Question

我有一个变量age，其中包含100名患者（年龄20到80岁）的年龄。

我想编写代码，从100名患者的样本中随机选择10名患者（一名患者可出现不止一次），以获得平均年龄45岁。

有谁知道我怎么能做到这一点？

Answer 1

我原本想要了解Matlab中是否有可用于解决此问题的函数。然而，没有一个与兰特相关的命令似乎做了我需要做的事情（如果我错过了，我很抱歉）。

但是，我编写了以下代码来解决我的问题。在循环中，我允许从所需平均值的0.1变化（可以根据需要进行调整）。

% Make an array of 100 ages between 20 and 80
min=20;
max=80;
AgeArray = min + (max-min).*rand(100,1);

% Set parameters for randomselection
samplenr=10;              % Want a 10 random samples from the original age array of 100
mean = 45;                % The mean of randomly selected 10 sample should be 45
half_nr=samplenr/2;       % Set the half-way point

% Get first random sample of 10 ages from the original array of 100 
RandSubSample = datasample(AgeArray,samplenr);

% Sort the array of 10 from low to high values
RandSubSample=sort(RandSubSample);

% Mean of this random sample
sumage=sum(RandSubSample);
meanage=sumage/samplenr;



% Loop to re-select 10 samples until mean matches 45 (with error allowed for +/-
% 0.1
while meanage<mean-0.1 || meanage>mean+0.1

if meanage<mean                 % If mean is too low we need to replace young ages with older aged

    index=randi([1 half_nr]);
    RandSubSample(index)=r(randi(numel(r)));

elseif meanage>mean             % If mean is too high we need to replace old age with younger age

    index=randi([half_nr nr]);
    RandSubSample(index)=r(randi(numel(r)));

end

RandSubSample=sort(RandSubSample);

sumage=sum(RandSubSample);
meanage=sumage/nr;

end
y=sort(y);

sumage=sum(y);
meanage=sumage/nr;

end

matlab随机抽样变量x次以得到y的均值

1 个答案: