Hello,
Can anyone explain why this error?
> library(splines)
>
> knots <- c(6, 12, 22, 30, 35)
> x <- c(0.0, .25, 1.0, 2.0, 3.0)
> SCurve <- bs(x = x, knots = knots, intercept = FALSE, Boundary.knots
= c(0,3.5))
> class(SCurve)
[1] "bs" "basis" "matrix"
>
> ?predict.bs
> predict.bs(SCurve, xnew = 40:45)
Error in predict.bs(SCurve, xnew = 40:45) :
could not find function "predict.bs"
Note that splines:::predict.bs(SCurve, xnew = 40:45) works.
> sessionInfo()
R version 3.4.0 (2017-04-21)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1
Matrix products: default
locale:
[1] LC_COLLATE=Portuguese_Portugal.1252
LC_CTYPE=Portuguese_Portugal.1252
[3] LC_MONETARY=Portuguese_Portugal.1252 LC_NUMERIC=C
[5] LC_TIME=Portuguese_Portugal.1252
attached base packages:
[1] splines stats graphics grDevices utils datasets methods
[8] base
loaded via a namespace (and not attached):
[1] compiler_3.4.0 tools_3.4.0
Rui Barradas
Em 28-05-2017 17:37, Rui Barradas escreveu:
Hello,
Since function splines::bs returns an object of class bs in order to
read the help page for function predict you need
?predict.bs
The syntax would be
## S3 method for class 'bs'
predict(object, newx, ...)
Please read that help page and maybe you'll get the answer you need.
Hope this helps,
Rui Barradas
Em 28-05-2017 15:51, Glenn Schultz escreveu:
If is specify a spline basis as follows
knots <- c(6, 12, 22, 30, 35)
x <- c(0.0, .25, 1.0, 2.0, 3.0)
SCurve <- bs(x = x, knots = knots, intercept = FALSE, Boundary.knots =
c(0,3.5))
I would like to now get the spline value for new values of x. However,
when I use predict the new basis is returned and I would like to get the
value. Nothing has worked so far and ?predict plus examples only show
prediction from linear models. Is there a way to extract the value from
a defined spline?
Best Regards,
Glenn
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