converting from cell 2 matrix
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i have a cell like
a{1,1}={160;45;58;89}% in the form of 1*50char a{2,1}={160;45;58;89}% in the form of 1*50char a{3,1}={160;45;58;89}% in the form of 1*50char a{4,1}={160;45;58;89}% in the form of 1*50char
like this i have a cellsize of 50000 rows(a{50000,1})
what is the best way to convert elements inside cell to matrix form
eg o/p : desired matrix to be
b(1,:)= 160 45 58 89 b(2,:)= 160 45 58 89 b(3,:)= 160 45 58 89 b(4,:)= 160 45 58 89
problem here is the forloop is taking lot of time to convert to serired matrix way
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Jan
el 15 de Nov. de 2012
Editada: Jan
el 15 de Nov. de 2012
I do not believe that the FOR loop is slow. If you post the code you are using, we will most likely find another problem. Without seeing the code, guessing the reasons is impossible.
testData = cell(1, 1e6);
testData(:) = {'12; 123; 45; 45; 0.1'};
dummy = sscanf(testData{1}, '%g;', [1, Inf]); % To get the width
width = numel(dummy);
Result = zeros(width, numel(testData)); % Pre-allocate!!!
for ii = 1:numel(testData)
Result(:, ii) = sscanf(testData{ii}, '%g;', width);
end
Another approach, which consumes more memory - as long as it fits in the processor cache, this could be faster than the loop:
dummy = sscanf(testData{1}, '%g;', [1, Inf]); % To get the width
width = numel(dummy);
dataStr = sprintf('%s;', testData{:});
Result = sscanf(dataStr, '%g;', [width, Inf]);
It looks like you need to transpose(Result) at the end.
The simple FOR-loop follows the important KISS programming paradigm:
Keep it simple stupid!
When the time required for programming and debugging exceeds the runtime, there is no better concept.
3 comentarios
Jan
el 15 de Nov. de 2012
This matrix requires 1e6*25*8 bytes of free contiguos RAM, otherwise swapping to disk will destroy any performance. This is only 200 MB and should be available on modern computers. Please explain exactly, which line of code creates which kind of "some memory issues".
Más respuestas (1)
Azzi Abdelmalek
el 15 de Nov. de 2012
b=cell2mat(a)
7 comentarios
Azzi Abdelmalek
el 15 de Nov. de 2012
Editada: Azzi Abdelmalek
el 15 de Nov. de 2012
b=cell2mat(cellfun(@(x) cellfun(@str2double, regexp(x,';','split')),a,'un',0))
Jan
el 15 de Nov. de 2012
@Azzi: str2double accepts cell strings as input. Therefore I'd omit at least the innermost cellfun. cell2mat contains an expensive loop. Therefore it is most likely naticably faster to create this loop manually, because this will reduce the amount of temporarily used memory.
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