传递时R向量松散一个组件

时间:2011-10-17 14:28:29

标签: r

recursiveCall <- function(x, N)
{
    cat("length = ", length(x))
    cat("vector x = ", x[1:2^N], "\n")
    return (x)
}
PaulLevyBrownianMotion <- function(N)
{
    cat("Paul Levy construction for N = ", N, "\n")
    W = c(rnorm(2^N+1, 0, 1))
    cat("length = ", length(W))
    cat("Wstandard = ", W, "\n")
    W <- recursiveCall(W[1:2^N+1], N)
    return (W) 
}

我的向量W在传递给另一个函数时似乎丢失了它的第一个组件。你能帮帮我吗?这是输出。

> W = PaulLevyBrownianMotion(2)
Paul Levy construction for N =  2 
length =  5Wstandard =  0.08641454 1.616638 -0.8747996 0.6149899 0.2689501 
length =  4vector x =  1.616638 -0.8747996 0.6149899 0.2689501 
> 

1 个答案:

答案 0 :(得分:2)

由于优先顺序,

W[1:2^N + 1]没有将您的想法编入索引。首先构造向量1:2^N,然后添加标量1(因此每个元素增加1),从而导致元素2到结尾被选中。