Structures with multiple, conntected anonymous functions

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I have a rather complicated workflow, however, I am having trouble identifying a way for it to work.
I start with a json of the form:
{
"x":3,
"y":6,
"fx":"@(a) a^2+y",
"fy":"@(a,b) y*fy(b)+b*x"
}
This gets read in and evaluated such that:
S.x = 3
S.y = 6
S.fx = @(a) a^2+y
S.fy = @(a,b) y*f(b)+b*x
I have tried both eval and str2funct to create the anonymous function.
Later in the workflow, this structure gets packaged into another structure
dS.S = S;
This is to help with passing a model around between functions.
The problem is, that later on, this command will give an error:
dS.S.fy(1,1)
The error states that the function fx cannot be found. However,
dS.S.fx(1)
does work, WITHOUT there being an "y" defined in the workspace.
Can anyone shed some light on this?
  3 个评论
Ameer Hamza
Ameer Hamza 2020-5-13
What is f(b) in this line
y*f(b)+b*x
is it supposed to be fx(b)?
Kyle Johnson
Kyle Johnson 2020-5-13
@James Tursa, they do not need to pick up S.x or S.y. These values will not change once they are read in.
@Ameer Hamza, yes that is a typo

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采纳的回答

Ameer Hamza
Ameer Hamza 2020-5-13
编辑:Ameer Hamza 2020-5-13
Ok. I know "eval is evil" but here it makes everything quite simple in this case. You can define a helper function like this
function s = helperFun(s)
names = fieldnames(s);
for i=1:numel(names)
eval(sprintf('%s=s.%s;', names{i}, names{i}));
end
for i=1:numel(names)
if ischar(s.(names{i}))
s.(names{i}) = eval(s.(names{i}));
end
end
end
and then call it on your json file like this
S = jsondecode(fileread('test.txt'));
S = helperFun(S)
ds.S = S;
Then call the functions
>> ds.S.fx(1)
ans =
7
>> ds.S.fy(1, 2)
ans =
246

更多回答(1 个)

the cyclist
the cyclist 2020-5-13
Are you sure that y didn't exist in the workspace when you defined the structures? Because in a fresh workspace,
S.x = 3
S.y = 6
S.fx = @(a) a^2+y
S.fy = @(a,b) y*f(b)+b*x
dS.S = S
dS.S.fx(1)
gives the error
Unrecognized function or variable 'y'.
Error in @(a)a^2+y
Does that shed light?
  9 个评论
James Tursa
James Tursa 2020-5-13
Yes, it appears you understand how function handles work now. Your current approach seems correct to me.

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