[Scilab-users] Advice from Scilab community

[Stefan Du Rietz]

Your second question:

-->(t == v)
  ans  =
   T T T T
-->and(t == v)
  ans  =
   T

-->(z == v)
  ans  =
   F T T F
-->and(z == v)
  ans  =
   F

PS  I didn't quite understand your first question, but you only have 
to use vectorfind with each possible permutation until it is found ;-)

/Stefan

On 2013-10-22 14:42, Stefan Du Rietz wrote:
--------------------
> You can use vectorfind() three times to get the line number.
>
> Regards
> Stefan.
>
>
>
> On 2013-10-22 13:58, Carrico, Paul wrote:
> --------------------
>> Dear All
>>
>> I’m asking here an advice from the community in order to find a (fast
>> and clean) solution ; I’ve the following issue :
>>
>> -I’ve a (n x 4) matrix where n may be huge (several hundred thousand
>> of lines is common )
>>
>> -I need to find and to remove the line wich has the same numbers (but
>> different in order)
>>
>> -I know that there’s only one occurrence … thus when it has been
>> found, not necessary to continue
>>
>> Example :
>>
>> A = [
>>
>> X  X  X      X
>>
>> X  X  X      X
>>
>> 1 9 1000 994423
>>
>> X  X  X      X
>>
>> X  X  X      X
>>
>> X  X  X      X
>>
>> 1000 994423 1 9  … or 994423 1 9 1000 … or 9 1000 994423 … only 3
>> possible permutations
>>
>> X  X  X      X
>>
>> X  X  X      X
>>
>> X  X  X      X
>>
>> ]
>>
>> Any suggestion that avoids the use of (intensive ugly) loops ?
>>
>> NB:
>>
>> additional question :
>>
>> -If t = [1 2 3 4]
>>
>> -If v = [1 2 3 4]
>>
>> -If z = [0 2 3 9]
>>
>> -Then (t == v) provides [T T T T]
>>
>> -Then (z == v) provides [F T T F]
>>
>> -…. Is it possible to have a single Boolean with all the cells met the
>> requirement (T) and when it’s not the case (F) ?
>>
>> Thanks in advance
>>
>> Paul
>>
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