how to mirror data and then plot it?
정보
This question is locked. 편집하거나 답변을 올리려면 질문을 다시 여십시오.
이전 댓글 표시
이 질문에 dpb
님이 플래그를 지정함
x =0:1:15;
y = tan(x);
plot(x,y)
i hv to mirror data and then plot also .how to do it
댓글 수: 9
ramya
2024년 8월 8일
So I understand that you want to mirror along the middle.
Thus
x = 0:1:15;
y = tan(x);
x_new = x;
x_mirrored = fliplr(x_new);
and
y_new = tan(x_mirrored);
Thus the plot obtained will be
figure;
plot(x_new, y_new, 'b', 'DisplayName', 'Symmetric Data');
legend;
xlabel('x');
ylabel('tan(x)');
title('Symmetric Data Plot');
Let me know if this is what was desired.
Aniket
2024년 8월 8일

Something like this ?
a= [1 2 3 4 5 6 7 8 9 10] ;
b=[0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.2 0.3] ;
ix=fix(numel(a)/2);
c=[b(1:ix) fliplr(b(1:ix))];
plot(a,c)
The above keeps the same number of elements as the original; to do your precise example reduces the number of points by one for an even number...your choice as to which would be the intent/need.
You still haven't precisely defined "symmetry about what?" for the original -- about the minimum, perhaps? That kind of thing may be very dependent upon just what the real application is, not just specifically for the one given example.
But a clear and unambiguous definition of just what "symmetrical" means to you is mandatory...
ramya
2024년 8월 8일
dpb
2024년 8월 8일
In that case, don't start a new and seemingly unrelated thread...what's wrong with the Answer @Cris LaPierre provided you there? Why don't you amplify in depth what "symmetry" is supposed to mean for the given data which look anything but as given?
It's not possible to answer a question when the specifications for the solution definition aren't clear...you may have a clear understanding of what you mean, but apparently nobody else here does...
I'm flagging this as duplicated to keep the conversation in one place and as a courtesy to those who responded there with best efforts...
답변 (0개)
This question is locked.
카테고리
도움말 센터 및 File Exchange에서 Programming에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!


