Question

我想生成满足以下等式的数据（w1，w2，w3，w4）：

w1 + w2 + w3 + w4 = 1使得对于所有i，wi> = 0。

那我该怎么办？

Answer 1

gend=function(n){#n is the number of such samples you would like to generate.
  D=matrix(NA,n,4)
  for(i in 1:n){
    a=runif(3)
    b=a[2];d=a[3];a=a[1]
    while((b+a)>1)b=runif(1)
    while((b+a+d)>1)d=runif(1)
    D[i,]=c(a,b,d,1-sum(a,b,d))
  }
 D
}
S=gend(10)
S
           [,1]       [,2]        [,3]         [,4]
 [1,] 0.8599033 0.04730250 0.008108873 0.0846853156
 [2,] 0.3181092 0.37239277 0.243917543 0.0655804751
 [3,] 0.4733250 0.07116504 0.145719661 0.3097903084
 [4,] 0.9482858 0.04967905 0.001082051 0.0009531255
 [5,] 0.7291447 0.16336750 0.106824697 0.0006631124
 [6,] 0.1896754 0.60669666 0.062790600 0.1408372964
 [7,] 0.8452493 0.12875891 0.022590556 0.0034012373
 [8,] 0.4452403 0.52793687 0.011378945 0.0154438773
 [9,] 0.4618185 0.23065407 0.224276892 0.0832504972
[10,] 0.2191794 0.09512862 0.110790973 0.5749009629

rowSums(S)
 [1] 1 1 1 1 1 1 1 1 1 1

Answer 2

没有循环你可以尝试这样的事情：

require(data.table)
n <- 200 # how much data you need
myFunc <- function(n) {
  v <- replicate(3, rnorm(1e6)) # from normal distribution simulate 3 variables
  v <- as.data.table(v)
  v <- v[V1 >= 0 & V2 >= 0 & V3 >= 0]
  v[, s := V1 + V2 + V3]
  v <- v[s <= 1]
  v[, V4 := 1 - s]
  v[, s := NULL]
  v <- v[1:n] # select needed count
  v[]
}
set.seed(32)
myFunc(n
#               V1         V2         V3          V4
#   1: 0.167742364 0.10237134 0.54388085 0.186005445
#   2: 0.268182684 0.15659897 0.23427071 0.340947639
#   3: 0.449311437 0.35508438 0.19126053 0.004343644
#   4: 0.248128979 0.18794309 0.20178770 0.362140224
#   5: 0.235724248 0.02147451 0.68609822 0.056703023
# ---                                              
# 196: 0.748858750 0.10044298 0.14843887 0.002259405
# 197: 0.527629846 0.15944044 0.11169268 0.201237029
# 198: 0.406024037 0.02289703 0.33207412 0.239004809
# 199: 0.024428994 0.31282766 0.61436293 0.048380418
# 200: 0.008766677 0.54692203 0.07853701 0.365774287

all(sapply(myFunc(n), function(x) all(x <= 1 & x >=0)))
# [1] TRUE
all.equal(rowSums(myFunc(n)), rep(1, n))
# [1] TRUE

速度比较：

microbenchmark::microbenchmark(gend(n),
                               myF(n))
# Unit: milliseconds
# expr      min       lq     mean   median       uq       max neval cld
# gend(n) 3.996232 7.034548 21.23260 10.32951 16.37486 272.49600   100   b
# myF(n) 7.723631 9.177185 11.00396 11.27347 13.05472  14.43866   100  a

模拟满足R中线性方程的数据？

2 个答案: