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Suppose observed X(1), X(2), ..., X(N) from a continuous time process. How can i discretize the time of this data on the grid {0,1/N,...,(N-1)/N,1} using R?

I really appreciate any help. Thanks.

Robervaldson Tigrilino
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    Welcome to SO. I'm afraid you'll have to explain this a bit more clearly... What does your data look like in R? What do you want it to look like? Take a read through [this post](http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example) while you're at it. – Justin Oct 18 '13 at 13:49
  • Hello, as @justin points out, we need a bit more information. To point you in the right direction though, you can look at `?cut`, `?cut2` (`library(Hmisc)`), and `?seq` will also be helpful. Depending on your application, the `zoo` package might be helpful as well – Ricardo Saporta Oct 18 '13 at 14:12

1 Answers1

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This would be the way to do it in continuous time:

x <- cumsum(abs(rnorm(20)))
n <- (x-min(x))/diff(range(x))
> n
 [1] 0.00000000 0.01884929 0.02874295 0.07230612 0.11253305 0.19770821 0.26356939
 [8] 0.33310811 0.36687944 0.47041629 0.53331128 0.61724640 0.72534086 0.74782335
[15] 0.79829820 0.83023417 0.85336221 0.85528100 0.90023497 1.00000000

To get a numeric vector analogous to what you might get from cut or Hmisc::cut2 you can use findInterval:

> findInterval(n, seq(0,1,length=length(n) )/length(n) )
 [1]  1  8 11 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20

And "normalizing to [0,1] is then simple, even trivial;

> findInterval(n, seq(0,1,length=length(n) ))/length(n)
 [1] 0.05 0.05 0.05 0.10 0.15 0.20 0.30 0.35 0.35 0.45 0.55 0.60 0.70 0.75 0.80 0.80 0.85
[18] 0.85 0.90 1.00
IRTFM
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