Storing data from a for loop into a structure
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I am trying to store all data into a single structure. The data is currently generated in a for loop cycling through various data sets (with the name of each set stored in names1 as 19x1 cell of strings). The code then runs a large number of functions to load in data and produce the desired outputs PreDonn, PostDonn etc. which I want to store in this single structure, with the field of the structure defined based on the name of each set from names1. I am certain I have used this approach in different code before and it has worked, but this time when the code runs my Output file only contains the values for the last loop, with the rest of the data missing. I believe it is writing over the Output file during each loop, but I don't know why. I have also attempted preallocating all of the desired fields of the structure but they got written over as well. Any thoughts?
for L=1:size(names1)
...
...
Output.(names1{L}).PreDonn=PreDonn;
Output.(names1{L}).PostDonn=PostDonn;
Output.(names1{L}).PreDoff=PreDoff;
Output.(names1{L}).PostDoff=PostDoff;
end
5 Comments
hello
we don't know if the rest of your code is faulty for that , but the basic operations work as you stated.
Now I would replace size with numel to avoid issues when names1 cell array has dimensions 1 x N instead of N x 1 .
names1 = {'aa','bb','cc'};
for L=1:numel(names1)
Output.(names1{L}).PreDonn=rand(1,1);
Output.(names1{L}).PostDonn=rand(1,1);
Output.(names1{L}).PreDoff=rand(1,1);
Output.(names1{L}).PostDoff=rand(1,1);
end
Output
Nicolaas Pickard
on 12 Dec 2024
Stephen23
on 12 Dec 2024
Note that rather than creating lots of nested structures inside one scalar structure and using dynamic fieldnames, you could easily flatten and simplify your data by using just one non-scalar structure with basic indexing:
for L = 1:numel(names1)
...
Output(L).PreDonn=PreDonn;
Output(L).PostDonn=PostDonn;
Output(L).PreDoff=PreDoff;
Output(L).PostDoff=PostDoff;
Output(L).Name = names1{L}; % much more robust!
end
Simpler, more robust, and generally more efficent this also has benefits when accessing your data, e.g.:
Mathieu NOE
on 12 Dec 2024
excellent suggestion !
Nicolaas Pickard
on 13 Dec 2024
Answers (1)
ScottB
on 12 Dec 2024
0 votes
size(names1) has 2 elements so I think that is the problem:
names = cell(1,3);
names(1,1) = 'Jim'
names(1,2) = 'Bill'
names(1,3) = 'Tyron'
PreDonn = eye(3)
sz_names = size(names)
for L = 1:sz_names(2)
Output.(names{L}).PreDonn=PreDonn;
end
whos
1 Comment
Nicolaas Pickard
on 12 Dec 2024
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