Nesting depth and the error "Expected one output from a curly brace or dot indexing expression, but there were x results."
11 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
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
Réponse acceptée
Bruno Luong
le 19 Août 2022
Modifié(e) : Bruno Luong
le 19 Août 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 commentaires
Plus de réponses (1)
Jan
le 19 Août 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 commentaires
Image Analyst
le 19 Août 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
le 19 Août 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
Voir également
Catégories
En savoir plus sur Matrix Indexing dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!