Filter löschen
Filter löschen

Calculate values of struct of struct

7 Ansichten (letzte 30 Tage)
Julian
Julian am 4 Sep. 2023
Kommentiert: Julian am 5 Sep. 2023
I have a struct with several fieldnames, each of them is again a struct and has a fieldname "P". This "P" is an array.
I want to calculate the sum of all "P "s like:
s.s1.P = (1:20).'; % For simplicity I have assigned dummy values to the "P "s.
s.s2.P = (1:20).';
s.s3.P = (1:20).';
% ...
result = s.s1.P + s.s2.P + s.s3.P % + ...
I have already a solution with a for loop:
field = fieldnames(s);
result = zeros(size(s.(field{1}).P));
for i = 1:length(field)
result = result +s.(field{i}).P;
end
I also tried to use "structfun" but were not able to get this result.
Is it possible to calculate this problem with "structfun" or another function or is the for loop the final solution for this problem?
  1 Kommentar
Stephen23
Stephen23 am 4 Sep. 2023
Rather than forcing pseudo-indices into fieldnames you should simply use a structure array, as dpb showed.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Bruno Luong
Bruno Luong am 4 Sep. 2023
Bearbeitet: Bruno Luong am 4 Sep. 2023
Single line (work for fields with struct NOT objects of class, struct is NOT some sort of class)
s.s1.P = (1:20).'; % For simplicity I have assigned dummy values to the "P "s.
s.s2.P = (1:20).';
s.s3.P = (1:20).';
sum([cat(1,struct('sx',struct2cell(s)).sx).P],2)
ans = 20×1
3 6 9 12 15 18 21 24 27 30

Weitere Antworten (2)

dpb
dpb am 4 Sep. 2023
Bearbeitet: dpb am 4 Sep. 2023
The problem of using sequentially-named fields in the struct instead of an array of struct; there's no good way to process the resulting variables or struct fields programmatically when do so.
structfun applies the function to each field of a scalar struct so it isn't the tool for the purpose here, anyway, you're trying to process across fields, not within each. And, arrayfun is for an array but you didn't use an array to hold the similar data, you chose to use multiple fields instead. A 2D array would have been an alternate choice that could have handled easily...
s.P = (1:20).';
s.P = [s.P (1:20).'];
s.P = [s.P (1:20).'];
result = sum(s.P,2)
result = 20×1
3 6 9 12 15 18 21 24 27 30
Or, alternatively, as a struct array
s.P = (1:20).';
s(2).P = (1:20).';
s(3).P = (1:20).';
result = sum([s.P],2)
result = 20×1
3 6 9 12 15 18 21 24 27 30
  7 Kommentare
Bruno Luong
Bruno Luong am 4 Sep. 2023
Bearbeitet: Bruno Luong am 4 Sep. 2023
No
D=delaunayTriangulation(rand(10,2))
D =
delaunayTriangulation with properties: Points: [10×2 double] ConnectivityList: [12×3 double] Constraints: []
structfun(@(D)D.Points, D)
Error using structfun
Inputs to STRUCTFUN must be scalar structures.
Sorry my first remark on structfun on class is not applicable int the present context, since s is still a struct with 3 fields. structun can be applied on s.
Julian
Julian am 5 Sep. 2023
Yes structfun was used on s, a struct and it's not posible to use structfun on class objects.

Melden Sie sich an, um zu kommentieren.


Bruno Luong
Bruno Luong am 4 Sep. 2023
Bearbeitet: Bruno Luong am 4 Sep. 2023
Honestly I prefer the for-loop
s.s1.P = (1:20).'; % For simplicity I have assigned dummy values to the "P "s.
s.s2.P = (1:20).';
s.s3.P = (1:20).';
c = struct2cell(structfun(@(sf) sf.P, s, 'UniformOutput', false));
sum(cat(2,c{:}),2)
ans = 20×1
3 6 9 12 15 18 21 24 27 30
  2 Kommentare
dpb
dpb am 4 Sep. 2023
Given the already bad idea of having created the sequentially-named fields, agree, the for loop is the way; as your example shows, to do anything else gets far too convoluted and undoubtedly would be slower besides.
At least with struct fields, one can programmatically access them w/o the eval abomination.
Julian
Julian am 4 Sep. 2023
Thanks for the solution.
Unfortunately, I already came up with this, but I didn't like this huge nesting of functions either.
I understand why you would prefer the for-loop.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Structures finden Sie in Help Center und File Exchange

Produkte


Version

R2023a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by