Borrar filtros
Borrar filtros

Repalce zero element by the number before it ?

1 visualización (últimos 30 días)
Willim
Willim el 17 de En. de 2019
Editada: madhan ravi el 17 de En. de 2019
If I have array called N,
N = [2 0 7 0 9 10 0 0 11 0 0 ];
I want to create new array M that have [2 2 7 7 9 10 10 10 11 11 11]
Here is other example
N=[ 0 1 4 5 6 7 0 10 0 0 0]
the one I would create is
M=[0 1 4 5 6 7 7 10 10 10 10].
I would replace each zero in array N by the number before it except the first zero.
Thank you in advance.

Respuesta aceptada

Image Analyst
Image Analyst el 17 de En. de 2019
Faraj:
Did you try the super obvious solution: a very simple "for" loop:
N = [2 0 7 0 9 10 0 0 11 0 0 ];
for k = 1 : length(N)
if N(k) == 0
N(k) = N(k-1);
end
end
Presumably so. Were you looking for some more compact vectorized or function-based solution instead?

Más respuestas (2)

KSSV
KSSV el 17 de En. de 2019
N=[ 0 1 4 5 6 7 0 10 0 0 0] ;
M = N ;
idx = find(N==0) ; % get the indices of zeros
while ~isempty(idx)
idx(idx==1) = [] ; % remove the index if zero is at one
M(idx) = M(idx-1) ; % replace zeros with previous values
idx = find(M==0) ;
idx(idx==1) = [] ;
end
  2 comentarios
Willim
Willim el 17 de En. de 2019
This works but for my entire code , my code keeps running until I pause the run and quit debugging. then type M in my command window then I got the right M.
could you please the algorithm code ?
KSSV
KSSV el 17 de En. de 2019
Give the value of N. Image Analyist solution is good....go ahead with that.

Iniciar sesión para comentar.


madhan ravi
madhan ravi el 17 de En. de 2019
Editada: madhan ravi el 17 de En. de 2019
EDITED
N = [2 0 7 0 9 10 0 0 11 0 0 ];
M=N;
M(M==0)=NaN;
M=fillmissing(M,'nearest')
Gives:
M =
2 2 7 7 9 10 10 10 11 11 11
N=[ 0 1 4 5 6 7 0 10 0 0 0];
M=N;
M(M==0)=NaN;
M=fillmissing(M,'previous');
M(1)=0
Gives:
M =
0 1 4 5 6 7 7 10 10 10 10
  5 comentarios
Willim
Willim el 17 de En. de 2019
great it works but the thing is that the frist element could be 1 not 0 all the time
madhan ravi
madhan ravi el 17 de En. de 2019
Editada: madhan ravi el 17 de En. de 2019
What? Did you see the first example in my answer?

Iniciar sesión para comentar.

Categorías

Más información sobre Matrix Indexing en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by