Add random numbess to matrix

Question

armin m 2021-11-30

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1599849-add-random-numbess-to-matrix

评论： armin m 2021-12-2

采纳的回答： DGM

Hi. I have 1×n matrix.i wanna add 1 to 5 percent of it,s actual value to it, randomly. Can any body help.me? Tnx

2 个评论
显示无隐藏无

dpb 2021-11-30

" add 1 to 5 percent of it,s actual value to it, randomly."

The above leaves a lot of uncertainty as to what you intend.

For starters what properties should the random variable have -- normally distributed, uniform, ...

Should the values added always be positive additions (which raises the overall mean of the observations) or zero mean or...

So many Q?, so little info...all we know for sure is you need n RVs...which is an easy thing to generate in MATLAB given a way to select the remaining undefined things like which distribution, what parameters for said distribution, ...

armin m 2021-11-30

Q is matrix, i wanna it be normal distribute.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

DGM 2021-11-30

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1599849-add-random-numbess-to-matrix#answer_844329

编辑：DGM 2021-11-30

Define "add 1-5% of its actual value to it"

Consider an array A. Does this mean

% random factor is scalar
B = A + (0.01+0.04*rand(1))*A;

or maybe

% random factor is an array 
% each element gets its own random factor
B = A + (0.01+0.04*rand(size(A))).*A;

Imnoise() allows the application of additive gaussian noise using intensity mapping of local noise variance. That might also apply.

16 个评论
显示 14更早的评论隐藏 14更早的评论

dpb 2021-12-1

编辑：dpb 2021-12-1

Another poster asked about a RNV within bounds yesterday...depending on how you set mu, sigma, the Answer I posted there may work for you here as well...

https://www.mathworks.com/matlabcentral/answers/1599699-how-can-i-randomize-target-positions-in-a-gaussian-distribution#answer_844364

In the above the desired limits were Z=+/-2 which is 95% so roughly with a small enough sample size that the likelihood of not exceeding the Z were good enough to probably get by. If your sample size goes way up, then the probability of drawing a value in the tails goes up until at some point it'll be highly unlikely to ever draw a sample that is within the bounds unless you do constrict the variance or truncate the distribution.

"There is no free lunch!" You simply can't have a normally-distributed variate and constrain the tails values and yet allow an arbitrary variance.

Alternatively, just use a uniform or if you do want a weighted to the center distribution, a triangular or other arbitrarily shaped distribution function that is bounded. But, they won't be normal -- can't have both, other than by the illustrated route of constraining the variance to make the probability of exceeding a tail value acceptably low such that it just doesn't occur. The downside to the latter is that you end up with effectively a much smaller error than the max almost always--in the above example the range didn't exceed 0.04 in the sample of (1000)^2 (randn(N) generates a NxN array, not a vector of length N) even though he set the tolerance for 6-sigma to be 5%, not 4% and the effective max range looking at the histogram tails is more nearly in the 3% range. Is that the behavior you're looking for?

DGM 2021-12-1

编辑：DGM 2021-12-1

Extending the same example as before, but with a wider scaling factor for sigma and truncating the distribution:

tol = 0.05;

k = 2;

pd = makedist('Normal','mu',0,'sigma',tol/k);

pdt = truncate(pd,-tol,tol);

R = random(pdt,1000);

histogram(R)

[min(R(:)) max(R(:))]

ans = 1×2

-0.0500 0.0500

I neglected to mention in the prior example that in order for R and A to be combined, you'd need to generate R such that it's the same size as A. In these examples, I'm just generating a large set so that the shape of the distribution is obvious.

As dpb mentions, I'm asserting that the tolerance is +/-5% for the sake of making a consistent example. That should be easy enough to change. I just don't want to chase moving targets.

With the prior example using an unconstrained random factor, you may choose to define the tolerance window as the interval in which a specified fraction of the random values will lie. For example, a scaling factor of 2 would be 95%

tol = 0.05;

k = 2;

s = diff(normcdf([-tol tol],0,tol/k))

s = 0.9545

R = tol/k*randn(1000);

histogram(R)

[min(R(:)) max(R(:))]

ans = 1×2

-0.1234 0.1152

3 would be 99.7% and so on...

k = 3;
s = diff(normcdf([-tol tol],0,tol/k))
s = 0.9973

Or you could find some other way to define the relationship between the tolerance window and the gaussian shape, e.g. you could say that the noise a zero-mean gaussian with 10% FWHM.

tol = 0.05;
k = sqrt(2*log(2))
k = 1.1774
s = normpdf([-tol 0],0,tol/k);
s(1)/s(2)
ans = 0.5000

DGM 2021-12-2

I was just showing how to find particular values of k. In that example, 95% of the noise lies within +/-tol when k = 2. If that's the desired scenario, then:

A = 1:10; % <-- a 1xn vector
tol = 0.05; % set to 4% or whatever you need
k = 2; % sigma scaling factor
R = tol/k*randn(size(A)); 
B = A + A.*R
B = 1×10
    1.0212    1.8844    2.9885    3.9351    4.9835    5.9440    6.6925    7.9764    8.9043   10.0414

armin m 2021-12-2

Thank you very much

请先登录，再进行评论。

Answer 2

dpb 2021-12-1

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1599849-add-random-numbess-to-matrix#answer_845295

If the desire is a bounded, symmetric, continuous distribution that approximates a normal, consider the beta with, eta,gamma equal. The pdf is then bounded between [0, 1] with mean gamm/(eta+gamma) --> gamma/(2*gamma) --> 0.5 for eta==gamma.

As for the normal, you can shift and scale the generated RNVs generated from random by whatever is needed to match the target range.

The 'pdf' normalization inside histogram results in the red overlaid normal; scaling the N() pdf to match the peak bin in the histogram results in the black overlay which emphasizes the extra weight of the beta towards the central tendency as compared to a normal. But, you can produce a bounded random variate this way that with the very nebulous requirements for the underlying error distribution could surely serve the purpose.

The above was generated by

rB55=random('beta',5,5,1e6,1);
histogram(rB55)
hold on
[mn,sd]=normfit(rB55)
pN55=normpdf(x,0.5,sd);
plot(x,pN55,'-r')
plot(x,pN55*2.48/2.64,'-k')
hLg=legend('pdf(\beta(5,5))','N(0.5,sd(\beta)','0.94*N(0.5,sd(\beta)');

where the magic constants were obtained by getting the maxima of the histogram binned values and the pdf peak

The above uses functions in the Statistics Toolbox...

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

dpb 2021-12-1

Illustrates can have very broad to quite narrow range depending on the input paramters. While not plotted, note that the B(1,1) case reduces to the uniform distribution.

请先登录，再进行评论。