Info

이 질문은 마감되었습니다. 편집하거나 답변을 올리려면 질문을 다시 여십시오.

Grouping and sharing of subcarriers in a random manner

조회 수: 1 (최근 30일)
Prabha Kumaresan
Prabha Kumaresan 2017년 12월 19일
마감: MATLAB Answer Bot 2021년 8월 20일
Could anyone tell me how to group rows(row 1 with row 2 or row 3 with row 5 or row2 and row 3 with row5) in a random manner such that their values should be shared for the matrix C =
0 0.4089 0 0 0 0 0 0.4725 0 0 0 0
0 0 0 0.0038 0.0023 0 0 0 0 0 0 0
0 0 0.0018 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0.0003 0 0 0 0 0 0
0.0565 0 0 0 0 0 0 0 0.0263 0.0523 0 0
0 0 0 0 0 0 0.0313 0 0 0 0.0394 0.0367
  댓글 수: 3
Prabha Kumaresan
Prabha Kumaresan 2017년 12월 19일
As i am unable to get the output in my code I am asking the question.
the final thing which i really required is for the above matrix i need user grouping with sharing of subcarriers and i want to get implemented in n number of rows and m number of columns.
Jan
Jan 2017년 12월 19일
@Prabha: Posting multiple questions concerning the same problem confuses the readers. Stay at one thread and discuss it there in one piece. Please follow KL's valuable advice carefully.
As i am unable to get the output in my code I am asking the question.
Asking many questions is not an efficient way to solve a problem. Better concentrate on one question and spend the time to formulate it clearly.
Please confirm that you have read Format questions and Ask good questions.

답변 (1개)

Vigneshwar Pesaru
Vigneshwar Pesaru 2018년 1월 3일
It seems to be trivial code,if I understood properly.And I don't want to share any code but algorithm or logic to get your desired o/p.
1.Generate a random number between 1 and n;here n is nothing but [n,m]=size(i/p_matrix)
2.Declare a new empty matrix(as many times as you need) in a loop.
3.Start copying the elements row vise(how many number of rows you copy to the new empty matrix depends on the random number you have generated instantly).
Hope the above one works.

이 질문은 마감되었습니다.

Community Treasure Hunt

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

Start Hunting!

Translated by