could anyone tell me how to replace the value of the given matrix

A=[0.01 0.02 0.03 0 0 0 0.02;
0.02 0.03 0.04 0 0 0 0.03;
0.03 0.02 0.02 0 0 0 0.04]

if i add all the numbers it comes 0.31. Could anyone tell me how to replace the number present not the zeros in such a way that if i add the numbers in the matrix it should be 0.35

[Stephen23]

@Prabha Kumaresan: are you still trying to generate random numbers that sum to some value?

https://www.mathworks.com/matlabcentral/answers/378681-could-anyone-help-me-to-fix-the-issue#answer_301472

Rather than asking the same question fifty times maybe you could give one clear description (and then it is more likely that you will get the results that you want).

[Rik]

@Stephen, If Prabha Kumaresan is so bad for your blood pressure, why don't you ignore his posts for a while? I did just that after I was about to post an angry and rude comment.

[Stephen23]

@Rik Wisselink: thank you for the concern. I actually want to help Prabha Kumaresan get a good solution... but it is not clear to me why Prabha Kumaresan is avoiding the use of the one function that easily solves this task. An explanation is what I am after!

[Jan]

@Stephen: Prabha does not give us explanations. It seems to be more a problem of the communication than of Matlab. Then this problem cannot be solved by communicating in the forum. We cannot help everybody, but all we can do is to suggest working solutions. We did this as good as we can, so it is time to be satisfied - even if the OP is not.

But driving crazy would be a reasonable reaction also. ;-)

Answer 1

[Stephen23]

This is trivially easy with randfixedsum, in just one line of code:

>> A = [0.01 0.02 0.03 0 0 0 0.02; 0.02 0.03 0.04 0 0 0 0.03; 0.03 0.02 0.02 0 0 0 0.04];
>> A(A~=0) = randfixedsum(nnz(A),1,0.35,0.01,0.09)
A =
   0.010084   0.024919   0.042718   0.000000   0.000000   0.000000   0.011433
   0.011970   0.010413   0.011182   0.000000   0.000000   0.000000   0.055572
   0.036536   0.016813   0.054944   0.000000   0.000000   0.000000   0.063416
>> sum(A(:))
ans =  0.35000

It is not clear why you are wasting so much of your time avoiding using randfixedsum.

Answer 2

[the cyclist]

A = 0.35*A/sum(A(:));

[Prabha Kumaresan]

ok.With respect to your answer A in numerator and denominator gets cancelled and it results in 0.35 But what i actually need is the numbers needs to be changed so that the sum of the matrix results in 0.35.But according to your answer when i add the matrix numbers the sum was not not equal to 0.35.

[Rik]

A doesn't cancel in this line of code. The denominator is the sum of all values of A. Dividing A by that sum sets the sum of the result to 1. Because your goal is a sum of 0.35, you multiply it with that number.

Shouldn't your call to randfixedsum have taken care of this?

[Prabha Kumaresan]

As both denominator and numerator has same value it will be 1.when 1 multipled by 0.35 gives 0.35.But the sum of the matrix is not actually 0.35.How to get the sum of the matrix to be 0.35.

[Rik]

Have you tried this code and checked with sum(A(:)) after this line? Because in my copy of Matlab the line of code the cyclist result in a matrix that sums to 0.35.

But again, shouldn't randfixedsum have solved this for you? See the link Stephen posted above.

[Prabha Kumaresan]

Could you tell me is there any other command other than randfixedsum can be used.

[Walter Roberson]

Yes, there a variety of ways to generate with a fixed sum. One of the ways is to generate numbers and then scale them so that the sum is correct -- just as you did here. There are other ways.

The difficulty is that a lot of ways give biased values. Roger's randfixedsum was written to carefully avoid biasing the values.

[Jan]

@Prabha: I do not see the problem.

A = [0.01 0.02 0.03 0 0 0 0.02; ...
     0.02 0.03 0.04 0 0 0 0.03; ...
     0.03 0.02 0.02 0 0 0 0.04]
A = 0.35 * A / sum(A(:));

Now sum(A(:)) is 0.35, exactly what you have asked for.

It is not useful to suggest another tool instead of randfixedsum, if this tool works perfectly.

[Walter Roberson]

The sum comes out as the closest representable floating point number to 35/100 which is the same floating point number that is used for the constant 0.35