How to copy every 500 rows to a new variable?
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akshay raj
le 16 Nov 2015
Réponse apportée : Walter Roberson
le 19 Juin 2019
Hi,
I have a [30000 x 57] matrix, I want to copy every new 500 rows to a new variable?
Thanks.
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Stephen23
le 16 Nov 2015
Modifié(e) : Stephen23
le 19 Juin 2019
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Ilham Hardy
le 16 Nov 2015
Wow, thanks for the concise explanation.
Now I am really convinced not to dynamically named variables.
Thanks Stephen :)
Plus de réponses (4)
Star Strider
le 16 Nov 2015
Modifié(e) : Star Strider
le 16 Nov 2015
Another approach:
M = randi(9, 30000, 57); % Create Data
Mr = reshape(M, 50, [], 57); % Reshape
Mr = permute(Mr, [1 3 2]); % Shuffle Dimensions
Mc = mat2cell(Mr, ones(size(Mr,1),1), size(Mr,2), size(Mr,3)); % Create Cell Array
Every element of ‘Mc’, from ‘Mc{1}’ to ‘Mc{50}’ is a different matrix variable. To convert them back into double arrays for calculations, use the cell2mat function.
2 commentaires
Star Strider
le 16 Nov 2015
Thank you!
My code does the same essential task yours does, just differently.
That brings to focus the observation that neither you nor I created 50 ‘new’ variables with our code, but instead created a single cell array with segments of the original matrix that are easier to access and work with, as ‘akshay raj’ essentially requested.
It would seem that reading and understanding the existing Answers would avoid much of the consternation that is too frequently apparent otherwise.
Thorsten
le 16 Nov 2015
Modifié(e) : Thorsten
le 16 Nov 2015
Why do you want to create 60 new variables and double the data? That's not a good idea. You can keep your original data and just address the i'th set of your 500 rows using
ind = [1:500] + (i-1)*500;
your_function(X(ind,:));
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Bish Erbas
le 16 Nov 2015
Do you really have to create individual variables for each 500 rows? This approach is not recommended because it is not using the array processing power of MATLAB. The preferred method is to store related data in a single array or a cell array. For example:
A = rand(30000,57);
N = (size(A,1)/500);
cellVar = cell(1,N);
for k = 1:N
cellVar{k} = A(500*(k-1)+1:500*(k-1)+500,:);
end
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Walter Roberson
le 19 Juin 2019
result = squeeze(cellfun(@transpose, num2cell(reshape(YourMatrix.', size(YourMatrix,2), 500, []), [1 2]), 'uniform', 0));
This will give you a 60 x 1 cell array, each entry of which is a 500 x 57 array.
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