vector index of consecutive gap (NaN) lengths?
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Hi
For a vector A with random, sometime consecutive gaps of NaN, I want to develop a vector B of same length A that will indicate the length of local consecutive gaps for every value in A. B would have zeros for non-NaN locations in A.
so for
A = [2 4 NaN 7 9 NaN NaN NaN 32 NaN NaN 8];
I'd get
B = [0 0 1 0 0 3 3 3 0 2 2 0];
Ideas? Speed is always a virtue.
Thanks!
Tom
2 commentaires
Réponse acceptée
Sean de Wolski
le 27 Juin 2012
And if you like Ryan's idea but don't like bwlabel because it's evil:
A = [2 4 NaN 7 9 NaN NaN NaN 32 NaN NaN 8];
CC= bwconncomp(isnan(A));
n = cellfun('prodofsize',CC.PixelIdxList);
b = zeros(size(A));
for ii = 1:CC.NumObjects
b(CC.PixelIdxList{ii}) = n(ii);
end
3 commentaires
Sean de Wolski
le 27 Juin 2012
BWCONNCOMP makes BWLABEL irrelevant for everything except LABEL2RGB! For that you have LABELMATRIX to convert from the output of BWCONNCOMP.
Anyway, yes, CC.PixelIdxList contains the indices you need to do most matrix manipulations easily and can be fed directly to REGIONPROPS, all while being faster!
Plus de réponses (3)
Ryan
le 27 Juin 2012
Thomas' answer is faster, but here is my go:
A = [2 4 NaN 7 9 NaN NaN NaN 32 NaN NaN 8];
idx = isnan(A);
[B n]= bwlabel(idx);
C = B;
prop = regionprops(idx,'Area');
area = cat(1,prop.Area);
for ii = 1:n
B(C == ii) = area(ii);
end
B
0 commentaires
Andrei Bobrov
le 28 Juin 2012
Modifié(e) : Andrei Bobrov
le 30 Juin 2012
A = [2 4 NaN 7 9 NaN NaN NaN 32 NaN NaN 8];
a = isnan(A);
t1 = find([true, diff(a)~=0]);
N = diff(t1);
out = zeros(size(A));
V = regionprops(a,'PixelIdxList');
out(cat(1,V.PixelIdxList)) = cell2mat(arrayfun(@(x)x*ones(x,1),N(a(t1))','un',0));
OR
A = [2 4 NaN 7 9 NaN NaN NaN 32 NaN NaN 8];
a = isnan(A);
n1 = regionprops(a,'Area');
out = a + 0;
out(a) = cell2mat(arrayfun(@(x)x*ones(1,x),[n1.Area],'un',0));
ADD variant
a = isnan(A);
t = [true,diff(a)~=0];
k = diff(find([t,true]));
k2 = k.*a(t);
out = k2(cumsum(t));
0 commentaires
Thomas
le 27 Juin 2012
A very crude way.. pretty sure can be done better...
A = [2 4 NaN 7 9 NaN NaN NaN 32 NaN NaN 8];
A(~isnan(A))=0;
A(isnan(A))=1;
c=diff(A);
start=find(c==1)+1;
stop=find(c==-1)+1;
out=stop-start;
for ii=1:length(out)
A(start(ii):(stop(ii)-1))=out(ii);
end
A
5 commentaires
Thomas
le 29 Juin 2012
another iteration here NaN can be first,last or anywhere in the middle.. 'hopefully'
A = [NaN 4 NaN 7 9 NaN NaN NaN 32 NaN NaN 8];
A(~isnan(A))=0;
A(isnan(A))=1;
c=diff(A);
start=find(c==1)+1;
stop=find(c==-1)+1;
if length(stop)<length(start)
stop=[stop start(end)+1];
end
if length(start)<length(stop)
start=[start(1)-1 start];
end
out=stop-start;
for ii=1:length(out)
A(start(ii):(stop(ii)-1))=out(ii);
end
A
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