Could anyone help me to fix the issue.

3 次查看（过去 30 天）

Prabha Kumaresan 2018-1-24

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/378681-could-anyone-help-me-to-fix-the-issue

评论： Rik 2018-1-26

If A=rand(5,10) how to have the values by fixing the minimum value to be 0.01 and maximum value to be 0.09 and by adding all the random values together it should result in 0.35. For example I have taken the minimum value and maximum value to be 0.01 and 0.09. And the addition of all the values should result in 0.35.Could anyone help me how to implement it.

2 个评论
显示无隐藏无

John D'Errico 2018-1-26

编辑：John D'Errico 2018-1-26

Why do you feel it necessary to ask the same question always at least TWICE, and often 3 or 4 times? You got an answer several times, with several ways to do it.

You have now asked something over 170 basic questions, and as has been pointed out, do not seem to be learning the basics of MATLAB yet. Surely it it time to read the documentation, rather than relying on Answers as your lazy manual? In this case, had you simply done a search on answers, you would have found the same question asked and answered repeatedly.

Stephen23 2018-1-26

See version number five:

https://www.mathworks.com/matlabcentral/answers/379141-could-anyone-help-me-to-solve-the-issue-in-the-following-code

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Birdman 2018-1-24

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/378681-could-anyone-help-me-to-fix-the-issue#answer_301471

在 MATLAB Online 中打开

One way:

a=0.01;b=0.09;N=10;
while true
    A=a+(b-a).*rand(N,1)
    if ismembertol(sum(A(:)),0.35,1e-2)
        break;
    else
    end
end

13 个评论
显示 11更早的评论隐藏 11更早的评论

Stephen23 2018-1-25

@Prabha Kumaresan: why are you wasting your time on this? I showed you a working solution.

Jan 2018-1-25

@Prabha: Birdman's code produced random numbers until the sum has the wanted value accidentally. You see that this takes a long time for 16 numbers. Note that this solution accepts a sum, which is not exactly 0.35, but uses a tolerance of 0.01 . For a reliable working code see Stephen's answer.

请先登录，再进行评论。

更多回答（1 个）

Stephen23 2018-1-24

3
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/378681-could-anyone-help-me-to-fix-the-issue#answer_301472

编辑：Stephen23 2018-1-25

在 MATLAB Online 中打开

"If A=rand(5,10) how to have the values by fixing the minimum value to be 0.01 and maximum value to be 0.09 and by adding all the random values together it should result in 0.35"

This is impossible: matrix A has size 5x10 with 5*10=50 elements, which means that the minimum possible sum of all of its values will be 0.01*50=0.5, which is greater than the requested sum of 0.35.

Once you select values or a matrix size that are actually possible, then the simplest solution is to download and use Roger Stafford's excellent randfixedsum:

https://www.mathworks.com/matlabcentral/fileexchange/9700-random-vectors-with-fixed-sum

Here is a complete working example for a more realistic matrix size of 2x10:

>> r = 2;
>> c = 10;
>> [x,v] = randfixedsum(r*c,1,0.35,0.01,0.09);
>> y = reshape(x,r,c)
y =
   0.018153   0.011913   0.025947   0.020434   0.010300   0.027032   0.017784   0.018632   0.025201   0.015418
   0.014428   0.013693   0.012165   0.012403   0.026608   0.015631   0.013894   0.012895   0.014529   0.022941
>> sum(y(:))
ans =  0.35000