Using rbinom might be more clear than my sample() solution. May I
speculate without actual testing that:
inv <- 2*rbinom(length(vec),1,0.5) - 1
might be more efficient than a 2-step rbinom / ifelse strategy?
--
David Winsemius
Heritage Labs
On Nov 18, 2008, at 5:21 PM, joris meys wrote:
The function rbinom might be a solution.
Try following simple program :
vec <- c(-1,1,-1,1,1,-1,-1,1,1)
inv <-rbinom(length(vec),1,0.5)
inv <-ifelse(inv==0,-1,1)
vec2 <- vec*inv #switches sign with p=0.5
In this, inv is a random binomial vector, where the probability for
being 1 is 0.5 in all positions. Changing the values of 0 in this
vector to -1, gives you a vector you can multiply with the original
one to change the sign with a probability of 0.5 for all positions
Kind regards
Joris
On Tue, Nov 18, 2008 at 6:11 PM, Salas, Andria Kay
<[EMAIL PROTECTED]> wrote:
To clear up a question regarding my earlier posting regarding
random changes in a vector:
Say you have a vector (1, -1, -1, 1, 1, -1). I want each value in
the vector to have a probability that it will change signs when
this vector is regenerated. For example, probability = 50%. When
the vector is regenerated, the first value in the vector (1) has a
50% chance of switching to -1. If I regenerated this vector 10
times, 5 of the times it would switch to -1. Similarly, I need
each value in the vector to have this same probability of switching
signs when the vector is regenerated, and each value's chances of
doing so is independent of the other values. So the second value
(-1) also has a 50% chance of switching to 1, and whether or not it
does so is independent of if the first value changes from 1 to -1
(also a 50% probability).
I hope this clears up the confusion, and I would appreciate any
help anyone can provide!
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