How can I split a table at given rows?

I have an n by m table, A, and a vector of indices, idx.
Is there a way to split my table into smaller tables, each beginning with the row indicated in idx?
For example:
A = ['str1', 'str2';
1, 2;
3, 4;
'str3', 'str4';
5, 6]
idx = [1; 4]
I would like:
A1 = ['str1', 'str2';
1, 2;
3, 4]
A2 = 'str3', 'str4';
5, 6]
Many thanks.

Respuestas (3)

Sad Grad Student
Sad Grad Student el 23 de Feb. de 2015
Try:
A1 = A(idx(1):idx(2)-1,:);
A2 = A(idx(2):end,:);
you can make that into a loop too if you have more values.
funkadelala
funkadelala el 23 de Feb. de 2015
Thank you. For looping, I have:
B = zeros(1,length(idx));
for i = 1:1:length(idx)
B(i) = A(idx(i):idx(i+1)-1,:);
end
But, this returns the error:
"The following error occurred converting from table to double: Error using double. Conversion to double from table is not possible."
I assume that this is because of mixed data types. How might I correct this?

3 comentarios

Sad Grad Student
Sad Grad Student el 23 de Feb. de 2015
Editada: Sad Grad Student el 23 de Feb. de 2015
I'm not too familiar with table in Matlab. But I believe the error is because you're declaring variable B as a vector (datatype: double) and trying to pass the table in it. Try using: table2array for conversion but since you have strings as your first row, it might give you issues. So along with table2array , look up: Access Data In Table
Some fidgeting around those commands might help you. Let me know what works! :)
funkadelala
funkadelala el 23 de Feb. de 2015
Editada: funkadelala el 23 de Feb. de 2015
Curly indexing did the trick. The working code is:
B = cell(length(idx),1); %preallocates dimensions
for i = 1:1:length(idx)
%creates cell array that contains tables; each cell beginning with
%index idx(i) of table A, ending on index idx(i+1)-1
if i ~= length(idx) %if not last pass
B{i,:} = A(idx(i):idx(i+1)-1,:);
else
B{i,:} = A(idx(i):end,:);
end
end
Thanks!
Sad Grad Student
Sad Grad Student el 23 de Feb. de 2015
Cool! Congrats :)

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Sad Grad Student
Sad Grad Student el 23 de Feb. de 2015
Solution to this question:
B = cell(length(idx),1); %preallocates dimensions
for i = 1:1:length(idx)
%creates cell array that contains tables; each cell beginning with
%index idx(i) of table A, ending on index idx(i+1)-1
if i ~= length(idx) %if not last pass
B{i,:} = A(idx(i):idx(i+1)-1,:);
else
B{i,:} = A(idx(i):end,:);
end
end

1 comentario

Sad Grad Student
Sad Grad Student el 23 de Feb. de 2015
Had to write it that way so someone having the same question can access the actual working solution! :)

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