How to find a value in a list?
42 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
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
6 Kommentare
Matt Fig
am 15 Aug. 2012
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.
Antworten (4)
Ilham Hardy
am 15 Aug. 2012
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
0 Kommentare
Andrei Bobrov
am 15 Aug. 2012
Bearbeitet: Andrei Bobrov
am 16 Aug. 2012
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
0 Kommentare
Jordan
am 15 Aug. 2012
4 Kommentare
Andrei Bobrov
am 16 Aug. 2012
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
Matt Fig
am 16 Aug. 2012
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
0 Kommentare
Siehe auch
Kategorien
Mehr zu Spatial Search finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!