Nested Structure in cellfun

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Ozge Moral
Ozge Moral am 28 Jul. 2016
Bearbeitet: Ozge Moral am 29 Jul. 2016
I have a structure that store lots of variable. I want to find function value of these x values. But there is wrong result. My variables :
StartingValue.x ={};
StartingValue.fn =[];
My function is
f = @(x) sin(5.1*pi.*x+0.5).^6;
I created a nested structure and i want to function value of this variable.
StartingValue(1).Next(1).x
= Columns 1 through 6
[0.0025] [0.0075] [0.0125] [0.0175] [0.0225] [0.0275]
Columns 7 through 10
[0.0325] [0.0375] [0.0425] [0.0475]
To find function values i wrote this code:
StartingValue(1).Next(1).fn=cellfun(f,StartingValue(1).Next(1).x)
ans =
Columns 1 through 8
0.0185 0.0385 0.0716 0.1213 0.1899 0.2781 0.3838 0.5023
Columns 9 through 10
0.6263 0.7464
This result is not correct according to mathematical calculation. I couldn't find what mistake i did. I used cellfun because of x variable is a cell array. I didn't understand whether problem is nested structure or cellfun. If my question is not clear, i can post all code.

Akzeptierte Antwort

Guillaume
Guillaume am 28 Jul. 2016
The result is entirely correct. Perhaps you're assuming that sin works in degrees? If that is the case, use sind instead of sin.
By the way, there is no reason to use a cell array for your example, a vector would make the code a lot simpler:
StartingValue(1).Next(1).x = [0.0025, 0.0075, 0.0125, 0.0175, 0.0225, 0.0275, 0.0325, 0.0375, 0.0425, 0.0475]
f = @(x) sin(5.1*pi.*x+0.5).^6;
StartingValue(1).Next(1).fn = f(StartingValue(1).Next(1).x)
Also note that the fn field name is misleading since its content is not a function but a vector.
  5 Kommentare
Guillaume
Guillaume am 29 Jul. 2016
There's a factor of 10 difference between the values generated by your initialization function, and the values in your original post:
0.025 vs 0.0025
This explains the vastly different results.
Ozge Moral
Ozge Moral am 29 Jul. 2016
Bearbeitet: Ozge Moral am 29 Jul. 2016
i'm going to show my eyes to doctor! Thanks a lot.

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