A basic question of matrix indexing can't get a proper output

Given matrix A, assign the second column of A to a variable v. Afterwards change each element of the last row of A to 0.

my code:

A = [1:5; 6:10; 11:15; 16:20];
v= A(1:4,2);

A(5, :) = zeros(1, 5);

[Walter Roberson]

The last row of A is not row 5.

https://www.mathworks.com/help/matlab/ref/end.html#bvih9bm

[Debaditya Chakraborty]

Thanks man it worked!!!

[Ayush Khandelwal]

Full answer dijiye

[Gaurav Saini]

It's not run on matlab

[Walter Roberson]

The posted code runs without difficulty on MATLAB. It is, however, not correct code for the Question. The Question requires that the last row of A be changed to 0, but the code posted here instead adds a new row.

[Ashim Bhat]

FOR ASSIGNING V

V = A(: , 2)

FOR GETTING LAST ROWS AS ZEROS

A(4, :) = 0

[Walter Roberson]

No, A(4,:)=0 changes the 4th row of A to 0 no matter how many rows A has.

[Adam Danz]

I'm baffled by the confusion in this thread. It's really quite simple.

a(end,:) 

selects the last row of 'a'.

a(end+1,:) = [. . .]

adds a row to the end of 'a'.

It's that simple.

[Walter Roberson]

"Given matrix A" and the example matrix happens to have 4 rows so people are customizing the code to the four row case.

[Lokesh Belekar]

V= A(1:4,2); A (4,:) = 0

///It's correct Answer.

[Walter Roberson]

No, Lokesh, the requirement is "Afterwards change each element of the last row of A to 0." You are changing the 4th row of A, not the last row of A. Suppose A has 7 rows, then you would have to change the 7th row not the 4th.

[Tridisha Gogoi]

A = [1:5; 6:10; 11:15; 16:20];
v=A(:,2)
A(4:end,:)=0

[Walter Roberson]

No, Tridisha, the requirement is not to set all rows from #4 onwards to zero, only to set the last row to 0.

[Mohd Suhel Ansari]

A = [1:5; 6:10; 11:15; 16:20];

v=A(:,2)


A(4,[1 2 3 4 5])=0;

[Walter Roberson]

No, Mohd Suhel Ansari, the requirement is not to set row #4 to zero: the requirement is to set the last row to zero.

[SHUBHAM HINGE]

A = [1:5; 6:10; 11:15; 16:20];
v=A(1:end,2)
A(4,1:end) = 0

[Adam Danz]

@SHUBHAM HINGE

1. "1:end" can and should be replaced by A(:,2) as my answer demonstrates.
2. avoid hard-coding indices in "A(4,1:end)". Instead, use A(end,:)=0 as my answer demonstrates.

[khaula]

it works

[Vishal]

make it -

......

A(5, :)= zeros(1,5);

[Walter Roberson]

@Vishal

Assigning to A(5,:) would be appropriate only for the cases where A happens to already have exactly 5 rows. The example A matrix has 4 rows, not 5. Assigning to A(5,:) for it would create a new row of zeros, whereas the requirements is that the last (existing) row of A is set to 0.

Answer 1

[Adam Danz]

Here are some improvements to your code so that it works no matter what size A is.

A = [1:5; 6:10; 11:15; 16:20]; 

v= A(:,2); 
A(end+1, :) = zeros(1, size(A,2)); 

Note that your instructions are to "change" the last row of A. That's not what your code is doing. You're adding a row of zeros. If you want to change the last row instead of adding another row,

A(end, :) = zeros(1, size(A,2));

Summary

Add a row of 0s to the end of matrix A

A(end+1,:) = 0; 

Replace the last row of matrix A with 0s.

A(end,:) = 0;

Answer 2

[amjad khan]

A = [1:5; 6:10; 11:15; 16:20];
v=A(:,2) % assigning variable v to the second column of matrix "A"
A(4,:)=0 % changing all the elements of row 4 to zeros

[Adam Danz]

This is essentially the same answer as mine except that you're replacing the 4th row with 0s whether or not the 4th row is the last row. This is why I suggest using A(end,:) = 0 so that it will work for all sizes of A.

Answer 3

[Badal Bhardwaj]

Answer is

A(4,1)=0 
A(4,2)=0 
A(4,3)=0 
A(4,4)=0 
A(4,5)=0

[Walter Roberson]

No, that only works if A happens to be 4 rows. The question requires that the last row of A be replaced with zeros -- no matter what size A is.

[Badal Bhardwaj]

I have do this my answer is correct At the end of these 5 steps the last row of A is 0 Firstly try then comment

[Adam Danz]

@Badal Bhardwaj , as Walter said, that only works if A happens to be 4 rows.

Example of it not working

A = [1 2 3 4 5;
    1 2 3 4 5;
    1 2 3 4 5;
    1 2 3 4 5;
    1 2 3 4 5;
    1 2 3 4 5];
A(4,1)=0 
A(4,2)=0 
A(4,3)=0 
A(4,4)=0 
A(4,5)=0

Result

A =
     1     2     3     4     5
     1     2     3     4     5
     1     2     3     4     5
     0     0     0     0     0
     1     2     3     4     5
     1     2     3     4     5

If you need help understanding why your solution only works when A has 4 rows, please let us know.

Also, your 5 lines of code can be reduced to 1 line (see my answer above).

[Badal Bhardwaj]

See the image 4th row is zero

[Badal Bhardwaj]

Do not use parenthesis at end of matrix I know this Is lengthy code but definitely correct

[Adam Danz]

This isn't a correct answer to the question. The question asks how to change values of the last row of a matrix to zero, not the 4th row.

[Walter Roberson]

See the image 4th row is zero

The question does not ask to make the 4th row zero: the question asks to make the last row zero.

Do not use parenthesis at end of matrix

? Where did Adam use parenthesis at the end of matrix?

Answer 4

[Ashim Bhat]

for assigning v

v = A(:,2);

for geeting zero values of last row of A

A(4,:) = 0

[Walter Roberson]

That assigns to the 4th row of A, which might not be the last row of A.

[Adam Danz]

Example 1 of this method failing:

A = [1 2 3; 
    1 2 3; 
    1 2 3; 
    1 2 3; 
    1 2 3;
    1 2 3];

A(4,:) = 0; 

% Result
A =
     1     2     3
     1     2     3
     1     2     3
     0     0     0  % <--- wrong row
     1     2     3
     1     2     3

Example 2 of this method failing

A = [1 2 3; 
    1 2 3]; 

A(4,:) = 0; 

% Result
A =
     1     2     3
     1     2     3
     0     0     0  
     0     0     0  % <--- Now matrix A has 4 rows

Answer 5

[Imane Tahar]

A = [1:5; 6:10; 11:15; 16:20];
v= A(:,2)
A(end,:)= 0

Answer 6

[Minal Kulkarni]

A=[1:5; 6:10; 11:15; 16:20];
v=[ A(1,2); A(2,2); A(3,2); A(4,2)]
A=[1:5; 6:10; 11:15; 0,0,0,0,0]

[Adam Danz]

This is not a solution.

You're overwriting A instead of replacing the last row.

Indexing in the second line is very inefficient.

And the 2nd and 3rd lines assume A has 4 columns.

Please consider taking the Matlab on-ramp.

https://www.mathworks.com/learn/tutorials/matlab-onramp.html

Answer 7

[Muniba]

A = [1:5; 6:10; 11:15; 16:20];
v=A(1:end,2)

v = 4×1

2 7 12 17

A(end:4,1:end)=0

A = 4×5

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 0 0 0 0

[Muniba]

This is the correct way.

[DGM]

This is arguably not the correct way, or at least not a robust way. Simpler and more robust answers have already been given.

It's unnecessary to do this. I think it just adds clutter that can make larger expressions less readable, but I suppose other opinions can exist.

%v=A(1:end,2) % is the same as
v=A(:,2) % just using :

The question asks to set the last row to zero. Your answer sets the fourth row to zero. There are four rows, so it's the same, right? Well, this only works if there are exactly four rows. If there are more than four rows, your method will fail silently, assigning nothing to zero. If there are fewer than four rows, it will set the last row (whichever that is) to zero, and add extra rows to the array such that it has four rows. If A had 3 rows, it will wind up with two rows of zeros.

%A(end:4,1:end)=0 % selects row 4, but only if there are exactly 4 rows!
A(end,:)=0 % select the _last row_!

There was no reason to add this extra complication and all its problems. The generalized solution is simpler to write and simpler to read.

Answer 8

[Haris]

Variable A must be of size [4 5]. It is currently of size [5 5]. Check where the variable is assigned a value.

[Dyuman Joshi]

@Haris, how exactly is your comment, which you have mistakenly posted as an answer, relate to this thread?

[DGM]

Nobody has given a clear description of the assignment, and almost all of the answers to this thread are either redundant, wrong, or both. As far as anyone here knows, there is no requirement that A is any particular size. In fact, it seems like a terrible idea to teach students to presume that inputs are a particular mystery size and then write code that relies on that presumption.

Of course, being a terrible idea makes it plausible that someone actually wrote an assignment like that, so if you are in a position to give a better description of the assignment, then by all means, go ahead.

Just because someone set the grader up so that it accepts bad code that panders to an overly simplistic test doesn't mean anybody should call the bad code "correct".