How to read elements from a computed matrix without define a new variable?

Dear all,
I am having a question about matrix computation. For example, a matrix A is calculated by an expression, . I am only interested in the element . Simply I can do and . I am wondering whether there is a way to do this without defining A. This line of code is not legal in MATLAB but it shows my goal clearly, .
Thanks!

7 件のコメント

This feature was asked for future versions, but the answer of the staff was for that you need to create the variable anyways, so you are only losing legilability, not wining performance.
Mathematically, you can just compute the first element as
a = exp(B(1,:)*C(:,1)+D(1));
Yize Wang
Yize Wang 2019 年 5 月 10 日
編集済み: Yize Wang 2019 年 5 月 10 日
Thank you for your fast answer. Loss of legilability is exactly what I am trying to avoid. Mathematical way works fine for me but may be confusing to other readers. Thanks!
Torsten
Torsten 2019 年 5 月 10 日
Of course,
a = exp(B(1,:)*C(:,1)+D(1));
is wrong in general.
It's true if, e.g., B*C+D is a diagonal matrix.
gonzalo Mier
gonzalo Mier 2019 年 5 月 10 日
編集済み: gonzalo Mier 2019 年 5 月 10 日
Eeeeeemmmm, Torsten... this equation is true always...
If you want to check:
A = rand(3);B = rand(3); C = A*B; D = A(1,:)*B(:,1); D==C(1,1)
Torsten
Torsten 2019 年 5 月 10 日
The matrix exponential of a matrix is not equal to the matrix where all of its elements are "exponentialized".
gonzalo Mier
gonzalo Mier 2019 年 5 月 10 日
Yize Wang, this is the easiest way and fastest to do it as I know. If you are concern about legibility, you can encapsulate this piece of code in a function called get_1_comp_system or some cool name (I'm not good at all naming).
Stephen23
Stephen23 2019 年 5 月 10 日
編集済み: Stephen23 2019 年 5 月 10 日
"Eeeeeemmmm, Torsten... this equation is true always..."
Except that you did not take into account the matrix exponential shown in the original question.

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 採用された回答

Stephen23
Stephen23 2019 年 5 月 10 日
編集済み: Stephen23 2019 年 5 月 10 日
MATLAB does not allow indexing into the results of operations. You can either use a temporary variable (which is probably the most efficient solution), or you could call subsref directly:
>> B = rand(3);
>> C = rand(3);
>> D = rand(3);
>> a = subsref(expm(B*C+D),substruct('()',{1,1}))
a = 8.6593
>> A = expm(B*C+D) % for comparison:
A =
8.6593 7.3898 6.2134
13.1945 16.0237 11.6736
13.1316 13.8705 11.8273

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