This illustrates why you need to post a reproducible example. You have a
number of confounding factors in your code.
First, "data" is a commonly-used function... avoid using it for variable
names.
Second, using the attach function this way leads to confusion... best to
forget this function until you start building packages.
Third, clka is linearly dependent on lka, so having them both in the
regression is not possible. In this case lm has chosen to ignore clka so
that bs("clka") is NA.
Fourth, curve expects you to give it a function, and instead you have
given it a vector.
Fifth, you are plotting versus lka, but attempting to vary clka in the
curve call.
There are a number of directions you could go with this to get a working
output... below is my version.
dta <- read.table(
"http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt";, header=TRUE )
fit2 <- lm( ruotsi.pist ~ mies + koulu*lka, data=dta )
bs <- coef( fit2 )
rpBylka <- function( lka ) {
kouluB <- factor( "B", levels = levels( dta$koulu ) )
newdta <- expand.grid( mies=0, koulu=kouluB, lka=lka )
predict( fit2, newdata = newdta )
}
dtaKouluB <- subset( dta, koulu == "B" )
varitB <- dtaKouluB$mies
varitB[ varitB == 0 ] <- 2
plot( dtaKouluB$lka
, dtaKouluB$ruotsi.pist
, col=varitB
, pch=16
, xlab='lka'
, ylab='ruotsi.pist'
, main='Lukio B'
)
curve( rpBylka, from = min( dta$lka ), max( dta$lka ), add=TRUE, col="red" )
On Sun, 25 Sep 2016, Matti Viljamaa wrote:
On 25 Sep 2016, at 19:37, Matti Viljamaa <mvilja...@kapsi.fi> wrote:
Okay here?s a pretty short code to reproduce it:
data <-
read.table("http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt";,
header=TRUE)
data$clka <- I(data$lka - mean(data$lka))
attach(data)
fit2 <- lm(ruotsi.pist ~ mies + koulu + lka + koulu*clka)
bs <- coef(fit2)
varitB <- c(data[koulu == 'B',]$mies)
varitB[varitB == 0] = 2
plot(data[data$koulu == 'B',]$lka, data[koulu == 'B',]$ruotsi.pist, col=varitB,
pch=16, xlab='', ylab='', main='Lukio B?)
curve(bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*x+bs["kouluB:clka"]*clka,
from=min(lka), to=max(lka), add=TRUE, col='red')
On 25 Sep 2016, at 19:24, Jeff Newmiller <jdnew...@dcn.davis.ca.us> wrote:
Go directly to C. Do not pass go, do not collect $200.
You think curve does something, but you are missing what it actually does.
Since you don't seem to be learning from reading ?curve or from our responses,
you need to give us an example you can learn from.
--
Sent from my phone. Please excuse my brevity.
On September 25, 2016 9:04:09 AM PDT, mviljamaa <mvilja...@kapsi.fi> wrote:
On 2016-09-25 18:52, Jeff Newmiller wrote:
You seem to be confused about what curve is doing vs. what you are
doing.
But my x-range in curve()'s parameters from and to should be the entire
lka vector, since they are from=min(lka) and to=max(lka). Then why does
this not span the entire of lka? Because of duplicate entries or what?
It seems like I cannot use curve(), since my x-axis must be exactly lka
for the function to plot the y value for every lka value.
A) Compute the points you want to plot and put them into 2 vectors.
Then figure out how to plot those vectors. Then (perhaps) consider
putting that all into one line of code again.
B) The predict function is the preferred way to compute points. It
may
be educational for you to do the computations by hand at first, but
in
the long run using predict will help you avoid problems getting the
equations right in multiple places in your script.
C) Learn what makes an example reproducible (e.g. [1] or [2]), and
ask
your questions with reproducible code and data so we can give you
concrete responses.
[1] http://adv-r.had.co.nz/Reproducibility.html
[2]
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
--
Sent from my phone. Please excuse my brevity.
On September 25, 2016 8:36:49 AM PDT, mviljamaa <mvilja...@kapsi.fi>
wrote:
On 2016-09-25 18:30, Duncan Murdoch wrote:
On 25/09/2016 9:10 AM, Matti Viljamaa wrote:
Writing:
bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*lka+bs["kouluB:clka"]*clka
i.e. without that being inside curve produces a vector of length
375.
So now it seems that curve() is really skipping some
lka-/x-values.
How could curve() know what the length of lka is? You're telling
it
to set x to a sequence of values of length 101 (the default) from
min(lka) to max(lka). You never tell it to set x to lka.
curve() is designed to plot expressions or functions, not vectors.
If
you actually want to plot line segments using your original data,
use
lines(). (You'll likely need to sort your x values into increasing
order if you do that, or you'll get a pretty ugly plot.)
Duncan Murdoch
I know that about curve(), but since this function uses lka as a
parameter, then how should I formulate it for curve so that I don't
get
the error about wrong lengths?
On 25 Sep 2016, at 16:01, Matti Viljamaa <mvilja...@kapsi.fi>
wrote:
I?m trying to plot regression lines using curve()
The way I do it is:
bs <- coef(fit2)
and then for example:
curve(bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*x+bs["kouluB:clka"]*clka,
from=min(lka), to=max(lka), add=TRUE, col='red')
This above code runs into error:
Error in curve(bs["(Intercept)"] + bs["mies"] * 0 + bs["kouluB"]
+
bs["lka"] * :
'expr' did not evaluate to an object of length 'n'
In addition: Warning message:
In bs["(Intercept)"] + bs["mies"] * 0 + bs["kouluB"] + bs["lka"]
*
:
longer object length is not a multiple of shorter object length
Which I?ve investigated might be related to the lengths of the
different objects being multiplied or summed.
Taking length(g$x) or length(g$y) of
g <-
curve(bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*x,
from=min(lka), to=max(lka), add=TRUE, col='red')
returns 101.
However length(lka) is 375. But perhaps these being different is
not
the problem?
I however do see that the whole range of lka is not plotted, for
some
reason. So how can I be sure
that it passes through all x-values in lka? And i.e. that the
lengths
of objects inside curve() are correct?
What can I do?
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