Nested struct arrays with variable format

What is the syntax for creating a nested struct array with the following format?

Varibale2_Name = struct('FieldName1',cell(Variable1_Name,1),'FieldName2',cell(Variable1_Name,1),'FieldName3',*struct*)

Where *struct* is meant to contain basically any other information/data returned from multiuple different functions called within a loop. I would like to be able to chuck a 1x7 struct, a 3x10 cell, etc. into that last field, essentailly treating it as a pointer/container. I am doing this specifically to avoid creating separate aggregator/container variables for each possible format of FieldName3 data, which I would have to concatenate with each iteration through a loop because, while Variable1_Name is known prior to the loop, the size of FieldName3 data cannot be known a-priori.

Answer 1

[Jan]

You can't.

S = struct('Name1', cell(n1, 1), ...
           'Name2', cell(n1, 1))

his creates a [n1 x 1] struct array with two fields. You cannot append an aribtrary struct magically to S. Struct array need to have the same fields for all elements.

Either you need another level of substructs or you can store the data of as arrays:

S.Value = struct('Name1', cell(n1, 1), 'Name2', cell(n1, 1))
S.Name3 = yourStruct;

% or:
S.Name1 = cell(n1, 1)
S.Name2 = cell(n1, 1)
S.Name3 = yourStruct;

[Gabriel Stanley]

So I think I did what you were talking about with the substructs thing without meaning to. I ran individual segments of the loop in the cmd window to get data sets, but with manual assignment was able to create the following:

AggregateData = struct('ConsistentInfo',cell(n1,1),'Semi-ConsistentInfo',cell(n1,1),'WildcardHolder',struct)

Fields ConsistentInfo Semi-ConsistentInfo WildcardHolder

1 1x5 cell [n2 x 2] double [1x16] struct

2 1x5 cell [n3 x 2] double [10x3] cell

3 1x5 cell [n4 x 2] double [1x7] struct

... ... ... ...

n1 1x5 cell [n{n1} x 2] double [1x1] struct

Which is what I was trying to do. Are there any potential consequences of what I've done here I should be warry of?

[Jan]

If it works, it works.