find length NaN segments

Question

Lieke Numan 2019-2-21

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https://ww2.mathworks.cn/matlabcentral/answers/446218-find-length-nan-segments

评论： madhan ravi 2019-2-22

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!

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Stephen23 2019-2-22

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. 

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Answer 1

madhan ravi 2019-2-21

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/446218-find-length-nan-segments#answer_362057

编辑：madhan ravi 2019-2-21

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

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Stephen23 2019-2-22

编辑：Stephen23 2019-2-22

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 2019-2-22

Turns out not possible :D.

madhan ravi 2019-2-22

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

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Answer 2

Stephen23 2019-2-21

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/446218-find-length-nan-segments#answer_362082

编辑：Stephen23 2019-2-22

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.

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madhan ravi 2019-2-22

Yes but in this thread :D.

Stephen23 2019-2-22

"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.

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Answer 3

Jan 2019-2-22

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/446218-find-length-nan-segments#answer_362216

编辑：Jan 2019-2-22

在 MATLAB Online 中打开

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);

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Answer 4

KSSV 2019-2-21

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/446218-find-length-nan-segments#answer_362061

Read about isnan and nnz

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madhan ravi 2019-2-21

编辑：madhan ravi 2019-2-21

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

KSSV 2019-2-21

he can use diff anddo that..

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