How to get a random vector whose elements arrangement is the same as that of my desired vector

If I have a desired vector u=[3 1 5 105 155 50].

Now I want to genrate 10 random vectors of the same size as above u. Then I want to get one of those random vectors which is nearly equal to u and has the same arrangment of elements as that of my desired vector u.

i.e. if one of those generated vector is say for example [1.001 2.99 4.999 154.999 105.001 49.999]. This is nearly equal to my above desired vector u but the arrangement of elements of this vector is not the same as that of my u vector. So I want to get it in the same arrangement of elements as that of my u vector. How can we do this?

[Sadiq Akbar]

Dear madhan ravi, the page shows me that you have edited the answer 11 minutes agao, but I cann't see it. What should I do now to see your answer? Regards

Answer 1

[madhan ravi]

Use the second output of sort() to the newly sorted generated vector as index.

 [~, ix] = sort(u);
 N = sort(newly_generated_vector);
 Wanted = N(ix)

[madhan ravi]

As requested, I don't see the difference to be honest.

>> u1=[1 3 5 10 30 50]; % Desired vector 1
u2=[3 1 5 30 10 50]; % Desired vector 2
u3=[5 1 3 50 10 30];  % Desired vector 3
 [~, ix] = sort(u2);
 N = sort(u3); 
 Wanted = N(ix)

Wanted =

     3     1     5    30    10    50

>> 

[Sadiq Akbar]

Thank you very much for your 2nd prompt response madhan ravi. As I told you that I tested your program for three cases. In each case I changed the arrangement in my u vector and then ran your program (considering the other u as my newly generated vector, so your program didn't give correct result in 3rd case. Let me re-phrase it. suppose I take 1st case in which my u is [1 3 5 10 30 50]; likewise let the "newly_generated_vector" is [5 1 3 50 10 30]; So when I run your program, I get the wanted vector whose elements are not in the same order(in arrangement) as that of my u vector. Actually the "newly_generated_vector" is a random vector in which the arrangement of the elements may not be the same as that of u. Suppose I generate 10 of those random numbers. Now I want to get one of them which is "nearly" the same as my u and the arrangement of elements in that random vector must be the same as that of my u vector.

[madhan ravi]

Here you go:

[~, ix] = sort(u3);
[U1, ix1(ix)] = sort(u1);  % If u1 is my newly generated vector for example
 Wanted = U1(ix1)

[Sadiq Akbar]

Dear madhan ravi thank you very much for your 3rd prompt response. My three u vectors were as follows:

u1=[1 3 5 10 30 50]; % Desired vector 1

u2=[3 1 5 30 10 50]; % Desired vector 2

u3=[5 1 3 50 10 30]; % Desired vector 3

You have taken u3 as my desired vector and u1 as the random vector. When I ran this program, the Matlab gives me this result:

Wanted =

5 3 1 50 30 10

As you can see the arrangement of elements in this Wanted vector is not the same as that of the u3 vector. In u3, the arragement of elements are 5 1 3 50 10 30 while in Wanted vector the arrangement of elements are 5 3 1 50 30 10. It ought to be like u3 i,e,

5 1 3 50 10 30.

So still it doesn't meet my requirement.

[madhan ravi]

What are you talking about? Did you even run my code??

 >> u1=[1 3 5 10 30 50]; % Desired vector 1
 u2=[3 1 5 30 10 50]; % Desired vector 2
 u3=[5 1 3 50 10 30];  % Desired vector 3

 [~, ix] = sort(u3);
[U1, ix1(ix)] = sort(u1);  % If u1 is my newly generated vector for example
  Wanted = U1(ix1)

 Wanted =

Columns 1 through 3

     5     1     3

Columns 4 through 6

    50    10    30

 >> isequal(u3, Wanted)

 ans =

logical

   1

[Sadiq Akbar]

Dear madhan ravi I am extremely thankful to you for your prompt responses. Yes, I ran your program.Just check the following yourself.

u1=[1 3 5 10 30 50]; % Desired vector 1
 u2=[3 1 5 30 10 50]; % Desired vector 2
 u3=[5 1 3 50 10 30];  % Desired vector 3

%1st check
[~, ix] = sort(u3);
[U1, ix1(ix)] = sort(u1);  % If u1 is my newly generated vector for example
Wanted = U1(ix1)   %-----------Correct
This gives correct result.

%2nd check
[~, ix] = sort(u1);
[U1, ix1(ix)] = sort(u2);  % If u1 is my newly generated vector for example
Wanted = U1(ix1)  % --------------Wrong
%This gives wrong result

% 3rd Check
[~, ix] = sort(u2);
 [U1, ix1(ix)] = sort(u3);  % If u1 is my newly generated vector for example
 Wanted = U1(ix1)   %---------------Wrong
% This also gives wrong result.

As I requested you that the arrangement of elements in Wanted vector must be the same as that of my u vector.

But it gives wrong result in the two cases.

[madhan ravi]

>> u1=[1 3 5 10 30 50]; % Desired vector 1
u2=[3 1 5 30 10 50]; % Desired vector 2
u3=[5 1 3 50 10 30];  % Desired vector 3

%1st check
[~, ix] = sort(u3);
[~, ix1(ix)] = sort(u1);
Wanted = u1(ix1);
Check_1 = isequal(u3, Wanted)

%2nd check
[~, ix] = sort(u1);
[~, ix1(ix)] = sort(u2);
Wanted = u2(ix1);
Check_2 = isequal(u1, Wanted)

% 3rd Check
[~, ix] = sort(u2);
[~, ix1(ix)] = sort(u3);
Wanted = u3(ix1);
Check_3 = isequal(u2, Wanted)

Check_1 =

  logical

   1


Check_2 =

  logical

   1


Check_3 =

  logical

   1

>> 

[Sadiq Akbar]

Thank you very much madhan ravi for your tireless efforts. Yes, you are right that we get its logical value 1, but as I have requested you I am not interested in its logical value. Rather I want to make the arrangement of elements in my random vector the same as that of my desired vector. If you run this program, the 1st check gives me CORRECT RESULT. But the 2nd and the 3rd check give me the following WRONG RESULTS respectively though their logical values are 1 .

Wanted= 5 1 3 50 10 30 % Result of check 2

Wanted= 5 1 3 50 10 30 % Result of check 3

Now if you compare arrangement of elements of Wanted vectors of these two results with u1 and u2 where they are our desired vectors respectively, you well see that its different because

u1=[1 3 5 10 30 50]; % Desired vector 1

u2=[3 1 5 30 10 50]; % Desired vector 2

As you can see elements in Wanted of check 2 are 5 1 3 50 10 30 while that of u1 are

1 3 5 10 30 50.

Likewise elements of Wanted of check3 are 5 1 3 50 10 30 while that of u2 are

3 1 5 30 10 50.

I emphasize that I am not interested in its logical values, rahter I want to make the arrangement of elements in the random vector the same as that of my desired vector.

[madhan ravi]

Man I get the correct results for God’s sake!

Do you know why isequal() was used or what the function actually does? You seem to be lacking basic MATLAB fundamentals. It means the arrangements are made according to the wanted vector. I have already answered your original question. Are you playing with me or what just to waste my time??

 >> u1=[1 3 5 10 30 50]; % Desired vector 1
 u2=[3 1 5 30 10 50]; % Desired vector 2
 u3=[5 1 3 50 10 30];  % Desired vector 3

 %1st check
 [~, ix] = sort(u3);
 [~, ix1(ix)] = sort(u1);
 Wanted = u1(ix1);
 u3
 Wanted

 %2nd check
 [~, ix] = sort(u1);
 [~, ix1(ix)] = sort(u2);
 Wanted = u2(ix1);
 u1
 Wanted

 % 3rd Check
 [~, ix] = sort(u2);
 [~, ix1(ix)] = sort(u3);
 Wanted = u3(ix1);

 u2
 Wanted

 u3 =

Columns 1 through 3

     5     1     3

Columns 4 through 6

    50    10    30

 Wanted =

Columns 1 through 3

     5     1     3

Columns 4 through 6

    50    10    30

 u1 =

Columns 1 through 3

     1     3     5

Columns 4 through 6

    10    30    50

 Wanted =

Columns 1 through 3

     1     3     5

Columns 4 through 6

    10    30    50

 u2 =

Columns 1 through 3

     3     1     5

Columns 4 through 6

    30    10    50

 Wanted =

Columns 1 through 3

     3     1     5

Columns 4 through 6

    30    10    50

If you can’t analyse the above that it gives the correct results. I don’t know what will make you realise.

[Sadiq Akbar]

I am extremely SORRY madhan ravi. No, I don't play with you. Even I don't want to waste anybody's time. Neither I can think like that. Its one's preciuos time that one spares for me. So why would i want to do like that when you are sparing your precious time for me. Actually its my fauklt as I am not as expert as you are. If I were expert, then why would I asked for help. But due to my weakness, I was doing a serious mistake which is that the program which I was running was as follows:

% Help by sir madhan ravi from Mathworks

u1=[1 3 5 10 30 50]; % Desired vector 1

u2=[3 1 5 30 10 50]; % Desired vector 2

u3=[5 1 3 50 10 30]; % Desired vector 3

%%%%%%%%%%%%%%%%%%%

%1st Try of sir madhan ravi

%%%%%%%%%%%%%%%%%%%

% % % [~, ix] = sort(u2);

% % % N = sort(u3);

% % % Wanted = N(ix)

%%%%%%%%%%%%%%%%%%%%%%%%%

% 2nd Try of sir madhan ravi

%%%%%%%%%%%%%%%%%%%%%%%%

% [~, ix] = sort(u3);

% [U1, ix1(ix)] = sort(u1); % If u1 is my newly generated vector for example

% Wanted = U1(ix1) %-----------Correct

% % My Change in u

[~, ix] = sort(u1);

[U1, ix1(ix)] = sort(u2); % If u1 is my newly generated vector for example

Wanted = U1(ix1) % --------------Wrong

% [~, ix] = sort(u2);

% [U1, ix1(ix)] = sort(u3); % If u1 is my newly generated vector for example

% Wanted = U1(ix1) %---------------Wrong

%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 3rd Try of sir madhan ravi

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

u1=[1 3 5 10 30 50]; % Desired vector 1

u2=[3 1 5 30 10 50]; % Desired vector 2

u3=[5 1 3 50 10 30]; % Desired vector 3

%%%%%%%%%%%%%%%%

% %1st check

%%%%%%%%%%%%%%%%%%

% [~, ix] = sort(u3);

% [~, ix1(ix)] = sort(u1);

% Wanted = u1(ix1);

% Check_1 = isequal(u3, Wanted)

%%%%%%%%%%%%%%%%%%%%%%%

%2nd check-----------------Wrong Result

%%%%%%%%%%%%%%%%%%%%%%%

[~, ix] = sort(u1);

[~, ix1(ix)] = sort(u2);

Wanted = u2(ix1);

Check_2 = isequal(u1, Wanted)

%%%%%%%%%%%%%%%%%%%%%%

% % 3rd Check

%%%%%%%%%%%%%%%%%%%%%%

% [~, ix] = sort(u2);

% [~, ix1(ix)] = sort(u3);

% Wanted = u3(ix1);

% Check_3 = isequal(u2, Wanted)

I didn't think that there was a semicolon at there end of the Wanted vector in your correct program.At the same time there was no semicolon at the end of the Wanted vector of the wrong program which was also running and was not commented by me. So its my fault and i am extremely sorry for that.

What should I do that your kind heart forgives me?

Answer 2

[the cyclist]

Do you want something like this, which adds some random "noise" around the elements of u?

u = [3 1 5 105 155 50];

m = 10;
n = size(u,2);

r = zeros(m,n);

for ii = 1:m
    r(ii,:) = u + 0.001*(rand(1,n)-0.5);
end

[Sadiq Akbar]

Thank you very much madhan ravi for your prompt response. I tested your above program for the following cases:

u1=[1 3 5 10 30 50]; % Desired vector 1

u2=[3 1 5 30 10 50]; % Desired vector 2

u3=[5 1 3 50 10 30]; % Desired vector 3

>>[~, ix] = sort(u1);

N = sort(u2); % If u2 is my newly generated vector for example

Wanted = N(ix)

Wanted =

1 3 5 10 30 50 ----------------------------(1)

>>[~, ix] = sort(u2);

N = sort(u1); % If u1 is my newly generated vector for example

Wanted = N(ix)

Wanted =

3 1 5 30 10 50 ----------------------------(2)

>>[~, ix] = sort(u3);

N = sort(u1); % If u1 is my newly generated vector for example

Wanted = N(ix)

Wanted =

3 5 1 30 50 10 --------------------------(3)

As you can see it gives correct result in case 1 and 2 above because the wanted vector is same as u, but in case of (3), wanted is not same as u.I want to get wanted vector whose element arrangement is same as that of my u in all cases.

[Sadiq Akbar]

Thank you very much the cyclist for your prompt response also. I checked your program also. But I don't want like this. I want as I said to madhan ravi above.