how to declare all the rows and all column in matlab in a matrix or array??

suppose i need to remove all the zeros from a dataset and within a if else condition if the i no rows found zeros with all its column then the entire row will be deleted .. i need to know how to declare the entire row with column in matlab

is this the way to write c(:)=[]

Answer 1

[dpb]

"if [...] rows found zeros with all its column then the entire row will be deleted"

If the array is x, then

idx=~any(x,2);  % logical vector T for rows all zero
x(idx,:)=[];    % remove those rows

NB: other scattered zeros will still be in the array and there's no way to remove them and keep a 2D array unless there's a fixed number in each row/column so the resulting length of each reduced row and column is a constant. This is an unlikely condition in general altho could be possible for specific cases.

[wasima tammi]

thnks a lot and i think dt will work for the huge dataset also ..:)

[wasima tammi]

can u please xplain that y we have used 2 in this segment

idx=~any(x,2);  % logical vector T for rows all zero
x(idx,:)=[];

it would be so much helpful

[dpb]

"... [e]xplain ... [wh]y we have used 2 in this segment idx=~any(x,2);"

See

doc any

for details but in general Matlab functions over arrays operate by default on columns. We wanted to operate by row in this case which is the second storage dimension.

Editorial aside--please use complete sentences and spelling to make interpretation more reliable and easier for potential respondents to decipher the question instead of having to try to interpret meanings. This one wasn't too bad, but I (for one at least) will quickly give up in more arcane cases thus depriving the poster of a potential answer to the problem.

Answer 2

[James Tursa]

c(:) is the syntax for the entire matrix, not a particular row. And when used on the rhs of a statement it means "reshape c as a column vector". For a 2D matrix, syntax for the k'th row is

c(k,:)

However, deleting the k'th row in a loop has the effect of resizing the array in a loop (can be very slow if many rows of a large matrix are deleted). Also, depending on how you do it, the indexing of your original problem can change if the k'th row suddenly disappears. Can you show more code for how you intend to do this, particularly if it is in a loop?

[wasima tammi]

thnks a lot .. this is some of y codes where i am facing problems..

for i=1:length(T)
  j=1;
  if find (T(i,j)==1)
    c1(i,:)=P(i,:);
%   if find(c1(i,:)==0)
%     c1(i,:)=[];
%   end
  end
end

Answer 3

[dpb]

The commented-out line deleting the row will, as James noted, break as the array will then have a size(C1,2) one less than the original length (which, btw, will only be correct to begin with if there are more rows than columns in the array) so you will then "run off" the end later on.

As he also said, to do this in a loop requires traversing from last to first as

for i=size(T,1):-1:1
  ...

to not have the problem. But again (as noted also by James) this will also cause a continual reallocation of memory every time it does occur and will be much less efficient than the illustrated use of any supplied function and logical addressing.

The portion

if find (T(i,j)==1)
  c1(i,:)=P(i,:);

is also amenable to logical addressing almost identically as to previous

idx=any(T==1),2);
c1(idx,:)=P(idx,:);

NB: the above two cases are two separate logical index vectors using the same variable over; it is also possible to write the expression entirely w/o the temporary variable but that means writing the logical test expression twice; once on each side of the assignment. I'm not sure if the JIT compiler is smart enough to recognize the duplicate operation and do the comparison only once or not; unfortunately TMW doesn't document behavior well enough to know so writing using the temporary ensures the one-time calculation at perhaps expense of some memory. It is also somewhat easier to see what's actually being done so there's some advantage in that, too, particularly for new users.