csiro.au> writes:
>
> y <- sort(rnorm(20)) # say...
>
> m <- s <- numeric(19)
>
> for(i in 2:20) {
> m[i-1] <- mean(y[1:i])
> s[i-1] <- sd(y[1:i])
> }
> -Original Message-
> On Behalf Of cvandy
> Subject: [R] Loop with variable index
> I have a list of 20 values. The fir
y <- sort(rnorm(20)) # say...
m <- s <- numeric(19)
for(i in 2:20) {
m[i-1] <- mean(y[1:i])
s[i-1] <- sd(y[1:i])
}
Easy peasy, ...
Bill Venables
CSIRO Laboratories
PO Box 120, Cleveland, 4163
AUSTRALIA
Office Phone (email preferred): +61 7 3826 7251
Fax (if absolutely necessa
a cleaner code would be:
sapply(2:20, function(i) c(mean=mean(x[1:i]), sd=sd(x[1:i])))
b
On Jan 30, 2008, at 3:16 PM, Henrique Dallazuanna wrote:
Try this:
x <- rnorm(20)
sapply(c("sd", "mean"), function(fun)lapply(lapply(lapply(2:20, seq,
from=1), function(.x)x[.x]), fun))
On 30/01/2008,
Try this:
x <- rnorm(20)
sapply(c("sd", "mean"), function(fun)lapply(lapply(lapply(2:20, seq,
from=1), function(.x)x[.x]), fun))
On 30/01/2008, cvandy <[EMAIL PROTECTED]> wrote:
>
> I have a list of 20 values. The first time through a loop I want to find the
> mean and stnd.dev. of the first t
On 31/01/2008, at 8:58 AM, cvandy wrote:
>
> I have a list of 20 values.
***NO***! You have (or should have) a *vector* of 20 values.
Vectors and lists are different concepts. Learn and understand
the difference, else the world will come to an end.
> The first time thr
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