On Mon, 28 Jan 2008, Waverley wrote:
> I have a data vector as following:
> > z
> [1] 183.1370 201.9610 113.7250 140.7840 156.2750 42.1569 42.1569 42.1569
> [9] 240.1960 308.4310 42.1569 42.1569 42.1569 42.1569 42.1569 42.1569
> [17] 42.1569 42.1569 42.1569 42.1569 279.8040 42.1569
that seems right
order() gives you the indexes "idx" such that x[idx] == sort(x)
> set.seed(123)
> x <- rnorm(10)
> idx <- order(x)
> identical(x[idx], sort(x))
[1] TRUE
best
b
On Jan 28, 2008, at 8:19 PM, Waverley wrote:
I have a data vector as following:
z
[1] 183.1370 201.9610 113.72
I have a data vector as following:
> z
[1] 183.1370 201.9610 113.7250 140.7840 156.2750 42.1569 42.1569 42.1569
[9] 240.1960 308.4310 42.1569 42.1569 42.1569 42.1569 42.1569 42.1569
[17] 42.1569 42.1569 42.1569 42.1569 279.8040 42.1569 42.1569
when I sort, it gave me the right ord
Hi,
I have an AR(1) series, so I thought that the order of the series should be
1. A simple lm fit with one period lag predicts the series pretty well. But
when I tried ar, I got different orders: ar.mle selected order 6,
ar.burgselected order 14, and
ar.yw selected order 6. So I am wondering mayb
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