how to make a function that calculate consective difference of elements of a vector

Hi every one. I am going to attempt that query.. Write a function called neighbor that takes as input a row vector called v and creates another row vector as output that contains the absolute values of the differences between neighboring elements of v. For example, if v == [1 2 4 7], then the output of the function would be [1 2 3]. Notice that the length of the output vector is one less than that of the input. Check that the input v is indeed a vector and has at least two elements and return an empty array otherwise. You are not allowed to use the diff built‐in function. I am trying that code, getting an error in for loop because its inner statement (vi+1) creating problem.To me instead of for loop while loop will be implement. But how i not know.Any correction will be highly appreciable..

function [A]=neighbor(v)
if isvector(v) && length(v)>=2 % checking whether v is a vector and have two elements
    for i=1:length(v)
        A=v(i+1)-v(i); % getting the absolute values of v(i+1)-v(i)   
    end
else
    A=[];
end
end

[Andrei Bobrov]

without diff function

conv2(v(:)',[1, -1],'valid')

Answer 1

[Thorsten]

Why not using your original solution with the changes that Yoav suggested, plus an additional abs() to get the absolute difference?

function [A]=neighbor(v)
if isvector(v) && length(v)>=2 % checking whether v is a vector and have two elements
    for i=1:length(v) - 1 % *** -1 otherwise i+1 becomes invalid 
        A(i)=abs(v(i+1)-v(i)); % getting the absolute values of v(i+1)-v(i)   
    end
else
    A=[];
end
end

Instead of the for loop, you can use

   A = abs(v(2:end) - v(1:end-1));

If you run into problems, please report input and the error it produces.

[Muhammad Usman Saleem]

@ Thorsten thanks.. Actually he write it very briefly.. that why i was not using this.. Your other code has already share by @Stephen Cobeldick .. please see his answer..

Answer 2

[Yoav Livneh]

If you insisnt on using for loops you just need to change the limit of the for loop to

length(v)-1

and of course add

A(i) = ...

so you get a vector and not just a single value.

But you can just use indexing instead of loops to solve this:

A = v(2:end) - v(1:end-1);