eleminating data from a long vector
显示 更早的评论
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 个评论
Walter Roberson
2013-2-19
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?
采纳的回答
Azzi Abdelmalek
2013-2-19
编辑：Azzi Abdelmalek
2013-2-19
A=rand(1,1000); % Example
index=[10 100 500]
idx=arrayfun(@(x) x:x+30,index,'un',0);
A(cell2mat(idx))=[]
2 个评论
Walter Roberson
2013-2-19
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
2013-2-19
编辑：Azzi Abdelmalek
2013-2-19
I don't think that what he meant by the sequence. I guess he want to remove them at the same time.
更多回答（1 个）
Andrei Bobrov
2013-2-19
A=1:1000; % Example
index=[10 100 500];
n = 30;
A(bsxfun(@plus,index,(0:n-1)'))=[];
3 个评论
Walter Roberson
2013-2-19
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
2013-2-19
But in general this method is faster than the arrayfun and cell2mat approach.
Rica
2013-2-19
类别
在 帮助中心 和 File Exchange 中查找有关 Data Type Conversion 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
