Replacing the zero entry of a vector with its nonzero entries

Hello,

Today I have another problem slightly different from my previous question, but they are of the same nature.

Suppose I have a vector A in which there are some zero and nonzero entries. How can I replace the nonzero entries of the vector in such a way that the replacement is done with the closet (left/right index) nonzero entry?

If 0 appears between 2 numbers then we replace it with leftmost nonzero number. For example, if A=[2;0;-1;-2;0;0;4;1], then the replacement is done as follows: the 0 in A(2) is replaced with the number in A(1) (i.e. 2) not A(3); the 0 in A(5) is replaced with the number in A(4) and the zero in A(6) is replaced with the number in A(7).

If it occurs that only 1 entry is nonzero, then we replaced all the zero entries with that nonzero entry. If we have only 2 nonzero entries in the middle of the vector, say k,k+1, then we replaced the 1,2,..,k-1 entries with nonzero entry k,and we replaced the k+2,k+3,...,n with the nonzero entry k+1.

Thank you.

[Jan]

The question is strange. You want to replace the nonzero elements by the closest nonzero elements? The same for: "only 1 entry is nonzero, then we replaced all the nonzero entries with that nonzero entry"?

Do you mean that you want to replace the zeros as in the given example? Please edit the question and fix this.

[Hassan]

@Jan Simon, yes you are right. I have edit the question. Thank you for the observation.

Answer 1

[Jan]

I guess, that you want to replace zeros by the nearest non-zero value. Then this is a job for interp1:

z    = (A == 0);
A(z) = interp1(find(~z), A(~z), find(z), 'nearest');

To consider leading and trailing zeros:

A(z) = interp1(find(~z), A(~z), find(z), 'nearest', 'extrap');

[Hassan]

@Jan Simon, I am analyzing your answer, I will get back to you.

[Hassan]

@Jan Simon, suppose A=[2;0], that is A is a vector of dimension 2. The second line of your code generate the following error message:

Error using griddedInterpolant Interpolation requires at least two sample points in each dimension.

Error in interp1 (line 149) F = griddedInterpolant(X,V,method);

Therefore, interp1 cannot work for an arbitrary vector.

[Hassan]

interp1 works only if there is at least 2 nonzero entry of a given vector.

[Jan]

@Hassan: At least interp1 performs the job for standard inputs already. Then it is easy to catch the exceptions:

z  = (A == 0);
nz = sum(z);  % [TYPO! Must be sum(~z), see below]
if nz == 1
   A(:) = A(~z);
elseif nz > 1
   A(z) = interp1(find(~z), A(~z), find(z), 'nearest', 'extrap');
 end

Does this catch all exceptions? If not, it is not hard to expand this.

[Hassan]

@Jan Simon. Let me analyze your answer. I will get back to you.

[Hassan]

@Jan Simon. The variable nz may sometimes be >1 and still the number of nonzero entries of A is 1. For example, try A=[3;0;0]. In this case, interp1 still fails.

[Jan]

@Hasan: Do you see, that this is a mistake? The number of non-zero elements must be 1 for the exception, but nz was the number of zeros. I forgot a ~ operator. Maybe you might be able to find this by your own.

z        = (A == 0);
nNonZero = sum(~z);  % Changed: sum(z) --> sum(~z)
if nNonZero == 1
   A(:) = A(~z);
elseif nNonZero > 1
   A(z) = interp1(find(~z), A(~z), find(z), 'nearest', 'extrap');
end

"nNonZero" is a better name, because "nz" can mean "non zero" or "number of zeros".

[Basil C.]

I guess your latest solution should solve the problem.

Nice approach by using interp1 @Jan Simon. Learnt something new.

[Hassan]

I hope so, but I have to analyze it critically.

Answer 2

[Basil C.]

• For your first part

 for(i=1:length(A)-1)
    if(A(i)~=0 && A(i+1)==0)
       A(i+1)=A(i);
    end
 end

However I wasn't able to figure out the pattern as to why do you want to set the zero in A(6) to the value in A(8)

• For the second part. If the matrix is B.

 % CASE 1
   if(find(B)== length(B)/2+0.5)
       B(:)=B(length(B)/2+0.5);
% CASE 2
   elseif(find(B)== [length(B)/2,length(B)/2+1] )
       B(1:length(B)/2)=B(length(B)/2);
       B(length(B)/2 +1:length(B))=B(length(B)/2+1);
  end

Hope this solves your doubt.

[Hassan]

@Basil, I mean A(7) not A(8), it was a typo, I have fix it.

[Hassan]

@Basil, the problem with your answer is that it contains looping. I prepare loop-free codes.