Re: [R] interpolation

[Moshe Olshansky]

If your data is in data.txt file you can do the
following:

x <- read.table(file="data.txt",header=TRUE)
t<-ISOdatetime(x[,1],x[,2],x[,3],x[,4],x[,5],x[,6])
secs <- as.numeric(t)

So now secs represents your time in seconds and you
can use any type of interpolation you wish to
interpolate co2obs.

--- Yogesh Tiwari <[EMAIL PROTECTED]> wrote:

> Hello R Users,
> 
> I am new to R and I have simple problem for R users.
> 
> I have CO2 observations defined on time
> axis(yr,mo,day,hr,min,sec). (DATA
> ATTACHED HERE)
> 
> First I want to convert time axis as one axis as
> 'hour' on regular interval
> as 1 hour. Say 00 hrs to 24hrs(jan1), 25hrs to
> 48hrs(jan2) and so on.
> 
> Then I want to interpolate CO2 at every hour.
> 
> Kindly anybody can help,
> 
> Many thanks,
> 
> Regards,
> Yogesh
> > yr  mo  dy  hr  min sec co2obs
> 1998  1   9   0   35  24  365.19
> 1998  1   9   1   17  39  363.54
> 1998  1   9   1   58  41  364.24
> 1998  1   9   2   39  42  364.88
> 1998  1   9   3   20  43  365.06
> 1998  1   9   4   1   44  364.75
> 1998  1   9   4   42  45  364.77
> 1998  1   9   5   23  46  364.87
> 1998  1   9   6   4   51  364.77
> 1998  1   9   6   45  52  364.73
> 1998  1   9   7   26  54  364.76
> 1998  1   9   8   7   55  363.49
> 1998  1   23  1   6   16  364.31
> 1998  1   23  1   48  32  364.38
> 1998  1   23  2   29  33  364.67
> 1998  1   23  3   10  35  365.53
> 1998  1   23  3   51  36  365.16
> 1998  1   23  4   32  37  364.56
> 1998  1   23  5   13  40  364.62
> 1998  1   23  5   54  41  365.05
> 1998  1   23  6   35  42  365.13
> 1998  1   23  7   16  44  365.45
> 1998  1   23  7   57  49  364.77
> 1998  1   23  8   38  49  364.65
> 1998  2   3   0   44  21  362.43
> 1998  2   3   1   26  36  363.96
> 1998  2   3   2   7   38  364.59
> 1998  2   3   2   48  40  364.62
> 1998  2   3   3   29  40  366.22
> 1998  2   3   4   10  41  365.4
> 1998  2   3   4   51  41  365.34
> 1998  2   3   5   32  42  365.31
> 1998  2   3   6   13  44  364.84
> 1998  2   3   6   54  46  365.07
> 1998  2   3   7   35  51  364.84
> 1998  2   3   8   16  50  364.81
> 1998  2   19  0   35  14  363.41
> 1998  2   19  1   17  31  362.93
> 1998  2   19  1   58  32  363.86
> 1998  2   19  2   39  33  364.87
> 1998  2   19  3   20  34  366.05
> 1998  2   19  4   1   36  364.84
> 1998  2   19  4   42  36  364.77
> 1998  2   19  5   23  37  365.01
> 1998  2   19  6   4   39  365
> 1998  2   19  6   45  39  365.9
> 1998  2   19  7   26  40  366.24
> 1998  2   19  8   7   41  366.55
> 1998  3   5   0   50  20  363.13
> 1998  3   5   1   32  37  363.6
> 1998  3   5   2   13  39  364.26
> 1998  3   5   2   54  40  364.26
> 1998  3   5   3   35  41  364.39
> 1998  3   5   4   16  42  365.24
> 1998  3   5   4   57  42  365.48
> 1998  3   5   5   38  43  365.01
> 1998  3   5   6   19  44  365.43
> 1998  3   5   7   0   45  365.11
> 1998  3   5   7   41  46  368.54
> 1998  3   5   8   22  48  364.96
> 1998  3   19  0   46  36  363.25
> 1998  3   19  1   28  51  363.8
> 1998  3   19  2   9   53  364.21
> 1998  3   19  2   50  55  364.46
> 1998  3   19  3   31  58  365.61
> 1998  3   19  4   12  59  365.57
> 1998  3   19  4   53  59  365.53
> 1998  3   19  5   35  0   365.38
> 1998  3   19  6   16  3   366.23
> 1998  3   19  6   57  4   364.28
> 1998  3   19  7   38  6   367.08
> 1998  3   19  8   19  5   369.19
> 1998  4   2   1   8   1   363.76
> 1998  4   2   1   50  17  365.14
> 1998  4   2   2   31  18  365.26
> 1998  4   2   3

Re: [R] Interpolation in time

[Gabor Grothendieck]

na.approx(zoo(ta[,-seq(2)], tt))

where tt is as before.


On 10/6/05, Anette Nørgaard <[EMAIL PROTECTED]> wrote:
> This is exactly what I requested, thank you!! However I do actually have
> several columns in my data sheet where I need to do the same thing, then
> how do I come about that?
>
> e.g.
>
> yr<-c(rep(2000,14))
> doy<-c(16:29)
> dat1<-c(3.2,NA,NA,NA,NA,NA,NA,5.1,NA,NA,NA,NA,NA,4.6)
> dat2<-c(2.2,NA,NA,NA,NA,NA,NA,6.1,NA,NA,NA,NA,NA,4.2)
> dat3<-c(3.4,NA,NA,NA,NA,NA,NA,4.1,NA,NA,NA,NA,NA,4.7)
> ta<-cbind(yr,doy,dat1,dat2,dat3)
>
>
>
>[[alternative HTML version deleted]]
>
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>

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Re: [R] Interpolation in time

[Gabor Grothendieck]

Is doy intended to represent the number of days since the beginning
of the year?  In that case convert the first two columns to class Date
and interpolate using approx.  See ?approx for variations:

tt <- as.Date(paste(yr, 1, 1, sep = "-")) + doy - 1
ta[,"dat"] <- approx(tt, dat, tt)$y

Even better would be to create an irregular time series object.

library(zoo)
tt <- as.Date(paste(yr, 1, 1, sep = "-")) + doy - 1
ta.z <- na.approx(zoo(dat, tt))

Now ta.z is a zoo object representing your time series. coredata(ta.z)
is the data and time(ta.z) are the dates.  See:

library(zoo)
vignette("zoo")

for more info.




On 10/6/05, Anette Nørgaard <[EMAIL PROTECTED]> wrote:
> Can anybody help me write a code on the following data example, which
> fills out all NA values by using a linear interpolation with the two
> closest values?
>
> Doy is day of year (%j).
>
> Code example:
> yr<-c(rep(2000,14))
> doy<-c(16:29)
> dat<-c(3.2,NA,NA,NA,NA,NA,NA,5.1,NA,NA,NA,NA,NA,4.6)
> ta<-cbind(yr,doy,dat)
>
> ta
>  yr doy dat
>  [1,] 2000  16 3.2
>  [2,] 2000  17  NA
>  [3,] 2000  18  NA
>  [4,] 2000  19  NA
>  [5,] 2000  20  NA
>  [6,] 2000  21  NA
>  [7,] 2000  22  NA
>  [8,] 2000  23 5.1
>  [9,] 2000  24  NA
> [10,] 2000  25  NA
> [11,] 2000  26  NA
> [12,] 2000  27  NA
> [13,] 2000  28  NA
> [14,] 2000  29 4.6
>
> Anette Norgaard
>
>[[alternative HTML version deleted]]
>
> __
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>

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Re: [R] Interpolation in time

[Achim Zeileis]

On Thu, 6 Oct 2005 16:10:15 +0200 Anette Nørgaard wrote:

> This is exactly what I requested, thank you!! However I do actually
> have several columns in my data sheet where I need to do the same
> thing, then how do I come about that?

Look at na.approx() in package zoo.
Best,
Z
  
> e.g.
>  
> yr<-c(rep(2000,14))
> doy<-c(16:29)
> dat1<-c(3.2,NA,NA,NA,NA,NA,NA,5.1,NA,NA,NA,NA,NA,4.6)
> dat2<-c(2.2,NA,NA,NA,NA,NA,NA,6.1,NA,NA,NA,NA,NA,4.2)
> dat3<-c(3.4,NA,NA,NA,NA,NA,NA,4.1,NA,NA,NA,NA,NA,4.7)
> ta<-cbind(yr,doy,dat1,dat2,dat3)
>  
>  
> 
>   [[alternative HTML version deleted]]
> 
> __
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> http://www.R-project.org/posting-guide.html
>

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Re: [R] Interpolation in time

[jim holtman]

Is this what you want?
 > yr<-c(rep(2000,14))
> doy<-c(16:29)
> dat<-c(3.2,NA,NA,NA,NA,NA,NA,5.1,NA,NA,NA,NA,NA,4.6)
> ta<-cbind(yr,doy,dat)
> ta
yr doy dat
[1,] 2000 16 3.2
[2,] 2000 17 NA
[3,] 2000 18 NA
[4,] 2000 19 NA
[5,] 2000 20 NA
[6,] 2000 21 NA
[7,] 2000 22 NA
[8,] 2000 23 5.1
[9,] 2000 24 NA
[10,] 2000 25 NA
[11,] 2000 26 NA
[12,] 2000 27 NA
[13,] 2000 28 NA
[14,] 2000 29 4.6
> good <- !is.na(ta[,'dat'])
> x.f <- approxfun(ta[good,'doy'], ta[good,'dat'], rule=2)
> ta[!good, 'dat'] <- x.f(ta[!good, 'doy'])
> ta
yr doy dat
[1,] 2000 16 3.20
[2,] 2000 17 3.471429
[3,] 2000 18 3.742857
[4,] 2000 19 4.014286
[5,] 2000 20 4.285714
[6,] 2000 21 4.557143
[7,] 2000 22 4.828571
[8,] 2000 23 5.10
[9,] 2000 24 5.016667
[10,] 2000 25 4.93
[11,] 2000 26 4.85
[12,] 2000 27 4.77
[13,] 2000 28 4.68
[14,] 2000 29 4.60
>


 On 10/6/05, Anette Nørgaard <[EMAIL PROTECTED]> wrote:
>
> Can anybody help me write a code on the following data example, which
> fills out all NA values by using a linear interpolation with the two
> closest values?
>
> Doy is day of year (%j).
>
> Code example:
> yr<-c(rep(2000,14))
> doy<-c(16:29)
> dat<-c(3.2,NA,NA,NA,NA,NA,NA,5.1,NA,NA,NA,NA,NA,4.6)
> ta<-cbind(yr,doy,dat)
>
> ta
> yr doy dat
> [1,] 2000 16 3.2
> [2,] 2000 17 NA
> [3,] 2000 18 NA
> [4,] 2000 19 NA
> [5,] 2000 20 NA
> [6,] 2000 21 NA
> [7,] 2000 22 NA
> [8,] 2000 23 5.1
> [9,] 2000 24 NA
> [10,] 2000 25 NA
> [11,] 2000 26 NA
> [12,] 2000 27 NA
> [13,] 2000 28 NA
> [14,] 2000 29 4.6
>
> Anette Norgaard
>
> [[alternative HTML version deleted]]
>
> __
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> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
>



--
Jim Holtman
Cincinnati, OH
+1 513 247 0281

What the problem you are trying to solve?

[[alternative HTML version deleted]]

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Re: [R] interpolation function

[Liaw, Andy]

I don't know if what Hao wanted to do is to hang himself, but if so, perhaps
here is one possible rope:

> idx <- which(m > 0, arr.ind=TRUE)
> m.nz <- m[idx]
> library(akima)
> (m.int <- interp.new(idx[,1], idx[,2], m.nz, xo=1:9, yo=1:9, extrap=TRUE))
$x
[1] 1 2 3 4 5 6 7 8 9

$y
[1] 1 2 3 4 5 6 7 8 9

$z
  [,1]  [,2]  [,3]   [,4]   [,5]   [,6]
[,7]
 [1,]   57.1356740  55.18522 38.413800  18.086957   2.30   0.737908  22.
544616
 [2,]   18.8241001  19.35348  3.10 -20.355403 -41.610566 -51.308028 -40.
090327
 [3,]3.2234619   8.63630 -5.312670 -29.221287 -53.732091 -69.487619 -67.
130408
 [4,]1.100  13.75521  4.290890 -17.935507 -43.566514 -63.244671 -67.
612515
 [5,]3.2013295  25.25298 22.408739   4.00 -20.615776 -42.081126 -51.
038588
 [6,]0.2005624  33.62766 39.538938  27.083292   5.618185 -15.498922 -26.
910566
 [7,]  -17.2478151  29.37731 46.179546  41.812431  25.633428   7.00  -4.
730390
 [8,]  -58.4893169   3.0 32.828625  38.685475  29.928013  15.913700   6.
00
 [9,] -130.5161684 -53.43667 -9.550716   8.258617   9.00   1.817369  -3.
604296
 [,8]   [,9]
 [1,]  76.7287988 170.407375
 [2,]   1.400  80.913745
 [3,] -37.3029968  28.876027
 [4,] -47.3125835   6.953079
 [5,] -38.1307000   6.00
 [6,] -19.2592860  16.750369
 [7,]  -0.2080329  29.451691
 [8,]   9.3350825  34.229953
 [9,]   0.8303142  21.211142

Note how wild the interpolated/extrapolated values can be...

Andy


> From: Adaikalavan Ramasamy
> 
> I do not understand your question. If this was not a sparse 
> matrix, then
> I would have asked you refer into the missing value literature. Even
> there, people generally remove any columns/rows that have too many
> missing values to avoid unreliable results. 
> 
> And since this is a sparse matrix, you are going to have too many
> missing values on all rows and columns. I could be wrong but if I am,
> someone will tell me that soon enough.
> 
> Regards, Adai
> 
> 
> 
> On Fri, 2005-08-05 at 12:12 +, 吴 昊 wrote:
> > Hi,
> > I have a sparse matrix.I want to fill values into the 
> entries whose value 
> > is 0.The new generated values should come from the 
> interpolation of the 
> > values have existed.Does R provide such interpolation 
> functions which 
> > operate on Matrix, for example ,such a matrix below
> >  0  0  0  0  2.3  0 0  0  0 
> >  0  0 3.1 0   0   0 0 1.4 0
> >  0  0  0  0   0   0 0  0  0
> > 1.1 0  0  0   0   0 0  0  0
> >  0  0  0  4   0   0 0  0  6
> >  0  0  0  0   0   0 0  0  0
> >  0  0  0  0   0   7 0  0  0
> >  0  3  0  0   0   0 6  0  0
> >  0  0  0  0   9   0 0  0  0
> > thanks a lot
> > hao wu
> > 
> > __
> > R-help@stat.math.ethz.ch mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide! 
> http://www.R-project.org/posting-guide.html
> 
> 
> __
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> http://www.R-project.org/posting-guide.html
>

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Re: [R] interpolation function

[tom wright]

On Fri, 2005-05-08 at 12:12 +, 吴 昊 wrote:
> Hi,
> I have a sparse matrix.I want to fill values into the entries whose value 
> is 0.The new generated values should come from the interpolation of the 
> values have existed.Does R provide such interpolation functions which 
> operate on Matrix, for example ,such a matrix below
>  0  0  0  0  2.3  0 0  0  0 
>  0  0 3.1 0   0   0 0 1.4 0
>  0  0  0  0   0   0 0  0  0
> 1.1 0  0  0   0   0 0  0  0
>  0  0  0  4   0   0 0  0  6
>  0  0  0  0   0   0 0  0  0
>  0  0  0  0   0   7 0  0  0
>  0  3  0  0   0   0 6  0  0
>  0  0  0  0   9   0 0  0  0
> thanks a lot
> hao wu

Does this look like an answer? If anyone knows a better way please tell
me.

> d<-c(0,2,3,2,0,3,4,0,0,0,0,0)
> d.mat<-matrix(data=d,nrow=4,ncol=3,byrow=TRUE)
> for(i in 1:length(d.mat[1,])){
+ d.mat[,i][d.mat[,i]==0]<-mean(d.mat[,i][d.mat[,i]>0])
+ }

Thanks
Tom

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Re: [R] interpolation function

[Spencer Graves]

  I hope you don't take offense at anything said on this list.  My 
philosophy about this is summarized in something I wrote last December 
that has since been immoratlized in the "fortunes" package:

 > library(fortunes)
 > fortune("Spencer Graves")

Our great-great grandchilren as yet unborn may read some of the stupid
questions and/or answers that I and perhaps others give from time to 
time. I'd
rather get flamed for saying something stupid in public on this list than to
continue to provide substandard service to the people with whom I work 
because
I perpetrated the same mistake in an environment in which no one 
questioned so
effectively my errors.
-- Spencer Graves (in a discussion on whether answers on R-help 
should be
   more polite)
   R-help (December 2004)

  Best Wishes,
  spencer graves

Rolf Turner wrote:

> What you intend strikes me as being pretty silly.  Do not expect R to
> work magic for you.  Even if there were such a function as you desire
> in R, the results it would give would be effectively meaningless for
> data such as you exhibited.
> 
>   cheers,
> 
>   Rolf Turner
>   [EMAIL PROTECTED]
> 
> __
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-- 
Spencer Graves, PhD
Senior Development Engineer
PDF Solutions, Inc.
333 West San Carlos Street Suite 700
San Jose, CA 95110, USA

[EMAIL PROTECTED]
www.pdf.com 
Tel:  408-938-4420
Fax: 408-280-7915

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Re: [R] interpolation function

[Adaikalavan Ramasamy]

I do not understand your question. If this was not a sparse matrix, then
I would have asked you refer into the missing value literature. Even
there, people generally remove any columns/rows that have too many
missing values to avoid unreliable results. 

And since this is a sparse matrix, you are going to have too many
missing values on all rows and columns. I could be wrong but if I am,
someone will tell me that soon enough.

Regards, Adai



On Fri, 2005-08-05 at 12:12 +, 吴 昊 wrote:
> Hi,
> I have a sparse matrix.I want to fill values into the entries whose value 
> is 0.The new generated values should come from the interpolation of the 
> values have existed.Does R provide such interpolation functions which 
> operate on Matrix, for example ,such a matrix below
>  0  0  0  0  2.3  0 0  0  0 
>  0  0 3.1 0   0   0 0 1.4 0
>  0  0  0  0   0   0 0  0  0
> 1.1 0  0  0   0   0 0  0  0
>  0  0  0  4   0   0 0  0  6
>  0  0  0  0   0   0 0  0  0
>  0  0  0  0   0   7 0  0  0
>  0  3  0  0   0   0 6  0  0
>  0  0  0  0   9   0 0  0  0
> thanks a lot
> hao wu
> 
> __
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Re: [R] interpolation function

[Rolf Turner]

What you intend strikes me as being pretty silly.  Do not expect R to
work magic for you.  Even if there were such a function as you desire
in R, the results it would give would be effectively meaningless for
data such as you exhibited.

cheers,

Rolf Turner
[EMAIL PROTECTED]

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Re: [R] interpolation function in R

[Sundar Dorai-Raj]

吴 昊 wrote:
> Hi
> does R provide some interpolation fucntions?
> thank
> 
> 
> 
> 
> __
> R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide!

help.search("interpolation")

returns

interpSpline(splines)   Create an Interpolation Spline
periodicSpline(splines)
Create a Periodic Interpolation Spline
NLSstClosestX(stats)Inverse Interpolation
approx(stats)   Interpolation Functions
spline(stats)   Interpolating Splines

for example. There may be others in packages I don't have installed as well.

HTH,

--sundar

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Re: [R] interpolation function in R

[Rolf Turner]

?approx

?splinefun

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Re: [R] Interpolation of azimuth values

[Prof Brian Ripley]

On Thu, 16 Oct 2003, temiz wrote:

> I will make an interpolation of data which represents azimuth direction
> ( angle from north in clockwise direction) values.
> But there is a problem.
> Say, for instance, while 1 and 359 indicate somewhat same direction, 
> interpolation puts  values
> in the range from 1 to 359.  What can I do to solve the problem ?
> 
> Anything you offer ?

The usual solution is to extend the data periodically, that is to put 
copies shifted by one rotation on each side.

-- 
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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Re: [R] interpolation methods

[Kris Nackaerts]

Dear,

Have a look at:

*akima : 
Interpolation of irregularly spaced data*

   Linear or cubic spline interpolation for irregular gridded data

Kris

 



--

http://perswww.kuleuven.ac.be/~u0027178/VCard/mycard.php?name=krisn

http://gloveg.kuleuven.ac.be/


Minds are like parachutes, they only work when open
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RE: [R] Interpolation

[Adaikalavan Ramasamy]

Here is another way of doing it:

x <- c(1990, 1994, 1995, 1997)
all.x <- seq(min(x), max(x))
complementary.x <- setdiff(all.x, x)



-Original Message-
From: Sundar Dorai-Raj [mailto:[EMAIL PROTECTED]] 
Sent: Wednesday, February 12, 2003 9:39 PM
To: Remigijus Lapinskas
Cc: [EMAIL PROTECTED]
Subject: Re: [R] Interpolation




Remigijus Lapinskas wrote:
> Many thanks to all who replied to my e-mail. My problem was that I had

> not known about the approx function.
> 
> By the way, if I have  x <- c(1990,1994,1995,1997), is there an 
> automated way to fill in the gaps, i.e., to get 
> c(1991,1992,1993,1996)?
> 

Try this:

R> x <- c(1990, 1994, 1995, 1997)
R> all.x <- seq(min(x), max(x))
R> missing.x <- all.x[!all.x %in% x]
R> missing.x
[1] 1991 1992 1993 1996
R>

Regards,
Sundar

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Re: [R] Interpolation

[Sundar Dorai-Raj]

Remigijus Lapinskas wrote:

Many thanks to all who replied to my e-mail. My problem was that I
had not known about the approx function.

By the way, if I have  x <- c(1990,1994,1995,1997), is there an
automated way to fill in the gaps, i.e., to get
c(1991,1992,1993,1996)?



Try this:

R> x <- c(1990, 1994, 1995, 1997)
R> all.x <- seq(min(x), max(x))
R> missing.x <- all.x[!all.x %in% x]
R> missing.x
[1] 1991 1992 1993 1996
R>

Regards,
Sundar

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Re: [R] Interpolation

[Ray Brownrigg]

> I have two vectors and connect the points by line segments:
> 
> x <- c(1990,1994,1995,1997)
> y <- c(30,40,80,20)
> plot(x,y,type="l")
> 
> I want to get the values of y's on these lines at missing x's,
> i.e., at 1991,1992,1993,and 1996. Is there a simple way to do this?
> 
Easy:
approx(x, y, c(1991:1993, 1996))$y
while
approx(x, y, 1990:1997)$y
will give you all the y values.

Hope this helps,
Ray Brownrigg <[EMAIL PROTECTED]>   http://www.mcs.vuw.ac.nz/~ray

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Re: [R] interpolation and 3D plots

[John Fox]

Dear Jonathan,

The all.effects function in the car package will identify the high-order 
terms in a linear or generalized-linear model and absorb terms marginal to 
them (e.g., for a two-way interaction, the main effects marginal to the 
interaction and the constant). The plot method for the object produced uses 
trellis graphics to plot terms, but for a two-way interaction, the 
information in the object could easily be plotted as an image (or other 3D) 
plot.

I hope that this helps,
 John

At 10:02 AM 2/3/2003 +, Davies, Jonathan wrote:
I've been told that you may be able to help.

I have a complex linear model with multiple two-way interaction terms. Is
there a way to view the interaction terms in a 3d plot equivalent to the S
functions:

> plan<-interp(x,y,z)
> image(plan)

where x and y are my explanatory variables and z my response variable.


-
John Fox
Department of Sociology
McMaster University
Hamilton, Ontario, Canada L8S 4M4
email: [EMAIL PROTECTED]
phone: 905-525-9140x23604
web: www.socsci.mcmaster.ca/jfox
-

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Re: [R] interpolation and 3D plots

[ripley]

library(akima)
plan <- interp.old(x,y.z)
image(plan)

is very close.

This is in MASS (the book) with examples.

On Mon, 3 Feb 2003, Davies, Jonathan wrote:

> I've been told that you may be able to help.
> 
> I have a complex linear model with multiple two-way interaction terms. Is
> there a way to view the interaction terms in a 3d plot equivalent to the S
> functions:
>  
> > plan<-interp(x,y,z)
> > image(plan)
> 
> where x and y are my explanatory variables and z my response variable.

-- 
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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