Split array into separate arrays at row with NaNs
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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
采纳的回答
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);
>> 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
>> 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
>> 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 个评论
Douglas Anderson
2017-4-14
Douglas Anderson
2017-4-14
编辑：Douglas Anderson
2017-4-14
Stephen23
2017-4-14
@Douglas Anderson: my pleasure. Remember to accept the answer that best resolves your original question. Accepting is an easy way to show us that you appreciate our (volunteer) effort, and shows other that your question has been resolved.
Peng Li
2020-3-31
@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
2023-1-21
Thank you very much! It's very helpful!
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