Yeah the TreeProxy won't work in our case, as this is a MLM tree which
means there is really no order. For example:
1
2 10
11 9 3 4
Etc, I tried to convince the CEO to use a balanced tree when we first
started programming but that wasn't an option he wants people/nodes in the
tree to determine where the new person goes and on which leg. So I am not
sure how to query all left and or right without using a loop/recursive
function.
On Sun, Apr 8, 2012 at 8:19 AM, Massimo Di Pierro <
[email protected]> wrote:
> The recipe book described a
>
> class TreeProxy(object): ...
>
> which implements unordered tree traversal. It is the most efficient way to
> store records in a tree and retrieve them efficiently in a single query.
> Anyway, if this is the bottle neck, you should definitively cache it
> somehow.
>
>
> On Saturday, 7 April 2012 10:42:03 UTC-5, Detectedstealth wrote:
>>
>> Well I know the major bottle neck in the site is the binary tree. I am
>> still trying to figure out how to do this the best and most efficient. If
>> this was a stand alone app in c/c++ this would all be in memory which
>> wouldn't be such a problem. However how this is implemented it needs to
>> query through the database to find nodes. As you can imagine as we get more
>> and more members this will take much longer to run.
>>
>> See bellow code, binary_tree is called from a action and distributor is a
>> database object IE: db(db.distributors.id == id).select().first()
>>
>> def binary_tree(self, original_dist, distributor, levels, counter=0,
>> month=None):
>>
>>
>> children = []
>> left_count = right_count = counter
>>
>> target_month = month
>>
>> if distributor.left_id and left_count < levels:
>>
>> left_count += 1
>> #print "left - A", distributor.left_id
>>
>> children.append(self.binary_**tree(original_dist,
>> distributor.left_id, levels, left_count, target_month))
>>
>>
>> if distributor.right_id and right_count < levels:
>>
>> right_count += 1
>> #print "right - B", distributor.right_id
>>
>> children.append(self.binary_**tree(original_dist,
>> distributor.right_id, levels, right_count, target_month))
>>
>>
>> if distributor.sponsor_id == original_dist:
>>
>> sponsored_by_original = True
>> else:
>>
>> sponsored_by_original = False
>>
>> tree = {
>>
>> 'id': ("B" if distributor.side else "A") + "-" +
>> str(distributor.id),
>>
>> 'name': "<img src='/static/images/" + \
>>
>> ("top_tree_icon.png" if distributor.id == original_dist else
>> ("sponsored_tree_icon.png" if sponsored_by_original else
>> "user_tree_icon.png")) + \
>>
>> "' id='img-" + ("B" if distributor.side else "A") + "-" +
>> str(distributor.id) + "' /><span class='idlabel'>" + \
>>
>> str("0"*(7-len(str(distributor**.id))))+str(distributor.id) +
>> "-" + ("B" if distributor.side else "A") + \
>>
>> "</span><div class='userinfo' id='id-" + ("B" if
>> distributor.side else "A") + "-" + \
>>
>> str(distributor.id) + "'><ul><li class='userinfotop'><h4>" +
>> distributor.display_name + "</h4><span>" + \
>>
>> self.T("A CV:") + "<b>" +
>> (str(distributor.left_id.binar**y_cv_total) if distributor.left_id else
>> str(0)) + \
>>
>> "</b></span><span>" + self.T("B CV:") + "<b>" +
>> (str(distributor.right_id.bina**ry_cv_total) if distributor.right_id else
>> str(0)) + \
>>
>> "</b></span></li><li class='userinfobottom'><p>" +
>> self.T("Sponsored by me ") + (self.T("YES") if sponsored_by_original else
>> self.T("NO")) + \
>>
>> "</p><p>" + (self.T("Has down lines") if distributor.right_id or
>> distributor.left_id else self.T("Has no down lines")) + "</p><p>" + \
>>
>> (self.T("Disqualified ") if not distributor.quali_bonus else
>> self.T("Qualified ")) + self.T("to get bonus") + "</p></li></ul></div>",
>>
>> 'children': children
>> }
>>
>> return tree
>>
>> *# This is the biggest bottle neck of our entire application. Need a
>> much better way to handle tree traversing.*
>>
>> def traversingTree(self, distributor, list):
>>
>> if distributor:
>> left_node = distributor.left_id
>>
>> right_node = distributor.right_id
>> if left_node:
>>
>> list.append(left_node)
>> self.traversingTree(left_node,**list)
>>
>>
>> if right_node:
>> list.append(right_node)
>>
>> self.traversingTree(right_node**,list)
>>
>>
>> def bottom_right_b(self, distributor):
>>
>> right_id = None
>> cur_dist = distributor
>> while cur_dist.right_id:
>>
>> right_id = cur_dist.right_id
>> cur_dist = cur_dist.right_id
>>
>> print "Right Bottom", right_id
>> return right_id
>>
>>
>> def bottom_left_a(self, distributor):
>>
>> left_id = None
>> cur_dist = distributor
>> while cur_dist.left_id:
>>
>> left_id = cur_dist.left_id
>> cur_dist = cur_dist.left_id
>>
>> print "Left Bottom", left_id
>> return left_id
>>
>>
>>
>> On Sat, Apr 7, 2012 at 8:16 AM, Massimo Di Pierro <
>> [email protected]> wrote:
>>
>>> Right....what? [changing color to red] Note to self. Never do math
>>> before fully waking up.
>>>
>>> Correction: 2 reqs/second/server. (1M/24/60/60/6 servers). Still ok from
>>> web2py prospective.
>>> If all requests hit the database once: 12 reqs/second. It really depends
>>> of what those requests do.
>>>
>>> Massimo
>>>
>>> On Saturday, 7 April 2012 09:33:38 UTC-5, Anthony wrote:
>>>>
>>>> 1M reqs/day is 7 reqs/minute.
>>>>>
>>>>
>>>> Now we know how Massimo gets so much done -- his days are 2400 hours
>>>> long. ;-)
>>>>
>>>
>>
>>
>> --
>> --
>> Regards,
>> Bruce Wade
>> http://ca.linkedin.com/in/**brucelwade<http://ca.linkedin.com/in/brucelwade>
>> http://www.wadecybertech.com
>> http://www.fittraineronline.**com <http://www.fittraineronline.com> -
>> Fitness Personal Trainers Online
>> http://www.warplydesigned.com
>>
>>
--
--
Regards,
Bruce Wade
http://ca.linkedin.com/in/brucelwade
http://www.wadecybertech.com
http://www.fittraineronline.com - Fitness Personal Trainers Online
http://www.warplydesigned.com
