Nesting depth and the error "Expected one output from a curly brace or dot indexing expression, but there were x results."
5 views (last 30 days)
Show older comments
Consider this toy example:
a(1).x.y=1
a(2).x.y=2
[a.x.y]
Why does this produce the infamous error
Expected one output from a curly brace or dot indexing expression, but there were 2 results.
..instead of just listing all values from across all indexed structure elements, as in this other example where the nesting is at level 2 instead of level 3:
a(1).x=1
a(2).x=2
[a.x]
>> [a.x]
ans =
1 2
Accepted Answer
Bruno Luong
on 19 Aug 2022
Edited: Bruno Luong
on 19 Aug 2022
a work around if you insist on oneline
a(1).x.y=1
a(2).x.y=2
axy = [struct([a.x]).y]
0 Comments
More Answers (1)
Jan
on 19 Aug 2022
R2022a creates a different error:
a(1).x.y=1;
a(2).x.y=2;
[a.x.y]
"instead of just listing all values from across all indexed structure elements"
Think twice. [a.x] is an array already with 2 elements. The dot operator cannot handle an array as input, but a scalar struct only. This is plausible. Consider, that there is no logical decision for the dimensions of the output. It is also unclear, what you call "just listing all values".
3 Comments
Image Analyst
on 19 Aug 2022
Is there a bracket/brace/parentheses solution to this, or is the only way a simple but intuitive for loop
a(1).x.y=1;
a(2).x.y=2;
all_y = zeros(numel(a), 1);
for k = 1 : numel(a)
all_y(k) = a(k).x.y;
end
or possibly a cryptic call to structfun or some other weird function
Stephen23
on 19 Aug 2022
"The dot operator cannot handle an array as input, but a scalar struct only."
???
a(1).x.y=1;
a(2).x.y=2;
tmp = [a.x] % array struct, not scalar struct
[tmp.y] % dot indexing accepts an array without any problem
See Also
Categories
Find more on Matrix Indexing in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!