How to find position of something

Hi, I am having a problem with part of my homework. I have a column with numbers and NaN-s. If I have 5 or more NaN-s in a row I need to know place/position where NaN start (j) and end (i) ( in all column). For example 7 6 5 NaN NaN NaN NaN NaN NaN 9 7

j=4 i=9 Thanks you very much :)

Answer 1

[Roger Stafford]

You can use the 'find' and 'diff' functions as well as 'isnan' to accomplish that. Let A be your column vector of elements with possible NaNs among them.

 m = 5; % The minimum length of consecutive NaNs to record
 f = find(diff([false;isnan(A);false]));
 B = [f(1:2:end-1),f(2:2:end)-1]; % Get all start and end indices for NaN consecutive sequences
 B = B(B(:,2)-B(:,1)+1>=m,:); % Eliminate all with fewer than m NaNs

The first column of array B contains the start indices and the second column the end indices for consecutive sequences of at least five NaNs.

Answer 2

[Mohammad Monfared]

Suppose you have a column or row vector named 'A',

 find(isnan(A))

[Martina]

But this gives all NaN-s position. I need positions only when I have 5 or more NaN-s. And only position of first and last NaN because I have vector with 60000 numbers. Thank you

[Mohammad Monfared]

If there are just one group of NaNs:

 j=find(isnan(A),1,'first')
 i=find(isnan(A),1,'last')

Is that your case?

[Martina]

But I have lot of groups :(

[Mohammad Monfared]

It's not a straightforward simple line job, to me. One trick is to use the diff(isnan(A)) to find the boundaries of NaN groups and one more 'diff' to get the length of each group. Then a loop might be used afterward. Think about it (since it is a homework ;) ) a bit. If no success, I'll give the answer.

Answer 3

[Image Analyst]

Really easy if you have the Image Processing Toolbox. Do you have that?

measurements = regionprops(isnan(vector), 'PixelIdx');
% just 2 or 3 lines more and you've got it.

By the way, you shoudl have tagged it as homework - I did this for you this time.

[Martina]

I don't have that toolbox.