Filter löschen
Filter löschen

Repalce zero element by the number before it ?

1 Ansicht (letzte 30 Tage)
Willim
Willim am 17 Jan. 2019
Bearbeitet: madhan ravi am 17 Jan. 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.

Akzeptierte Antwort

Image Analyst
Image Analyst am 17 Jan. 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?

Weitere Antworten (2)

KSSV
KSSV am 17 Jan. 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 Kommentare
Willim
Willim am 17 Jan. 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 am 17 Jan. 2019
Give the value of N. Image Analyist solution is good....go ahead with that.

Melden Sie sich an, um zu kommentieren.


madhan ravi
madhan ravi am 17 Jan. 2019
Bearbeitet: madhan ravi am 17 Jan. 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 Kommentare
Willim
Willim am 17 Jan. 2019
great it works but the thing is that the frist element could be 1 not 0 all the time
madhan ravi
madhan ravi am 17 Jan. 2019
Bearbeitet: madhan ravi am 17 Jan. 2019
What? Did you see the first example in my answer?

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by