Related to matrix partitioning?
1 回表示 (過去 30 日間)
古いコメントを表示
hi, I have a matrix of dimension (1*n) where n is a multiple of 4. I have to recursively partition it into four equal parts until n=4. How can I do it in matlab? Also I want all the partitioned matrices generated.
0 件のコメント
回答 (2 件)
David Sanchez
2014 年 3 月 4 日
Try this:
M=rand(1,32);
n=length(M);
A=zeros(4,n/4);
for k=1:4%k=k+1
A(k,:) = M((1+(k-1)*n/4):n*k/4);
end
Access your new matrices with A(k,:),for example:
>> A
A =
0.2599 0.8001 0.4314 0.9106 0.1818 0.2638 0.1455 0.1361
0.8693 0.5797 0.5499 0.1450 0.8530 0.6221 0.3510 0.5132
0.4018 0.0760 0.2399 0.1233 0.1839 0.2400 0.4173 0.0497
0.9027 0.9448 0.4909 0.4893 0.3377 0.9001 0.3692 0.1112
>> A(1,:)
ans =
0.2599 0.8001 0.4314 0.9106 0.1818 0.2638 0.1455 0.1361
0 件のコメント
Image Analyst
2014 年 3 月 4 日
Why recursively? Why mess with that complication? Unless you did some tricky stuff the 2,nd, 3rd, and 4th parts won't be the same height. Because you could extract the top 1/4 and then call the function on the remainder - the remaining 3/4. But 1/4 of 4/3 is 3/16, not one quarter, so the second time through it's shorter. Is that what you want? You didn't really say how n was getting changes at each level of recursion. Let's say it starts out at 32. How does n eventually get to 4? Do you divide by 4? How are the parts going to be equal each time? Let's say n=32 and you get 4 arrays of 8 rows each and then n now equals 32/4 = 8. Do you then recurse again and make 4 additional arrays of 2 rows each???
Why not simply get your 4 arrays like this:
m1 = m(1:rows/4,:);
m2 = m(rows/4+1:2*rows/4,:);
m3 = m(rows/2+1:3*rows/4,:);
m4 = m(3*rows/4+1:end,:);
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Logical についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!