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Update field value in structures?

Najmeh Eskandari さんによって質問されました 2019 年 2 月 10 日
最新アクティビティ Walter Roberson
さんによって コメントされました 2019 年 2 月 10 日
Hi.
I have structure L with field L0. L0 has two fields:H0 and H1.
H0=struct('h',[0,0],'Q_centroids',[-5,0,5,0],'Q_actions',[15,7,10],'Q_table',[0]);
H1=struct('h',[0,2],'Q_centroids',[1,0,pi/2,0],'Q_actions',[2,7,8],'Q_table',[2]);
L0=struct('H0',H0,'H1',H1);
L=struct('L0',L0);
relevant_L=L.L0;
relevant_L.H0.Q_centroids=[-pi/2,0,pi/2,0;1 5 4 2];
I want to update H0 and then save it in L. so i load L0(relevant_L=L.L0) then update it through the last line but new strcture is created(name:relevant_L).
i know we can do this by:
L.L0.H0.Q_centrids
but i want to change Q_centroids first by load L.L0 in another variable(relevant_L) and then change it.
Is it possible?
Thanks.

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1 件の回答

Walter Roberson
回答者: Walter Roberson
2019 年 2 月 10 日
 採用された回答

If the question is whether it is possible to create an "alias" to something inside a structure, such that assigning to the name will have the effect of changing the part inside the structure -- if that is the question, then the answer is NO, that cannot be done in MATLAB with struct
That said, something that would look like that can be done in MATLAB. It would require that L0 would be an object of a class derived from handle and that the class had a H0 property. In that particular configuration, then updating relevant_L.H0 would have the effect of updating L.L0.H0 .

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Najmeh Eskandari 2019 年 2 月 10 日
Ok.
i use fieldname to find fields name in L:
H0=struct('h',[0,0],'Q_centroids',[-5,0,5,0],'Q_actions',[15,7,10],'Q_table',[0]);
H1=struct('h',[0,2],'Q_centroids',[1,0,pi/2,0],'Q_actions',[2,7,8],'Q_table',[2]);
L0=struct('H0',H0,'H1',H1);
L=struct('L0',L0);
FieldNames=fieldnames(L);
FieldNames={'L0'}.
Now can i for example define
a={'L0'};
then use
L.(a{1});
?
Walter Roberson
2019 年 2 月 10 日
Yes, you can use
STRUCTURENAME.(CHARACTER_VECTOR)
to access the field in STRUCTURENAME that the CHARACTER_VECTOR evaluates to. Your example of A={'L0'} and L.(a{1}) is fine, and is the same as L.('L0') which will evaluate as if you had used L.L0

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