make a delay in my vector, matlab [closed] - matlab

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I have vector for example. Ones and zeros are representing traffic and idle states for base station traffic.
u=[1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 ]
I know how to calculate number of ones and zeros in it after each break.
But I need help for delay. Zeros are telling me when I can put my base station in sleep mode.
Is there a way to make a delay for example that, base station does not go to sleep after first zero, instead of that that it goes to sleep after third zero, that means with some kind o delay.

Answer 1

Not a simple solution but yet I think you should find it interesting. You can optimize from here.
First of all I assumed you wanted to go to sleep after the third consecutive 0.
Check the example.
CODE:
u=[1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 ];
bw_u=bwlabel(u==0);
get_breaks=bsxfun(#eq, bw_u, unique(bw_u)');
pos_break=cumsum(get_breaks,2).*get_breaks;
third_0=pos_break(2:end,:)==3;
[~,indx_third_0]=find(third_0)
OUTPUT:
indx_third_0 =
8
20
32
43

Answer 2

Again assuming ASantosRibeiro interpreted your question correctly, you can do it a little more simply as follows. Let n denote the desired number of zeros. In your case, n=3;
>> ind = find(diff(conv(2*u-1, repmat(-1, [1 n])) == n) == 1) + 1
ind =
8 20 32 43

Answer 3

Assuming ASantosRibeiro interpreted your question correctly, you could do this to find when to "go to sleep":
delay = 3;
idx = find(diff(u)==-1) + delay
idx =
8 20 32 43
This is assuming you want a delay of 3.
diff(u) returns the difference between consecutive elements in u. Finding when the difference equals -1 is equivalent to finding the indices of when the light changes from 1 to 0. By adding the delay, you have your desired indices.