help with making my code more efficient

[Larry.Martell at gmail.com]

On Friday, December 21, 2012 10:57:19 AM UTC-7, Larry.... at gmail.com wrote:
> On Thursday, December 20, 2012 8:31:18 PM UTC-7, Dave Angel wrote:
> > On 12/20/2012 08:46 PM, Larry.Martell at gmail.com wrote: 
> > > On Thursday, December 20, 2012 6:17:04 PM UTC-7, Dave Angel wrote: 
> > >> <snip>
 
> > > Of course it's a fragment - it's part of a large program and I was just showing the relevant parts. 
 
> > But it seems these are methods in a class, or something, so we're 
> > missing context.  And you use self without it being an argument to the 
> > function.  Like it's a global.

> I didn't show the entire method, only what I thought was relevant to my issue. The method is declared as:
>  
>     def generate_data(self):

> > > <snip>
 
> > > Yes, the code works. I end up with just the rows I want.
  
> > >> Are you only concerned about speed, not fixing features?  
 
> > > Don't know what you mean by 'fixing features'. The code does what I want, it just takes too long.
 
> > >> As far as I can tell, the logic that includes the time comparison is bogus.  
 
> > > Not at all. 

> > >> You don't do  anything there to worry about the value of tup[2], just whether some
> > >> item has a nearby time.  Of course, I could misunderstand the spec.

> > > The data comes from a database. tup[2] is a datetime column. tdiff comes from a datetime.timedelta() 
 
> > I thought that tup[1] was the datetime.  In any case, the loop makes no
> > sense to me, so I can't really optimize it, just make suggestions.
 
> Yes, tup[1] is the datetime. I mistyped last night. 

> > >> Are you making a global called 'self' ?  That name is by convention only
> > >> used in methods to designate the instance object.  What's the attribute
> > >> self?

> > > Yes, self is my instance object. self.message contains the string of interest that I need to look for. 

> > >> Can cdata have duplicates, and are they significant? 

> > > No, it will not have duplicates.

> > >> Is the list sorted in any way?
 
> > > Yes, the list is sorted by tool and datetime.

> > >> Chances are your performance bottleneck is the doubly-nested loop.  You

> > >> have a list comprehension at top-level code, and inside it calls a

> > >> function that also loops over the 600,000 items.  So the inner loop gets

> > >> executed 360 billion times.  You can cut this down drastically by some

> > >> judicious sorting, as well as by having a map of lists, where the map is
> > >> keyed by the tool.

> > > Thanks. I will try that.

> > So in your first loop, you could simply split the list into separate 
> > lists, one per tup[0] value, and the lists as dictionary items, keyed by
> > that tool string.

> > Then inside the determine() function, make a local ref to the particular 
> > list for the tool.
> >    recs = messageTimes[tup[0]]

> I made that change ant went from taking over 2 hours to 54 minutes. A dramatic improvement, but still not adequate for my app. 

> > Instead of a for loop over recs, use a binary search to identify the 
> > first item that's >= date_time-tdiff.  Then if it's less than 
> > date_time+tdiff, return True, otherwise False.  Check out the bisect 
> > module.  Function bisect_left() should do what you want in a sorted list.

> Didn't know about bisect. Thanks. I thought it would be my savior for sure. But unfortunaly when I added that, it blows up with out of memory. 

The out of memory error had nothing to do with using bisect. I had introduced a typo that I really though would have caused a variable referenced before assignment error. But it did not do that, and instead somehow caused all the memory in my machine to get used up. When I fixed that, it worked really well with bisect. The code that was taking 2 hours was down to 20 minutes, and even better, a query that was taking 40 minutes was down to 8 seconds. 

Thanks very much for all your help. 




> This was the code I had:
>  
> times = messageTimes[tup[0]]
> 
> le = bisect.bisect_right(times, tup[1])
> 
> ge = bisect.bisect_left(times, tup[1])
> 
> return (le and tup[1]-times[le-1] <= tdiff) or (ge != len(times) and times[ge]-tup[1] <= tdiff)
> 
> 
> 
> > >>> cdata[:] = [tup for tup in cdata if determine(tup)]
> 
> > 
> 
> > >> As the code exists, there's no need to copy the list.  Just do a simple 
> 
> > >> bind.
> 
> > 
> 
> > > This statement is to remove the items from cdata that I don't want. I don't know what you mean by bind. I'm not familiar with that python function. 
> 
> > 
> 
> > Every "assignment" to a simple name is really a rebinding of that name.
> 
> > cdata = [tup for tup in cdata if determine(tup)]
> 
> > 
> 
> > will rebind the name to the new object, much quicker than copying.  If
> 
> > this is indeed a top-level line, it should be equivalent.  But if in
> 
> > fact this is inside some other function, it may violate some other
> 
> > assumptions.  In particular, if there are other names for the same
> 
> > object, then you're probably stuck with modifying it in place, using
> 
> > slice notation.
> 
> 
> 
> The slice notation  was left over when when cdata was a tuple. Now that it's a list I don't need that any more. 
> 
> 
> 
> > BTW, a set is generally much more memory efficient than a dict, when you
> 
> > don't use the "value".  But since I think you'll be better off with a
> 
> > dict of lists, it's a moot point.
> 
> 
> 
> I'm going back to square 1 and try and do all from SQL.