Re: Yet Another Vectorization Quiz

[c.]

On 14 Dec 2010, at 23:44, Judd Storrs wrote:

> On Tue, Dec 14, 2010 at 5:12 PM, Carlo de Falco <address@hidden> wrote:
>> Hi all,
>> 
>> I need to shift all the zeros in a (possibly very large) matrix
>> to the end of each column without changing the ordering of the
>> remaing elements, e.g.:
>> 
>> a = [ 1 2 4 0         [ 1 2 4 6
>>      2 3 5 6    ==>    2 3 5 7
>>      3 0 0 7           3 4 6 0
>>      0 4 6 0]          0 0 0 0]
>> 
>> with a for loop this can be done as
>> 
>> for j = 1:columns (a)
>>  v = zeros (rows (a), 1);
>>  v(1:nnz (a(:, j))) = a(a(:, j) != 0, j);
>>  a(:, j) = v;
>> endfor
>> 
>> I personally see no obvious way to improve this,
>> does anyone have any idea?
> 
> idx = find(a) ;
> idx2 = (0:numel(idx)-1)' + ceil(idx/size(a,1)) ;
> b = zeros(size(a)) ;
> b(idx2) = a(idx) ;
> 
> But, I'm not sure this is so hot for a large a. In that case you could
> maybe save some memory by processing a smaller set of columns at a
> time. Maybe something like (not tested)
> 
> function b = sortzeros(a)
>   idx = find(a) ;
>   idx2 = (0:numel(idx)-1)' + ceil(idx/size(a,1)) ;
>   b = zeros(size(a)) ;
>   b(idx2) = a(idx) ;
> endfunction
> 
> n = size(a,2) ;
> m = 8 ;   # process in chunks of 8 columns at a time
> for i=1:m:n
>   j = min(i+m-1,n) ;
>   a(:,i:j) = sortzeros(a(:,i:j)) ;
> endfor
> 
> 
> --judd

Judd,
Thanks, your solution also works but I went with the one suggested by Ben which 
is more fun ;)
c.