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How to obtain variables from a structure inside a for loop?

Latest activity Commented on by Rik
on 26 Aug 2019
Hello everyone, I have a code to call intensity values which are stored in a structure 's'. The structure 's' has various fields and the field(or variable, I'm not sure what it's called) signal2 has the values I need to access. When I use the following code for just one of the signal2 data's I get the desired output :
int_val= s.cellList.meshData{1,19}{1,1}.signal2;
list of numbers in an excel sheet
({1,19} is a cell inside the field meshData containing the desired values).
However, when I try to create a loop to access each of the cells(1:25),
for j=1:25
int_val= 's_',num2str(i),'.cellList.meshData{1,',num2str(j),'}{1,1}.signal2';
%this ends up creating a character array.
How do I get the values from the structure?


Your examples are inconsistent: your first example shows a variable named s, then your example
shows a variable named s_X for some integer X.: which of these is correct?
Also, why are you converting perfectly good indices into strings?
's' is a structure that I've created. I'm accessing variables in the structure.
I realized I could directly write the command as:
int_val= s.cellList.meshData{1, j}{1,1}.signal2;
instead of this:
int_val= 's_',num2str(i),'.cellList.meshData{1,',num2str(j),'}{1,1}.signal2';

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1 Answer

Answer by Bjorn Gustavsson on 26 Aug 2019
 Accepted Answer

It seems that you read in entire data-sets into a number of numbered struct-variables (presumably s_1, s_2,..., s_25). Dont do that (search for "dynamically generating variable names" or something like that and read the reasons). Instead
read everything into an array of structs. Then you'll have something like this:
s = % Your data-reading-function/script, now returning a struct array
for i1 = numel(s):-1:1
for i2 = size(s(i1).cellList.meshData,1):-1:1
int_val(i1,i2) = s(i1).cellList.meshData{1,i2}{1,1}.signal2;
int_val(int_val(:)<=threshold) = 0;
int_val(int_val(:)>threshold) = 1;

  1 Comment

Here is the thread you mentioned.

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