Find the rth "0" in specific columns in a matrix
1 visualización (últimos 30 días)
Mostrar comentarios más antiguos
monkey_matlab
el 27 de Oct. de 2017
Editada: Andrei Bobrov
el 27 de Oct. de 2017
Hello,
In the code below, thanks to the solution by Cedric Wannaz posted here , I am able to efficiently find the first and last "0" in an n x m matrix.
n = 10;
m = 20;
M = randi([0 1], n,m);
%%Start and Goal Points
[r1,c1] = ind2sub( size( M ), find( M(:) == 0, 1, 'first' )); % Start Point
[r2,c2] = ind2sub( size( M ), find( M(:) == 0, 1, 'last' )); % Goal Point
How do I go about finding the 3rd or 4th or the rth "0" in any column. For example, how to find the index of the 3rd "0" in the 6th column?
Thanks.
0 comentarios
Respuesta aceptada
Andrei Bobrov
el 27 de Oct. de 2017
Editada: Andrei Bobrov
el 27 de Oct. de 2017
cols = [1 3 4 6 7]; % selected columns.
num_zeros = [3,4];% The ordinal number "0" in each selected column.
M_nozeros = cumsum(~M).*~M;
[~,ii] = ismember(M_nozeros(:,cols),num_zeros);
[k,jj] = ndgrid(1:size(M,1),cols);
idx = [k(:),jj(:)];
out = accumarray(ii(:) + 1,1:numel(ii),[],@(x){idx(x,:)});
out = out(2:end);
out - cell array, here each cell corresponds to the ordinal number "0" ( num_zeros), each cell contains a double matrix: rows are the ordinal number of the row containing zero, the columns correspond to cols.
Use:
>> cols = [1 3 4 6 7]; % selected columns.
num_zeros = [3,4];% The ordinal number "0" in each selected column.
M = rand(15,10) < .25
M =
15×10 logical array
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 1 0 0 1
1 0 0 1 1 0 0 0 0 1
0 0 0 1 0 0 1 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 1 1 0 0 0 0 0
0 1 0 0 0 1 0 0 0 0
0 1 0 0 0 0 0 0 1 1
1 0 1 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 1 0 0 0
0 0 0 1 1 1 1 1 0 1
0 0 0 0 0 0 1 0 0 0
0 0 1 0 0 1 1 0 0 0
1 0 0 0 0 0 1 0 0 0
>>
>> M_nozeros = cumsum(~M).*~M;
[~,ii] = ismember(M_nozeros(:,cols),num_zeros);
[k,jj] = ndgrid(1:size(M,1),cols);
idx = [k(:),jj(:)];
out = accumarray(ii(:) + 1,1:numel(ii),[],@(x){idx(x,:)});
out = out(2:end);
>>
>> out{:}
ans =
4 1
3 3
3 6
5 4
5 7
ans =
4 6
5 1
6 7
4 3
7 4
>>
0 comentarios
Más respuestas (1)
Image Analyst
el 27 de Oct. de 2017
In general:
rows = 10;
columns = 16;
M = randi([0 1], rows,columns)
r = 2 % Whatever...
% Start and Goal Points
linearIndexes = find( M(:) == 0, r, 'first' )
[r1,c1] = ind2sub( size( M ), linearIndexes(end)) % Start Point
linearIndexes = find( M(:) == 0, r, 'last' )
[r2,c2] = ind2sub( size( M ), linearIndexes(1)) % Goal Point
0 comentarios
Ver también
Productos
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!