Functions defined by "inline" with vector input in MATLAB

Question

Defining a function with

f = inline('x+P1*P2-P3',3);

one can calculate f(1,2,3,4), f(0,1,2,1), etc.

How should I write the function f so that I can use vectors such as 1:4 or [2,3,6,4] as an input?

Do you mean f = inline('x+P1*P2-P3',3);? What behavior do you expect? What happens now if you try f(1:4,2,3,4)? Also, you should be using Anonymous Functions instead of inline which is deprecated. — jodag
– jodag, Commented Dec 14, 2017 at 4:23
@jodag: Thanks for that! I put a 4 there to make sure the code works. 3 should work as well. I was trying to fix some old implementation of numerical algorithms, which used that function. — user758077
– user758077, Commented Dec 14, 2017 at 12:56
@jodag: you are right. One should have 3 instead of 4 there if f has only 4 arguments. Fixed. (I didn't notice this problem until I tried to find my own answer.) — user758077
– user758077, Commented Dec 21, 2017 at 14:28

TroyHaskin · Accepted Answer · 2017-12-14 04:26:59Z

5

The posted code works because of the extremely rigid structure allowed by the deprecated inline:

inline(expr,n) where n is a scalar, constructs an inline function whose input arguments are x, P1, P2, ... .

Note: "inline will be removed in a future release. Use Anonymous Functions instead."

Noting the note, you can duplicate the behavior of the posted code by doing:

f = @(x,P1,P2,P3) x+P1*P2-P3;

You can also get your desired behavior by just having an x and indexing it within the body of the anonymous function:

f = @(x) x(1)+x(2)*x(3)-x(4);

answered Dec 14, 2017 at 4:26

TroyHaskin

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Comments

score 0 · Accepted Answer · 2017-12-21 14:46:17Z

0

I have just learned from these two answers:

that the keyword is Comma-Separated Lists. One can simply use

f = inline('x+P1*P2-P3',3);
a = [1,2,3,4]; % or a = 1:4;
c = num2cell(a); 
f(c{:})

answered Dec 21, 2017 at 14:30

user758077