Well, you can have exactly 70:30%, i.e. 70% 1s and 30% 0s, but
in random order. For example:
Popn <- c(rep(1,70),rep(0,30))
Samp <- sample(Pop)
(see '?sample' for this usage -- the result of sample(x) is a
random permutation of the elements of x).
In probabilistic terms, this is a "condition
On 05/07/2011 2:25 PM, Joshua Wiley wrote:
Hi Ana,
Look at the documentation for ?sample, specifically, the "prob"
argument. In your case this should work:
sample(c(0,1), 100, replace = TRUE, prob = c(.3, .7))
note that you may not have *exactly* 70% 1 and 30%, in any given sample.
And if y
Hi Ana,
Look at the documentation for ?sample, specifically, the "prob"
argument. In your case this should work:
sample(c(0,1), 100, replace = TRUE, prob = c(.3, .7))
note that you may not have *exactly* 70% 1 and 30%, in any given sample.
HTH,
Josh
On Tue, Jul 5, 2011 at 11:21 AM, Ana Kolar
Hi there,
I guess this is an easy one, but still:
I would like to randomly sample 0s and 1s but in a way that I end up having for
example 70% of 1s and the rest of 0s and not 50:50 as this function
does: sample(c(0,1), 100, replace = TRUE)
Any recommendations?
Many thanks!
Ana
[[al
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