Hi there,
That's exactly what I want.
I have checked ?combn out,
but I could get the following,
suppose that I want ALL possible combinations of them,
as this
==
apply(
combn(paste('x', 1:4, sep =""), 2), 2,
function(v) get(v[1])*get(v[2])
An aside to the main question:
I don't think that
i+1:3
is doing what you think it is.
On 26/03/2010 23:01, casperyc wrote:
Hi,
I am tring to write a loop to compute this,
==
x1=c(
rep(-1,4),
rep(1,4)
)
x2=c(
rep(c(-1,-1,1,1),2)
Hi casperyc,
Here is a suggestion:
# all at once
apply(combn(paste('x', 1:3, sep =""), 2), 2,
function(v) get(v[1])*get(v[2]) )
# step by step
thex <- paste('x', 1:3, sep ="")
thex
combs <- combn(thex, 2)
combs
apply(combs, 2, function(v) get(v[1])*get(v[2]) )
x1, x2 and x3 correspo
Hi,
I am tring to write a loop to compute this,
==
x1=c(
rep(-1,4),
rep(1,4)
)
x2=c(
rep(c(-1,-1,1,1),2)
)
x3=c(
rep(c(-1,1),4)
)
x1*x2
x1*x3
x2*x3
suppose i have x1,x2,x3
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