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Pick a value with some probability

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Francesco Pio
Francesco Pio 2023 年 5 月 21 日
編集済み: John D'Errico 2023 年 5 月 22 日
Hello everyone. Let's suppose we have a Gaussian distribution on people's height. there will be an average value with higher probability and values ​​that deviate with lower probability. In the example, x contains some height samples and y contains the probability with which that element occurs.
x = [1.50 1.60 1.70 1.70 1.70 1.80 1.90];
mu = mean(x);
s = std(x);
y = normpdf(x, mu, s);
Let's suppose we want to take a random value from the Gaussian distribution keeping in mind the various different probabilities. How can i do that?
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John D'Errico
John D'Errico 2023 年 5 月 21 日
編集済み: John D'Errico 2023 年 5 月 21 日
I think you don't understand what a gaussian distribution means. The mean and standard deviation of those points do not imply a Gaussian that has the same distribution as that set of heights. The mean and variance will be known, but those points do not follow a Gaussian.
As such, you are not taking those probabilities into account, IF you use that normal PDF in y.
If instead, what you really want to do is sample from the given distribution, then you could use a discrete distribution, with the specific probabilities.

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回答 (2 件)

Image Analyst
Image Analyst 2023 年 5 月 22 日
Try randn with your desired mean and std
r = mu + s * randn(1000, 1); % 1000 random numbers

John D'Errico
John D'Errico 2023 年 5 月 22 日
編集済み: John D'Errico 2023 年 5 月 22 日
Please stop asking the same question. There is NO way to know what distribution any set of data comes from. You can use tools to fit a distribution to some data. But that does not prove it is the true distribution. And even when you do, for example, fit a normal distribution to your data, that won't insure that the random samples from that normal distribution have the same distribution as your data.
You CAN decide to use a discrete distribution. Here, for example, you have a sample where 1.70 arises 3 times as often as the others.
help datasample
DATASAMPLE Randomly sample from data, with or without replacement. Y = DATASAMPLE(DATA,K) returns K observations sampled uniformly at random, with replacement, from the data in DATA. If DATA is a vector, then Y is a vector containing K elements selected from DATA. If DATA is a matrix, then Y is a matrix containing K rows selected from DATA. If DATA is an N-D array, DATASAMPLE samples along its first non-singleton dimension. DATA may be a dataset array or table. Because the sample is taken with replacement, the observations that DATASAMPLE selects from DATA may be repeated in Y. Y = DATASAMPLE(DATA,K,DIM) returns a sample taken along dimension DIM of DATA. For example, if DATA is a matrix and DIM is 2, Y contains a selection of DATA's columns. If DATA is a dataset array or table and DIM is 2, Y contains a selection of DATA's variables. Use DIM to ensure sampling along a specific dimension regardless of whether DATA is a vector, matrix or N-dimensional array. Y = DATASAMPLE(DATA,K, 'PARAM1',val1, 'PARAM2',val2, ...) or Y = DATASAMPLE(DATA,K,DIM, 'PARAM1',val1, 'PARAM2',val2, ...) specifies optional parameter name/value pairs to control how DATASAMPLE creates the sample. Parameters are: 'Replace' - select the sample with replacement if REPLACE is true (the default), or without replacement if REPLACE is false. When sampling without replacement, the observations that DATASAMPLE selects from DATA are unique. 'Weights' - create a weighted sample using the positive weights in the vector W. [Y,I] = DATASAMPLE(...) returns an index vector indicating which values were sampled from DATA. For example, Y = DATA(I) if DATA is a vector, Y = DATA(I,:) if DATA is a matrix, etc. DATASAMPLE uses RANDPERM and RANDI to generate random values and therefore changes the state of MATLAB's global random number generator. Control that generator using RNG. Y = DATASAMPLE(S,...) uses the random number stream S for random number generation. Examples: Draw five unique values from the integers 1:10. y = datasample(1:10,5,'Replace',false) Generate a random sequence of the characters ACGT, with replacement, according to specified probabilities. seq = datasample('ACGT',48,'Weights',[0.15 0.35 0.35 0.15]) Select a random subset of columns from a data matrix. X = randn(10,1000); Y = datasample(X,5,2,'Replace',false) Resample observations from a dataset array to create a bootstrap replicate dataset. load hospital y = datasample(hospital,size(hospital,1)) Use the second output to sample "in parallel" from two data vectors. x1 = randn(100,1); x2 = randn(100,1); [y1,i] = datasample(x1,10) y2 = x2(i) See also RAND, RANDI, RANDPERM, RNG. Documentation for datasample doc datasample Other uses of datasample tall/datasample
x = [1.50 1.60 1.70 1.70 1.70 1.80 1.90];
xsam = datasample(x,20)
xsam = 1×20
1.7000 1.7000 1.7000 1.8000 1.5000 1.8000 1.7000 1.8000 1.8000 1.8000 1.7000 1.7000 1.6000 1.6000 1.9000 1.7000 1.7000 1.6000 1.7000 1.9000
So the vector x contains ONLY elments which lie in the original data set. And they will have the same relative frequency. It is my guess this what you want.
histogram(datasample(x,1000000),'norm','pdf')
But if you use a normal approximation to that distribution, the relative frequencies of the data will not match the original data set at all well.
[muhat,sigmahat] = normfit(x)
muhat = 1.7000
sigmahat = 0.1291
fplot(@(x) normpdf(x,muhat,sigmahat),[1.4,2])
hold on
histogram(x,5,'norm','pdf')
As you can see, a normal pdf fits that data like I fit into the suit I wore when I got married.

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