Modify array zeros detect

Hello Community,

there is a vector lets say:

A = [1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0];

What I want to know are the last indices when it is going to switch to 1. Only this, is not a problem. Additionally the indices should be read out only if the length of the zeros are longer than lets say 12 positions.

In this array for example: 19, 58 etc.

Tried it with combination of for, if loop and find (A), but it was not 100% the expected value.

Thanks a lot!

[Image Analyst]

Can you fill out the "etc." in "19, 58 etc." - what are all the indices in your example. And clarify "last indices when it is going to switch to 1." So does that mean the last index that is a zero right before the array starts being 1's as long as that prior stretch of zeros was 13 or more elements long?

In your example, element 19 is a zero right in the middle of a long stretch of 0's. It is not the last zero before it switches to 1, so can you explain why 19 is part of your desired answer?

Answer 1

[Walter Roberson]

A = [1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0];
idx = strfind(A, [zeros(1, 12), 1]) + 11;

[N/A]

Thanks Walter. To learn from this, could you pls explain this strfind..? An open point, the last index (92) is not mentioned in the solution, although it has 12 zeros.

[Walter Roberson]

"the last index (92) is not mentioned in the solution, although it has 12 zeros"

It does not switch to 1 there, which was a requirement of your question.

The strfind() used here is the same as for strings https://www.mathworks.com/help/matlab/ref/strfind.html . It is not documented there, but it happens that strfind works with numeric values as well as characters, For example if you had done

S = sprintf('%d',A);

then you could clearly use

strfind(S, '0000000000001')

to find the position of a 1 that follows 12 zeros.

[N/A]

Is it possible to modify this

    idx = strfind(A, [zeros(1, 12), 1]) + 11;

Instead of '1', there could be every other decimal variable.

"It does not switch to 1 there, which was a requirement of your question." That is right, but maybe there are some arrays, which end with zero. So if there is a possibility to cover also this case, it would be great.

[Walter Roberson]

If you want to match on any non-zero value, then

A = [1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 7 1 1 1 1 1 1 1 1 1 1 1 5 1 1 19 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0];

idx = strfind(A ~= 0, [zeros(1, 12), 1]) + 11;

If you want to match on a specific non-zero value then

A = [1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 7 1 1 1 1 1 1 1 1 1 1 1 5 1 1 19 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0];

target = 8;
idx = strfind(A, [zeros(1, 12), target]) + 11;

If you want to find a match at the end as well,

A = [1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 7 1 1 1 1 1 1 1 1 1 1 1 5 1 1 19 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -4 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0];

target = 8;
idx = strfind([A, target], [zeros(1, 12), target]) + 11;

Answer 2

[Roger Stafford]

You could also use the following method. A is a row vector with 1’s and 0’s, and m is the least number of successive 0’s before switching to a 1 that have that last 0 position recorded in idx.

f = find(diff([1,A,0]~=0));
f1 = f(1:2:end);
f2 = f(2:2:end);
idx = f2(find(f2-f1(1:end-1)>=m))-1; % <-- The desired indices