Turning doubles of different dimensions in cells

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Sven Krasel 2020-5-5

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评论： Sven Krasel 2020-5-6

temp = [time, Dynamics];
temp = num2cell([temp{:}]);

Hey guys,

I have a 1x6 cell called temp. It contains doubles with different dimensions:

double 222x1 double 222x1 double 222x1 double 222x1 double 222x1 double 151x1

Is there a way to 'trick' the num2cell function to the make the temp cell look like being made of consistent dimensions?

Thanks a lot for your help! :-)

Best,

Sven

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Answer 1

Image Analyst 2020-5-5

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I can think of two ways

Create a matrix with a height of the tallest vector, and then paste the vectors in. Short vectors will have zeros below them in their column.
Interpolate the short vectors to be the same length as the taller ones using interp1() spline() or similar functions.

Would any of those methods suffice?

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Image Analyst 2020-5-5

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Sven, try this and see if either one of these two ways meets your needs. I did it both ways for you, but it really seems that you should know in advance which way meets your needs since they give totally different results.

% Create sample data.
temp = {rand(222,1), rand(222,1), rand(222,1), rand(222,1), rand(222,1), rand(151,1)}
% Get height
maxHeight = -inf;
for k = 1 : length(temp)
	maxHeight = max(maxHeight, size(temp{k}, 1))
end
%------------------------------------------------------------------------------------------
% Method 1 : past columns onto canvass matrix.
% Make matrix to hold values.
m = zeros(maxHeight, size(temp, 2));
% Stuff each cell of temp into a column of m.
for col = 1 : length(temp)
	thisCellContents = temp{col};
	m(1 : size(thisCellContents, 1), col) = thisCellContents;
end
m % Show in command window.
%------------------------------------------------------------------------------------------
% Method 2 : interpolate instead of pasting.
% Make matrix to hold values.
m2 = zeros(maxHeight, size(temp, 2));
% Stuff each cell of temp into a column of m.
for col = 1 : length(temp)
	thisCellContents = temp{col};
	thisMatrixHeight = size(thisCellContents, 1);
	interpolatedColumn = interp1((1:length(thisCellContents))', thisCellContents, linspace(1, thisMatrixHeight, maxHeight)');
	m2(:, col) = interpolatedColumn;
end
m2 % Show in command window.

Image Analyst 2020-5-6

If you're going to get each max, and compare it to the current max, you need to have a current max for the very first time you do that. So for the very first time, before you've even looked at any of your variable values, what would be a good value to compare it against such that your first variable value will be guaranteed to be the max so far? Minus infinity of course. Any value you have is certainly guaranteed to be greater than minus infinity, right?

Sven Krasel 2020-5-6