Find at least 4 consecutive values less than 1 in an array

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Neha
Neha el 29 de En. de 2014
Comentada: Jos (10584) el 29 de En. de 2014
I have an array in which I need to find four or more consecutive values that are less than one. For example: M = [1 0 .3 .5 .2 .1 6 7 .3 .5 10 1 .8 .9 .7 .2 .1 .3]
The program should return the indexes 3-6 and 13-18.
How can I go about solving this?

Respuestas (6)

Roger Stafford
Roger Stafford el 29 de En. de 2014
Editada: Roger Stafford el 29 de En. de 2014
Here's another one-liner if you accept initial indices as an answer.
f = find(all(hankel(M(1:4),M(4:end))<1,2));

Azzi Abdelmalek
Azzi Abdelmalek el 29 de En. de 2014
Editada: Azzi Abdelmalek el 29 de En. de 2014
Edit
M= [1 0 .3 .5 .2 .1 6 7 .3 .5 10 1 .8 .9 .7 .2 .1 .3]
if diff(size(M))<0
M=M'
end
a=M<1;
b=strrep(num2str(a),' ','');
[ii,jj]=regexp(b,'1+','start','end')
d=jj-ii>=4
out=[ii;jj];
out=out(:,d)
  3 comentarios
Azzi Abdelmalek
Azzi Abdelmalek el 29 de En. de 2014
Editada: Azzi Abdelmalek el 29 de En. de 2014
If you copy and paste the code, there is no error, unless you are using M that is not a 1xn array. What is the size of M
size(M)
Azzi Abdelmalek
Azzi Abdelmalek el 29 de En. de 2014
If M is a column array, just add at the beginning of your code
M=M'

Iniciar sesión para comentar.


Azzi Abdelmalek
Azzi Abdelmalek el 29 de En. de 2014
Editada: Azzi Abdelmalek el 29 de En. de 2014
Edit
M= [1 0 .3 .5 .2 .1 6 7 .3 .5 10 1 .8 .9 .7 .2 .1 .3];
if diff(size(M))<0
M=M'
end
a=[0 M<1 0];
ii=strfind(a,[0 1]);
jj=strfind(a,[1 0])-1;
d=jj-ii>=4;
out=[ii;jj];
out=out(:,d)
  3 comentarios
Azzi Abdelmalek
Azzi Abdelmalek el 29 de En. de 2014
It doesn't matter if the length changes. Maybe you are asking if the size changes, from 1xn to nx1. In this case we can edit our code. Look at edited answesrs
Neha
Neha el 29 de En. de 2014
Okay ,yes that works. Thank you very much.

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José-Luis
José-Luis el 29 de En. de 2014
M = [1 0 .3 .5 .2 .1 6 7 .3 .5 10 1 .8 .9 .7 .2 .1 .3];
numVal = 4;
your_vals = conv(double(M<1),ones(1,numVal)/numVal,'valid') == 1;
start = find(diff([0 your_vals]) == 1)
finish = fliplr(numel(M) + 1 - find(diff([0 fliplr(your_vals)]) == 1))

Andrei Bobrov
Andrei Bobrov el 29 de En. de 2014
t = M<1&M>0;
t1 = [true;diff(t(:))~=0];
idx=accumarray(cumsum(t1),(1:numel(M))',[],@(x){x});
ii = idx(t(t1));
out = ii(cellfun(@numel,ii)>=4);

Jos (10584)
Jos (10584) el 29 de En. de 2014
Editada: Jos (10584) el 29 de En. de 2014
Here is a relatively simple one-liner:
M = [1 0 .3 .5 .2 .1 6 7 .3 .5 10 1 .8 .9 .7 .2 .1 .3] % example data
[startIDX, endIDX] = regexp(char((M(:).'< 1)+'0'), '1111+')

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