Find a series of consecutive numbers in a vector

Hello, I have a small problem I am trying to solve on Matlab, but I am a stuck.

I have a vector containing timestamps: [34 35 36 78 79 80 81 82 84 85 86 102 103 104 105 106 107 201 202 203 204 ...]

And I would like to find the timestamp which is followed by at least 5 consecutive numbers after it. So in my example, the answer would be 102, because it is the first number which is followed by 5 consecutive numbers.

I tried many things using diff(), but I cannot find a simple way to get that result.

If anyone can help, it would be greatly appreciated.

Thank you!

Answer 1

[Roger Stafford]

Yet another method. Let t be your timestamp row vector.

 N = 5; % Required number of consecutive numbers following a first one
 x = diff(t)==1;
 f = find([false,x]~=[x,false]);
 g = find(f(2:2:end)-f(1:2:end-1)>=N,1,'first');
 first_t = t(f(2*g-1)); % First t followed by >=N consecutive numbers

[Caixia Liu]

Thanks for your codes. But it fails when the first index is actually the result.

[Nurul Ain Basirah Zakaria]

hi, what if, i want something like this:

to find the consecutive value below -1

2.0 2.5 2.1 1.3 1.4 -1.0 -2.1 -1.2 -1.5 -2.1 2.0 3.2 3.0 -1.0 -4.0 -2.1 -1.45 -1.20 -2.0 3.0 2.5 1.2

the first consecutive negative value is categorize as first event and second consecutive negative value is the second event,

then i want to calculate the mean value for each event;

sum of -1.0 -2.1 -1.2 -1.5 -2.1, divide by 5, as there are 5 num of -ve value for first event,

and the second event,

sum of -1.0 -4.0 -2.1 -1.45 -1.20 -2.0, divide by 6, as the are 6 num of -ve value for second event

can i do this in matlab?

Answer 2

[Laurent]

The following will give the lengths of the consecutive sequences of your vector:

q=yourvector;

a=diff(q);

b=find([a inf]>1);

c=diff([0 b]); length of the sequences

d=cumsum(c); endpoints of the sequences

[Image Analyst]

Not quite - try it and see.

[Laurent]

This is what I get:

>> c

c =

       3     5     3     6     4

which are the lengths of the sequences.

>> d

d =

       3     8    11    17    21

which is where the sequences end.

Then with a 'find(c>5)' you will know the location of the sequences larger than 5. Then from d you can deduce where the start of this sequence is.

Or did I misunderstand the question?

[Image Analyst]

No, sorry, I misunderstood your comment. I thought your endpoints was both endpoints - the starting and stopping indexes, but it's only where they stop.

Answer 3

[Jan]

This is almost a run-length problem:

x = [34 35 36 78 79 80 81 82 84 85 86 102 103 104 105 106 107 201 202 203 204];
[b, n, idx] = RunLength(x - (1:length(x)));
match  = (n > 5);
result = x(idx(match));

With http://www.mathworks.com/matlabcentral/fileexchange/41813-runlength.

Answer 4

[David Sanchez]

my_array =  [34 35 36 78 79 80 81 82 84 85 86 102 103 104 105 106 107 201 202 203 204];
my_num = 0;
consec = 1;
for k= 1:(numel(my_array)-1)
    if ( my_array(k+1) == (my_array(k) + 1) )
        consec = consec +1;
               if consec > 5
            my_num = my_array(k-4);
                     break
               end
    else 
                consec = 1;
    end
end

my_num =

   201

[David Sanchez]

I used this array instead:

my_array =  [34 35 36 78 79 80 81 82 84 85 86 102 103 104 105 108 109 201 202 203 204 205 206 207 210];
my_num =

     201

With the code above, the answer will be:

my_num =

     102

Answer 5

[JEM]

Easier like this

t=[34 35 36 78 79 80 81 82 84 85 86 102 103 104 105 106 107 201 202 203 204];
% search for the derivative 1 1 1 1 1 corresponding to 5 consecutive values
result = t(findstr(diff(t),[1 1 1 1 1]));

[dpb]

The difference vector of ones will be N-1 length to be found, not N, though. Five differences==1 will correspond to six consecutive values incremented by one.

[Sinem Balta Beylergil]

One correction and it works perfectly:

t = find(overlapR==1);
result = t(intersect(diff(t),[1 1 1 1 1]))

PS [1 1 1 1 1] can be written as ones(1,5) and 5 can be any number of repetitions you want.

Answer 6

[Andrei Bobrov]

a = [34 35 36 78 79 80 81 82 84 85 86 102 103 104 105 106 107 201 202 203 204]
n = 6 %  number consecutive numbers

k = [true;diff(a(:))~=1 ];
s = cumsum(k);
x =  histc(s,1:s(end));
idx = find(k);
out = a(idx(x==n))

[Andrew Newell]

Doesn't work if a contains negative numbers.

[Martti K]

To me it seems to work well with negative numbers also, but not with decreasing sequencies. In order to take the decreasing sequencies, use

    k = [true;diff(a(:))~=-1];

In order to take at_least n consecutive numbers, use

    out = a(idx(x>=n))

Answer 7

[Fahd Elabd]

% Here I have get the first and last numbers of consecutive group in iOnes array,

iOnes =  [34 35 36 78 79 80 81 82 84 85 86 102 103 104 105 106 107 201 202 203 204];
k=1;
j=1;
for i=1:length(iOnes)-1
    if  iOnes(i+1)-iOnes(i)==1  % means the next point is follwoing the current point
        firstOnes(k) = iOnes(j);
        lastOnes(k) = iOnes(i+1);
    else
        j=i+1;
        k=k+1;
    end
end