vector index of consecutive gap (NaN) lengths?

Asked by tom on 27 Jun 2012
Latest activity Commented on by Thomas on 29 Jun 2012

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 Comments

Ryan on 27 Jun 2012

do you have the image processing toolbox?

tom on 27 Jun 2012

yes.

tom

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4 Answers

Answer by Sean de Wolski on 27 Jun 2012
Accepted answer

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 Comments

Ryan on 27 Jun 2012

And without the evil function it is faster! I have never used bwconncomp (mostly because I am used to matrices and vectors and not cells), but this may be the start of it.

Sean de Wolski on 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!

tom on 28 Jun 2012

Thanks Sean, this works great, even with NaNs at edges. -Tom

Sean de Wolski
Answer by Ryan on 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 Comments

Ryan
Answer by Thomas on 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 Comments

Show 2 older comments
Ryan on 28 Jun 2012

If you're going for speed here is how they stack up:

Ryan's Answer: 0.014344 sec
Andrei's Answer: 0.05369 sec
Sean's Answer: 0.003353 sec
Thomas' Answer: 0.000045 sec
tom on 29 Jun 2012

This still doesn't work if A(1)=NaN; Thanks for continuing to work on it!

Thomas on 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
Thomas
Answer by Andrei Bobrov on 28 Jun 2012
Edited by Andrei Bobrov on 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 Comments

Andrei Bobrov

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