4D matrix manipulation

9 ビュー (過去 30 日間)
Asliddin Komilov
Asliddin Komilov 2019 年 8 月 16 日
コメント済み: Star Strider 2019 年 8 月 17 日
I have a 4D matrix G size of [k*j*i*l] so its G(k,j,i,l) ,
the sizes are:
k=1x365,
j=1x31,
i=1x24,
l=1x91.
For estimation purposes I need a 3D matrix G(k,j-i,l), which I can later also plot.
I did not try anything, since I have no clue how to do it.
Thanks
  4 件のコメント
Asliddin Komilov
Asliddin Komilov 2019 年 8 月 17 日
編集済み: Asliddin Komilov 2019 年 8 月 17 日
sorry for the confusion, the last dimension is not 1 (one) but l (L), j-i is a new dimension that is equal to j-i, but because j-i is for each values of j and i it becomes a matrix j-i(j,i) instead of an array. Here my imagination stops working, I cannot figure out how it should be. Or maybe I am wrong and the solution looks otherwise.
thanks
Asliddin Komilov
Asliddin Komilov 2019 年 8 月 17 日
I did not remove or edit it since the first posted, it must have been a system fault maybe.

サインインしてコメントする。

回答 (1 件)

Star Strider
Star Strider 2019 年 8 月 16 日
I am not certain what you want.
Try this:
G = rand(365,31,24,91); % Create ‘G’
Gnew = reshape(G, 365, [], 91); % Desired Result (?)
Experiment to get the result you want.
  2 件のコメント
Asliddin Komilov
Asliddin Komilov 2019 年 8 月 17 日
thanks, I am on it.
Star Strider
Star Strider 2019 年 8 月 17 日
My pleasure.
If my Answer helped you solve your problem, please Accept it!

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeMatrix Indexing についてさらに検索

製品


リリース

R2016a

Community Treasure Hunt

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

Start Hunting!

Translated by