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
采纳的回答
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
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!
更多回答(3 个)
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
0 个评论
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));
0 个评论
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
5 个评论
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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