On Apr 10, 2012, at 8:59 AM, David Winsemius wrote:
On Apr 10, 2012, at 3:16 AM, aajit75 wrote:
I have got solution using within function as below
dd$Seg <- 1
dd <- within(dd, Seg[x2> 0 & x3> 200] <- 1)
In this instance the first of your assignments appears superfluous.
dd <- within(dd, Seg[x2> 100 & x3> 300] <- 2)
dd <- within(dd, Seg[x2> 200 & x3> 400] <- 3)
dd <- within(dd, Seg[x2> 300 & x3> 500] <- 4)
I sthere any better way of doing it!!
dd <- with(dd, Seg <- min(findInterval(x2, c(-Inf ,100, 200, 300,
Inf) ),
findInterval(x3, C(-Inf ,300, 400, 500,
Inf) )
) )
Should have labeled that as an untested guess. A couple of typoes. And
should have assigned to $Seg since I was using `with` instead within
as you used AND the cap C is wrong AND should have used `pmin` instead
of 'min'
dd$Seg <- with(dat, pmin( findInterval(x2, c(-Inf,100, 200, 300,
Inf) ) ,
findInterval(x3, c(-Inf ,300, 400, 500,
Inf) ) ) )
Tested with the dataset offered in another question yesterday:
with(dat, table( findInterval(x, c(-Inf, -2, -1.5, -1, Inf) ) ,
findInterval(y, c(-Inf ,0, .05, .1, Inf) ) ) )
#---------
1 2 3 4
1 0 3 9 40
2 11 9 11 5
3 21 5 7 0
4 6 2 1 0
#---------
with(dat, pmin( findInterval(x, c(-Inf, -2, -1.5, -1, Inf) ) ,
findInterval(y, c(-Inf ,0, .05, .1,
Inf) ) ) )
#---------
[1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
[48] 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 1 1 1 1 3
3 2 2 2 2 2 1 1 1 1 1 1 3 3 3
[95] 2 2 2 1 1 1 1 1 1 1 3 3 2 2 1 1 1 1 1 1 1 1 1 3 2 2 1 1 1 1 1 1
1 1 1 1
David Winsemius, MD
West Hartford, CT
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