Split array into separate arrays at row with NaNs

21 vues (au cours des 30 derniers jours)
Douglas Anderson
Douglas Anderson le 14 Avr 2017
Commenté : Andrew Sol le 21 Jan 2023
Hello,
Probably a simple problem? I have been sent some data in an Excel file that has several sets of data on one sheet. The data are separated into the individual sets with a blank line, that becomes a row of "NaN"s. How can I make separate arrays from this big one? I didn't see the answer to this in documentation.
Thanks!
Doug Anderson

Réponse acceptée

Stephen23
Stephen23 le 14 Avr 2017
Modifié(e) : Stephen23 le 14 Avr 2017
First lets define some trial data:
>> M = [0,1;2,3;NaN,NaN;4,5;6,7;NaN,NaN;8,9]
M =
0 1
2 3
NaN NaN
4 5
6 7
NaN NaN
8 9
>> idx = all(isnan(M),2);
Method One: accumarray:
>> idy = 1+cumsum(idx);
>> idz = 1:size(M,1);
>> C = accumarray(idy(~idx),idz(~idx),[],@(r){M(r,:)});
>> C{:}
ans =
0 1
2 3
ans =
4 5
6 7
ans =
8 9
Method Two: mat2cell:
>> idr = diff(find([1;diff(idx);1]));
>> D = mat2cell(M,idr(:),size(M,2));
>> D{1:2:end}
ans =
0 1
2 3
ans =
4 5
6 7
ans =
8 9
You can access these data easily and efficiently using cell indexing, e.g.:
>> C{2}
ans =
4 5
6 7
Warning: Whatever you do, do NOT try to create lots of arrays in the workspace. This will be slow, buggy, obfuscated, and makes working with the data difficult:
  6 commentaires
Peng Li
Peng Li le 31 Mar 2020
@Stephen Cobeldick
Just a comment that the accumarray solution doesn't correctly deal with contiguous rows of NaNs, as as long as there are more than 1 contiguous rows of NaNs, cumsum will pick additional NaN rows, which results in a jump in idy(~idx). Therefore, the final C will contains cell elements that are empty. So additional steps are required to filter them. See an example modified based on yours:
M = [0,1;2,3;NaN,NaN;nan nan; nan nan; 4,5;6,7;NaN,NaN;8,9];
idx = all(isnan(M),2);
idy = 1+cumsum(idx);
idz = 1:size(M,1);
C = accumarray(idy(~idx),idz(~idx),[],@(r){M(r,:)})
C =
5×1 cell array
{2×2 double}
{0×0 double}
{0×0 double}
{2×2 double}
{1×2 double}
Andrew Sol
Andrew Sol le 21 Jan 2023
Thank you very much! It's very helpful!

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