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

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?

Answers (6)

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));
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 Comments

I got an error when I tried this:
Error using strrep Input strings must have one row.
Error in accelfilter (line 636) b=strrep(num2str(a),' ','');
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)
If M is a column array, just add at the beginning of your code
M=M'

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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 Comments

suppose that the array M length could change? How can I make the code still work for a matrix of size (1,n)?
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
Okay ,yes that works. Thank you very much.

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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))
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);
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+')

Asked:

on 29 Jan 2014

Commented:

on 29 Jan 2014

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