finding nearest vlaue
5 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I have an matrix
A=[1 2 8
7 9 6
10 14 89]
now i want to find the nearest value for example,lets ur assume 2,the nearest values are 1,7,9,8,6
for 1 it is 2,7,9
please tell how to find the nearest value
0 commentaires
Réponse acceptée
Jonathan Sullivan
le 6 Avr 2012
Kash,
You might want to try this. It'll allow for you to do the neighbors of more than 1 number at a time. Just change the variable val to be what you want (either a scalar, or a vector).
A = [1 2 8
7 9 6
10 14 89];
val = [1 2];
[y x] = find(ismember(A,val));
ys = unique(bsxfun(@plus,y,-1:1));
xs = unique(bsxfun(@plus,x,-1:1));
ys(ys <= 0 | ys > size(A,1)) = [];
xs(xs <= 0 | xs > size(A,1)) = [];
A(ys,xs)
0 commentaires
Plus de réponses (1)
Andrei Bobrov
le 6 Avr 2012
A=[1 2 8
7 9 6
10 14 89]
a = zeros(3);
out = cell(numel(A),2);
for i1 = 1: numel(A)
b = a;
b(i1) = 1;
out(i1,:) = {A(i1),A(bwdist(b,'chessboard') == 1)};
end
OR
B = nan(size(A)+2);
B(2:end-1,2:end-1) = A;
m1 = bsxfun(@plus,(1:3).',(0:2)*size(B,1));
i1 = bsxfun(@plus,m1(:),m1(:).'-1);
B1 = B(i1);
ic = ceil(size(i1,1)/2);
out = [B1(ic,:)' sort(B1([1:ic-1,ic+1:end],:).',2)];
With show?
B = nan(size(A)+2);
B(2:end-1,2:end-1) = A;
m1 = bsxfun(@plus,(1:3).',(0:2)*size(B,1));
i1 = bsxfun(@plus,m1(:),m1(:).'-1);
B1 = B(i1);
ic = ceil(size(i1,1)/2);
B2 = arrayfun(@(ii)reshape(B1(:,ii),size(A,1),[]),(1:size(B1,2))','un',0);
t2 = cellfun(@(x)~isnan(x),B2,'un',0);
out = [num2cell(B1(:,ic)),cellfun(@(x,y)x(any(y,2),any(y)),B2,t2,'un',0)]
3 commentaires
Voir également
Catégories
En savoir plus sur Matrix Indexing dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!