I think so, but it neither take long time
julia> let
A = rand(4,4);
@time for i = 1:10^6;A - 5.0*eye(4);end
end;
elapsed time: 0.177621144 seconds (534 MB allocated, 5.71% gc time in 25
pauses with 0 full sweep)
julia> let
A = rand(4,4);
@time for i = 1:10^6;A - 5.0*I;end
end;
elapsed time: 0.053749097 seconds (167 MB allocated, 5.04% gc time in 8
pauses with 0 full sweep)
2015-02-01 11:49 GMT-05:00 paul analyst <paul.anal...@mail.com>:
> Thx, nice. Is the fast way in Julia ?
> Paul
>
> W dniu niedziela, 1 lutego 2015 17:18:03 UTC+1 użytkownik Andreas Noack
> napisał:
>>
>> This is cheaper: rand(4,4) - 5*I because 5I is a special type that
>> doesn't store all elements.
>>
>> 2015-02-01 11:03 GMT-05:00 paul analyst <paul.a...@mail.com>:
>>
>>> I have somethink like this :
>>>
>>> rand(4,4).-eye(4,4)*5
>>>
>>> Qestion : how to simpler subtracted from the diagonal value(5) ,
>>> Without multiplying matrices .... I have great and I have to simplify
>>> the code.
>>>
>>> Paul
>>>
>>
>>
