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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Réponse acceptée
Andrei Bobrov
le 18 Oct 2016
Modifié(e) : Andrei Bobrov
le 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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Plus de réponses (4)
KSSV
le 18 Oct 2016
Modifié(e) : KSSV
le 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 ;
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Gareth Lee
le 18 Oct 2016
Modifié(e) : Gareth Lee
le 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 commentaire
Jan
le 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
le 18 Oct 2016
Modifié(e) : Walter Roberson
le 18 Oct 2016
Odd how two different people have such similar tasks for the same data...
0 commentaires
Jan
le 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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