eleminating data from a long vector

Hi all!
i have an index vector:
%
index=[id1;id2;id3.....idn]
and i have a long vector: A
i want to eleminate the elment of the vector a which have the index in index in this way:
%
A(id1:id1+30)=[]
A(id2:id2+30)=[]
.
.
.A(idn:idn+30)=[]
how could i write this in matlab in a compact manner?
thank you

1 comentario

Walter Roberson
Walter Roberson el 19 de Feb. de 2013
Are you certain this is what you want to do? After the first removal, everything from id1+30 onward in the vector would "fall down" 31 places. Does id2 take that renumbering into account?

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 Respuesta aceptada

Azzi Abdelmalek
Azzi Abdelmalek el 19 de Feb. de 2013
Editada: Azzi Abdelmalek el 19 de Feb. de 2013
A=rand(1,1000); % Example
index=[10 100 500]
idx=arrayfun(@(x) x:x+30,index,'un',0);
A(cell2mat(idx))=[]

2 comentarios

Walter Roberson
Walter Roberson el 19 de Feb. de 2013
This is not equivalent to the original sequence in that the original sequence does not delete id2:id2+30 until after id1:id1+30 are gone, so in absolute locations it might need to add 31 more for the 2nd index, 62 for the 3rd, and so on... depending on the sort order of the indices.
Azzi Abdelmalek
Azzi Abdelmalek el 19 de Feb. de 2013
Editada: Azzi Abdelmalek el 19 de Feb. de 2013
I don't think that what he meant by the sequence. I guess he want to remove them at the same time.

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Más respuestas (1)

Andrei Bobrov
Andrei Bobrov el 19 de Feb. de 2013
A=1:1000; % Example
index=[10 100 500];
n = 30;
A(bsxfun(@plus,index,(0:n-1)'))=[];

3 comentarios

Walter Roberson
Walter Roberson el 19 de Feb. de 2013
Should either be n=31 or run from 0:n instead of 0:n-1 as the original question asks to delete id1:id1+30 which is 31 locations.
Jan
Jan el 19 de Feb. de 2013
But in general this method is faster than the arrayfun and cell2mat approach.
Rica
Rica el 19 de Feb. de 2013
Thank you all

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