lm combined with splines

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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'.

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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