Grouping elements by conditions

1 visualización (últimos 30 días)
Amro Lulu
Amro Lulu el 21 de Ag. de 2018
Comentada: dani elias el 19 de Nov. de 2020
I have an array [ 1 1.2 1.5 1.9 5 8 8.1 10 12 12.3 12.5]. I'm trying to group the elements that are within a window of length 1, so I get [ 1 1.2 1.5 1.9] [8 8.1] [12.3 12.5] How can this be done efficiently
  2 comentarios
Matt J
Matt J el 21 de Ag. de 2018
Will the grouped elements always be consecutive?
dani elias
dani elias el 19 de Nov. de 2020
how can I use the same concept to group every 4 binary digit, for the case I have an array of 256 binary number,
example
a=[1 1 1 1 1 1 1 1 1 0 1 1 1 0 0 0];
to be in a form of
b=[1111;1111;1011;1000]

Iniciar sesión para comentar.

Respuesta aceptada

Matt J
Matt J el 21 de Ag. de 2018
Editada: Matt J el 21 de Ag. de 2018
You could try this file
x=[ 1 1.2 1.5 1.9 5 8 8.1 10 12 12.3 12.5];
label = graph_connected_components(abs(x.'-x)<=1)
groupcell = splitapply(@(g){g},x,label)
  5 comentarios
Amro Lulu
Amro Lulu el 29 de Ag. de 2018
Hey Matt, I have a question related to this, what if I have a matrix x=[ 1 1.2 1.5 1.9 5 8 8.1 10 12 12.3 12.5; 1 5.3 2.1 5.4 1 6 2.1 45 23 14.3 13.4] And I need to group them the same way, however, only based on the first row, and the resulting group will include the corresponding elements from the second row, as
[ 1 1.2 1.5 1.9; 1 5.3 2.1 5.4] ..etc..
Thank you and I really appreciate it
Matt J
Matt J el 29 de Ag. de 2018
Adapting Yuvaraj's answer,
x=[ 1 1.2 1.5 1.9 5 8 8.1 10 12 12.3 12.5; 1 5.3 2.1 5.4 1 6 2.1 45 23 14.3 13.4];
a=x(1,:);
Len=diff([0,find(diff(a)>1),numel(a)]);
S=mat2cell(x,2,Len)

Iniciar sesión para comentar.

Más respuestas (1)

Yuvaraj Venkataswamy
Yuvaraj Venkataswamy el 21 de Ag. de 2018
Editada: Matt J el 21 de Ag. de 2018
This is your answer.
X=[ 1 1.2 1.5 1.9 5 8 8.1 10 12 12.3 12.5];
Len=diff([0,find(diff(a)>1),numel(a)]);
S=mat2cell(X,1,Len);

Categorías

Más información sobre Data Type Conversion en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by