Grouping sequence within interval

2 visualizaciones (últimos 30 días)
Benu
Benu el 17 de Mayo de 2020
Comentada: Benu el 5 de Jun. de 2020
I got stuck in grouping my sequence.
How can I automatically grouping element of sequence? within interval for this case maybe 0.5
suppose I have 3 sequence
V1 = [0.73 0.74 0.77 0.78 0.79 11.17 11.18 11.19 11.20 11.22 11.23];
V2 = [0.71 0.78 0.76 6.04 6.08 6.01 11.38 11.4 11.1] ;
V3 = [0.73 0.74 0.75 4.29 4.33 7.83 7.85 11.38 11.34]
My expectation is to have a group as follows
G1 = {[0.73 0.74 0.77 0.78 0.79] ; [11.17 11.18 11.19 11.20 11.22 11.23]}
G2 = {[0.71 0.78 0.76] ; [6.04 6.08 6.01] ; [11.38 11.4 11.1] }
G3 = {[0.73 0.74 0.75] ; [4.29 4.33] ; [7.83 7.85] ; [11.38 11.34]}

Respuesta aceptada

Thiago Henrique Gomes Lobato
Thiago Henrique Gomes Lobato el 17 de Mayo de 2020
Something like this should do the trick for you
V1 = [0.73 0.74 0.77 0.78 0.79 11.17 11.18 11.19 11.20 11.22 11.23];
V2 = [0.71 0.78 0.76 6.04 6.08 6.01 11.38 11.4 11.1] ;
V3 = [0.73 0.74 0.75 4.29 4.33 7.83 7.85 11.38 11.34]
Vsorted = sort(V3); % Works also with V1, V2
NewSeqLocations = find(diff(Vsorted)>0.5); % 0.5 is your interval, i.e, maximum difference between adjacent numbers
NewSeqLocations = [0,NewSeqLocations,length(Vsorted)]; % Last sequence ends at last array element and starts at first element
Seq = cell(length(NewSeqLocations)-1,1); % Save sequences in cell array
for idx=1:length(Seq)
Seq{idx} = Vsorted(NewSeqLocations(idx)+1:NewSeqLocations(idx+1));
end
Seq{:}
ans =
0.730000000000000 0.740000000000000 0.750000000000000
ans =
4.290000000000000 4.330000000000000
ans =
7.830000000000000 7.850000000000000
ans =
11.340000000000000 11.380000000000001
  1 comentario
Benu
Benu el 5 de Jun. de 2020
I never think this kind of a way. this is really help me.
wonderful ! Thank you so much !

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Matrix Indexing en Help Center y File Exchange.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by