efficient way to process data

[Mark Lawrence]

On 13/01/2014 18:27, Larry Martell wrote:
> On Mon, Jan 13, 2014 at 1:09 AM, Chris Angelico <rosuav at gmail.com> wrote:
>> On Mon, Jan 13, 2014 at 2:35 PM, Larry Martell <larry.martell at gmail.com> wrote:
>>> Thanks for the reply. I'm going to take a stab at removing the group
>>> by and doing it all in python. It doesn't look too hard, but I don't
>>> know how it will perform.
>>
>> Well, if you can't switch to PostgreSQL or such, then doing it in
>> Python is your only option. There are such things as GiST and GIN
>> indexes that might be able to do some of this magic, but I don't think
>> MySQL has anything even remotely like what you're looking for.
>>
>> So ultimately, you're going to have to do your filtering on the
>> database, and then all the aggregation in Python. And it's going to be
>> somewhat complicated code, too. Best I can think of is this, as
>> partial pseudo-code:
>>
>> last_x = -999
>> x_map = []; y_map = {}
>> merge_me = []
>> for x,y,e in (SELECT x,y,e FROM t WHERE whatever ORDER BY x):
>>      if x<last_x+1:
>>          x_map[-1].append((y,e))
>>      else:
>>          x_map.append([(y,e)])
>>      last_x=x
>>      if y in y_map:
>>          merge_me.append((y_map[y], x_map[-1]))
>>      y_map[y]=x_map[-1]
>>
>> # At this point, you have x_map which is a list of lists, each one
>> # being one group, and y_map which maps a y value to its x_map list.
>>
>> last_y = -999
>> for y in sorted(y_map.keys()):
>>      if y<last_y+1:
>>          merge_me.append((y_map[y], last_x_map))
>>      last_y=y
>>      last_x_map=y_map[y]
>>
>> for merge1,merge2 in merge_me:
>>      merge1.extend(merge2)
>>      merge2[:]=[] # Empty out the list
>>
>> for lst in x_map:
>>      if not lst: continue # been emptied out, ignore it
>>      do aggregate stats, get sum(lst) and whatever else
>>
>> I think this should be linear complexity overall, but there may be a
>> few aspects of it that are quadratic. It's a tad messy though, and
>> completely untested. But that's an algorithmic start. The idea is that
>> lists get collected based on x proximity, and then lists get merged
>> based on y proximity. That is, if you have (1.0, 10.1), (1.5, 2.3),
>> (3.0, 11.0), (3.2, 15.2), they'll all be treated as a single
>> aggregation unit. If that's not what you want, I'm not sure how to
>> handle it.
>
> Thanks. Unfortunately this has been made a low priority task and I've
> been put on to something else (I hate when they do that). I'll revive
> this thread when I'm allowed to get back on this.
>

I've not followed this thread closely but would this help 
http://pandas.pydata.org/ ?  When and if you get back to it, that is!!!

-- 
My fellow Pythonistas, ask not what our language can do for you, ask 
what you can do for our language.

Mark Lawrence