How to convert a layered structure into arrays.

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John Petersen
John Petersen 2021-3-18
编辑： Matt J 2021-3-18
I have a structure like this
A.a.b.first, A.a.b.second, A.a.b.third
A.c.d.first, A.c.d.second, A.c.d.third
A.e.f.first, A.e.f.second, A.e.f.third
I want to extract the fields "first", "second" and "third" into arrays so that one array has all the firsts, another has all the seconds, and a thrid and all the thirds. Is there a simulink method to use or a matlab function? I am currently brute forcing it in matlab code. That is,
B(1) = A.a.b.first
B(2) = A.c.d.first
B(3) = A.e.f.first
and so forth. So it would be much better to have a loop or something even more streamlined.
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Matt J
Matt J 2021-3-18
编辑：Matt J 2021-3-18
Are the desired fields always at a fixed depth (4 fields deep in your example)? Also, if the field names are not a simple alphabetical progression, what decides the order of the elements in B(i)? We need to know the general pattern before a solution can be recommended.
John Petersen
John Petersen 2021-3-18
编辑：John Petersen 2021-3-18
No. Most are four, but some are only three deep. There is a bus in simulink that defines the order. Is it possible to use that? There is also an enum that is used in the code that defines the same order.

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采纳的回答

Matt J
Matt J 2021-3-18
编辑：Matt J 2021-3-18
Ran in:
Here's a way you can recursively search down through a tree of nested structs. Since I don't know how you want to handle multiple occurrences of the target field name (e.g. "first") on the same branch of the treee, I propose one possibility:
A.b.c.d.e.first=3;
A.g.h.first=4;
out = findField(A,'first')
out = 1×2 cell array
{[3]} {[4]}
function out=findField(S,tfield)
if ~isstruct(S)
out=[];
elseif isfield(S, tfield)
out=S.(tfield); %found a match
else%recurse
f=fieldnames(S);
if numel(f)==1
out=findField( S.(f{1}) , tfield);
else
for i=1:numel(f)
out{i}=findField( S.(f{i}) , tfield);
end
end
end
end

更多回答（1 个）

Matt J
Matt J 2021-3-18
编辑：Matt J 2021-3-18
A more automated option,
f=string(num2cell('a':'z'));
for i=1:13
B(i)=A.( f(2*i-1) ).( f(2*i) ).first;
C(i)=A.( f(2*i-1) ).( f(2*i) ).second;
D(i)=A.( f(2*i-1) ).( f(2*i) ).third;
end
  1 个评论
John Petersen
John Petersen 2021-3-18
编辑：John Petersen 2021-3-18
I was looking for a more general approach that was able to extract the last fields in the structure. The fields are not actually 'a', 'b', 'c'....

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