随机抽样以给出准确的总和

Question

我想对 1000 到 100000 之间的 140 个数字进行采样，使这 140 个数字的总和约为 200 万（2000000）：

sample(1000:100000,140)

这样：

sum(sample(1000:100000,140)) = 2000000

有什么方法可以实现吗？

Answer 1

这里有一些获得接近 200 万的 hacky 方法。希望有人会发布更聪明的解决方案。

在这个选项中，我们使用prob 参数来使更小的值更有可能，并且我们通过反复试验来选择指数。此方法严重倾向于在 OP 中指定的范围内选择较低的值。

x1 = sample(1000:100000,140, prob=(1e5:1e3)^5.5)
mean(replicate(100, sum(sample(1000:100000,140, prob=(1e5:1e3)^5.5))))

[1] 2015620

在此选项中，我们从截断的法线（在您给定的边界处截断）进行采样。我们最初将平均值设置为 2e6/140=14285.71。但是，如果标准差大到足以导致下边界附近有很多值，则截断会使均值偏高，因此我们添加了通过反复试验选择的校正。

library(truncnorm)
x2 = rtruncnorm(140, 1e3, 1e5, mean=0.82*2e6/140, sd=1e4)
mean(replicate(1000, sum(rtruncnorm(140, 1e3, 1e5, mean=0.82*2e6/140, sd=1e4))))

[1] 2008050

如果您设置的标准差较小，则无需进行校正。但是，通过这种方式，您得到的远离均值的值会更少。

mean(replicate(1000, sum(rtruncnorm(140, 1e3, 1e5, mean=2e6/140, sd=0.5e4))))

[1] 2008494

在任何一种情况下，sample 方法的指数或对截断法线的校正都可以通过自动搜索来选择，该搜索具有平均和与 200 万之间的差异的容差。

以下是输出的一些典型分布：

Answer 2

这是一个尝试，试图改变上键。这个想法是当总和越来越高时减少上限。

sup<- 100000
tir <- vector(length = 140)
for(i in 1:140){
  print(i)
  tir[i] <- sample(1000:sup,1)
  sup <- max(1001,min(sup,abs(2000000 - sum(tir,na.rm = T))/(140-i)*2))
}
sum(tir)
[1] 2001751

Answer 3

这是一种命中注定的方法。基本思想是找到 140 个总和为 2000000 的数字与将 1:2000000 分成 140 个部分相同，需要 139 个切点。另外，请注意，最小值 1000 有点烦人。只需从所有问题数据中减去它并在最后添加回来：

rand.nums <- function(a,b,n,k){
  #finds n random integers in range a:b which sum to k
  while(TRUE){
    x <- sample(1:(k - n*a),n-1, replace = TRUE) #cutpoints
    x <- sort(x)
    x <- c(x,k-n*a) - c(0,x)
    if(max(x) <= b-a) return(a+x)
  }
}

然后rand.nums(1000,100000,140,2000000) 计算出给定范围内的 140 个整数，总和为 2000000。对于这些参数选择，函数几乎立即返回。对于参数的其他选择，解决方案可能是不可能的，或者受到如此严格的限制，以至于偶然找到一个实际上是不可能的。因此，在使用该功能时需要谨慎。可以通过添加 maxtrials 参数来修改它，如果超过了 maxtrials 却没有找到解决方案，则返回 NA。

Answer 4

存在一种生成此类随机数的算法。

最初为MATLAB创建，有一个R实现：

Surrogate::RandVec

来自 MATLAB 脚本注释的引用：

%   This generates an n by m array x, each of whose m columns
% contains n random values lying in the interval [a,b], but
% subject to the condition that their sum be equal to s.  The
% scalar value s must accordingly satisfy n*a <= s <= n*b.  The
% distribution of values is uniform in the sense that it has the
% conditional probability distribution of a uniform distribution
% over the whole n-cube, given that the sum of the x's is s.
%
%   The scalar v, if requested, returns with the total
% n-1 dimensional volume (content) of the subset satisfying
% this condition.  Consequently if v, considered as a function
% of s and divided by sqrt(n), is integrated with respect to s
% from s = a to s = b, the result would necessarily be the
% n-dimensional volume of the whole cube, namely (b-a)^n.
%
%   This algorithm does no "rejecting" on the sets of x's it
% obtains.  It is designed to generate only those that satisfy all
% the above conditions and to do so with a uniform distribution.
% It accomplishes this by decomposing the space of all possible x
% sets (columns) into n-1 dimensional simplexes.  (Line segments,
% triangles, and tetrahedra, are one-, two-, and three-dimensional
% examples of simplexes, respectively.)  It makes use of three
% different sets of 'rand' variables, one to locate values
% uniformly within each type of simplex, another to randomly
% select representatives of each different type of simplex in
% proportion to their volume, and a third to perform random
% permutations to provide an even distribution of simplex choices
% among like types.  For example, with n equal to 3 and s set at,
% say, 40% of the way from a towards b, there will be 2 different
% types of simplex, in this case triangles, each with its own
% area, and 6 different versions of each from permutations, for
% a total of 12 triangles, and these all fit together to form a
% particular planar non-regular hexagon in 3 dimensions, with v
% returned set equal to the hexagon's area.
%
% Roger Stafford - Jan. 19, 2006

例子：

test <- Surrogate::RandVec(a=1000, b=100000, s=2000000, n=140, m=1, Seed=sample(1:1000, size = 1))
sum(test$RandVecOutput)
# 2000000
hist(test$RandVecOutput)