Hello,
I have a mxn matrix of binary data. I want to write a code where I find the rows in column n which have 1s in them and move them to bottom of the row. I want a generalised code for this as it will be in a loop where multiple iterations of this will be executed.
%For example, assume this is my matrix
% a b c d e f g h i j k l
H = [0 0 1 1 0 1 0 1 0 1 1 0; %1
0 0 0 1 0 1 1 0 1 0 1 0; %2
0 1 1 0 0 1 1 1 0 0 0 1; %3
0 1 1 0 1 0 1 0 1 0 0 1; %4
0 0 0 1 1 0 0 1 1 0 0 0; %5
0 1 0 0 1 0 0 0 0 0 1 0]; %6
%In column 'l' there are 2 rows with 1s in them (row 3 and 4).
% I want to move rows 3 and 4 to the bottom. My final matrix should look like this.
% a b c d e f g h i j k l
H = [0 0 1 1 0 1 0 1 0 1 1 0; %1
0 0 0 1 0 1 1 0 1 0 1 0; %2
0 0 0 1 1 0 0 1 1 0 0 0; %5
0 1 0 0 1 0 0 0 0 0 1 0; %6
0 1 1 0 0 1 1 1 0 0 0 1; %3
0 1 1 0 1 0 1 0 1 0 0 1]; %4
%Any help is appreciated. I thank in advance.

 採用された回答

Ive J
Ive J 2020 年 12 月 23 日

0 投票

H = randi([0 1], 6, 10)
H =
0 1 0 1 1 0 0 0 1 0
1 1 1 0 1 1 0 1 1 1
0 0 0 0 1 0 0 1 0 1
1 1 0 0 1 1 1 0 0 1
1 1 0 0 1 1 1 1 0 0
1 0 1 1 1 1 1 0 1 1
col = 9;
H = [H(H(:, col) < 1, :); H(H(:, col) > 0, :)]
0 0 0 0 1 0 0 1 0 1
1 1 0 0 1 1 1 0 0 1
1 1 0 0 1 1 1 1 0 0
0 1 0 1 1 0 0 0 1 0
1 1 1 0 1 1 0 1 1 1
1 0 1 1 1 1 1 0 1 1

5 件のコメント

Rishi Balasubramanian
Rishi Balasubramanian 2020 年 12 月 23 日
That works perfectly. Thank you.
Rishi Balasubramanian
Rishi Balasubramanian 2020 年 12 月 23 日
Hey I have another question. How would I choose that over a range?
% a b c d e f g h i j k l
H = [0 0 1 1 0 1 0 1 0 1 1 0; %1
0 0 0 1 0 1 1 0 1 0 1 0; %2
0 1 1 0 0 1 1 1 0 0 0 1; %3
0 1 1 0 1 0 1 0 1 0 0 1; %4
0 0 0 1 1 0 0 1 1 0 0 0; %5
0 1 0 0 1 0 0 0 0 0 1 0]; %6
%For ex in my matrix if I only wanted to look at column 'k' over rows 1 to 5, how can I use that?
%I want my final matrix to be something like this.
% a b c d e f g h i j k l
H = [0 1 1 0 0 1 1 1 0 0 0 1; %3
0 1 1 0 1 0 1 0 1 0 0 1; %4
0 0 0 1 1 0 0 1 1 0 0 0; %5
0 1 0 0 1 0 0 0 0 0 1 0]; %6
0 0 1 1 0 1 0 1 0 1 1 0; %1
0 0 0 1 0 1 1 0 1 0 1 0; %2
Ive J
Ive J 2020 年 12 月 23 日
You can also do that with some tiny modifications
H =
0 1 0 1 1 0 0 0 1 0
1 1 1 0 1 1 0 1 1 1
0 0 0 0 1 0 0 1 0 1
1 1 0 0 1 1 1 0 0 1
1 1 0 0 1 1 1 1 0 0
1 0 1 1 1 1 1 0 1 1
col = 9;
rowsOnly = 1:5; % i.e. do not change row 6
idxRows = ismember(1:size(H, 1), rowsOnly)'; % ~idxRows: row 6
idx0 = H(:, col) < 1 & idxRows; % 3, 4 and 5
idx1 = H(:, col) > 0 & idxRows; % 1 and 2
H = [H(idx0, :); H(~idxRows, :); H(idx1, :)]
0 0 0 0 1 0 0 1 0 1
1 1 0 0 1 1 1 0 0 1
1 1 0 0 1 1 1 1 0 0
1 0 1 1 1 1 1 0 1 1
0 1 0 1 1 0 0 0 1 0
1 1 1 0 1 1 0 1 1 1
Rishi Balasubramanian
Rishi Balasubramanian 2020 年 12 月 23 日
Wonderful, that works perfectly. Thanks man.
Rishi Balasubramanian
Rishi Balasubramanian 2020 年 12 月 23 日
Hello, One more question. Hopefully its the last. Say I want to move a single row to a specific postion? How do I do that?
For ex
% a b c d e f g h i j k l
H = [0 0 1 1 0 1 0 1 0 1 1 0; %1
0 0 0 1 0 1 1 0 1 0 1 0; %2
0 1 1 0 0 1 1 1 0 0 0 1; %3
0 1 1 0 1 0 1 0 1 0 0 1; %4
0 0 0 1 1 0 0 1 1 0 0 0; %5
0 1 0 0 1 0 0 0 0 0 1 0]; %6
% I want to move row 2 to row index 5 such that my final matrix is -
% a b c d e f g h i j k l
H = [0 0 1 1 0 1 0 1 0 1 1 0; %1
0 1 1 0 0 1 1 1 0 0 0 1; %3
0 1 1 0 1 0 1 0 1 0 0 1; %4
0 0 0 1 1 0 0 1 1 0 0 0; %5
0 0 0 1 0 1 1 0 1 0 1 0; %2
0 1 0 0 1 0 0 0 0 0 1 0]; %6

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

ヘルプ センター および File ExchangeCreating and Concatenating Matrices 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!

Translated by