How to compare rows of an array with other rows?
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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 个评论
Jan
2018-1-19
编辑:Jan
2018-1-19
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.
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Jan
2018-1-19
编辑:Jan
2018-1-19
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
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更多回答(1 个)
Birdman
2018-1-19
编辑:Birdman
2018-1-19
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 个评论
Jan
2018-1-19
编辑:Jan
2018-1-19
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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