Concatenate arrays of different length into a matrix
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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 件のコメント
Stephen23
2018 年 8 月 29 日
編集済み: Stephen23
2018 年 8 月 29 日
"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
採用された回答
Stephen23
2018 年 8 月 29 日
編集済み: Stephen23
2018 年 8 月 29 日
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 件のコメント
Stephen23
2018 年 8 月 29 日
編集済み: Stephen23
2018 年 8 月 29 日
@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,:)
その他の回答 (1 件)
Yang Liu
2024 年 1 月 26 日
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.
0 件のコメント
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