Matti Viljamaa
2016-Sep-25 17:21 UTC
[R] curve() doesn't seem to use the whole range of x? And Error: longer object length is not a multiple of shorter object length
> On 25 Sep 2016, at 19:37, Matti Viljamaa <mviljamaa at 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 <jdnewmil at 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 <mviljamaa at 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 <mviljamaa at 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 <mviljamaa at 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? >>>>>>> >>>>>>> ______________________________________________ >>>>>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >>>>>>> https://stat.ethz.ch/mailman/listinfo/r-help >>>>>>> PLEASE do read the posting guide >>>>>>> http://www.R-project.org/posting-guide.html >>>>>>> and provide commented, minimal, self-contained, reproducible code. >>>>>>> >>>>> >>>>> ______________________________________________ >>>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >>>>> https://stat.ethz.ch/mailman/listinfo/r-help >>>>> PLEASE do read the posting guide >>>>> http://www.R-project.org/posting-guide.html >>>>> and provide commented, minimal, self-contained, reproducible code. >> >
Jeff Newmiller
2016-Sep-25 18:23 UTC
[R] curve() doesn't seem to use the whole range of x? And Error: longer object length is not a multiple of shorter object length
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$lka ), add=TRUE,
col="red" )
On Sun, 25 Sep 2016, Matti Viljamaa wrote:
>
>> On 25 Sep 2016, at 19:37, Matti Viljamaa <mviljamaa at kapsi.fi>
wrote:
>>
>> Okay here?s a pretty short code to reproduce it:
>>
>> data <-
read.table("http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt",
header=TRUE)
>
> data$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 <jdnewmil at
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 <mviljamaa at
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
<mviljamaa at 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
<mviljamaa at 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?
>>>>>>>>
>>>>>>>> ______________________________________________
>>>>>>>> R-help at r-project.org mailing list -- To
UNSUBSCRIBE and more, see
>>>>>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>>>>>> PLEASE do read the posting guide
>>>>>>>> http://www.R-project.org/posting-guide.html
>>>>>>>> and provide commented, minimal, self-contained,
reproducible code.
>>>>>>>>
>>>>>>
>>>>>> ______________________________________________
>>>>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE
and more, see
>>>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>>>> PLEASE do read the posting guide
>>>>>> http://www.R-project.org/posting-guide.html
>>>>>> and provide commented, minimal, self-contained,
reproducible code.
>>>
>>
>
>
---------------------------------------------------------------------------
Jeff Newmiller The ..... ..... Go Live...
DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
Matti Viljamaa
2016-Sep-26 16:41 UTC
[R] curve() doesn't seem to use the whole range of x? And Error: longer object length is not a multiple of shorter object length
Thank you. However, I?m having some trouble converting your code to use clka, because the model I was using was: fit2 <- lm(ruotsi.pist ~ mies + koulu + clka + koulu*clka, data=dta)> On 25 Sep 2016, at 21:23, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > > 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$lka ), add=TRUE, col="red" ) > > On Sun, 25 Sep 2016, Matti Viljamaa wrote: > >> >>> On 25 Sep 2016, at 19:37, Matti Viljamaa <mviljamaa at kapsi.fi> wrote: >>> >>> Okay here?s a pretty short code to reproduce it: >>> >>> data <- read.table("http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt", header=TRUE) >> >> data$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 <jdnewmil at 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 <mviljamaa at 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 <mviljamaa at 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 <mviljamaa at 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? >>>>>>>>> >>>>>>>>> ______________________________________________ >>>>>>>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >>>>>>>>> https://stat.ethz.ch/mailman/listinfo/r-help >>>>>>>>> PLEASE do read the posting guide >>>>>>>>> http://www.R-project.org/posting-guide.html >>>>>>>>> and provide commented, minimal, self-contained, reproducible code. >>>>>>>>> >>>>>>> >>>>>>> ______________________________________________ >>>>>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >>>>>>> https://stat.ethz.ch/mailman/listinfo/r-help >>>>>>> PLEASE do read the posting guide >>>>>>> http://www.R-project.org/posting-guide.html >>>>>>> and provide commented, minimal, self-contained, reproducible code. >>>> >>> >> >> > > --------------------------------------------------------------------------- > Jeff Newmiller The ..... ..... Go Live... > DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live Go... > Live: OO#.. Dead: OO#.. Playing > Research Engineer (Solar/Batteries O.O#. #.O#. with > /Software/Embedded Controllers) .OO#. .OO#. rocks...1k > ---------------------------------------------------------------------------