How can i define with variable

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fatih ayça
fatih ayça 2021 年 3 月 6 日
編集済み: fatih ayça 2021 年 3 月 6 日
COORD = [0 0;2 0;4 0;2 2;0 2];
CON = [2 1;3 2;3 4;2 4;4 1;4 5];
for a = 1:6
i = CON(a,1);
j = CON(a,2);
COORD(j,1);
COORD(i,1);
dx = COORD(j,1)-COORD(i,1);
dy = COORD(j,2)-COORD(i,2);
lambdax(a) = dx ;
lambday(a) = dy ;
end
for a = 1:6
K = [lambdax(a)^2 lambdax(a)*lambday(a) -lambdax(a)^2 -lambdax(a)*lambday(a);lambdax(a)*lambday(a) lambday(a)^2 -lambdax(a)*lambday(a) -lambday(a)^2;-lambdax(a)^2 -lambdax(a)*lambday(a) lambdax(a)^2 lambdax(a)*lambday(a);-lambdax(a)*lambday(a) -lambday(a)^2 lambdax(a)*lambday(a) lambday(a)^2]
end
Hello everyone, this is my script it is working now but i need K in the form of K(a) or K(b), if i write K(a) it is not working. How can i write like K(a)?

採用された回答

Star Strider
Star Strider 2021 年 3 月 6 日
I still do not understand what you want to do, however it is straightforward to create ‘K’ as an anonymous function:
K = @(a) [lambdax(a).^2 lambdax(a)*lambday(a) -lambdax(a).^2 -lambdax(a).*lambday(a);lambdax(a).*lambday(a) lambday(a).^2 -lambdax(a).*lambday(a) -lambday(a).^2;-lambdax(a).^2 -lambdax(a).*lambday(a) lambdax(a).^2 lambdax(a).*lambday(a);-lambdax(a).*lambday(a) -lambday(a).^2 lambdax(a).*lambday(a) lambday(a).^2];
That should then produce whatever it is that you want from calling ‘K’ as a function.
See the documentation section on Anonymous Functions for details on how they work and how to use them.
I also vectorised ‘K’. See Array vs. Matrix Operations for those details.
.
  6 件のコメント
fatih ayça
fatih ayça 2021 年 3 月 6 日
Thank you so much.
Star Strider
Star Strider 2021 年 3 月 6 日
As always, my pleasure!

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