How to call element of matrix of symbolic variables

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Hazim Nasir
Hazim Nasir 2018-7-16
编辑： Peyman Amiri 2024-5-15
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

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采纳的回答

Walter Roberson
Walter Roberson 2018-7-17
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.
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显示 8更早的评论
Walter Roberson
Walter Roberson 2022-10-20
@Fenglei Gu
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.

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更多回答（3 个）

Nathan Hardenberg
Nathan Hardenberg 2022-5-19
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 个评论
Stu
Stu 2022-10-19
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.

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João Vitor Fraga
编辑：João Vitor Fraga 2023-2-1
Ran in:
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.
Peyman Amiri
Peyman Amiri 2024-5-15
编辑：Peyman Amiri 2024-5-15
Ran in:
Even we can make the brilliant soultion of @Nathan Hardenberg better by the following improvement (this solution does not change anything about the initial tc1 matrix):
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];
tc1=tc1(t);
pc1 = tc1(1:3,4)
pc1 = 

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