Create a new array over each iteration

Question

Riccardo Tronconi 2021-6-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/855985-create-a-new-array-over-each-iteration

编辑： Rik 2021-6-16

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,:);

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Riccardo Tronconi 2021-6-14

I need it parametric so Its functioning is still valid if max(indices) is either lower or greater.

Stephen23 2021-6-15

"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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Answer 1

Rik 2021-6-14

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/855985-create-a-new-array-over-each-iteration#answer_724640

在 MATLAB Online 中打开

ex[1].xlsx

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]}

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Riccardo Tronconi 2021-6-16

编辑：Rik 2021-6-16

在 MATLAB Online 中打开

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 2021-6-16

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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