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Richard Nixon

2003-Apr-02 16:49 UTC

[R] lm with an arbitrary number of terms

Hello folks,

Any ideas how to do this?

data.frame is a data frame with column names "x1",...,"xn"
y is a response variable of length dim(data.frame)[1]

I want to write a function

function(y, data.frame){
    lm(y~x1+...+xn)
}

This would be easy if n was always the same.
If n is arbitrary how could I feed the x1+...+xn terms into lm(response~terms)?

Thanks
Richard

--
Dr. Richard Nixon
MRC Biostatistics Unit, Cambridge, UK
http://www.mrc-bsu.cam.ac.uk/personal/richard
Tel: +44 (0)1223 330382, Fax: +44 (0)1223 33038

Spencer Graves

2003-Apr-02 21:44 UTC

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[R] lm with an arbitrary number of terms

The following might work:

	mdl <- paste("y~", paste(names(data.frame),
collapse="+"))
	lm(mdl, ...)

If y = "y" is a column of your data.frame, you can delete it be 
selecting "names(data.frame)[!is.element(y, names(data.frame)]"

Can you solve the problem from here?
Best Wishes,
Spencer Graves

Richard Nixon wrote:> Hello folks,
> 
> Any ideas how to do this?
> 
> data.frame is a data frame with column names
"x1",...,"xn"
> y is a response variable of length dim(data.frame)[1]
> 
> I want to write a function
> 
> function(y, data.frame){
>     lm(y~x1+...+xn)
> }
> 
> This would be easy if n was always the same.
> If n is arbitrary how could I feed the x1+...+xn terms into
lm(response~terms)?
> 
> Thanks
> Richard
> 
> --
> Dr. Richard Nixon
> MRC Biostatistics Unit, Cambridge, UK
> http://www.mrc-bsu.cam.ac.uk/personal/richard
> Tel: +44 (0)1223 330382, Fax: +44 (0)1223 33038
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://www.stat.math.ethz.ch/mailman/listinfo/r-help

Vadim Ogranovich

2003-Apr-02 21:58 UTC

head link

[R] lm with an arbitrary number of terms

I think you can do it like this

lm(y~., data=data.frame) # note the dot to the right of ~
> -----Original Message-----
> From: Richard Nixon [mailto:richard.nixon at mrc-bsu.cam.ac.uk]
> Sent: Wednesday, April 02, 2003 8:50 AM
> To: r-help at stat.math.ethz.ch
> Subject: [R] lm with an arbitrary number of terms 
> 
> 
> Hello folks,
> 
> Any ideas how to do this?
> 
> data.frame is a data frame with column names
"x1",...,"xn"
> y is a response variable of length dim(data.frame)[1]
> 
> I want to write a function
> 
> function(y, data.frame){
>     lm(y~x1+...+xn)
> }
> 
> This would be easy if n was always the same.
> If n is arbitrary how could I feed the x1+...+xn terms into 
> lm(response~terms)?
> 
> Thanks
> Richard
> 
> --
> Dr. Richard Nixon
> MRC Biostatistics Unit, Cambridge, UK
> http://www.mrc-bsu.cam.ac.uk/personal/richard
> Tel: +44 (0)1223 330382, Fax: +44 (0)1223 33038
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://www.stat.math.ethz.ch/mailman/listinfo/r-help
> 
-------------------------------------------------- 
DISCLAIMER\ This e-mail, and any attachments thereto, is intende... {{dropped}}

Peter Dalgaard BSA

2003-Apr-02 22:21 UTC

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[R] lm with an arbitrary number of terms

Richard Nixon <richard.nixon at mrc-bsu.cam.ac.uk> writes:
> Hello folks,
> 
> Any ideas how to do this?
> 
> data.frame is a data frame with column names
"x1",...,"xn"
> y is a response variable of length dim(data.frame)[1]
> 
> I want to write a function
> 
> function(y, data.frame){
>     lm(y~x1+...+xn)
> }
> 
> This would be easy if n was always the same.
> If n is arbitrary how could I feed the x1+...+xn terms into
lm(response~terms)?
> 
You could build up the formula in a for loop; something like

n <- names(my.data.frame)
f <- as.symbol(n[1])
for (i in 2:length(n))
  f <- substitute(f+a,list(a=as.symbol(n[i])) 
f <- substitute(y~f)

(I didn't try it. Most likely it doesn't quite work.)

However, what's wrong with 

lm(y ~ ., data=my.data.frame)

??

-- 
   O__  ---- Peter Dalgaard             Blegdamsvej 3  
  c/ /'_ --- Dept. of Biostatistics     2200 Cph. N   
 (*) \(*) -- University of Copenhagen   Denmark      Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)             FAX: (+45) 35327907

John Fox

2003-Apr-02 22:21 UTC

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[R] lm with an arbitrary number of terms

Dear Richard,

At 05:49 PM 4/2/2003 +0100, Richard Nixon wrote:
>Any ideas how to do this?
>
>data.frame is a data frame with column names
"x1",...,"xn"
>y is a response variable of length dim(data.frame)[1]
>
>I want to write a function
>
>function(y, data.frame){
>     lm(y~x1+...+xn)
>}
>
>This would be easy if n was always the same.
>If n is arbitrary how could I feed the x1+...+xn terms into 
>lm(response~terms)?
If y contains the *name* of the response variable, which is also in the 
data frame, then the following should work:

fun <- function(y, df){
     lm(as.formula(paste(y, "~.")), data=df)
     }

Alternatively, if y is a vector and is not in the data frame, then you 
might try

fun <- function(y, df){
     df <- data.frame(y, df)
     lm(y~., data=df)
     }

I hope that this helps,
  John
____________________________
John Fox
Department of Sociology
McMaster University
email: jfox at mcmaster.ca
web: http://www.socsci.mcmaster.ca/jfox

Roger Peng

2003-Apr-02 23:48 UTC

head link

[R] lm with an arbitrary number of terms

You need to paste() together a formula.  There's an example in ?formula.
Try,

n <- 10
rhs <- paste("x", 1:n, collapse = "+", sep =
"")
lhs <- "y ~"
f <- as.formula(paste(lhs, rhs))

Then pass `f' into lm().

-roger
_______________________________
UCLA Department of Statistics
rpeng at stat.ucla.edu
http://www.stat.ucla.edu/~rpeng

On Wed, 2 Apr 2003, Richard Nixon wrote:
> Hello folks,
> 
> Any ideas how to do this?
> 
> data.frame is a data frame with column names
"x1",...,"xn"
> y is a response variable of length dim(data.frame)[1]
> 
> I want to write a function
> 
> function(y, data.frame){
>     lm(y~x1+...+xn)
> }
> 
> This would be easy if n was always the same.
> If n is arbitrary how could I feed the x1+...+xn terms into
lm(response~terms)?
> 
> Thanks
> Richard
> 
> --
> Dr. Richard Nixon
> MRC Biostatistics Unit, Cambridge, UK
> http://www.mrc-bsu.cam.ac.uk/personal/richard
> Tel: +44 (0)1223 330382, Fax: +44 (0)1223 33038
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://www.stat.math.ethz.ch/mailman/listinfo/r-help
>

Bill.Venables@csiro.au

2003-Apr-03 04:31 UTC

head link

[R] lm with an arbitrary number of terms

What are you using the data frame argument for?

If y, x1, x2, ..., xn were all the variables in the data frame all you need
do is

myfun <- function(y, dat) {
	lm(y ~ ., dat) 
}

but your question is very vague and ill-posed.


-----Original Message-----
From: Richard Nixon [mailto:richard.nixon at mrc-bsu.cam.ac.uk]
Sent: Thursday, April 03, 2003 2:50 AM
To: r-help at stat.math.ethz.ch
Subject: [R] lm with an arbitrary number of terms 


Hello folks,

Any ideas how to do this?

data.frame is a data frame with column names "x1",...,"xn"
y is a response variable of length dim(data.frame)[1]

I want to write a function

function(y, data.frame){
    lm(y~x1+...+xn)
}

This would be easy if n was always the same.
If n is arbitrary how could I feed the x1+...+xn terms into
lm(response~terms)?

Thanks
Richard

--
Dr. Richard Nixon
MRC Biostatistics Unit, Cambridge, UK
http://www.mrc-bsu.cam.ac.uk/personal/richard
Tel: +44 (0)1223 330382, Fax: +44 (0)1223 33038

______________________________________________
R-help at stat.math.ethz.ch mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help

Darryl

2003-Apr-03 06:31 UTC

head link

[R] lm with an arbitrary number of terms

How about the following?

lm(as.formula(paste("y~",paste(names(data.frame),
collapse="+"),sep=""),cbind(data.frame,y))

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch]On Behalf Of Richard Nixon
Sent: 02 April 2003 19:50
To: r-help at stat.math.ethz.ch
Subject: [R] lm with an arbitrary number of terms


Hello folks,

Any ideas how to do this?

data.frame is a data frame with column names "x1",...,"xn"
y is a response variable of length dim(data.frame)[1]

I want to write a function

function(y, data.frame){
    lm(y~x1+...+xn)
}

This would be easy if n was always the same.
If n is arbitrary how could I feed the x1+...+xn terms into
lm(response~terms)?

Thanks
Richard

--
Dr. Richard Nixon
MRC Biostatistics Unit, Cambridge, UK
http://www.mrc-bsu.cam.ac.uk/personal/richard
Tel: +44 (0)1223 330382, Fax: +44 (0)1223 33038

______________________________________________
R-help at stat.math.ethz.ch mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help

Richard Nixon

2003-Apr-03 10:04 UTC

head link

[R] lm with an arbitrary number of terms

Thanks to James Holtman, Matt Wiener, Douglas Bates, Spencer Graves, John Fox,
Roger Peng, Bill Venable, Darryl Greig, Vadim Ogranovich and Peter Dalgaard

There are several ways to tackel this problem, and one easy one which I missed
:-) 


----------------------------------------------------------------------
Question
data.frame is a data frame with column names "x1",...,"xn"
y is a response variable of length dim(data.frame)[1]

I want to write a function

function(y, data.frame){
    lm(y~x1+...+xn)
}

This would be easy if n was always the same.
If n is arbitrary how could I feed the x1+...+xn terms into
lm(response~terms)?

----------------------------------------------------------------------
y <- c(67,76,56,48,10,43,12,33,88,63,75,14,58,19,14)
x1 <- as.factor(c(2,2,2,2,2,1,1,1,1,1,1,1,1,1,1))
x2 <- c(25,40,35,34,51,24,37,31,26,21,41,45,24,26,29)

data<-data.frame(x1,x2)
----------------------------------------------------------------------
1) James Holtman

fun <- function(y, data.frame){
    .vars <- paste(names(data.frame), collapse="+")
    eval(parse(text=paste('lm(y~',.vars,',data.frame)')))
}

fun(y,data)

Call:
lm(formula = y ~ x1 + x2, data = data.frame)

Coefficients:
(Intercept)          x12           x2  
     84.565       18.763       -1.403  
----------------------------------------------------------------------
2) Matt Wiener

fun <- function(y, data.frame){
    lm(y ~ ., data = data.frame)
}

fun(y,data)

Call:
lm(formula = y ~ ., data = data.frame)

Coefficients:
(Intercept)          x12           x2  
     84.565       18.763       -1.403  

----------------------------------------------------------------------
3) Matt Wiener

fun <- function(y, data.frame){
    .vars <- paste(names(data.frame), collapse="+")
     lm(as.formula(paste('y~',.vars)),data.frame)
}

fun(y,data)

Call:
lm(formula = as.formula(paste("y~", .vars)), data = data.frame)

Coefficients:
(Intercept)          x12           x2  
     84.565       18.763       -1.403  


----------------------------------------------------------------------
Douglas Bates: same as (2)
Spencer Graves: similar to (3)
John Fox: mixture of (2) and (3) 
Roger Peng: similar to (3)
Bill Venable: same as (2)
Darryl Greig: simmilar to (3)
Vadim Ogranovich: same as (2)

Peter Dalgaard
n <- names(my.data.frame)
f <- as.symbol(n[1])
for (i in 2:length(n))
  f <- substitute(f+a,list(a=as.symbol(n[i])) 
f <- substitute(y~f)
(I didn't try it. Most likely it doesn't quite work.)
Neither did I - Richard ;)




--
Dr. Richard Nixon
MRC Biostatistics Unit, Cambridge, UK
http://www.mrc-bsu.cam.ac.uk/personal/richard
Tel: +44 (0)1223 330382, Fax: +44 (0)1223 33038

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