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vector index of consecutive gap (NaN) lengths?

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tom
tom 2012-6-27
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

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采纳的回答

Sean de Wolski
Sean de Wolski 2012-6-27
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 个评论
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Sean de Wolski
Sean de Wolski 2012-6-27
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!
tom
tom 2012-6-28
Thanks Sean, this works great, even with NaNs at edges. -Tom

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更多回答（3 个）

Ryan
Ryan 2012-6-27
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
Andrei Bobrov
Andrei Bobrov 2012-6-28
编辑：Andrei Bobrov 2012-6-30
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));
Thomas
Thomas 2012-6-27
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
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tom
tom 2012-6-29
This still doesn't work if A(1)=NaN; Thanks for continuing to work on it!
Thomas
Thomas 2012-6-29
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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