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

[amjad khan]

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

v=A(:,2)

A(4,:)=0

[amjad khan]

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 2

[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 3

[John Cyruz Las Piñas]

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

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

[Walter Roberson]

Afterwards change each element of the last row of A to 0.

That does not say to add a new row to A: it says to change the last row of A.

[Walter Roberson]

Adam gave the full command in https://www.mathworks.com/matlabcentral/answers/463130-a-basic-question-of-matrix-indexing-can-t-get-a-proper-output#answer_375887

[Walter Roberson]

Ayush Khandelwal, what code did you end up with when you followed Adam's guideance ?

[Ayush Khandelwal]

OK done

Answer 4

[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 5

[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 6

[MICHAEL]

v=A(:,2);

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

this is the write answer

[Walter Roberson]

that is not valid syntax for using zeros : the semicolon is not correct

Answer 7

[Imane Tahar]

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

Answer 8

[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 9

[Idowu]

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

[Walter Roberson]

The requirements are to change the last row to 0, not to change the 4th row specifically to 0.

Your code is the same as the code posted at https://www.mathworks.com/matlabcentral/answers/463130-a-basic-question-of-matrix-indexing-can-t-get-a-proper-output#comment_938612