Reputation: 1610
I want to define a piecewise function using R, however, my R code goes wrong. Any suggestion is welcome.
x<-seq(-5, 5, by=0.01)
for (x in -5:5){
if (-0.326 < x < 0.652) fx<- 0.632
else if (-1.793<x<-1.304) fx<- 0.454
else if (1.630<x<2.119) fx<-0.227
else fx<- 0 }
Upvotes: 6
Views: 39267
Reputation: 23975
I'm a little late to the party, but I couldn't resist posting a couple more ways to do this. Both take advantage of R capabilities for working with intervals on the real line.
If you define your cut points and function values in the vectors cuts
and vals
like so:
cuts <- c( -Inf, -1.793, -1.304, -0.326, 0.625, 1.630, 2.119 )
vals <- c( 0, 0.454, 0, 0.632, 0, 0.227, 0 )
Then you can use findInterval
to efficiently look up the values of x
in your cutpoints:
fx <- vals[findInterval(x, c(-Inf, cuts))]
If this function needed to do fancier stuff than just look up a constant value, you can put expressions or functions or whatever you want in vals
, possibly using a list
if you want.
Alternatively, since this function is a step function, you can use stepfun
:
f <- stepfun(cuts[-1], vals)
fx <- f(x)
Then you also get to use the nice plotting methods of stepfun
too.
Upvotes: 13
Reputation: 121077
Yet another option, this time using cut
.
regions <- c(-Inf, -1.793, -1.304, -0.326, 0.652, 1.63, 2.119, Inf)
group <- cut(x, regions)
f_values <- c(0, 0.454, 0, 0.632, 0, 0.227, 0)
(fx <- f_values[group])
Upvotes: 2
Reputation: 21502
Unless you have varying cutoff points, I'd use switch
. Here is an example w/ simplified cut values.
xcuts<-1:10 #the values at which you change fx assignment
xx<- seq(1.5,10,5, by =10) #vector of fx values to be selected
switch(max(which(x>xcuts)),
1= fx<-xx[1],
2= fx<-xx[2],
..."et cetera"...
)
Loop over x.
Upvotes: 1
Reputation: 269556
Try this:
x <- seq(-5, 5, 0.01)
fx <- (x > -0.326 & x <0.625) * 0.632 +
(x > -1.793 & x < -1.304) * 0.454 +
(x > 1.630 & x < 2.119) * 0.227
plot(x, fx)
Upvotes: 17
Reputation: 70643
Or you could use ifelse
.
fx <- ifelse(x > -0.326 & x <0.625, 0.632,
ifelse(x > -1.793 & x < -1.304, 0.454,
ifelse(x > 1.630 & x < 2.119, 0.227, 0)))
Upvotes: 12
Reputation: 18420
Maybe if you split the conditions
if((-1.793<x) & (x < 0.652)) ...
EDIT: This seems not to be all, here is a different approach:
x<-seq(-5, 5, by=0.01)
fx <- function(x) {
res <- rep(0, length(x))
res[(-0.326 < x) & (x < 0.652)] <- 0.632
res[(-1.793<x) & (x < (-1.304))] <- 0.454
res[(1.630<x) & (x <2.119)] <- 0.227
return(res)
}
fx(x)
Upvotes: 2