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bbb_bbb
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Writing data to a structure array without loop

Asked by bbb_bbb
on 5 Jul 2019
Latest activity Answered by Jan
on 5 Jul 2019
Accepted Answer by Jan
% Let an array of axes be given
ax(1)=axes;
ax(2)=axes;
ax(3)=axes;
% Let's write something to axes user data in the "data" field
ax(1).UserData.data=1;
ax(2).UserData.data=1;
ax(3).UserData.data=1;
% Now change "data" in all axes to "2"
for i=1:numel(ax)
ax(i).UserData.data=2;
end
How can this be done without a loop?
I could write somethng like this:
s.data=2;
[ax.UserData] = deal(s);
But if there were another user data in these axes (e.g. ax(1).UserData.data1='something';), they will be gone. I need to change only one field, without affecting the other fields.

  3 Comments

"How can this be done without a loop?"
There is no easy way to do this without a loop. A loop is the easiest way to access the data of nested non-scalar structures. Why do you need to avoid using a loop?
Stephen +1
Thank you for the answer.
I always wonder how some operations can be done without a loop. This time I killed half a day for this, and decided to ask the community.

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2 Answers

Answer by Jan
on 5 Jul 2019
 Accepted Answer

As written by Stephen before:
There is no vectorized way to set the value of sub-fields of struct arrays.
The expression
[ax(1:3).UserData.data] = ...
cannot work, because [ax(1:3).UserData] creates a comma separated list, but you cannot obtain a subfield of such a list by the dot operator.
A loop is a simple and efficient solution. If you have spent a half day for searching a vectorized version, you are a victim of the "premature optimization" programming pattern. In many cases the programming and debug time exceeds the run time massively, such that the total time to solve a problem does not profit from too tricky improvements in lines of code, which do not consume a remarkably part of the total processing time. So start with writing dull loops and if the program is running and tested successfully, use the profiler to find the most time consuming parts. Optimize these bottlenecks only.

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Answer by KSSV
on 5 Jul 2019

  6 Comments

KSSV
on 5 Jul 2019
T = num2cell(1:2);
ax = struct('UserData',T)
This gives
ax(1).UserData
ans =
1
But it should be:
ax(1).UserData
ans =
data: 2
KSSV
on 5 Jul 2019
Yes......UserData.Data is must?

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