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How to find a value in a list?

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Jordan
Jordan 2012-8-15
I have a m-by-3 matrix of numbers (as below) and I would like to find a specific number in a row and then find the other numbers in that same row then write these in a seperate matrix. For example I want to look for the numbers connected to the number 4, in row 1 these will be 15 and 12 and in row 15 these will be 12 and 3. I then want a matrix called 4 to contain the values 15, 12 and 3. I do not want any repeats in this new matrix, hence 12 only appears once in the new matrix.
15 4 12
5 7 2
22 23 21
18 11 10
10 8 14
16 12 11
12 3 6
8 7 5
8 5 9
14 8 9
7 1 2
5 13 9
5 2 13
7 6 1
12 4 3
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Matt Fig
Matt Fig 2012-8-15
Questions:
Do you want all of the numbers in a row that has the given number in it, or just those connected to it (adjacent neighbors)? Take the number 15 in your given list. Do you want to include both the 4 and the 12 from row 1 or just the 4?
What if we had a row like this:
4 4 6
Would we count 4 as a neighbor to 4 (Andrei's code will not)?
Will the real array you need to work with have numbers only on [1,23] or what range?
Thanks.
Jordan
Jordan 2012-8-16
@Matt I want to get all the numbers in the same row as a given number, so in the example you gave, I would want 4 and 12. Also you will never have a row where a number would be repeated as the numbers represent points in space and the 3 columns give three points which make a triangle. I want the code to be compatible with any range of numbers as the range will change depending on the input, so [1:i]. Thanks.

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回答（4 个）

Ilham Hardy
Ilham Hardy 2012-8-15
Unchecked..
test_val = [15 4 12;5 7 2;22 23 21;18 11 10;10 8 14;16 12 11;12 3 6;8 7 5;8 5 9;14 8 9;7 1 2;5 13 9;5 2 13;7 6 1;12 4 3];
base_num = 4;
[r4 c4] = find(test_val==base_num);
matrx4 = unique(test_val(r4,:));
matrx4(matrx4==base_num)=[];
HTH, IH
Andrei Bobrov
Andrei Bobrov 2012-8-15
编辑：Andrei Bobrov 2012-8-16
z = [...
15 4 12
5 7 2
22 23 21
18 11 10
10 8 14
16 12 11
12 3 6
8 7 5
8 5 9
14 8 9
7 1 2
5 13 9
5 2 13
7 6 1
12 4 3];
out = setdiff(unique(z(any(z == 4,2),:)),4);
ADD ( EDIT )
c = unique(z);
n = numel(c);
out = cell(1,n);
for j1 = 1:n
out{j1} = unique(z(conv2(+(z==c(j1)),[1 0 1],'same')>0));
end
Jordan
Jordan 2012-8-15
Both of these solutions work for a single number, i.e 4, but is there a way which I can automate it so it creates a seperate new matrix for each value, 1-23. Could I possibly use a for loop?
Cheers
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Jordan
Jordan 2012-8-16
@Andrei, when I run this I get the error 'Cell contents assignment to a non-cell array object.'
Thanks
Andrei Bobrov
Andrei Bobrov 2012-8-16
Hi Jordan! Please use:
c = unique(z);
n = numel(c);
out = cell(1,n);
for j1 = 1:n
out{j1} = unique(z(conv2(+(z==c(j1)),[1 0 1],'same')>0));
end

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Matt Fig
Matt Fig 2012-8-16
For large data sets, this is the fastest I could come up with:
A = ceil(rand(19e4,3)*255); % Large data set on [1 255]
% Begin Code, store results in cell array T.
m = max(A(:));
T = cell(1,m);
for ii = 1:m
T{ii} = A(any(A==ii,2),:);
T{ii} = sort(T{ii}(:).');
T{ii} = T{ii}([~isempty(T{ii}) diff(T{ii})~=0]);
T{ii} = T{ii}(T{ii}~=ii);
end

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