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Concatenate arrays of different length into a matrix

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Stefano Grillini
Stefano Grillini 2018-8-29
回答： Yang Liu 2024-1-26
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
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Stephen23
Stephen23 2018-8-29
编辑：Stephen23 2018-8-29
"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
Stefano Grillini
Stefano Grillini 2018-8-29
Apologise for the misunderstanding Stephen. The arrays are column vectors of the form
A = [NaN; 2; 3; 4; 5; 6; 7; NaN];
B = [5; NaN; 6; 7; NaN; 8; 9; 10; 11; 12];
Therefore dimensions are inconsistent for horzcat.
Thanks jonas. I'll have a look at synchronize(). By the way I need to run the MS_Regress_Fit function where the dependent variable is a matrix of two columns. My imported data are all vectors of size 187x1 with NaN. The problem is that MS_Regress_Fit does not accept NaN in the time-series. Therefore, before concatenating, I need to
milliq1(isnan(milliq1))=[];
lnavilliq(isnan(lnavilliq))=[];
But this command reduces the dimensions according to the number of NaNs so I'm unable to concatenate the two arrays.
Thanks
S

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Stephen23
Stephen23 2018-8-29
编辑：Stephen23 2018-8-29
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
Pad to longest length using padcat:
>> padcat(A,B)
ans =
NaN 2 3 4 5 6 7 NaN NaN NaN
5 NaN 6 7 NaN 8 9 10 11 12
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Stephen23
Stephen23 2018-8-29
编辑：Stephen23 2018-8-29
@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,:)
Stefano Grillini
Stefano Grillini 2018-8-30
Thank you very much @Stephen! It's actually the only choice

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Yang Liu
Yang Liu 2024-1-26
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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