How to call element of matrix of symbolic variables

197 views (last 30 days)
I have two symbolic variables
syms th1(t) Lc1
and generated a matrix in terms of th(t) and Lc1
tc1(t) =
[ cos(th1(t)), -sin(th1(t)), 0, Lc1*cos(th1(t))]
[ sin(th1(t)), cos(th1(t)), 0, Lc1*sin(th1(t))]
[ 0, 0, 1, 0]
[ 0, 0, 0, 1]
now when I try to call the first three elements of the last column
pc1 = tc1(1:3,4);
I got this error:
Error using symfun/subsref (line 169)
Symbolic function expected 1 inputs and received 2.
Error in Fk (line 10)
pc1 = tc1(1:3,4);
thanks

Accepted Answer

Walter Roberson
Walter Roberson on 17 Jul 2018
It is not possible to index the entries of a symbolic function that computes a matrix.
The closest you can get is to call the function passing in variables for the parameters, get an array output, index into the array, and use the resulting expression or turn it into a function if needed.
  10 Comments
Walter Roberson
Walter Roberson on 20 Oct 2022
Using () indexing to access the components of a symbolic function that defines an array, is something that is not going to be permitted. The () indexing method for symbolic function is already overloaded to evaluate the function, and there is no way to say "when I write these () brackets I mean indexing instead of evaluation".
There is already formula to retrieve the array expression in a way that can be later indexed.

Sign in to comment.

More Answers (2)

Nathan Hardenberg
Nathan Hardenberg on 19 May 2022
Since the code answer is a bit hidden in the comments I'm posting the solution as full code here:
syms th1(t) Lc1
tc1(t) = [
cos(th1(t)), -sin(th1(t)), 0, Lc1*cos(th1(t));
sin(th1(t)), cos(th1(t)), 0, Lc1*sin(th1(t));
0, 0, 1, 0;
0, 0, 0, 1;]
pc1 = tc1(t); % call the function with t first
pc1 = pc1(1:3,4) % now you can index
  2 Comments
Stu
Stu on 19 Oct 2022
Wow, thanks for that. It's beautiful.
Now I can pass a vector (which itself contains symbolic variables) into another symbolic function which requires that vector input.

Sign in to comment.


João Vitor Fraga
João Vitor Fraga on 1 Feb 2023
Edited: João Vitor Fraga on 1 Feb 2023
Here's a simple solution that doesn't turn your symfun into a sym: simply left-multiply and right-multiply your symfun by appropriate matrices whose entries are either zero or one, depending on which elements you wish to extract. Your symfun will remain a symfun. In your case:
syms th1(t) Lc1
tc1(t) = [ cos(th1(t)), -sin(th1(t)), 0, Lc1*cos(th1(t))
sin(th1(t)), cos(th1(t)), 0, Lc1*sin(th1(t))
0, 0, 1, 0
0, 0, 0, 1];
Lmatrix = [1 0 0 0
0 1 0 0
0 0 1 0];
Rmatrix = [0
0
0
1];
pc1 = Lmatrix*tc1*Rmatrix
pc1(t) = 
class(pc1)
ans = 'symfun'
This can easily be condensed into a simple function you could call whenever needed.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!