how can I write a code segment that flips an N*M array up to down *Do not use loops or flip function

2 件のコメント

James Tursa
James Tursa 2020 年 12 月 26 日
What have you done so far? What specific problems are you having with your code?
Stephen23
Stephen23 2020 年 12 月 27 日

サインインしてコメントする。

 採用された回答

Image Analyst
Image Analyst 2020 年 12 月 25 日

4 投票

What did your course tell you about indexing? Like
indexes = startingValue : stepValue : endingValue;
If you still need help, reply only after reading the link below:

5 件のコメント

omar khasawneh
omar khasawneh 2020 年 12 月 25 日
sorry , is there any mistake with my question ? !
Image Analyst
Image Analyst 2020 年 12 月 25 日
Do you want us to do your homework for you so that you can turn in our code to your class while telling your professor that you wrote the code yourself? Please explain what you want us to do for you and what you will do with our answer.
omar khasawneh
omar khasawneh 2020 年 12 月 26 日
Listen , I never allow you to accuse me of lying to my professor, if you also lie, that's your problem.
I asked because I really need help, I don't need to manipulate as you pretend, and I didn't ask you to solve the question
Image Analyst
Image Analyst 2020 年 12 月 26 日
Read my comment again - there is no accusation. Just a question as to explain how we can help you with this question that can be solved in one line of code. Since it's only one line of code, it's hard to help you without doing the whole thing for you. That's why I gave the hint I did. If I give any more than that, then it's completely done and there is nothing more for you to do. At least this way, I think you can change variable names and be able to say you did it yourself.
Jan
Jan 2020 年 12 月 27 日
@omar khasawneh: There is no accusation. Image Analyst gave you a useful hint.

サインインしてコメントする。

その他の回答 (1 件)

omar khasawneh
omar khasawneh 2020 年 12 月 26 日

0 投票

okay , I got it ...
I solved like this A = [1 2 3 ; 4 5 6; 7 8 9];
A = A' % transpos
A = rot180(A) %rotation

1 件のコメント

Image Analyst
Image Analyst 2020 年 12 月 26 日
OK, that works as long as you have written a special rot180() function. I'm not seeing it in base MATLAB. If I instead use imrotate() to do that part:
A = [1 2 3 ; 4 5 6; 7 8 9];
A = A' % transpos
% A = rot180(A2) %rotation
A = imrotate(A, 180)
I get
A =
1 4 7
2 5 8
3 6 9
A =
9 6 3
8 5 2
7 4 1
Which is not flipped up to down. It also flips left to right.
The solution most MATLABers would use is a single line of code using the hint I gave. So here is a little bit more:
A = A(startingValue : stepValue : endingValue, :);
That's it. One single line of code. See if you can supply the values.

サインインしてコメントする。

カテゴリ

ヘルプ センター および File ExchangeMatrix Indexing についてさらに検索

製品

リリース

R2015a

質問済み:

2020 年 12 月 25 日

コメント済み:

Jan
2020 年 12 月 27 日

Community Treasure Hunt

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

Start Hunting!

Translated by