The function move_me is defined like this: function w = move_me(v,a). The first input argument v is a row-vector, while a is a scalar. The function moves every element of v that is equal to a to the end of the vector.

function w=move_me(v,a)
if
w_1=v(v>a);
w_2=v(v<a);
w=[w_1 w_1 a];
end

It does work for function when variable 'a' has some value but when 'a' is empty it doesn't work. Actually in the question it is also written that If 'a' is omitted, the function moves occurrences of zeros. could you please help me I am trying to figure out this problem but it has not been done. which function can be used to shift 'a' in vector 'v' to the end of vector 'v'. Thanks in advance

[KSSV]

When a is empty obviously it will not work, because there is no number to split/ divide. What you expect actually?

[Wasi von Deutschland]

Actually I expect how one can make a function when 'a' is omitted then 'a' should be equal to zero in this function. Means by using 'if statement' or 'for loop' This is my full question ''The function move_me is defined like this: function w = move_me(v,a). The first input argument v is a row-vector, while a is a scalar. The function moves every element of v that is equal to a to the end of the vector. For example, the command >> x = move_me([1 2 3 4],2); makes x equal to [1 3 4 2]. If a is omitted, the function moves occurrences of zeros.''

Answer 1

[Rik]

function w=move_me(v,a)
if ~exist('a','var'),a=0;end
w=[v(v~=a) v(v==a)];
end

If this is homework, try to think of how this works and why. Add comments you wrote yourself to explain this code. Teachers know of this forum as well, so don't blindly copy stuff; it will harm you in the long run.

[Wasi von Deutschland]

As far as I got it. if variable 'a' doesn't exist a=0 and if it's not the case w=[v(v¬=a) v(v==a)] in vector 'v' if 'a' exists it should be ignored and in the second in vector 'v' if 'a' exists it should be placed on the second position. Am I right? and I really appreciate your work thanks

[Rik]

You are close. You need to do some Googling (or Yahooing or Binging) on the topic of logical indexing (which is what is happening here). Also, it isn't really a second position, but a concatenation, so I would advise you to search for that as well.

[Rahul Sen]

I think its combination of logical indexing and stacking. First it checks for the input "a". if it doesn't exist it assign a value "0". Next it creates two row vectors using logical indexing. First one not equals to the value "a". second one, which equals to "a". Is my understanding is right?

Answer 2

[Jorge Briceño]

Hi,

This answer might be helpful:

function w= move_me(v,a)
% The first input argument v is a row-vector, while a is a scalar. 
% The function moves every element of v that is equal to "a" to the end of the vector.
% If a is omitted, the function moves occurrences of zeros.
% Use nargin or exist to evaluate the existence of the second argument.
if nargin<2
% If a is omitted, replace it by zero.
% Function horzcat concatenates the answer.
  a=0;
  w=horzcat(v(v~=a),v(v==a));
else
  w=horzcat(v(v~=a),v(v==a));
end
end

The built-in function "horzcat" concatenates the column vectors. In case you are not using a row vector as an input, you just need to add something like this:

v=v(:)'

As Rik Wisselink mentioned, try to think and understand what you are being asked and how the code is working. In the end, you need to develop the logic as well as programming skills.

I hope it helps.

[Javed Ahmad]

Thanks sir, your answer is correct

Answer 3

[Vaibhav Sharma]

Answer 4

[Vaibhav Sharma]

This is the answer

Answer 5

[Lars Wolff]

Hi together!

It is me again....

I tried:

function [w] = move_me(v,a)
isscalar(a)
a
v(:) = v 
w = [v(v~=a)  v(v==a)]
if a ~= v
    w = (a==v)
end

but i get the message:

Problem 1 (move_me):
    Feedback: Your function performed correctly for argument(s) [1 2 3 4], 2
    Feedback: Your program made an error for argument(s) [0 1 2 3]

      Your solution is _not_ correct.

Thanks for your help and ideas in advance!

[Walter Roberson]

Your code is testing whether a is a scalar, and it displays the result of that, but that makes no difference to the rest of the calculation.

The problem description in the original question does not say what should happen if the function is not passed any arguments, or is only called with one argument, or is called with a 2D array for the first argument, or is called with a non-scalar for the second argument.

The grading system is giving you the feedback that it passed in a single argument containing [0 1 2 3] with no second argument, and that your routine is not doing what the grading system expects in response. Your code would be failing with an error message when it tried to access the a that was not being passed in.