vector index of consecutive gap (NaN) lengths?

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

[Ryan]

do you have the image processing toolbox?

[tom]

yes.

Answer 1

[Sean de Wolski]

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

[Ryan]

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]

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]

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

Answer 2

[Ryan]

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

Answer 3

[Andrei Bobrov]

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));

Answer 4

[Thomas]

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

[tom]

Thanks. This works nicely unless NaNs are on the ends, which is something it needs to be able to handle. Ideas?

[Thomas]

Based on your request

A = [2 4 NaN 7 9 NaN NaN NaN 32 NaN NaN 8 NaN];
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
out=stop-start;

for ii=1:length(out)
A(start(ii):(stop(ii)-1))=out(ii);
end

A

[Ryan]

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]

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

[Thomas]

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