How to compare rows of an array with other rows?
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lucksBi
el 19 de En. de 2018
Comentada: lucksBi
el 24 de En. de 2018
Hey.. How to compare rows of an array with rows defined in x?
array={[1,2,3,4,6,13,18,9];[13];[12,1,2,5];[3,4];[1,5,6]}
x=[1;3]
comparison will be based on x. e.g. first element in x is 1 so it will compare 1st row of array with all other rows. 2nd element is 3 so 3rd row will be compared will all other rows. Comparison will get intersection of elements like:
result{1,1}= {[13;[1,2];[3,4];[1,6]]}
result{2,1}= {[[1,2];0;0;[1,6];[1,5]]}
0 means no common element was found.
please help.
4 comentarios
Jan
el 19 de En. de 2018
Editada: Jan
el 19 de En. de 2018
The question is not clear. The elements of the cell array called "array" have one row only. Do you mean, that their first columns are compared? Or do you mean the 1st and 3rd element of the cell array?
Note that there is a missing ] in
result{1,1}= {[13;[1,2];[3,4];[1,6]]}
and this is not valid also:
result{2,1}= {[[1,2];0;0;[1,6];[1,5]]}
because you cannot concatenate vectors of different width vertically.
Jos (10584)
el 19 de En. de 2018
and you do realise that you only have to process the unique values in x?
Respuesta aceptada
Jan
el 19 de En. de 2018
Editada: Jan
el 19 de En. de 2018
The question is not clear, but I guess a code which produces the output:
array = {[1,2,3,4,6,13,18,9];[13];[12,1,2,5];[3,4];[1,5,6]}
x = [1;3];
Result = cell(1, numel(x));
n = numel(array);
for ix = 1:numel(x)
index = x(ix);
A = array{index};
B = cell(1, n - 1);
iB = 0;
for ia = 1:n
if ia ~= index
iB = iB + 1;
Inter = intersect(A, array{ia});
if isempty(Inter)
Inter = 0;
end
B{iB} = Inter;
end
end
Result{ix} = B;
end
It might help to save time, if you sort the arrays at first:
for k = 1:numel(array)
array{k} = sort(array{k});
end
4 comentarios
Más respuestas (1)
Birdman
el 19 de En. de 2018
Editada: Birdman
el 19 de En. de 2018
Another approach for your problem, which is shorter:
for i=1:numel(x)
temp=setdiff(1:size(array,1),x(i));
for j=1:numel(temp)
result{j,i}=intersect(array{x(i),1},array{temp(j),1});
end
end
2 comentarios
Jan
el 19 de En. de 2018
Editada: Jan
el 19 de En. de 2018
Yes, it is shorter. But it omits the pre-allocation of result and using 0 as output, if no intersections are found. It is slower, because array{x(i),1} is addressed in each iteration.
Another idea, which avoids setdiff, which is an overkill for a scalar input:
n = size(array, 1);
v = 1:n;
result = cell(n-1, numel(x));
result(:) = {0};
for i = 1:numel(x)
temp = v(v ~= i); % Short form of: setdiff(v, i)
ai = array{x(i)}; % Or maybe: ai = unique(array{x(i)})
for j = 1:n-1
r = intersect(ai, array{temp(j)});
if ~isempty(r)
result{j,i} = r;
end
end
end
This might be an efficient mix of our solutions.
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