How to find position of something
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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 :)
采纳的回答
Roger Stafford
2013-12-7
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.
更多回答（2 个）
Mohammad Monfared
2013-12-7
Suppose you have a column or row vector named 'A',
find(isnan(A))
4 个评论
Martina
2013-12-7
Mohammad Monfared
2013-12-7
编辑：Mohammad Monfared
2013-12-7
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
2013-12-7
Mohammad Monfared
2013-12-7
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.
Image Analyst
2013-12-7
编辑：Image Analyst
2013-12-7
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.
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