Re: vectorization quiz

[Francesco Potortì]

>
>On Fri, Sep 10, 2010 at 10:42 AM, Francesco Potortì <address@hidden> wrote:
>>>> I thought a little, but I do not see the way of vectorising this.  It's
>>>> a loop that is run only few times, but is called very frequently.
>>>> Essentially, there are few 2-D maps stacked one over the other which
>>>> must be shifted each by a variable amount and then summed to obtain a
>>>> new 2-D map.  Here is the code:
>>>>
>>>>    for mi = 1:m
>>>>      llka += llkm(by+offm(mi,1), bx+offm(mi,2), mi);
>>>>    endfor
>>>>
>>>> where:
>>>> m          ==>     5
>>>> size(llka) ==>     156 222
>>>> size(llkm) ==>     186 252 5
>>>> by         ==>     16:172
>>>> bx         ==>     16:238
>>>> size(offm) ==>     5 2
>>>> offm(m,i)  ==>     a scalar in the range [-15;15]
>>>>
>>>> Thanks in advance to those wishing to try :)
>>>>
>>>I don't think there's a good way. You can, of course, extract the
>>>submatrices via cellfun and then use plus, but it's not going to be
>>>significantly better.
>>
>> I think the slow thing is the index computation, so maybe I could try to
>> compute them all at once and then index, reshape and sum.  Maybe I'll try.
>>
>
>Is it? range + scalar should result in a range when possible. I think
>this already worked in Octave 3.2.4.
>Indexing with ranges is generally the fastest option.

I am referring to the fact that the same thing is done several times
because of the loop.  But in fact, if you precompute indexes then you
lose the advantage of having ranges as indexes.  Mh.

>> With the new diag implementation now we have a fast way of multiplying
>> each row of a matrix by a different number.  But I think we have no
>> euivalently fast way of summing a different number to each row.  Am I
>> wrong?
>>
>In 3.3.52+, you can use bsxfun (@plus, a, b).

It also works in 3.2.4:

octave> bsxfun(@plus,1:7,(1:3)')
ans =

    2    3    4    5    6    7    8
    3    4    5    6    7    8    9
    4    5    6    7    8    9   10

At last I understood what's the use of bsxfun!  Thanks :)

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