Find a series of consecutive numbers in a vector
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Julia
el 5 de Sept. de 2013
Comentada: Nurul Ain Basirah Zakaria
el 12 de Jun. de 2021
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!
0 comentarios
Respuesta aceptada
Roger Stafford
el 6 de Sept. de 2013
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 comentarios
Caixia Liu
el 10 de Abr. de 2018
Editada: Caixia Liu
el 10 de Abr. de 2018
Thanks for your codes. But it fails when the first index is actually the result.
Nurul Ain Basirah Zakaria
el 12 de Jun. de 2021
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?
Más respuestas (6)
Laurent
el 5 de Sept. de 2013
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 comentarios
Laurent
el 5 de Sept. de 2013
Editada: Laurent
el 5 de Sept. de 2013
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
el 5 de Sept. de 2013
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.
David Sanchez
el 5 de Sept. de 2013
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 comentario
David Sanchez
el 5 de Sept. de 2013
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
JEM
el 30 de Mayo de 2017
Editada: Walter Roberson
el 5 de Jun. de 2021
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 comentarios
dpb
el 16 de Oct. de 2018
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
el 5 de Jun. de 2021
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.
Andrei Bobrov
el 5 de Sept. de 2013
Editada: Andrei Bobrov
el 5 de Sept. de 2013
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 comentarios
Martti K
el 27 de Oct. de 2015
Editada: Martti K
el 27 de Oct. de 2015
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))
Fahd Elabd
el 30 de Dic. de 2020
% 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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