Why does the anonymous function not evaluate correctly
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I posted this in stack exchange comp science site 2 days back but couldn't get a response. Will restate with a simplified example and hope I get some directions.
Consider the Rosenbrock function in http://www.mathworks.com/help/optim/ug/writing-objective-functions.html#brhkghv-6
Defining the following in a script (modifying matlab example for how x is inputted) and run it,
clc; clear;
anonrosen = @(x1,x2)(100*(x2 - x1^2)^2 + (1-x1)^2);
and then typing the following in command window,
anonrosen(-1,2)
yields,
ans =
104
But if I define the following in a script and run it,
clc; clear;syms x1 x2;
p = (100*(x2 - x1^2)^2 + (1-x1)^2);
anonrosen = @(x)p;
and then typing the following in command window,
anonrosen(-1,2)
yields,
ans =
(x1 - 1)^2 + 100*(- x1^2 + x2)^2
Why don't I get 104 as answer in 2nd approach?
The reason I want to make the 2nd approach work is the functional form "(100*(x2 - x1^2)^2 + (1-x1)^2)" in my application is being generated on the fly during application run depending on user inputs using symbolic toolbox so I cannot hard-code it the way it is done in matlab example code.
Thanks in advance Hari
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采纳的回答
Star Strider
2014-6-29
编辑:Star Strider
2014-6-29
You have to define functions differently in the Symbolic Math Toolbox:
syms x1 x2
anonrosen = symfun(100*(x2 - x1^2)^2 + (1-x1)^2, [x1 x2])
anonrosen(-1,2)
produces:
anonrosen(x1, x2) =
(x1 - 1)^2 + 100*(- x1^2 + x2)^2
ans =
104
11 个评论
Star Strider
2014-7-5
My pleasure!
I wanted to be sure I continued to monitor your question in the event we had not resolved it.
I’ll look for your next Question. Good call on creating it separately, since once an Answer is Accepted (THANKS!) it tends to not be followed-up.
John D'Errico
2014-7-5
Don't close the question. Don't delete what you wrote. Don't do anything of the sort.
Leave the question there, for others to read and learn from it.
Then accept the answer if it solved your problem. Vote it up.
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