This is really a question for R-help: it is not about R development
nor software development in R.
The answer to the only question I see is simple: your model depends on
'fb3' and you supplied 'height'.
On Tue, 9 Feb 2010, Randriamiharisoa Alex wrote:
Hello,
In the following I tried 3 versions of an example in R help. Only
the two first predict command work.
Which 'example in R help'? If you mean that from ?bs, then the
comment about 'safe prediction' is why your third approach (even if
done something like
ph3 <- predict(fm3, data.frame(fb3 = I(bs(ht3, df = 5))))
) would not be correct. You need to predict using the spline with the
knots computed by the original set of data, and predict() is smart
enough to figure that out when it sees a formula involving bs() (look
at str(fm1) to see where the information is stored, and hence how you
could mimic this).
After :
library(splines)
require(stats)
1)
fm1 <- lm(weight ~ bs(height, df = 5), data = women)
ht1 <- seq(57, 73, len = 200)
ph1 <- predict(fm1, data.frame(height=ht1)) # OK
plot(women, xlab = "Height (in)", ylab = "Weight (lb)")
lines(ht1, ph1)
2)
height <- women$height # 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72
weight <- women$weight # 115 117 120 123 126 129 132 135 139 142 146 150
154 159 164
fm2 <- lm(weight ~ bs(height, df = 5))
ht2 <- seq(57, 73, len = 200)
ph2 <- predict(fm2, data.frame(height=ht2)) # OK
plot(women, xlab = "Height (in)", ylab = "Weight (lb)")
lines(ht2,ph2)
3)
height <- women$height # 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72
weight <- women$weight # 115 117 120 123 126 129 132 135 139 142 146 150
154 159 164
fb3 <- bs(height, df = 5)
fm3 <- lm(weight ~ fb3)
ht3 <- seq(57, 73, len = 200)
ph3 <- predict(fm3, data.frame(height=ht3)) # Error message about newdata. Why
?
plot(women, xlab = "Height (in)", ylab = "Weight (lb)")
lines(ht3,ph3) # no line
Thanks for the reason of this message.
Alex Randria
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