How to Convert vector elements to zero for certain N length when its values gets negative?
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Sagar Dhage
am 6 Aug. 2014
Kommentiert: Sagar Dhage
am 7 Aug. 2014
I have a column vector F= (0; 0; 0; 0; 5; 7; 12; 11; 23; 32; 22; 10; 8; -6 0; -8; 1; 4; 8; -5; 0; 0; 0; 6; 8; 9; 13; 14; 12; 23; 34; 22; 16; 17; 4; -5 ; -6; 5; 7; 0; 0; 0)
I want to convert values of F when it gets negative (in this case -6 and -5) till N (lets N=8)
so result would be F= (0; 0; 0; 0; 5; 7; 12; 11; 23; 32; 22; 10; 8; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 6; 8; 9; 13; 14; 12; 23; 34; 22; 16; 17; 4; 0; 0; 0; 0; 0; 0; 0) how to do this?
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Akzeptierte Antwort
Azzi Abdelmalek
am 6 Aug. 2014
F(F<0 & F>=-8)=0
3 Kommentare
Azzi Abdelmalek
am 6 Aug. 2014
Bearbeitet: Azzi Abdelmalek
am 6 Aug. 2014
F= [0; 0; 0; 0; 5; 7; 12; 11; 23; 32; 22; 10; 8; -6 ;0; -8; 1; 4; 8; -5; 0; 0; 0; 6; 8; 9; 13; 14; 12; 23; 34; 22; 16; 17; 4; -5 ; -6; 5; 7; 0; 0; 0]
idx=find(F<0,1)
n=numel(F);
while ~isempty(idx)
F(idx:min(idx+7,n))=0;
idx=find(F'<0,1);
end
Weitere Antworten (3)
Yu Jiang
am 6 Aug. 2014
Bearbeitet: Yu Jiang
am 6 Aug. 2014
while ~isempty(find(F<0))
id1 = find(F<0,1);
id2 = min(id1 + 7, length(F));
F(id1:id2) = 0;
end
2 Kommentare
Yu Jiang
am 6 Aug. 2014
How about this?
while ~isempty(find(F<0))
id1 = find(F<0,1);
id2 = min(id1 + 7, length(F));
F(id1:id2) = 0;
end
Roger Stafford
am 6 Aug. 2014
I am not sure what you mean in your remark "till N (lets N=8)". Do you mean that a maximum of N successive negative values in F are to be converted to 0 with all the possible remaining negative values, if any, unaffected? If so, the example you should have used ought to have demonstrated that behavior. In any case, here is code that would accomplish such a task:
F(find(F<0,N)) = 0;
Roger Stafford
am 6 Aug. 2014
I think I understand you now. How about this:
n = length(F);
for k = 1:n
if F(k) < 0
F(k:min(k+N-1,n)) = 0;
end
end
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