find length NaN segments

I have large vectors, containing quite a lot of NaN samples. I want to know the length of each array of NaNs within this vector, even when this equals 1. So I want to have the lenght of all NaN segments.

Thanks in advance!

[Stephen23]

Note to future readers: the accepted answer fails for many simple cases:

Only NaN

>> A = [NaN,NaN,NaN,NaN];
>> index=find(isnan(A));
>> idx=find(diff(index)~=1);
>> R=[idx(1) diff(idx) numel(index)-idx(end)]
Attempted to access idx(1); index out of bounds because numel(idx)=0.

Only Numbers

>> A = [1,2,3,4];
>> index=find(isnan(A));
>> idx=find(diff(index)~=1);
>> R=[idx(1) diff(idx) numel(index)-idx(end)]
Attempted to access idx(1); index out of bounds because numel(idx)=0.

Empty Vector

>> A = [];
>> index=find(isnan(A));
>> idx=find(diff(index)~=1);
>> R=[idx(1) diff(idx) numel(index)-idx(end)]
Attempted to access idx(1); index out of bounds because numel(idx)=0.

Leading Numbers

>> A = [1,NaN,NaN,NaN];
>> index=find(isnan(A));
>> idx=find(diff(index)~=1);
>> R=[idx(1) diff(idx) numel(index)-idx(end)]
Attempted to access idx(1); index out of bounds because numel(idx)=0.

Trailing Numbers

>> A = [NaN,NaN,NaN,1];
>> index=find(isnan(A));
>> idx=find(diff(index)~=1);
>> R=[idx(1) diff(idx) numel(index)-idx(end)]
Attempted to access idx(1); index out of bounds because numel(idx)=0. 

Answer 1

[madhan ravi]

https://www.mathworks.com/matlabcentral/answers/444595-count-the-occurence-of-a-number-in-between-other-numbers - replace the first line with isnan().

[madhan ravi]

A = [NaN NaN NaN 1 2 3 4 NaN 3 44 NaN] ; % example data
index=find(~isnan(A)); % will give you the length of numbers inbetween nan , if you remove ~ it will give the length of nans inbetween numbers
idx=find(diff(index)~=1);
R=[idx(1) diff(idx) numel(index)-idx(end)]

[Lieke Numan]

Thanks a lot!

[madhan ravi]

Jos did you read the comment first? Just remove ~ infront of isnan() please examine before flagging.

[Jos (10584)]

My apologies, I will remove the flag and comment.

[Stephen23]

Note that this answer fails in many simple cases, e.g.:

All NaN

>> A = [NaN,NaN,NaN,NaN];
>> index=find(isnan(A));
>> idx=find(diff(index)~=1);
>> R=[idx(1) diff(idx) numel(index)-idx(end)]
Attempted to access idx(1); index out of bounds because numel(idx)=0.

No NaN

>> A = [1,2,3,4];
>> index=find(isnan(A));
>> idx=find(diff(index)~=1);
>> R=[idx(1) diff(idx) numel(index)-idx(end)]
Attempted to access idx(1); index out of bounds because numel(idx)=0.

Empty Vector

>> A = [];
>> index=find(isnan(A));
>> idx=find(diff(index)~=1);
>> R=[idx(1) diff(idx) numel(index)-idx(end)]
Attempted to access idx(1); index out of bounds because numel(idx)=0.

Leading Numbers

>> A = [1,NaN,NaN,NaN];
>> index=find(isnan(A));
>> idx=find(diff(index)~=1);
>> R=[idx(1) diff(idx) numel(index)-idx(end)]
Attempted to access idx(1); index out of bounds because numel(idx)=0.

Trailing Numbers

>> A = [NaN,NaN,NaN,1];
>> index=find(isnan(A));
>> idx=find(diff(index)~=1);
>> R=[idx(1) diff(idx) numel(index)-idx(end)]
Attempted to access idx(1); index out of bounds because numel(idx)=0.

Its use of indexing means that any special cases will need to be handled separately.

See my answer for simpler code that works for all of these cases.

[madhan ravi]

Just an if condition should suffice that need.

[Stephen23]

"Just an if condition should suffice that need. "

Please show exactly how one simple IF statement would fix those cases.

[madhan ravi]

Turns out not possible :D.

[madhan ravi]

@Lieke there are so many limitations to this answer , you have accepted the wrong answer ;-)

Answer 2

[Stephen23]

This is simpler and actually works for all horizontal vectors (unlike the accepted answer):

>> A = [NaN NaN NaN 1 2 3 4 NaN 3 44 NaN];
>> D = diff([false,isnan(A),false]);
>> find(D<0)-find(D>0)
ans =
     3     1     1

For a slightly faster version you can call find once:

>> F = find(diff([false,isnan(A),false]));
>> F(2:2:end)-F(1:2:end)
ans =
     3    1     1

EDIT: uses Jan's logical vector suggestion.

[madhan ravi]

Even the accepted answer works as mentioned exactly in the comment.

[madhan ravi]

I accept it's simpler than the accepted answer but I don't think it's faster:

>> A = [NaN NaN NaN 1 2 3 4 NaN 3 44 NaN];
tic
for k=1:1e5
D = diff([0,isnan(A),0]);
find(D<0)-find(D>0);
end
toc
tic
for k=1:1e5
index=find(isnan(A)); 
idx=find(diff(index)~=1);
R=[idx(1) diff(idx) numel(index)-idx(end)];
end
toc
Elapsed time is 1.533159 seconds.
Elapsed time is 0.308740 seconds.
>> 

[Jos (10584)]

A diff-find-diff solution :-D

D = diff(find(diff([0,isnan(A),0])))
R = D(1:2:end)

[Stephen23]

"Even the accepted answer works as mentioned exactly in the comment."

Except when it doesn't work:

https://www.mathworks.com/matlabcentral/answers/446218-find-length-nan-segments#comment_673961

[madhan ravi]

"faster too"

how?https://www.mathworks.com/matlabcentral/answers/446218-find-length-nan-segments#comment_673784

[Stephen23]

Faster, although not as fast as Jan's STRFIND solution:

https://www.mathworks.com/matlabcentral/answers/444595-count-the-occurence-of-a-number-in-between-other-numbers#comment_673969

[madhan ravi]

Yes but in this thread :D.

[Stephen23]

"Yes but in this thread "

The main difference is swapping == for isnan (which everyone used). I doubt that it makes much difference, but you are welcome to do some tests and post the results here.

Answer 3

[Jan]

And again: With FEX: RunLength :

[B, N] = RunLength(A);
Result = N(isnan(B));

Or:

y      = [false, isnan(A), false];
Result = strfind(y, [true, false]) - strfind(y, [false,true])

For this test vector:

A = ones(1, 1e5);
A(randperm(1e5, 5e4)) = NaN;

the method posted by Jos (link) is about 10% faster:

D = diff(find(diff([false, isnan(A), false])));
R = D(1:2:end);

Answer 4

[KSSV]

Read about isnan and nnz

[madhan ravi]

nnz will give you the total number of elements not the length of each array of nans

[KSSV]

A = [NaN NaN NaN 1 2 3 4 NaN] ;
nnz(isnan(A))

There are four NaN's.

[madhan ravi]

OP wants it like 3 nans and 1 nan not the total number of nans or the length of each numbers inbetween nans.

[KSSV]

he can use diff anddo that..