Grouping and sharing of subcarriers in a random manner
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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
3 Comments
is it really a different question? You seem to have asked the very same question around 10-15 times. It's a waste of time for everyone. Just ask once
- provide enough information of what you have and what you want
- try to "understand" the answers given to you (this means not just copy-paste the answer inside your original code)
- read the documentation of matlab to know how your actual code works.
Unfortunately your questions give an impression that you're badly trying to do some project with a code you've got from somewhere. Take some time to understand it and I'm sure it will be more efficient than throwing 10 questions each day on an online forum.
Prabha Kumaresan
on 19 Dec 2017
Jan
on 19 Dec 2017
@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.
Answers (1)
Vigneshwar Pesaru
on 3 Jan 2018
0 votes
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.
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