- Concatenate all the vectors into one column vector.
- Sort it.
- Find indexes where the difference is smaller than 3 (some threshold).
- Find the corresponding indexes in the original vector.
Approximate matching of numbers across 20 Matrices
3 views (last 30 days)
Show older comments
I have 20 matrices which house multiple columns of data. The first column is the datenum. I need to essentially look through all 20 matrices to find approximate datenum matches according to a threshold (threshold probably = 3 or 4). The threshold is needed due to a spatial lag between locations (each location is a different matrix with different dates). Example:
A=(712549, 712658, 715188, 715935, 716613, 716948, 717319, 717764, 718091, 718484),
B = (714401, 716219, 717321, 717716, 718506, 718802, 722427, 722853, 723546, 723554)
and C = (718802, 722427, 722853, 723546, 723554, 724633, 726821, 731624, 731958, 733730).
If a datenum in A is within 3 of a datenum in B and/or C, I want to identify that datenum as well as tag which matrices were matches and the corresponding datenum. An example output for one approx. match between matrix B and C would be: 723554 | B | 723555 | C (with the | symbol meaning separated by columns and the letter belonging to the datenum on its left). Any help would be great! Thanks.
0 Comments
Accepted Answer
michio
on 28 Sep 2016
If all the 20 datenum vectors has the same length, the following (though not straight forward) could work.
A=[712549, 712658, 715188, 715935, 716613, 716948, 717319, 717764, 718091, 718484];
B =[714401, 716219, 717321, 717716, 718506, 718802, 722427, 722853, 723546, 723554];
C =[718802, 722427, 722853, 723546, 723554, 724633, 726821, 731624, 731958, 733730];
All = [A', B', C'];
% 1: Concatenate all the vectors into one column vector.
Alldata = All(:);
% 2: Sort it. (index refers to the original location in Alldata)
[sorted,index] = sort(Alldata);
% 3: Find indexes where the difference is smaller than 3 (some threshold).
idx_close = find(diff(sorted) < 3);
% 4: Find the corresponding indexes in the original vector.
a0 = Index(idx_close);
a1 = Index(idx_close+1);
[I0,J0] = ind2sub([10,3],a0); % 10 : length of the datenum vector
[I1,J1] = ind2sub([10,3],a1); % 3 : number of variables (A, B, C)
% Now we know that All(I0,J0) is close to All(I1,J1). Let's display it.
variable_name = ['A','B','C'];
for ii=1:length(I0)
str0 = [variable_name(J0(ii)), ' ', num2str(All(I0(ii),J0(ii)))];
str1 = [variable_name(J1(ii)), ' ', num2str(All(I1(ii),J1(ii)))];
disp(['match : ', str0, ' and ', str1]);
end
More Answers (1)
José-Luis
on 28 Sep 2016
Edited: José-Luis
on 28 Sep 2016
A = [712549, 712658, 715188, 715935, 716613, 716948, 717319, 717764, 718091, 718484];
B = [714401, 716219, 717321, 717716, 718506, 718802, 722427, 722853, 723546, 723554];
tol = 3;
[idxB, idxA] = ind2sub([numel(A),numel(B)],find(abs(bsxfun(@minus,A,B')) < tol));
3 Comments
José-Luis
on 28 Sep 2016
You could solve that with a loop. Give it a shot and I'll be happy to help if you get stuck.
There are other alternatives as well.
See Also
Categories
Find more on Creating and Concatenating Matrices in Help Center and File Exchange
Products
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!