Find a series of consecutive numbers in a vector

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Julia
Julia 2013-9-5
评论： Nurul Ain Basirah Zakaria 2021-6-12
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!

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采纳的回答

Roger Stafford
Roger Stafford 2013-9-6
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
  2 个评论
Caixia Liu
Caixia Liu 2018-4-10
编辑：Caixia Liu 2018-4-10
Thanks for your codes. But it fails when the first index is actually the result.
Nurul Ain Basirah Zakaria
Nurul Ain Basirah Zakaria 2021-6-12
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?

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更多回答（6 个）

Laurent
Laurent 2013-9-5
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
  3 个评论
显示 1更早的评论
Laurent
Laurent 2013-9-5
编辑：Laurent 2013-9-5
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
Image Analyst 2013-9-5
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.

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Jan
Jan 2013-9-8
编辑：Jan 2013-9-8
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.
David Sanchez
David Sanchez 2013-9-5
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
  1 个评论
David Sanchez
David Sanchez 2013-9-5
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

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JEM
JEM 2017-5-30
编辑：Walter Roberson 2021-6-5
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]));
  2 个评论
dpb
dpb 2018-10-16
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
Sinem Balta Beylergil 2021-6-5
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.

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Andrei Bobrov
Andrei Bobrov 2013-9-5
编辑：Andrei Bobrov 2013-9-5
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))
  2 个评论
Martti K
Martti K 2015-10-27
编辑：Martti K 2015-10-27
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))

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Fahd Elabd
Fahd Elabd 2020-12-30
% 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

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