row[1 : ]):
print row[0], col, val
and the dictionary?
it is the ultimate goal of what I am intending...
Thank you
--
http://mail.python.org/mailman/listinfo/python-list
Re: Building a dict from a tuple of tuples
vsoler wrote:
On Feb 20, 7:00 pm, MRAB pyt...@mrabarnett.plus.com wrote:
vsoler wrote:
Hello everyone!
I have a tuple of tuples, coming from an Excel range, such as this:
((None, u'x', u'y'),
(u'a', 1.0, 7.0),
(u'b', None, 8.0))
I need to build a dictionary that has, as key, the row and column
header.
For example:
d={ (u'a',u'x'):1.0, (u'a',u'y'): 7.0, (u'b',u'y'):8.0 }
As you can see, if the value in the matrix is None, no key has to be
added to the dictionary.
Of course, my tuple of tuples is a lot bigger.
How can I possibly do this?
Thank you
Does this help?
matrix = ((None, u'x', u'y'),
(u'a', 1.0, 7.0),
(u'b', None, 8.0))
for row in matrix[1 : ]:
for col, val in zip(matrix[0][1 : ], row[1 : ]):
print row[0], col, val
and the dictionary?
it is the ultimate goal of what I am intending...
Thank you
The difficult bit is working out how to produce the keys and values for
the dict from the tuple of tuples, and I've shown you that. The rest is
straightforward.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Building a dict from a tuple of tuples
On Feb 20, 8:54 pm, MRAB pyt...@mrabarnett.plus.com wrote:
vsoler wrote:
On Feb 20, 7:00 pm, MRAB pyt...@mrabarnett.plus.com wrote:
vsoler wrote:
Hello everyone!
I have a tuple of tuples, coming from an Excel range, such as this:
((None, u'x', u'y'),
(u'a', 1.0, 7.0),
(u'b', None, 8.0))
I need to build a dictionary that has, as key, the row and column
header.
For example:
d={ (u'a',u'x'):1.0, (u'a',u'y'): 7.0, (u'b',u'y'):8.0 }
As you can see, if the value in the matrix is None, no key has to be
added to the dictionary.
Of course, my tuple of tuples is a lot bigger.
How can I possibly do this?
Thank you
Does this help?
matrix = ((None, u'x', u'y'),
(u'a', 1.0, 7.0),
(u'b', None, 8.0))
for row in matrix[1 : ]:
for col, val in zip(matrix[0][1 : ], row[1 : ]):
print row[0], col, val
and the dictionary?
it is the ultimate goal of what I am intending...
Thank you
The difficult bit is working out how to produce the keys and values for
the dict from the tuple of tuples, and I've shown you that. The rest is
straightforward.
I'll try. Thank you very much MRAB
--
http://mail.python.org/mailman/listinfo/python-list
