Making each element of a row vector equal to zero

Hi Everyone, I have a row vector (size 1*100) which contains randomly distributed 1s and 0s. How can i make each element of this row vector equal to zero using for loop ?

Sorry, if this is a very basic question !!

Answer 1

[James Kristoff]

There are many ways to modify arrays in MATLAB. First, let's look at the for loop method:

given:

% this could be any size row vector
foo = [0,1,0,0,1,1,0];

% loop through all elements of foo
for( i = 1:length(foo) )
    % set each element to 0
    foo(i) = 0;
end

Alternatively you could use logical indexing, e.g.

% this could be any size row vector
foo = [0,1,0,0,1,1,0];

% set foo equal to zero where foo equals 1
foo(foo == 1) = 0;

Also, in this simple case, you could just create a new row vector of all zeros:

% this could be any size row vector
foo = [0,1,0,0,1,1,0];

foo = zeros(size(foo));

[Aftab Ahmed Khan]

Hi, thank you for the reply. But i am trying to achieve something else. I have to use a for loop as i am using a "find function" to first find the location of 1 in the matrix and then replace it with zero. let say this is my matrix, BS_channeltable = [1 1 0 1; 0 0 1 1; 1 0 1 0; 0 1 0 1];

but when i use this command, find(BS_channeltable(1,:))~= 0; it answers me 1 1 1, but i want to find their vertices in the matrix.

I hope i have explained it well for you.

[James Kristoff]

There are also several options when dealing with Matrices. First, let me explain why you are seeing 1 1 1 when you execute the commands you mentioned.

% creates a 4x4 matrix
BS_channeltable = [1 1 0 1; 0 0 1 1; 1 0 1 0; 0 1 0 1];

let's split up this command to understand it better:

find(BS_channeltable(1,:))~= 0;

working from the innermost expression we have:

BS_channeltable(1,:)

which returns the first row (a 1x4 row vector) of the BS_channeltable matrix:

[1, 1, 0, 1]

then, that value is sent to the find function.

find([1, 1, 0, 1])

which returns the indices of any non-zero elements specifically:

[1, 2, 4]

then you are comparing these indices with 0 using not-equals

[1, 2, 4] ~= 0

and since none of these indices are zero, you get an array of true values, represented as:

[1, 1, 1]

This means that you could either use the indices returned by the find function directly i.e.

% get the indices of any non-zero values
indices = find(BS_channeltable(1,:));
% replace these specific values with 0
BS_channeltable(1, indices) = 0

Alternatively you could use logical indexing to get around using the find function i.e.

% set the any values in the first row of BS_channeltable
% that are not equal to zero, to zero
BS_channeltable(1, BS_channeltable(1,:) ~= 0) = 0;

[Aftab Ahmed Khan]

Thank you so much, Done.

Answer 2

[James Tursa]

You don't need a for loop. You can just do this:

row_vector(:) = 0; % Set all elements to 0, keep original variable type the same

[Aftab Ahmed Khan]

Hi, thank you for the reply. But i am trying to achieve something else. I have to use a for loop as i am using a "find function" to first find the location of 1 in the matrix and then replace it with zero. let say this is my matrix, BS_channeltable = [1 1 0 1; 0 0 1 1; 1 0 1 0; 0 1 0 1];

but when i use this command, find(BS_channeltable(1,:))~= 0; it answers me 1 1 1, but i want to find their vertices in the matrix.

I hope i have explained it well for you.

[James Tursa]

Not clear yet what you want. Are you trying to do this operation for only certain rows of a 2D matrix? If so, you can still do this without find and a for loop. E.g.,

BS_channeltable(1,:) = 0;

Is there some reason you need the indexes of these locations, other than to set their locations equal to 0?

Answer 3

[George Papazafeiropoulos]

You can create a new vector with all zeros by typing the command:

new=zeros(1,100)

or by using a for loop:

for I=1:100
    if v(I)==1
    v(I)=0;
    end
end