R 随机数生成序列
我需要创建一个包含五个变量的函数R 随机数生成序列,r,R,我需要创建一个包含五个变量的函数 a(乘数) n(样本量) c(以默认值0递增) m(模数) x0(初始种子值) 我需要用这个方程生成一个随机数序列 xi=(a*xi-1+c)(mod m),i=1,2,…,n 如向量x=(x1,…,xn)所示 我的尝试: my.unif1(0){ x[n]听起来您想了解更多关于线性同余生成器的信息。以下资源可能会帮助您解决代码问题: lcg可以帮助: my.fct.1 <- function(x, multiplier, increment,
- a(乘数)
- n(样本量)
- c(以默认值0递增)
- m(模数)
- x0(初始种子值)
- xi=(a*xi-1+c)(mod m),i=1,2,…,n
my.unif1(0){
x[n]听起来您想了解更多关于线性同余生成器的信息。以下资源可能会帮助您解决代码问题:
lcg可以帮助:
my.fct.1 <- function(x, multiplier, increment, modulus){
increment <- ifelse(missing(increment), 0, increment) # setting the default increment to 0
newval <- (multiplier*x + increment) %% modulus
return(newval)
}
my.fct.2 <- function(x0, n, multiplier, increment, modulus){
if(n == 1){
val <- my.fct.1(x = x0, multiplier = multiplier, increment = increment, modulus = modulus)
vec <- c(x0, val)
return(vec)
}
if(n > 1){
vec <- my.fct.2(x = x0, n = n-1, multiplier = multiplier, increment = increment, modulus = modulus)
val <- vec[length(vec)]
newval <- my.fct.1(x = val, multiplier = multiplier, increment = increment, modulus = modulus)
newvec <- c(vec, newval)
return(newvec)
}
}
my.fct.1 <- function(x, multiplier, increment, modulus){
increment <- ifelse(missing(increment), 0, increment) # setting the default increment to 0
newval <- (multiplier*x + increment) %% modulus
return(newval)
}
my.fct.2 <- function(x0, n, multiplier, increment, modulus){
if(n == 1){
val <- my.fct.1(x = x0, multiplier = multiplier, increment = increment, modulus = modulus)
vec <- c(x0, val)
return(vec)
}
if(n > 1){
vec <- my.fct.2(x = x0, n = n-1, multiplier = multiplier, increment = increment, modulus = modulus)
val <- vec[length(vec)]
newval <- my.fct.1(x = val, multiplier = multiplier, increment = increment, modulus = modulus)
newvec <- c(vec, newval)
return(newvec)
}
}
> my.fct.2(3, 9, 7, -1, 4)
[1] 3 0 3 0 3 0 3 0 3 0
> my.fct.2(1, 9, 2, 1, 13)
[1] 1 3 7 2 5 11 10 8 4 9
> my.fct.2(0, 17, 5, 3, 7)
[1] 0 3 4 2 6 5 0 3 4 2 6 5 0 3 4 2 6 5
# and here the arguments set to cross check it against @mysteRious's answer
> my.fct.2(5, 20, 6, 7, 23)
[1] 5 14 22 1 13 16 11 4 8 9 15 5 14 22 1 13 16 11 4 8 9
U <- my.fct.2(5, 20, 6, 7, 23)/23
> U
[1] 0.21739130 0.60869565 0.95652174 0.04347826 0.56521739 0.69565217 0.47826087 0.17391304
[9] 0.34782609 0.39130435 0.65217391 0.21739130 0.60869565 0.95652174 0.04347826 0.56521739
[17] 0.69565217 0.47826087 0.17391304 0.34782609 0.39130435