Summing a 2D list

[Paddy]

On Jun 13, 1:12 pm, Karsten Heymann <karsten.heym... at blue-cable.net>
wrote:
> Hi Mark,
>
>
>
> Mark <markjtur... at gmail.com> writes:
> > I have a scenario where I have a list like this:
>
> > User            Score
> > 1                 0
> > 1                 1
> > 1                 5
> > 2                 3
> > 2                 1
> > 3                 2
> > 4                 3
> > 4                 3
> > 4                 2
>
> > And I need to add up the score for each user to get something like
> > this:
>
> > User            Score
> > 1                 6
> > 2                 4
> > 3                 2
> > 4                 8
>
> > Is this possible? If so, how can I do it? I've tried looping through
> > the arrays and not had much luck so far.
>
> Although your problem has already been solved, I'd like to present a
> different approach which can be quite a bit faster. The most common
> approach seems to be using a dictionary:
>
> summed_up={}
> for user,vote in pairs:
>   if summed_up.has_key(user):
>     summed_up[user]+=vote
>   else:
>     summed_up[user]=vote
>
> But if the list of users is compact and the maximum value is known
> before, the using a list and coding the user into the list position is
> much more elegant:
>
> summed_up=list( (0,) * max_user )
> for user,vote in pairs:
>   summed_up[user] += vote
>
> I've run a quick and dirty test on these approaches and found that the
> latter takes only half the time than the first. More precisely, with
> about 2 million pairs, i got:
>
> * dict approach: 2s
>         (4s with "try: ... except KeyError:" instead of the "if")
> * list approach: 0.9s
>
> BTW this was inspired by the book "Programming Pearls" I read some
> years ago where a similar approach saved some magnitudes of time
> (using a bit field instead of a list to store reserved/free phone
> numbers IIRC).
>
> Yours,
> Karsten

How does your solution fare against the defaultdict solution of:

d = collections.defaultdict(int)
for u,s in zip(users,score): d[u] += s


- Paddy