Replace rows with NaN only if there are more than two continous zero values in the same column
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I am a begginer in Matlab. I need to replace with NaN all rows which contain more than two continous zero values in the same column (second column). Look at the next example:
Input=
124.2 0 7.2 -4.8
131.1 1.8 1.4 -1.2
131.9 0 1 3
123 0 4 5
987 0 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 2 7
4454 0 3 4
3 0 0 2
434 0 2 0
Output=
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
Thanks for your help fellows!
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Accepted Answer
Andrei Bobrov
on 18 Oct 2016
Edited: Andrei Bobrov
on 18 Oct 2016
t = Input(:,2) == 0;
[ii,c] = bwlabel(t);
x = accumarray(ii+1,1);
x = x(2:end);
z = 1:c;
Input(ismember(ii,z(x > 2)),2) = nan;
for all Input
t = Input == 0;
z = cumsum(diff([zeros(1,size(Input,2));t]) == 1);
ii = bsxfun(@plus,z,cumsum([0,z(end,1:end-1)])).*t;
b = accumarray(ii(:)+1,1);
Output = Input;
Output(ismember(ii,find(b(2:end)>2 ))) = nan;
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More Answers (4)
KSSV
on 18 Oct 2016
Edited: KSSV
on 18 Oct 2016
A = [124.2 0 7.2 -4.8
131.1 1.8 1.4 -1.2
131.9 0 1 3
123 0 4 5
987 0 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 2 7
4454 0 3 4
3 0 0 2
434 0 2 0];
c2 = A(:,2) ; % pick second column
temp = diff(c2 == 0 ) ;
bs = find( temp == 1 ) + 1 ; % start of zero
be = find( temp == -1 ) ; % end of zero
be = [be(2:end) ; length(c2)] ;
%%replace with Nan's
for i = 1:length(bs)
c2(bs(i):be(i)) = NaN ;
end
A(:,2) = c2 ;
0 Comments
Gareth Lee
on 18 Oct 2016
Edited: Gareth Lee
on 18 Oct 2016
First, you can find the column with continous zeros(more than two), then find the index for replacement with nan. for example:
a =(A==0);
a = double(a); % you can loop to find the number of 1 (more than 2)
m = a(:,2);
[cc,dd] = regexp(sprintf('%d', m), '1{3,}', 'start', 'end'); % find the index of continous ones
Next, replace the zeros between cc and dd with nan.
1 Comment
Jan
on 18 Oct 2016
The conversion between DOUBLEs and CHARs is time consuming and prone to errors and rounding for floating point values. Better stay at the same type.
Walter Roberson
on 18 Oct 2016
Edited: Walter Roberson
on 18 Oct 2016
Odd how two different people have such similar tasks for the same data...
0 Comments
Jan
on 18 Oct 2016
[B, N] = RunLength(Data(:, 2));
B(B == 0 & N >= 3) = NaN;
Data(:, 2) = reshape(RunLength(B, N), [], 1);
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