Concatenate arrays of different length into a matrix
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Stefano Grillini
el 29 de Ag. de 2018
Respondida: Yang Liu
el 26 de En. de 2024
Assume I have two arrays (time-series) of the form:
A = [NaN, 2, 3, 4, 5, 6, 7, NaN]
B = [5, NaN, 6, 7, NaN, 8, 9, 10, 11, 12]
Since two arrays of different length can not be horzcat (obviously), how can I combine them as to obtain a 8x2 matrix where available data match. I have long time-series, so this is just an example, but it points out how crucial it is to have matching observations. Ideally, the output should be:
C = [NaN, 2, 3, 4, 5, 6, 7, NaN; 5, NaN, 6, 7, NaN, 8, 9, 10]
Thanks
Stefano
3 comentarios
Stephen23
el 29 de Ag. de 2018
Editada: Stephen23
el 29 de Ag. de 2018
"Since two arrays of different length can not be horzcat (obviously),"
I didn't have any problems using horzcat:
>> A = [NaN, 2, 3, 4, 5, 6, 7, NaN];
>> B = [5, NaN, 6, 7, NaN, 8, 9, 10, 11, 12];
>> horzcat(A,B)
ans =
NaN 2 3 4 5 6 7 NaN 5 NaN 6 7 NaN 8 9 10 11 12
Respuesta aceptada
Stephen23
el 29 de Ag. de 2018
Editada: Stephen23
el 29 de Ag. de 2018
Truncate to shortest length using indexing:
>> N = min(numel(A),numel(B));
>> [A(1:N);B(1:N)]
ans =
NaN 2 3 4 5 6 7 NaN
5 NaN 6 7 NaN 8 9 10
>> padcat(A,B)
ans =
NaN 2 3 4 5 6 7 NaN NaN NaN
5 NaN 6 7 NaN 8 9 10 11 12
5 comentarios
Stephen23
el 29 de Ag. de 2018
Editada: Stephen23
el 29 de Ag. de 2018
@Stefano Grillini: you really have two choices: either interpolate to fill in the NaN data, or remove the entire row from your data wherever there is a NaN. Judging by your data interpolation does not make much sense, however removing the rows is easy:
idx = any(isnan(c),2);
new = c(~idx,:)
Más respuestas (1)
Yang Liu
el 26 de En. de 2024
I'd say, the most straight forward method would be using cell to combine whatever dimension you have, and use Cell{a,b}(x,y) to access the elements.
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