Identify first and last indicies of numbers in an array containing NaN's

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Nev Pires 2020-8-6

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评论： Stephen23 2020-8-7

I have an array containing numbers and NaN's. The numbers and NaN's are interspersed with each other:

M = [NaN, 1.3, 4, 5.2, 3, NaN, NaN, NaN, 4.6, 2, 6.2, 3, 2, NaN, 7, 3.2, 5, NaN, NaN, NaN, 12.1 ,6.8];

I want to extract the first and last indicie of each section containing real numbers. In this example it would be

startIndex = [2, 9, 15, 21];
endIndex = [5, 13, 17, 22];

I know I can find the index of each non NaN value by:

index = find(~isnan(M));
index = [2, 3, 4, 5, 9, 10, 11, 12, 13, 15, 16, 17, 21, 22];

The first derivative would help me identify the break points in the index array:

idx = find(diff(index));
idx = [1, 1, 1, 4, 1, 1, 1, 1, 2, 1, 1, 4, 1];

I'm now stuck on how to use that to get the data that I want.

Thanks in advance!

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Answer 1

Stephen23 2020-8-6

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Your original idea of using diff is exactly the simple and efficient solution that experienced MATLAB users would use:

>> M = [NaN,1.3,4,5.2,3,NaN,NaN,NaN,4.6,2,6.2,3,2,NaN,7,3.2,5,NaN,NaN,NaN,12.1,6.8];
>> D = diff([true;isnan(M(:));true]);
>> B = find(D<0)
B =
    2
    9
   15
   21
>> E = find(D>0)-1
E =
    5
   13
   17
   22

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Nev Pires 2020-8-6

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This definitely helps a lot. Could you please explain how this line works?:

D = diff([true;isnan(M(:));true]);

Thank you!

Stephen23 2020-8-7

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"Could you please explain how this line works?"

Break it down into its constituent pieces:

D = diff([true;isnan(M(:));true]);
                     M(:)          % convert M to column vector (optional)
               isnan(    )         % logical vector of NaN locations
         [true;           ;true]   % concatenate TRUE to ensure edge-cases are detected
    diff(                       )  % differences between adjacent logical values

A simple example should help too:

>> M = [1;2;NaN;3;NaN];
>> D = diff([true;isnan(M(:));true])
D =
    -1
     0
     1
    -1
     1
     0

Note that the start of the first number sequence would not be detected without the concatenated true values.

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Answer 2

Sudheer Bhimireddy 2020-8-6

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Try this:

% Instead of diff, just loop through the indices to see the order and group them
your_nan_indices = [2, 3, 4, 5, 9, 10, 11, 12, 13, 15, 16, 17, 21, 22];
nInd = numel(your_nan_indices);
start_ind(1) = your_nan_indices(1);
j = 2; k = 1; 
for i = 2:nInd-1
    if your_nan_indices(i) ~= your_nan_indices(i+1)-1
        start_ind(j) = your_nan_indices(i+1); j = j + 1;
        end_ind(k)   = your_nan_indices(i);   k = k + 1;
    else
        end_ind(k)   = your_nan_indices(i+1);
    end
end

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Sudheer Bhimireddy 2020-8-6

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I ran the code on my system and it gave me the answer you mentioned in your question. i.e., start_ind(1) = 2.

Writing two expressions seperated by a semi-colon is same as writing them in two lines. I usually do this when the second expression is a trivial one like in this example.

% Matlab takes ; as end of current expression
a = 1;
b = 2; 
% the above two lines are same as
a = 1; b = 2;
% removing the semi-colon in between will give you error
a = 1 b = 2; % <- this is not an acceptable syntax
% This is perfectly fine
a = 1 
b = 2  

numel vs length:

numel gives the number of elements, while length gets the maximum of the array size. If you have a 1D array, then both give the same answer. It is reported that numel is faster than length for a 1D array of larger size.

Stephen23 2020-8-6

编辑：Stephen23 2020-8-6

"is there any advantage of using numel vs length?"

length changes the dimension that it measures depending on the size of the provided array, which can lead to unexpected bugs. The robust alternatives are to use numel for the total number of elements in an array, and size for the size of any one particular dimension.

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