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.

 Akzeptierte Antwort

Ive J
Ive J am 23 Dez. 2020

0 Stimmen

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 Kommentare

Rishi Balasubramanian
Rishi Balasubramanian am 23 Dez. 2020
That works perfectly. Thank you.
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
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 am 23 Dez. 2020
Wonderful, that works perfectly. Thanks man.
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

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