Converting a vector to multiple variables required as input for 'matlabFunction'.
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I am converting symbolic function to a Matlab function using 'matlabFunction'.
For e.g.
syms x y
g=matlabFunction(x+y-1)
Output
g = @(x,y)x+y-1.0
The returned matlab function 'g' accepts individual variables like g(1,1). However, I want to give inputs as a vector, like,
input = [1 1]
g(input)
I get error 'Not enough input arguments.' I want to give input as a single vector because number of input in my actual function vary during run-time. Any idea how to convert a vector to multiple variable at run-time efficiently or any other solution.
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채택된 답변
Andrei Bobrov
2012년 9월 24일
input = {1 1}
g(input{:})
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Leo Simon
2012년 11월 6일
Thanks Walter
I could do everything without anonymous functions, but would prefer not to because of another infuriating matlab property which is that I can't define a function within an m-file, have to do it within a function. Should I just give up on passing an anonymous function to fsolve? I'm willing to do it the hard way if there is a hard way!
Walter Roberson
2012년 11월 6일
I do not understand what you mean about can't define a function within an m-file ?
The code I show above involves passing anonymous functions to fsolve: it just uses a helper function which would be the same for all cases. The "g" that I reference in the expression call_on(g, x) is the anonymous function constructed via matlabFunction. The version that does not use this (invariant) helper function would need to use some nasty MATLAB behavior to fudge the temporary variable.
추가 답변 (2개)
Brad Carman
2018년 1월 26일
The matlab documentation fails to mention that it's possible to set this up with matlabFunciton() itself. I've discovered that how you input the 'Vars' argument adjusts how the function is built. Using a vector input of variables for 'Vars' will give you what you want. See the following example:
syms x y
z = x+y;
Without vector input:
matlabFunction(z, 'Vars', {x, y})
function_handle with value:
@(x,y)x+y
With vector input:
matlabFunction(z, 'Vars', {[x, y]})
function_handle with value:
@(in1)in1(:,1)+in1(:,2)
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Ti Miller-Jackson
2019년 8월 21일
Oh my gosh this works so well! You just saved me many hours. God bless you sir!
Graham Rowe
2020년 11월 7일
This should be the accepted answer. So much more elegant than defining temporary variables that get created and then removed at run time.
Azzi Abdelmalek
2012년 9월 24일
편집: Azzi Abdelmalek
2012년 9월 24일
g = @(x,y)x+y-1.0
%g is a function of two inputs x and y, if you want one vector input
g=@(x) x(1)+x(2)-1.0
%then
g([2 3])
=4
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