Replace with the mean if there are at least six continous NaN in the same column

Hello fellows. I am looking for help. In the second column of a matrix, if there are at least six continous NaN, I need to replace it by the mean, calculated from the previous and the last numerical data of the NaN interval. Please look at the next example:

Input:

124.2  0  7.2  -4.8
131.1  1.8  1.4  -1.2
131.9  NaN  1  3
123  NaN  4  5
987  NaN  0  9
2323  3  3  3
2323  0  3  2
3  0  6  0
221  3.1  4  5.6
57  1  231  122
987  NaN  2  7
4454  NaN  3  4
3  NaN  0  2
434  NaN  2  0
123  NaN  4  5
987  NaN  0  9
3  0  6  0
987  NaN  2  7
4454  NaN  3  4
3  NaN  0  2
434  NaN  2  0
123  NaN  4  5
987  NaN  0  9
987  NaN  0  9
124.2  0  7.2  -4.8

Output:

124.2  0  7.2  -4.8
131.1  1.8  1.4  -1.2
131.9  2.4  1  3
123  2.4  4  5
987  2.4  0  9
2323  3  3  3
2323  0  3  2
3  0  6  0
221  3.1  4  5.6
57  1  231  122
987  0.5  2  7
4454  0.5  3  4
3  0.5  0  2
434  0.5  2  0
123  0.5  4  5
987  0.5  0  9
3  0  6  0
987  NaN  2  7
4454  NaN  3  4
3  NaN  0  2
434  NaN  2  0
123  NaN  4  5
987  NaN  0  9
987  NaN  0  9
124.2  0  7.2  -4.8

Thanks in advance for your help!

[Jan]

And again the same data as in Answers: 307774 and Answers: 307780

Answer 1

[KSSV]

Let I be your Input matrix.

c2 = I(:,2) ;   % pick second column 
temp = diff(isnan(c2 )) ;
bs = find( temp == 1 ) + 1 ;   % start of NaN
be   = find( temp == -1 ) ;    % end of NaN
%%replace with mean 
for i = 1:length(be)
    m = (c2(bs(i)-1)+c2(be(i)+1))/2 ;  % get mean 
    I(bs(i):be(i),2) = m ;
end

Answer 2

[Walter Roberson]

tf = [false, isnan(YourMatrix(:,2)) .', false];
start_of_block = strfind(tf, [false, true(1,6)]);
end_of_block = strfind(tf, [true(1,6), false]) + 4;

Now for each pair start_of_block(K), end_of_block(K), there is a run of at least 6 NaN occupying YourMatrix(start_of_block(K):end_of_block(K),2) .

meanvals = (YourMatrix(start_of_block-1, 2) + YourMatrix(end_of_block+1, 2));

for K = 1 : length(start_of_block)
  YourMatrix(start_of_block(K):end_of_block(K), 2) = meanvals(K);
end

Answer 3

[Andrei Bobrov]

t = isnan(A(:,2));
tt = diff(t);
n = mean([A(tt == 1,2),A([false;tt == -1],2)],2);
ii = t.*cumsum([false;tt == 1]);
n(accumarray(ii(t),1) > 6) = nan;
A(t,2) = n(ii(t));

[Walter Roberson]

This appears to replace all groups of NaNs, not just groups that are at least 6 long.

[Andrei Bobrov]

Opp! Thank you Walter!

[Andrei Bobrov]

Corrected.