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 Kommentare
Akzeptierte Antwort
Sean de Wolski
am 27 Jun. 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 Kommentare
Sean de Wolski
am 27 Jun. 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!
Weitere Antworten (3)
Ryan
am 27 Jun. 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 Kommentare
Andrei Bobrov
am 28 Jun. 2012
Bearbeitet: Andrei Bobrov
am 30 Jun. 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 Kommentare
Thomas
am 27 Jun. 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 Kommentare
Thomas
am 29 Jun. 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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