v(n:1)=1 where n>1

Hello! I have some old code from a colleague who is unavailable to answer Qs. There is a case where:

v(n:1)=1 where n>1

The fact that n>1 is the part that is confusing me. I understand that the : acts as an indices range assigning the value 1 to all values in vector v.

What am I missing? Why would 1 be at the end? Does this reverse the order of the array or continue back to the first element after reaching the end?

Thanks!

[Image Analyst]

Please copy and paste the exact line of code. I don't understand the "where n>1" part. How are you determining that from the code? Or was it in a comment? Or do you have a for loop like "for k = 1 : n" and then the "v" line of code inside the for loop? Don't make us guess.

Answer 1

[Amir]

Matlab doesn't reverse the order. Try this code:

A=1:10;
n=3;
A(n:1)=1

or run this:

3:1

answer is:

ans =

     Empty matrix: 1-by-0

[AnnaB]

Thanks for the very quick response :) and self help way of approaching this. I noticed that nothing actually happens to A. So does that mean that code is flawed or am I missing something again?

[AnnaB]

So basically when A(n,m)=1 where n>m nothing happens where as if n<m then 1 is assigned to elements n to m. I just want to be sure that I'm not missing a mathlab function that I didn't know about.

[AnnaB]

Looking at the values for v in the matlab workspace, the code has actually populated the value '1' for positions n to end. This must be another way of writing v(n:end)=1. Im just guessing though.

[Amir]

AnnaB, when does it give 1 from n to end? v(n:m)? or v(n:end)?

[Amir]

Are you sure that the value of n has not been changed before this was run?

[AnnaB]

I haven't used (n:end) in the code. It has worked with v(n:m) where n>m. I'm not sure how it works. You could be right that it has been successfully run in the past. However, if that was the case then v(n:1) would return an error, right?

It is so strange, there are 4 instances of this in the code and I don't know how it works.

for n>1
  v(n:1)=1 
  t2=t(n:1)
  d2=d1(n:1)-r(n:1)

[Andy L]

just swap n:1 with 1:n if that is the range that you want to make == 1, i.e

v(n:1) == 1; % becomes...
v(1:n) == 1;

[Julia]

From the Matlab help:

j:k is the same as [j,j+1,j+2,...,j+m], where m = fix(k-j). In the case where both j and k are integers, this is simply [j,j+1,...,k]. This syntax returns an empty matrix when j > k

So I think your guess that nothing happens is right.

[Amir]

I will ask this from my friends. If you find the reason please share it here.