Create a new array over each iteration

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Riccardo Tronconi
Riccardo Tronconi 2021 年 6 月 14 日
編集済み: Rik 2021 年 6 月 16 日
Hi guys! Starting from one table (A) I would divide it in n-table according to the max value of the A(:,4). At this point every each iteration I must create another table as it follows:
function [T_num2str(i)] = par(A)
n= max(A{:,4});
for i=1:n
search = find(A{:,4}==i);
eval(['T_' num2str(i) ' = table;']);
T_num2str(i) = position(search,:);
end
end
I have two problem:
1- How to define output values if I do not know how many they would be?
2- How to solve this indexing problem ?
T_num2str(i) = position(search,:);
  7 件のコメント
Riccardo Tronconi
Riccardo Tronconi 2021 年 6 月 14 日
I need it parametric so Its functioning is still valid if max(indices) is either lower or greater.
Stephen23
Stephen23 2021 年 6 月 15 日
"I need it parametric so Its functioning is still valid if max(indices) is either lower or greater. "
Sure, but you did not answer Rik's question.
So far there is no obvious reason why you cannot use simpler indexing, rather than your complex and inefficient approach.

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Rik
Rik 2021 年 6 月 14 日
The code below gives you what you want. The only difference is that you need to write my_data{1} instead of my_data1.
[~,~,data]=xlsread('ex[1].xlsx')
data = 12×6 cell array
{'Location'} {[1.6149e+18]} {'lab'} {[1]} {[2.6936]} {[3.5776]} {'Location'} {[1.6149e+18]} {'lab'} {[2]} {[2.5642]} {[3.5736]} {'Location'} {[1.6149e+18]} {'lab'} {[3]} {[2.6610]} {[3.5411]} {'Location'} {[1.6149e+18]} {'lab'} {[2]} {[2.6049]} {[3.5502]} {'Location'} {[1.6149e+18]} {'lab'} {[3]} {[2.6080]} {[3.5852]} {'Location'} {[1.6149e+18]} {'lab'} {[3]} {[2.6523]} {[3.5085]} {'Location'} {[1.6149e+18]} {'lab'} {[3]} {[2.6343]} {[3.5355]} {'Location'} {[1.6149e+18]} {'lab'} {[2]} {[2.6998]} {[3.5359]} {'Location'} {[1.6149e+18]} {'lab'} {[2]} {[2.6149]} {[3.5874]} {'Location'} {[1.6149e+18]} {'lab'} {[1]} {[2.6527]} {[3.4991]} {'Location'} {[1.6149e+18]} {'lab'} {[1]} {[2.6670]} {[3.5399]} {'Location'} {[1.6149e+18]} {'lab'} {[1]} {[2.6206]} {[3.5023]}
indices=cell2mat(data(:,4));
my_data=cell(max(indices),1)
for n=1:max(indices)
my_data{n}=data(indices==n,:);
end
my_data
my_data = 3×1 cell array
{4×6 cell} {4×6 cell} {4×6 cell}
my_data{1}
ans = 4×6 cell array
{'Location'} {[1.6149e+18]} {'lab'} {[1]} {[2.6936]} {[3.5776]} {'Location'} {[1.6149e+18]} {'lab'} {[1]} {[2.6527]} {[3.4991]} {'Location'} {[1.6149e+18]} {'lab'} {[1]} {[2.6670]} {[3.5399]} {'Location'} {[1.6149e+18]} {'lab'} {[1]} {[2.6206]} {[3.5023]}
  4 件のコメント
Riccardo Tronconi
Riccardo Tronconi 2021 年 6 月 16 日
編集済み: Rik 2021 年 6 月 16 日
for i=1:length(my_data{1})
if my_data{1}{i,5} >= 2
% do some calculation
end
end
Doing so It returns me an error
Rik
Rik 2021 年 6 月 16 日
You made the mistake of using length and assuming it would return the number of rows. The length function does not guarantee that. Use size(my_data{1},1) instead if you want the number of rows. You might also want to learn about the numel function if you want to loop over all elements of an array.

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