date:20091127

On Fri, Nov 27, 2009 at 1:13 AM, Kjetil Halvorsen
kjetilbrinchmannhalvor...@gmail.com wrote:
 I think the subject says it all. I want to make a simple lattice plot,
 using xyplot with the
 argument     type=c(l,a).
 The problem then is that in the resulting plot it is
 difficult/impossible to see which plot corresponds to the average
 and which to the individual profiles. I triedthings like extra
 arguments  lwd=c(1,3)   or   col=c(blue,red)
 hoping this would be interpreteded parallely to the type=  argument,
 but no.  Like:

 xyplot(response ~ time|group, repa0, groups=~Participant, type=c(b,
 a), lwd=c(1, 3),
       ylim=c(0, 10))

Are you looking for per-group average or overall average? All your
attempts will do per-group average, which is rather pointless here.
For a global average you need to call panel.average with all the data,
not through panel.superpose:

mypanel - function(x, y, ..., groups, type){
panel.xyplot(x, y, ..., groups = groups, type=type)
panel.average(x, y, horizontal = FALSE,
  col = black, lwd = 3)
}

xyplot(response ~ time|group, repa0, groups=~Participant, type=l,
   ylim=c(0, 10), panel=mypanel)


-Deepayan


 and many other variants ...

 Then I vent for writing my own panel functions:
 mypanel - function(x, y, ..., type){
    panel.average(x, y, ...,  horizontal=FALSE)
    panel.xyplot(x, y, ..., type=type)
  }
 xyplot(response ~ time|group, repa0, groups=~Participant, type=l,
 ylim=c(0, 10),
       panel=panel.superpose, panel.groups=mypanel)

 (which doesn't work)
 ???
 Kjetil Halvorsen
 To recreate the data:

 repa0 -
 structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L,
 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
 2L, 2L, 2L, 2L, 2L), .Label = c(1, 2), class = factor),
    Participant = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L, 13L, 14L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L, 13L, 14L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L, 13L, 14L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L, 13L, 14L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L, 13L, 14L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L, 13L, 14L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L, 13L, 14L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L, 13L, 14L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L, 13L, 14L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L, 13L, 14L), time = c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
    2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
    3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
    4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L,
    5L, 5L, 5L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
    1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
    2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
    4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L,
    5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), response = c(2L,
    10L, 8L, 4L, 0L, 4L, 10L, 2L, 3L, 4L, 1L, 2L, 3L, 3L, 5L,
    2L, 10L, 3L, 0L, 3L, 10L, 1L, 7L, 3L, 2L, 10L, 8L, 5L, 8L,
    0L, 10L, 2L, 0L, 2L, 10L, 10L, 3L, 2L, 10L, 4L, 6L, 0L, 5L,
    0L, 6L, 2L, 0L, 2L, 10L, 5L, 3L, 2L, 6L, 4L, 5L, 0L, 5L,
    0L, 5L, 0L, 0L, 0L, 10L, 0L, 0L, 2L, 5L, 0L, 5L, 0L, 3L,
    8L, 6L, 1L, 3L, 0L, 5L, 4L, 10L, 0L, 3L, 3L, 6L, 5L, 2L,
    0L, 2L, 0L, 4L, 0L, 5L, 4L, 5L, 1L, 0L, 2L, 0L, 5L, 3L, 1L,
    8L, 2L, 4L, 0L, 0L, 0L, 10L, 4L, 3L, 0L, 0L, 10L, 3L, 3L,
    6L, 1L, 4L, 0L, 0L, 0L, 10L, 3L, 3L, 2L, 2L, 10L, 2L, 3L,
    5L, 0L, 1L, 0L, 0L, 0L, 10L, 2L, 3L, 2L, 0L, 6L)), .Names = c(group,
 Participant, time, response), row.names = c(1.1, 2.1,
 3.1, 4.1, 5.1, 6.1, 7.1, 8.1, 9.1, 10.1, 11.1,
 12.1, 13.1, 14.1, 1.2, 2.2, 3.2, 4.2, 5.2, 6.2,
 7.2, 8.2, 9.2, 10.2, 11.2, 12.2, 13.2, 14.2,
 1.3, 2.3, 3.3, 4.3, 5.3, 6.3, 7.3, 8.3, 9.3,
 10.3, 11.3, 12.3, 13.3, 14.3, 1.4, 2.4, 3.4,
 4.4, 5.4, 6.4, 7.4, 8.4, 9.4, 10.4, 11.4, 12.4,
 13.4, 14.4, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5,
 8.5, 9.5, 10.5, 11.5, 12.5, 13.5, 14.5, 1.11,
 2.11, 3.11, 4.11, 5.11, 6.11, 7.11, 8.11, 9.11,
 10.11, 11.11, 12.11, 13.11, 14.11, 1.21, 2.21,
 3.21, 4.21, 5.21, 6.21, 7.21, 8.21, 9.21, 10.21,
 11.21, 12.21, 13.21, 14.21, 1.31, 2.31, 3.31, 4.31,
 5.31, 6.31, 7.31, 8.31, 9.31, 10.31, 11.31, 12.31,
 13.31, 14.31, 1.41, 2.41, 3.41, 4.41, 5.41, 6.41,
 7.41, 8.41, 9.41, 10.41, 11.41, 12.41, 13.41, 14.41,
 1.51, 2.51, 3.51, 4.51, 5.51,

Re: [R] order of panels in xyplots

On Thu, Nov 26, 2009 at 12:09 AM, Titus Malsburg malsb...@gmail.com wrote:
 The documentation of xyplot could be improved here.  It says:

  If 'index.cond' is a list, it has to be as long as the number of 
 conditioning
   variables, and the 'i'-th component has to be a valid indexing vector for 
 the
   integer vector '1:nlevels(g_i)' (which can, among other things, repeat some
   of the levels or drop some altogether).

 It should make explicit that nlevels is the number of levels actually
 used in the data and not length(levels(f)).

Actually, it should say that whether unused levels are counted depends
on the value of the 'drop.unused.levels' argument. I will update the
documentation.

-Deepayan

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Re: [R] Concave hull

2009-11-27 Thread baptiste auguie

Thanks for the interesting reference to alphahull. It might be a good
starting point for placing e.g. a legend in a plot (I think the usual
techniques for this (gregmisc?) are a bit more brute-force.)


baptiste

2009/11/27 Kjetil Halvorsen kjetilbrinchmannhalvor...@gmail.com:
 There is a package on CRAN implementing such an idea:
 alphahull, phull is other package,

 kjetil

 On Thu, Nov 26, 2009 at 6:11 PM, baptiste auguie
 baptiste.aug...@googlemail.com wrote:
 2009/11/26 Ted Harding ted.hard...@manchester.ac.uk:
 Raising a rather general question here.

 This is a tantalising discussion, but the notion of concave hull
 strikes me as extremely ill-defined!

 I'd like to see statement of what it is (generically) supposed to be.

 I'm curious too, but I can imagine the following definition,

 Consider a sphere (n-dimensional maybe) that we let come in contact
 with the scatter of points from outside. The set of points that the
 sphere can attain may define unambiguously (I think) a concave hull,
 for a specified sphere radius. The convex hull is obtained in the
 limit of infinite radius (plane).

 It's probably not exactly this, but I guess that's the rough idea.

 Just a thought,

 baptiste

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Re: [R] Adding text in the panels for Trellis plot ...

On Thu, Nov 26, 2009 at 6:22 AM, ychu066 ychu...@aucklanduni.ac.nz wrote:

 i was trying to do a for loop for plotting the histograms , but it doesnt
 work properly

 library(lattice)
 columns - 8:153
 plots - vector(list, length(columns))
 j - 0
 for (i in columns)
 + {
 +   plots[[ j - j+1 ]] - histogram( ~ data[,i] | data[,2],ylab =
 Frequency,
 +   xlab = Score, xlim = c(1,5), ylim = c(0,100),layout=c(3,1),
 +   main = c(colnames(data)[i],index,j+1),mycolors =
 colors()[c(536,552,652,254,26)],
 +   panel = function(..., col, mycolors) {
 +   panel.histogram(..., col = mycolors[panel.number()])})
 + trellis.focus('strip', 1, 1, highlight=FALSE)
 + ltext(0.60, -0.25, 'PPM', col='blue', pos=3)
 + trellis.unfocus()
 + }
 Error in grid.Call.graphics(L_downviewport, name$name, strict) :
  Viewport 'plot1.strip.1.1.vp' was not found

 print(plots[[1]])


 I am not sure what was happeneing. . . can anyone help me ?

Panel functions are for controlling the panel display. If you want to
control the strip, use the 'strip' argument to provide a custom strip
function.

-Deepayan

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Re: [R] barchart() {Lattice} help.

On Fri, Nov 27, 2009 at 9:21 AM, Peng Cai pengcaimaill...@gmail.com wrote:
 @ Peter: I got it, thanks a lot for all your help! And yes, as you said the
 title option in auto.key is redundant.


 @ All, Hi: I need to add percentage sign to y-axis labels (like 0%, 20%,
 ..., 100%). How can I get it. I'm using barchart() function as given below
 along with the data set.

Read ?barchart, specifically the entry for 'scales'. Hint: figure out
what the line

 scales = list(y = list(at = myYscale)),

does.

-Deepayan



 Data:
 Sample    Col1    Col2    Col3
 Row1    -20    40    -10
 Row2    30    -20    40
 Row3    30    10    -20
 Row4    20    20    -10

 R code:

 library(lattice)
 dta-read.table(data.txt, header=TRUE, row.names=Sample)
 myYscale - seq(-140,140,20)
 barchart(data.matrix(dta),
    horizontal=FALSE,
      stack=TRUE,
    par.settings = simpleTheme(col = c(2:4)),
    auto.key=list(space = 'right', rows = 3, rectangles = TRUE,
                  points = FALSE ),
    scales = list(y = list(at = myYscale)),
    panel=function(x,y,...){
        panel.abline(h=c(myYscale), col.line=gray)
          panel.barchart(x,y,...)}
 )

 Any help would be greatly appreciated,
 Peng

 On Thu, Nov 26, 2009 at 10:38 PM, P Ehlers ehl...@ucalgary.ca wrote:

 As I wrote earlier:


 I had to add the rectangles= and points= arguments to

 auto.key to get the same key as you had earlier.

 and the relevant line in the code was:


  auto.key = list(space = 'right', rectangles=TRUE, points=FALSE)

 -Peter Ehlers

 Peng Cai wrote:

 Hello Peter and David,

 Thanks for your help. I have added what you suggested and its working
 perfectly fine except:

 When I add the panel function, the legend changes. In the sense without
 the
 panel function the column names are shown with small colored rectangles
 (on
 right), but after adding it the rectangles change to tiny un-filled
 diamonds. Any suggestions?

 My current code and data is below,

 Thanks a lot,
 Peng


 Data:


 Sample Col1 Col2 Col3
 Row1 -2 4 -1
 Row2 3 -2 4
 Row3 3 5 -2
 Row4 4 1 -1

 Code:







 dta-read.table(data.txt, header=TRUE, row.names=Sample)
 coltemp=c(619,376,497)
 myYscale - seq(-10, 10, 1)
 barchart(data.matrix(dta),
       horizontal=FALSE,
       stack=TRUE,
     par.settings = simpleTheme(col = colors()[coltemp]),
       auto.key=list(space=right),
     border=NA,
     panel=function(x,y,...){
              panel.abline(h=c(myYscale), col.line=gray)
              panel.barchart(x,y,...)
       },
     scales = list(y = list(at = myYscale))
 )












 Thanks,


 Peng


 On Thu, Nov 26, 2009 at 7:23 PM, Peng Cai pengcaimaill...@gmail.com
 wrote:



 Hi Again,

 Before I start getting into what you just suggested, let me confirm if I
 made my point clear previously. I'm looking for horizontal lines similar
 to
 one on the following link (It has parallel lines for each y=200,
 y=400,...):

 http://pfiles.5min.com/images/176735/176734313.jpg

 What you just suggested can solve this purpose? Thanks,

 Peng


 On Thu, Nov 26, 2009 at 7:09 PM, Peter Ehlers ehl...@ucalgary.ca
 wrote:



 Peng Cai wrote:



 Thanks David, I tried panel.abline(h=somevalue) -- both inside and
 outside
 of barchart() function but its not working. Any suggestions?

 Peng



 Here's some code related to the data you posted earlier.


 barchart(data.matrix(dta), horizontal = FALSE, stack = TRUE,
            par.settings = simpleTheme(col = 2:4),
            panel=function(x,y,...){
              panel.abline(h=c(-2,0,3,4), col.line=gray)
              panel.barchart(x,y,...)

            },
            scales = list(y = list(at = -2:8)),
            auto.key = list(space = 'right', rectangles=TRUE,
                points=FALSE)
 )

 If you want the gray lines in front of the bars, switch the
 order of the panel functions. With lattice, it's all about
 what goes into each panel (you have only one panel here).
 If you want more than one thing in a panel, you have to set
 up a function to do those things.

 I had to add the rectangles= and points= arguments to
 auto.key to get the same key as you had earlier.

  -Peter Ehlers





 On Thu, Nov 26, 2009 at 6:42 PM, David Winsemius 
 dwinsem...@comcast.netwrote:

  On Nov 26, 2009, at 6:12 PM, Peng Cai wrote:


  Thanks a lot Peter! One more help, is there a similar function
 abline()



 for
 barchart().

  ?panel.abline



  I'm trying to add a (light gray colored) horizontal lines, one for


 each
 y-value.

 Peng

 On Thu, Nov 26, 2009 at 5:59 PM, Peter Ehlers ehl...@ucalgary.ca
 wrote:

  Peng Cai wrote:



  Hi Peter,



 I'm not sure but it seems scales command works only with integer
 values.

 If the y-axis values are very small (such as -0.03, -0.02, -0.01,
 0,
 0.01,..., 0.08). My current plot has values 0, 0.05, and 0.10
 only.
 But
 I
 need it to extend it to negative numbers and reduce the scale
 width
 (like
 -0.04, -0.02, 0, 0.02,...).

 Can I change these too? Thanks!


  Use, e.g.


 myYscale - seq(-0.04,

[R] plotting two surfaces simultaneously in a single panel

2009-11-27 Thread Umesh Srinivasan

Hi,

I have recently begun using the lattice package, and have been using
the wireframe command to visualise matrices which are model outputs. I
have been trying to plot two surfaces (from two matrices)
simultaneously in one panel, to visualise intersections etc., but
neither my attempts or trawling the net are helping me find how to do
this.

Can someone help? Thanks

Cheers,
Umesh Srinivasan
NCBS-TIFR,
Bangalore, India

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Re: [R] If condition using accessors

2009-11-27 Thread Peter Ehlers


Vitória Magalhães Piai wrote:

Hi,

I'm quite new using R and have got no one to help me get through it. 
Hopefully someone can help me with one problem I've been struggling with 
for the last hours!!

(Sorry if I'm using the wrong terminology as well!)

I have a data matrix in which SIE is one of my variables. What I need to 
do now is to create a new variable (group) if a certain condition in SIE 
is met.

I tried:

if (data2$SIE  50.646){
data2$group=as.factor(small)
}
which gives me In if (data2$SIE  50.646) { :  the condition has length 
  1 and only the first element will be used


and
ifelse (data2$SIE  50.64593, data2$group = 
as.factor(small),data2$group = as.factor(large))
with Error: unexpected '=' in ifelse (data2$SIE  50.64593, 
data2$group =


I tried all other kinds of variations and so far, nothing!!


Do all of the following:

?=
?==
?if
?ifelse

 -Peter Ehlers



Is there a good way to do this that someone could teach me?
Thank you so much,
Vitória

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R-help@r-project.org mailing list
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Re: [R] If condition using accessors

2009-11-27 Thread Peter Ehlers



Peter Ehlers wrote:

Vitória Magalhães Piai wrote:

Hi,

I'm quite new using R and have got no one to help me get through it. 
Hopefully someone can help me with one problem I've been struggling 
with for the last hours!!

(Sorry if I'm using the wrong terminology as well!)

I have a data matrix in which SIE is one of my variables. What I need 
to do now is to create a new variable (group) if a certain condition 
in SIE is met.

I tried:

if (data2$SIE  50.646){
data2$group=as.factor(small)
}
which gives me In if (data2$SIE  50.646) { :  the condition has 
length   1 and only the first element will be used


and
ifelse (data2$SIE  50.64593, data2$group = 
as.factor(small),data2$group = as.factor(large))
with Error: unexpected '=' in ifelse (data2$SIE  50.64593, 
data2$group =


I tried all other kinds of variations and so far, nothing!!


Do all of the following:

?=
?==
?if
?ifelse

 -Peter Ehlers


Actually, I assumed without checking that ?ifelse would
have an example of assignment of the result of the test.
(Maybe it could be added.)

yourVariable - ifelse(...)

 -Peter Ehlers





Is there a good way to do this that someone could teach me?
Thank you so much,
Vitória

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Re: [R] Barplot with confidence intervals

2009-11-27 Thread Marc Schwartz


On Nov 26, 2009, at 3:23 PM, Jonhnny Weslley wrote:


I have a file in the following format:

Scenario1 Scenario1CIL Scenario1CIU Scenario2 Scenario2CIL  
Scenario2CIU

60 57 62 45 48 50
110 101 111 51 50 52
120 117 122 64 62 66
192 190 194 79 75 79

where:
First  column  = Scenario1 mean value
Second  column  = Scenario1 Low Confidence Interval
Third  column  = Scenario1 Upper Confidence Interval
Fourth  column  = Scenario2 mean value
Fifth  column  = Scenario2 Low Confidence Interval
Sixth  column  = Scenario2 Upper Confidence Interval

Then, I tried this:

library(gplots)
data - read.table(data.file, header=T, sep= )
legend - c(line1,line2,line3,line4)

ci.l - as.matrix(c(c(data$Scenario1CIL), c(data$Scenario2CIL)))
ci.u - as.matrix(c(c(data$Scenario1CIU), c(data$Scenario2CIU)))

barplot2(as.matrix(c(c(data$Scenario1), c(data$Scenario2))),  
beside=TRUE,

   legend=legend, ylim=c(0, 200), main=Experiment X, ylab=Total
size,
   font.main=4, cex.axis=1.2, cex.lab=1.5, cex.names=1.5,
   plot.ci=TRUE, ci.l=ci.l, ci.u=ci.u, plot.grid=TRUE)

But this code doesn't group the values in Scenario1 and Scenario2, as
expected.
All plotted bars are joined side by side.

Considering the letter 'H' a bar, the expected result was: '   
'

but the result of the above code was: '' (no space between the
Scenario1's values and Scenario2'values)  and no legends (Scenario1  
and

Scenario2).

How I must to do?

Thanks in advance!



The data structures that you are passing to barplot2() are single  
column matrices. They need to be multicolumn matrices, where each  
column is a group and as per ?barplot2, the ci.l and ci.u arguments  
need to have the same structure as the height argument.


Thus:


 DF
  Scenario1 Scenario1CIL Scenario1CIU Scenario2 Scenario2CIL
160   57   6245   48
2   110  101  11151   50
3   120  117  12264   62
4   192  190  19479   75
  Scenario2CIU
1   50
2   52
3   66
4   79


height - as.matrix(DF[, c(1, 4)])
ci.l - as.matrix(DF[, c(2, 5)])
ci.u - as.matrix(DF[, c(3, 6)])

 height
 Scenario1 Scenario2
[1,]6045
[2,]   11051
[3,]   12064
[4,]   19279

 ci.l
 Scenario1CIL Scenario2CIL
[1,]   57   48
[2,]  101   50
[3,]  117   62
[4,]  190   75

 ci.u
 Scenario1CIU Scenario2CIU
[1,]   62   50
[2,]  111   52
[3,]  122   66
[4,]  194   79



library(gplots)

legend - c(line1, line2, line3, line4)

barplot2(height, plot.ci = TRUE, ci.l = ci.l, ci.u = ci.u,
 beside = TRUE, legend = legend, ylim = c(0, 200),
 main = Experiment X, ylab = Total size,
 font.main = 4, cex.axis = 1.2, cex.lab = 1.5,
 cex.names = 1.5, plot.grid = TRUE)


BTW, once you get the plot created, you may want to note some errors  
in your CI values.


HTH,

Marc Schwartz

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[R] how to analyze this design using lmer

2009-11-27 Thread Adaikalavan Ramasamy


Dear all,

A friend of mine requested me to analyze some data she has generated. I 
am hoping for some advice on best way of properly analyzing the data as 
I have never worked with such complicated or nested designs.


Here is the setup. She has taken material from 5 animals and each 
material is subdivided into 6 plate (30 plates in total). Each plate is 
then assigned as either a control or a treated with a chemical AND kept 
at one of three concentrations. A sample is taken daily from each plate 
for six continuous days and measured (180 measurement in total). Her 
main question is whether treatment has an effect.


Here is a simulated dataset:

 df - expand.grid( animal=LETTERS[1:5], group=c(Control, Treated),
conc=c(X, Y, Z), day=1:6 )
 df$plate - as.numeric(factor(apply(df[ ,1:3], 1, paste, collapse=)))
 df - df[ order(df$plate), ]
 df$plate - as.factor(df$plate)
 rownames(df) - NULL

 set.seed(1066)
 df$value - runif(90, 1, 2)*(df$group==Control) +
 c(0, -0.5, -0.20)[as.numeric(df$conc)] +
 rnorm(30)[ as.numeric(df$plate) ] +
 runif(180, 0.9, 1.1)*df$day + rnorm(180, sd=0.5)

 df[1:10, ]
 animal   group conc day plate value
 1A ControlX   1 1 3.3403510
 2A ControlX   2 1 5.1042965
 3A ControlX   3 1 5.4003462
 ...
 ...
 178  E TreatedZ   430 2.8558186
 179  E TreatedZ   530 4.4567206
 180  E TreatedZ   630 5.4542460


I have tried analyzing the data as follows:
library(lme4)
lmer( value ~ group + day + conc + (1 | animal/plate), data=df )
lmer( value ~ group + day + conc + (1 | animal), data=df )
lmer( value ~ group + day + conc + (1 | plate), data=df )

BUT I am not sure which of the models above is appropriate. Any advice 
would be very useful. Many thanks in advance.


Regards, Adai

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Re: [R] layers in xYplot of Hmisc

I took the OP's variables, reversed the direction of the second y,  
bundled them up in a dataframe and then assigned the grp. So starting  
with the OP's code:


 x-seq(1,10,1)
y-seq(1,10,1)
ci-y*.10
ciupper-y+ci
cilower-y-ci
x2-seq(1,5,.5)
y2-seq(5, 1, -.5) #reverse second variable
ci2-y2*.10
ciupper2-y2+ci2
cilower2-y2-ci2
dfci - data.frame(x =c(x,x2), y=c(y,y2), ciupper=c(ciupper,ciupper2),  
cilower=c(cilower,cilower2) )
 dfci$grp - 1:19  # groupify, needed to do it by hand since  
lengths were different

 dfci$grp[1:10] - 1
 dfci$grp[11:19] - 2
 dfci$grp - factor(dfci$grp)
dfci
dput(dfci)



On Nov 27, 2009, at 6:13 AM, butter wrote:



do you mind me asking what code you used to create that data frame  
and name

the groups 1 and 2?

David Winsemius wrote:



On Nov 26, 2009, at 10:26 PM, Joe King wrote:


In the filled bands part of xYplot of the Hmisc package, is there
a way to
have multiple bands with multiple lines? or does it just allow one
for now?


No problem as long as you use groups... which means you probably  
ought

to be providing a dataframe format with a grouping factor. (At least
that was how I read the help page saying that:
-
Details
Unlike xyplot, xYplot senses the presence of a groups variable and
automatically invokes panel.superpose instead of panel.xyplot. The
same is true for Dotplot vs. dotplot.

-


I would give it a dataset that does not have lines coincident if you
want to see them.

(You should also get rid of that space in your color parameter.

dfci - structure(list(x = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 1.5,  
2,

2.5, 3, 3.5, 4, 4.5, 5), y = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 5, 4.5,
4, 3.5, 3, 2.5, 2, 1.5, 1), ciupper = c(1.1, 2.2, 3.3, 4.4, 5.5, 6.6,
7.7, 8.8, 9.9, 11, 5.5, 4.95, 4.4, 3.85, 3.3, 2.75, 2.2, 1.65, 1.1),
cilower = c(0.9, 1.8, 2.7, 3.6, 4.5, 5.4, 6.3, 7.2, 8.1, 9, 4.5,  
4.05,

3.6, 3.15, 2.7, 2.25, 1.8, 1.35, 0.9), grp = structure(c(1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L), .Label = c(1, 2), class = factor)), .Names = c(x, y,
ciupper, cilower, grp), row.names = c(NA, -19L), class =
data.frame)

xYplot(Cbind(y,cilower,ciupper)~x, groups=grp, data=dfci,
method=filled bands,col.fill=lightgrey, type=c(b))

--
David






So I had an example bit ago had a made up line and CI, now if I
wanted to
make a second line with a CI filled in can I put them on the same
plot?



x-seq(1,10,1)

y-seq(1,10,1)

ci-y*.10

ciupper-y+ci

cilower-y-ci



xYplot(Cbind(y,cilower,ciupper)~x, method=filled bands,
col.fill=light
grey, type=c(b))



x2-seq(1,5,.5)

y2-seq(1,5,.5)

ci2-y2*.10

ciupper2-y2+ci2

cilower2-y2-ci2



xYplot(Cbind(y2,cilower2,ciupper2)~x2, method=filled
bands,col.fill=light
grey, type=c(b))



---

Joe King, M.A.

Ph.D. Student

University of Washington - Seattle

206-913-2912

j...@joepking.com

---

Never throughout history has a man who lived a life of ease left a
name
worth remembering. --Theodore Roosevelt




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David Winsemius, MD
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--
View this message in context: 
http://old.nabble.com/layers-in-xYplot-of-Hmisc-tp26537542p26541016.html
Sent from the R help mailing list archive at Nabble.com.

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David Winsemius, MD
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Re: [R] If condition using accessors



On Nov 27, 2009, at 5:34 AM, Vitória Magalhães Piai wrote:


Hi,

I'm quite new using R and have got no one to help me get through it.  
Hopefully someone can help me with one problem I've been struggling  
with for the last hours!!

(Sorry if I'm using the wrong terminology as well!)

I have a data matrix in which SIE is one of my variables. What I  
need to do now is to create a new variable (group) if a certain  
condition in SIE is met.

I tried:

if (data2$SIE  50.646){
data2$group=as.factor(small)
}
which gives me In if (data2$SIE  50.646) { :  the condition has  
length  1 and only the first element will be used



If ( cond) is not vectorised. Ifelse is.


and
ifelse (data2$SIE  50.64593, data2$group =  
as.factor(small),data2$group = as.factor(large))
with Error: unexpected '=' in ifelse (data2$SIE  50.64593,  
data2$group =


Almost... try:

data2$group - ifelse (data2$SIE  50.64593,  as.factor(small),  
as.factor(large) )


... but that looks wrong to my wetware R interpreter, too, so could  
also try:


data2$group - factor( ifelse (data2$SIE  50.64593,  small, large )

(And of course this would have been tested had you supplied a small  
test dataset, but this week I am being passive aggressive toward  
persons who don't follow the Posting Guide.)


--
David



I tried all other kinds of variations and so far, nothing!!

Is there a good way to do this that someone could teach me?
Thank you so much,
Vitória

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[R] post

2009-11-27 Thread Anastasia

Dear all,

I have the following problem which I cannot solve:

data - numeric (100)
for (i in 1:100){
  p - runif(1,min=0,max=1)
  data[i] - rnorm(1,mean=2,sd=1)
}

ke - density(data,bw=sj,n=61)

How can I now find the value of this density function for example ke(0),
ke(2) etc.
Maybe this is an easy question but I cannot resolve it.

Thank you very much for the help.



On Fri, Nov 27, 2009 at 3:34 PM, r-help-requ...@r-project.org wrote:

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[R] Time Series Data

2009-11-27 Thread chris carleton


Hi All,

I'm trying to analyze some time series data and I have run into difficulty. I 
have decadal sun spot data and I want to separate the very regular periodic 
function from the trend and noise. I looked into using stl(), but the frequency 
of the time series data must be greater than 1 for stl(). My data covers a 1000 
year interval from 9095 BP to 8095BP and the frequency is, therefore, 0.1 
(because the data are decadal). I've tried changing the frequency, but the only 
frequency that creates a plot of the time series data which matches the raw 
data is 0.1. Is there anything I can do to the data that will make it more 
amenable to stl(), or is there another package that I could use for decomposing 
the signal that does not require that the frequency of the time series to be 
greater than 1? Thanks,

Chris
  
_

 Facebook.

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[R] problem tick marker and text

2009-11-27 Thread Duijvesteijn, Naomi


   Hi R-ers,


   I am struggling with my x-axis in a association plot. What I would like is
   to place the labels of the x-axis between the tick markers and normally the
   labels are printed at the place where the tick marker is placed. I donât
   want to move the tick marker (it gives the switch between one chromosome and
   the next) but I just want to put the chromosome number in between the two
   tick markers.


   This example from mhtplot has the chromosome number at the place of the
   tickmarker. How can I change this to âin betweenâ?


   Thanks!,


   Naomi
 
# fake example with Affy500k data
affy -c(40220, 41400, 33801, 32334, 32056, 31470, 25835, 27457, 22864, 28501, 
26273, 
 24954, 19188, 15721, 14356, 15309, 11281, 14881, 6399, 12400, 7125, 62
07)
CM - cumsum(affy)
n.markers - sum(affy)
n.chr - length(affy)
test - data.frame(chr=rep(1:n.chr,affy),pos=1:n.markers,p=runif(n.markers))
 
oldpar - par()
par(las=1,cex=0.6)
colors - rep(c(blue,green),11)
par(cex.axis=1.3)
mhtplot(test,usepos=TRUE,colors=colors,gap=1,pch=19,bg=colors)
box()
par(oldpar)

   
   
   
   Disclaimer:  De  informatie opgenomen in dit bericht (en bijlagen) kan
   vertrouwelijk zijn en is uitsluitend bestemd voor de geadresseerde(n).
   Indien u dit bericht ten onrechte ontvangt, wordt u geacht de inhoud niet te
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Re: [R] post

2009-11-27 Thread Peter Dalgaard

Anastasia wrote:
 Dear all,
 
 I have the following problem which I cannot solve:
 
 data - numeric (100)
 for (i in 1:100){
   p - runif(1,min=0,max=1)
   data[i] - rnorm(1,mean=2,sd=1)
 }

Er what is p doing in there. As far as I can see, you might as well do

data - rnorm(200, 2, 1)

 ke - density(data,bw=sj,n=61)

 How can I now find the value of this density function for example ke(0),
 ke(2) etc.
 Maybe this is an easy question but I cannot resolve it.

approxfun is your friend, e.g.:

 d - density(rnorm(200,2,1))
 dd - approxfun(d)
 dd(3)
[1] 0.2267764
 dd(5)
[1] 0.006181384
 dd(20) # from default rule=1 in approxfun
[1] NA

 
 Thank you very much for the help.
 
 


And, BTW, please don't hijack old mails for new topics (and in
particular, don't show your password to the world):
 On Fri, Nov 27, 2009 at 3:34 PM, r-help-requ...@r-project.org wrote:
 
 Welcome to the R-help@r-project.org mailing list!

 To post to this list, send your email to:

  r-help@r-project.org
-- 
   O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - (p.dalga...@biostat.ku.dk)  FAX: (+45) 35327907

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[R] problem with dynformula from plm package

Hello list,

I'm following the paper (http://www.jstatsoft.org/v27/i02/paper) on how to
use plm to run panel regressions, and am having trouble with what I
believe should be something very basic.

When I run the command (p.9 in the paper):

R
dynformula(emp~wage+capital,log=list(capital=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))

I see:

emp ~ wage + capital

rather than the complete model that is given in the paper:

log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(wage), 2) +
lag(log(wage), 3) + diff(capital, 2) + diff(capital, 3)

And indeed, when I try to run a regression using that formula, it appears to
not contain any lags or logs (output below).

Any ideas? Thanks in advance,

~Owen

-- 
Owen Powell
http://center.uvt.nl/phd_stud/powell

R library(plm)
R data(EmplUK, package=plm)
R a =
dynformula(emp~wage+capital,log=list(capital=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))
R testModel - plm(formula = a,data=EmplUK,model=within)
[1] 10312
R summary(testModel)
Oneway (individual) effect Within Model

Call:
plm(formula = a, data = EmplUK, model = within)

Unbalanced Panel: n=140, T=7-9, N=1031

Residuals :
Min.  1st Qu.   Median  3rd Qu. Max.
-17.1000  -0.3060   0.0137   0.3070  27.3000

Coefficients :
 Estimate Std. Error t-value  Pr(|t|)
wage-0.143626   0.032790 -4.3802 1.186e-05 ***
capital  0.801495   0.064088 12.5062  2.2e-16 ***
---
Signif. codes:  0 *** 0.001 ** 0.01 * 0.05 . 0.1   1

Total Sum of Squares:5030.6
Residual Sum of Squares: 4207.8
F-statistic: 86.9179 on 2 and 889 DF, p-value:  2.22e-16

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Re: [R] Time Series Data



On Nov 27, 2009, at 9:55 AM, chris carleton wrote:



Hi All,

I'm trying to analyze some time series data and I have run into  
difficulty. I have decadal sun spot data and I want to separate the  
very regular periodic function from the trend and noise. I looked  
into using stl(), but the frequency of the time series data must be  
greater than 1 for stl(). My data covers a 1000 year interval from  
9095 BP to 8095BP and the frequency is, therefore, 0.1 (because the  
data are decadal). I've tried changing the frequency, but the only  
frequency that creates a plot of the time series data which matches  
the raw data is 0.1. Is there anything I can do to the data that  
will make it more amenable to stl(), or is there another package  
that I could use for decomposing the signal that does not require  
that the frequency of the time series to be greater than 1? Thanks,


I do not understand why you are not multiplying by 10 and rounding.

--
David





David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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[R] problem with dynformula from plm package [RE-POST]

Hello list,

I'm following the paper (http://www.jstatsoft.org/v27/i02/paper) on
how to use plm to run panel regressions, and am having trouble with
what I believe should be something very basic.

When I run the command (p.9 in the paper):

R
dynformula(emp~wage+capital,log=list(capital=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))

I see:

emp ~ wage + capital

rather than the complete model that is given in the paper:

log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(wage), 2) +
lag(log(wage), 3) + diff(capital, 2) + diff(capital, 3)

And indeed, when I try to run a regression using that formula, it
appears to not contain any lags or logs (output below).

Any ideas? Thanks in advance,

~Owen

-- 
Owen Powell
http://center.uvt.nl/phd_stud/powell

R library(plm)
R data(EmplUK, package=plm)
R a =
dynformula(emp~wage+capital,log=list(capital=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))
R testModel - plm(formula = a,data=EmplUK,model=within)
[1] 10312
R summary(testModel)
Oneway (individual) effect Within Model

Call:
plm(formula = a, data = EmplUK, model = within)

Unbalanced Panel: n=140, T=7-9, N=1031

Residuals :
Min.  1st Qu.   Median  3rd Qu. Max.
-17.1000  -0.3060   0.0137   0.3070  27.3000

Coefficients :
 Estimate Std. Error t-value  Pr(|t|)
wage-0.143626   0.032790 -4.3802 1.186e-05 ***
capital  0.801495   0.064088 12.5062  2.2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Total Sum of Squares:5030.6
Residual Sum of Squares: 4207.8
F-statistic: 86.9179 on 2 and 889 DF, p-value:  2.22e-16

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Re: [R] problem with dynformula from plm package [RE-POST]



On Nov 27, 2009, at 10:25 AM, Owen Powell wrote:


Hello list,

I'm following the paper (http://www.jstatsoft.org/v27/i02/paper) on
how to use plm to run panel regressions, and am having trouble with
what I believe should be something very basic.

When I run the command (p.9 in the paper):

R
dynformula(emp~wage 
+ 
capital 
,log 
= 
list 
(capital 
=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))




Perhaps you could have read the help page for the current version of  
the package which says the argument have been modified. Using the  
current arguments:


dynformula(emp~wage 
+ 
capital 
,log 
.form 
= 
list 
(capital 
= 
FALSE 
,TRUE),lag.form=list(emp=2,c(2,3)),diff.form=list(FALSE,capital=TRUE))


log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(emp),
1) + lag(log(emp), 2) + lag(log(wage), 2) + lag(log(wage),
3) + diff(capital, 2) + diff(capital, 3)

--
David Winsemius, MD


I see:

emp ~ wage + capital

rather than the complete model that is given in the paper:

log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(wage), 2) +
lag(log(wage), 3) + diff(capital, 2) + diff(capital, 3)

And indeed, when I try to run a regression using that formula, it
appears to not contain any lags or logs (output below).

Any ideas? Thanks in advance,

~Owen

--
Owen Powell
http://center.uvt.nl/phd_stud/powell

R library(plm)
R data(EmplUK, package=plm)
R a =
dynformula(emp~wage 
+ 
capital 
,log 
= 
list 
(capital 
=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))


snipped


David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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[R] matching and extracting data

2009-11-27 Thread ram basnet

Dear all,
I have querry on how to extract the data by matching between two data set where 
one has the same elements multiple times?
 
For example, I have two matrix X and Y.
 X
   [,1]    [,2]   [,3]
1  A     5  P
2  B  6     P
3  C 7     P
4  D 5     Q
5  E  6     Q
6  F  7     Q
7  G     5  R
8  H     6  R
9  I   7  S
10 J  5     S
11 K    6  T
12 L     7  T
 
and 
 
Y   [,1]
1  P
2  Q
3  R
4  S
 
Now, I want to select and extract all the data of P, Q, R and S elements of 
column 3 of X matrix by matching with column 1 of Y matrix like below:
 
 [,1]   [,2]   [,3]
1  A     5  P
2  B  6     P
3  C 7     P
4  D 5     Q
5  E  6     Q
6  F  7     Q
7  G     5  R
8  H     6  R
9  I   7  S
10 J  5     S
 
I guess, the answer might be simple but i am not getting way to figure out. 
And, i have to select subset from very huge data set. So, i need some kinds of 
automated procedure.
If some one can help me, it will be great
 
Thanks in advance.
 
Sincerely,
Ram Kumar Basnet
 


  
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Re: [R] matching and extracting data

2009-11-27 Thread Chuck Cleland

On 11/27/2009 10:25 AM, ram basnet wrote:
 Dear all,
 I have querry on how to extract the data by matching between two data set 
 where one has the same elements multiple times?
  
 For example, I have two matrix X and Y.
 X
[,1][,2]   [,3]
 1  A 5  P
 2  B  6 P
 3  C 7 P
 4  D 5 Q
 5  E  6 Q
 6  F  7 Q
 7  G 5  R
 8  H 6  R
 9  I   7  S
 10 J  5 S
 11 K6  T
 12 L 7  T
  
 and 
  
 Y   [,1]
 1  P
 2  Q
 3  R
 4  S
  
 Now, I want to select and extract all the data of P, Q, R and S elements of 
 column 3 of X matrix by matching with column 1 of Y matrix like below:
  
 [,1]   [,2]   [,3]
 1  A 5  P
 2  B  6 P
 3  C 7 P
 4  D 5 Q
 5  E  6 Q
 6  F  7 Q
 7  G 5  R
 8  H 6  R
 9  I   7  S
 10 J  5 S
  
 I guess, the answer might be simple but i am not getting way to figure out. 
 And, i have to select subset from very huge data set. So, i need some kinds 
 of automated procedure.
 If some one can help me, it will be great

 subset(X, X[,3] %in% Y[,1])
  [,1] [,2] [,3]
 [1,] A  5  P
 [2,] B  6  P
 [3,] C  7  P
 [4,] D  5  Q
 [5,] E  6  Q
 [6,] F  7  Q
 [7,] G  5  R
 [8,] H  6  R
 [9,] I  7  S
[10,] J  5  S

 Thanks in advance.
  
 Sincerely,
 Ram Kumar Basnet
  
 
 
   
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-- 
Chuck Cleland, Ph.D.
NDRI, Inc. (www.ndri.org)
71 West 23rd Street, 8th floor
New York, NY 10010
tel: (212) 845-4495 (Tu, Th)
tel: (732) 512-0171 (M, W, F)
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[R] Simple Function doesn't work?

2009-11-27 Thread Anastasia

Hello,

I am new to R program, therefore, I am sorry if this is a really stupid
question.
I wrote a simple function and for some reason it doesn't work

ReturnsGrid = function(x,y,m){
for (i in 1:m){
   grid[i] - x + (i-1)*(y-x)/m
}
grid
}

xx=ReturnsGrid(0,9,3)

Thanks a lot!

[[alternative HTML version deleted]]

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Re: [R] matching and extracting data

2009-11-27 Thread David B. Thompson, Ph.D., P.E., D.WRE, CFM



On Nov 27, 2009, at 10:25 AM, ram basnet wrote:


Dear all,
I have querry on how to extract the data by matching between two  
data set where one has the same elements multiple times?


For example, I have two matrix X and Y.

X

   [,1][,2]   [,3]
1  A 5  P
2  B  6 P
3  C 7 P
4  D 5 Q
5  E  6 Q
6  F  7 Q
7  G 5  R
8  H 6  R
9  I   7  S
10 J  5 S
11 K6  T
12 L 7  T

and


Y   [,1]

1  P
2  Q
3  R
4  S

Now, I want to select and extract all the data of P, Q, R and S  
elements of column 3 of X matrix by matching with column 1 of Y  
matrix like below:



[,1]   [,2]   [,3]

1  A 5  P
2  B  6 P
3  C 7 P
4  D 5 Q
5  E  6 Q
6  F  7 Q
7  G 5  R
8  H 6  R
9  I   7  S
10 J  5 S


Perhaps (untested):

X[ X[,3] %in% Y[,1] , ]

Would have been tested if you had used dput or dump on your matrices.

--
David.



I guess, the answer might be simple but i am not getting way to  
figure out. And, i have to select subset from very huge data set.  
So, i need some kinds of automated procedure.

If some one can help me, it will be great

Thanks in advance.

Sincerely,
Ram Kumar Basnet




[[alternative HTML version deleted]]

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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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[R] Force a variable substitution in expression

2009-11-27 Thread Saptarshi Guha

Hello,
I have a function that creates an expression object with some variables
substituted e.g

foo - function(s){
  expression({
v - s
print(v)
  })
}

Thus foo returns an expression, however the expression has the symbol 's'
contained within it and thus returns an error when eval'd e.g

x - foo(10)
eval(x)
Error in eval(expr, envir, enclos) : object 's' not found

Q: How do I force a substitution so that the returned expression has the value
of 's' 'hardcoded'  in it e.g

 foo(10)

returns

expression({
  v - 10
  print(v)
})

Regards
Saptarshi

--
saptarshi guha | http://www.stat.purdue.edu/~sguha
It is when I struggle to be brief that I become obscure.
-- Quintus Horatius Flaccus (Horace)

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Re: [R] Simple Function doesn't work?

2009-11-27 Thread Ista Zahn

Hi,
You need to create the grid object before you can assign values to it. Try

ReturnsGrid = function(x,y,m){
grid - numeric()
  for (i in 1:m){
  grid[i] - x + (i-1)*(y-x)/m
}
grid
}

On Fri, Nov 27, 2009 at 11:00 AM, Anastasia nast...@gmail.com wrote:
 Hello,

 I am new to R program, therefore, I am sorry if this is a really stupid
 question.
 I wrote a simple function and for some reason it doesn't work

 ReturnsGrid = function(x,y,m){
 for (i in 1:m){
   grid[i] - x + (i-1)*(y-x)/m
 }
 grid
 }

 xx=ReturnsGrid(0,9,3)

 Thanks a lot!

        [[alternative HTML version deleted]]

 __
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-- 
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org

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Re: [R] problem tick marker and text

2009-11-27 Thread Peter Ehlers


Duijvesteijn, Naomi wrote:

   Hi R-ers,


   I am struggling with my x-axis in a association plot. What I would like is
   to place the labels of the x-axis between the tick markers and normally the
   labels are printed at the place where the tick marker is placed. I donâ€™t
   want to move the tick marker (it gives the switch between one chromosome and
   the next) but I just want to put the chromosome number in between the two
   tick markers.


   This example from mhtplot has the chromosome number at the place of the
   tickmarker. How can I change this to â€˜in betweenâ€™?


   Thanks!,


   Naomi
 

Kudos for including reproducible code, BUT
1. do you really consider a dataframe of
   dimension ~500K by 3 to be minimal?
2. there are over 2000 R-packages on CRAN;
   wouldn't it be polite to say which one contains mhtplot()?


# fake example with Affy500k data
affy -c(40220, 41400, 33801, 32334, 32056, 31470, 25835, 27457, 22864, 28501, 
26273, 
 24954, 19188, 15721, 14356, 15309, 11281, 14881, 6399, 12400, 7125, 62

07)
CM - cumsum(affy)


is CM used anywhere?


n.markers - sum(affy)
n.chr - length(affy)
test - data.frame(chr=rep(1:n.chr,affy),pos=1:n.markers,p=runif(n.markers))
 
oldpar - par()

par(las=1,cex=0.6)
colors - rep(c(blue,green),11)
par(cex.axis=1.3)
mhtplot(test,usepos=TRUE,colors=colors,gap=1,pch=19,bg=colors)
box()
par(oldpar)


Here is one way, using about half the data.
I use mtext() to place the labels, since axis= seems
to be hard-coded in mhtplot.

# make a vector of labels to place between ticks
textvec - letters[1:5]

# plot
mhtplot(test[1:2e5,],usepos=TRUE,colors=colors,gap=1,pch=19,bg=colors)

# now figure out where to put the labels. This would be
# easy with seq() if the scale were uniform but it's not.
# So we use the locator() function.

xat - locator()$x
xat - round(xat)
xat
#[1]  34316  85096 132458 173961 216929

# place the labels; adjust line= if you wish
mtext(textvec, side=1, line=0, at=xat)

If I had to do this frequently, I would hack the
mhtplot() function.

 -Peter Ehlers

   
   
   
   Disclaimer:  De  informatie opgenomen in dit bericht (en bijlagen) kan

   vertrouwelijk zijn en is uitsluitend bestemd voor de geadresseerde(n).
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Re: [R] Simple Function doesn't work?

2009-11-27 Thread baptiste auguie

Hi,

The error message,

Error in grid[i] - x + (i - 1) * (y - x)/m :
  object of type 'closure' is not subsettable

indicates that grid is actually known to R as a function (type grid
to see its definition). You can define your own variable with the same
name, but that needs to be done before the assignment in the for loop,


ReturnsGrid = function(x,y,m){

grid - vector(length = m)

for (i in 1:m){
  grid[i] - x + (i-1)*(y-x)/m
}
grid
}

ReturnsGrid(0,9,3)


HTH,

baptiste
2009/11/27 Anastasia nast...@gmail.com:
 Hello,

 I am new to R program, therefore, I am sorry if this is a really stupid
 question.
 I wrote a simple function and for some reason it doesn't work

 ReturnsGrid = function(x,y,m){
 for (i in 1:m){
   grid[i] - x + (i-1)*(y-x)/m
 }
 grid
 }

 xx=ReturnsGrid(0,9,3)

 Thanks a lot!

        [[alternative HTML version deleted]]

 __
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 https://stat.ethz.ch/mailman/listinfo/r-help
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Re: [R] Force a variable substitution in expression

2009-11-27 Thread Duncan Murdoch


On 27/11/2009 11:04 AM, Saptarshi Guha wrote:

Hello,
I have a function that creates an expression object with some variables
substituted e.g

foo - function(s){
  expression({
v - s
print(v)
  })
}

Thus foo returns an expression, however the expression has the symbol 's'
contained within it and thus returns an error when eval'd e.g

x - foo(10)
eval(x)
Error in eval(expr, envir, enclos) : object 's' not found

Q: How do I force a substitution so that the returned expression has the value
of 's' 'hardcoded'  in it e.g

 foo(10)

returns

expression({
  v - 10
  print(v)
})
  


The substitute() function can do it, but the bquote() function is the 
most convenient:


 foo - function(s) {
+   bquote(expression({
+v - .(s)
+print(v)
+   }))
+ }
 foo(10)
expression({
   v - 10
   print(v)
})

Duncan Murdoch

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Re: [R] Simple Function doesn't work?

2009-11-27 Thread Alain Guillet


Hi,

If you execute the following code it works but I wouldn't use grid if I 
were you as a vector as this name is already used by R (check 
help(grid)) and it explains why you have to define it in the function.


ReturnsGrid = function(x,y,m){
grid - numeric(m)
for (i in 1:m){
  grid[i] - x + (i-1)*(y-x)/m
}
grid
}

xx=ReturnsGrid(0,9,3)

Regards,
Alain


Anastasia wrote:

Hello,

I am new to R program, therefore, I am sorry if this is a really stupid
question.
I wrote a simple function and for some reason it doesn't work

ReturnsGrid = function(x,y,m){
for (i in 1:m){
   grid[i] - x + (i-1)*(y-x)/m
}
grid
}

xx=ReturnsGrid(0,9,3)

Thanks a lot!

[[alternative HTML version deleted]]

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--
Alain Guillet
Statistician and Computer Scientist

SMCS - Institut de statistique - Université catholique de Louvain
Bureau c.316
Voie du Roman Pays, 20
B-1348 Louvain-la-Neuve
Belgium

tel: +32 10 47 30 50

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Re: [R] If condition using accessors

2009-11-27 Thread Vitória Magalhães Piai


Thanx.
I'll just post in here the way I solved it in case someone happens to 
need something similar in the future. It may not be the most elegant 
way, but it worked!


yes = small
no = large
data3$group = ifelse(data3$SIE  50.64593,yes,no)

Peter Ehlers wrote:


Peter Ehlers wrote:

Vitória Magalhães Piai wrote:

Hi,

I'm quite new using R and have got no one to help me get through it. 
Hopefully someone can help me with one problem I've been struggling 
with for the last hours!!

(Sorry if I'm using the wrong terminology as well!)

I have a data matrix in which SIE is one of my variables. What I 
need to do now is to create a new variable (group) if a certain 
condition in SIE is met.

I tried:

if (data2$SIE  50.646){
data2$group=as.factor(small)
}
which gives me In if (data2$SIE  50.646) { :  the condition has 
length   1 and only the first element will be used


and
ifelse (data2$SIE  50.64593, data2$group = 
as.factor(small),data2$group = as.factor(large))
with Error: unexpected '=' in ifelse (data2$SIE  50.64593, 
data2$group =


I tried all other kinds of variations and so far, nothing!!


Do all of the following:

?=
?==
?if
?ifelse

 -Peter Ehlers


Actually, I assumed without checking that ?ifelse would
have an example of assignment of the result of the test.
(Maybe it could be added.)

yourVariable - ifelse(...)

 -Peter Ehlers





Is there a good way to do this that someone could teach me?
Thank you so much,
Vitória

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Re: [R] Time Series Data

2009-11-27 Thread chris carleton


I'm not sure what you mean by 'multiplying by 10 and rounding'. I've tried to 
re-organize the data and then create a ts object that is analogous to the 
various sunspot datasets provided in the data package with R, but the matrix I 
generate by binning the data into centuries (which would make the frequency in 
stl() 10) is being read by plot() and stl() as a multivariate time series 
dataset (stl() requires that the data be univariate). For those that haven't 
read stl() docs, the frequency is the number of observations per unit of time. 
In the case of quarterly economic data the frequency is 4 (4 observations over 
a year), and in the case of decadal sunspot numbers the frequency is .1 (0.1 
observations per year). I would have thought that after I binned the data into 
centuries (now ten observations per unit of time) that it would be sufficient, 
but I don't know why stl() interprets the matrix as a multivariate dataset when 
a similar matrix (sunspots - measured each month for a !
 frequency of 12; although from what I can tell the sunspots data in R is not a 
ts object anyhow) encounters no such difficulty. Any suggestions would be 
appreciated,

Chris

 CC: r-help@r-project.org
 From: dwinsem...@comcast.net
 To: w_chris_carle...@hotmail.com
 Subject: Re: [R] Time Series Data
 Date: Fri, 27 Nov 2009 10:10:36 -0500
 
 
 On Nov 27, 2009, at 9:55 AM, chris carleton wrote:
 
 
  Hi All,
 
  I'm trying to analyze some time series data and I have run into  
  difficulty. I have decadal sun spot data and I want to separate the  
  very regular periodic function from the trend and noise. I looked  
  into using stl(), but the frequency of the time series data must be  
  greater than 1 for stl(). My data covers a 1000 year interval from  
  9095 BP to 8095BP and the frequency is, therefore, 0.1 (because the  
  data are decadal). I've tried changing the frequency, but the only  
  frequency that creates a plot of the time series data which matches  
  the raw data is 0.1. Is there anything I can do to the data that  
  will make it more amenable to stl(), or is there another package  
  that I could use for decomposing the signal that does not require  
  that the frequency of the time series to be greater than 1? Thanks,
 
 I do not understand why you are not multiplying by 10 and rounding.
 
 -- 
 David
 
 
 
 David Winsemius, MD
 Heritage Laboratories
 West Hartford, CT
 
  
_

 Facebook.

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[R] changing titlessymbols in plot.spm

2009-11-27 Thread Assa Yeroslaviz

Hi all,

I'm trying my best in changing the names of my experiments in the spectral
map.

My experiments are from six different time points, two as control and two
treated experiments.
I set have for each time point a different color .
Now I want to change the symbols so that I have for the treated experiments
one and for the controls a different one. But I want only two different
symbols which.
The best option will be to have a C for the controls and a T for the
treated Experiments. I than won't be needing the title of the experiments.
But a different symbol will also be good.
I tried it with this command:

 stress - c(control , control , treated , treated , control ,
control , treated , treated , control , control , treated ,
treated , control , control , treated , treated , control ,
control , treated , treated , control , control , treated ,
treated)

# each four exp. are in one time point, therefore I am using six different
colors for the experiments.

 stress.Factor - factor(stress)

 plot(r.sma,label.tol=0,scale=uvc,col.group=(ramgrp[,2]),colors =
c(grey90,gray5,green3,gold1,dodgerblue2,indianred3,lightskyblue,hotpink4),
col.areas
= FALSE, col.symbols =unclass(stress.Factor) ,zoom=c(1,5),col.size=2)

Unfortunately it's not working. What i get is to see in the attachment. i
have two symbols but for the wrong experiments.
I would happy for any suggestions which help me solve this problem.

Thanks

Assa


-- 
Assa Yeroslaviz
Kockelsberg 22
51371 Leverkusen
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Force a variable substitution in expression

2009-11-27 Thread Saptarshi Guha

 The substitute() function can do it, but the bquote() function is the most
 convenient:

 foo - function(s) {
 +   bquote(expression({
 +    v - .(s)
 +    print(v)
 +   }))
 + }
 foo(10)
 expression({
   v - 10
   print(v)
 })

 Duncan Murdoch


Thanks, after sending this email, i used substitute, but first eval'ing
the formals (since they are delayed assignments).
I see bquote does this for me.


Regards
Saptarshi

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Re: [R] Save workspace with ff objects

2009-11-27 Thread Jens Oehlschlägel

 My script generates a mixture of normal and ff objects. 
 Very often I would like to save the workspace for 
 each parameter setting, so that I can get back to it later on.
 Is there an easy way to do this, 
 instead of needing to save individual ff objects separately? 

With one save() you can store as many ff objects as you like. 
However, this does not save the ff files to a different location.

 I've tried the naive way of just saving the workspace, only to find that ff 
 objects are empty.

When loading the ff objects, the ff files need to be in their original 
locations.
You need to make sure that you do not overwrite those 
and they survive finalizer and tempdir remove at rm(ff) or q() time.
Do read the ff help on parameters 'filename', 'pattern', 'finalizer', 
'finonexit'.

The next version of ff will have ffsave() 
which will store a mixture of normal and ff objects 
*and* all ff-files into a ffarchive, i.e. two files 
ffarchive.RData and ffarchive.ffData 
from which you can restore all or a selection of 
ff objects / files using the ffload() command.

Regards
Jens Oehlschlägel

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Re: [R] questions on the ff package

2009-11-27 Thread Jens Oehlschlägel

 I wonder how efficiently it is to do the following command on a frequent 
 basis.
 nrow(matFF) - nrow(matFF)+1

Obviously there is overhead (closing file, enlarging file, openeing file). 
I recommend you measure yourself whether this is acceptable for you.

 no large file copying is needed each time the nrow is changed?

With a decent filesystem there is *no* copying from smaller to larger file.

 would you think I can open 2000 large matrices and leave them open or I
need to close each after it is opened and used?

Not tested yet. 
I guess the number of open files can be configured when compiling your OS. 
Please test and let us know your experience.

Regards
Jens Oehlschlägel

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Re: [R] Simple Function doesn't work?

2009-11-27 Thread Colin Millar

Hi,

You would also make your code more efficient and possible more readable
by doing

ReturnsGrid - 
function(x, y, m)
{
  x + (seq.int(m) - 1) * (y - x) / m
}

(xx - ReturnsGrid(0, 9, 3))
#[1] 0 3 6

And if you want to supply vector x and y you could do something like
(there are probably better ways..)

ReturnsGrid - 
function(x, y, m)
{
  if (length(x) != length(y)  (length(x)==1 | length(y) == 1)) stop
(inputs not compatible) # or something
  n - max(length(x), length(y))
  out - sapply(seq.int(n), function(i) x[i] + (1:m - 1) * (y[i] - x[i])
/ m)
  
  drop(out)
}

(xx  - ReturnsGrid(0, 9, 3))
#[1] 0 3 6

(xx  - ReturnsGrid(0:2, 9:11, 3))
#[1,]012
#[2,]345
#[3,]678


But it seems like you could also do it using sequence ...

seq(x, y-1, by = m)

HTH,
Colin 



-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Anastasia
Sent: 27 November 2009 16:01
To: r-help@r-project.org
Subject: [R] Simple Function doesn't work?

Hello,

I am new to R program, therefore, I am sorry if this is a really stupid
question.
I wrote a simple function and for some reason it doesn't work

ReturnsGrid = function(x,y,m){
for (i in 1:m){
   grid[i] - x + (i-1)*(y-x)/m
}
grid
}

xx=ReturnsGrid(0,9,3)

Thanks a lot!

[[alternative HTML version deleted]]

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Re: [R] Simple Function doesn't work?

2009-11-27 Thread Colin Millar


Erm... Maybe the sequence bit wont work ... A bit hasty there

And also the length check should be

if (length(x) != length(y)  !(length(x) == 1 | length(y) == 1)) stop
(inputs not compatible) # or something
 
I missed out the not!

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Colin Millar
Sent: 27 November 2009 16:41
To: Anastasia; r-help@r-project.org
Subject: Re: [R] Simple Function doesn't work?

Hi,

You would also make your code more efficient and possible more readable
by doing

ReturnsGrid -
function(x, y, m)
{
  x + (seq.int(m) - 1) * (y - x) / m
}

(xx - ReturnsGrid(0, 9, 3))
#[1] 0 3 6

And if you want to supply vector x and y you could do something like
(there are probably better ways..)

ReturnsGrid -
function(x, y, m)
{
  if (length(x) != length(y)  (length(x)==1 | length(y) == 1)) stop
(inputs not compatible) # or something
  n - max(length(x), length(y))
  out - sapply(seq.int(n), function(i) x[i] + (1:m - 1) * (y[i] - x[i])
/ m)
  
  drop(out)
}

(xx  - ReturnsGrid(0, 9, 3))
#[1] 0 3 6

(xx  - ReturnsGrid(0:2, 9:11, 3))
#[1,]012
#[2,]345
#[3,]678


But it seems like you could also do it using sequence ...

seq(x, y-1, by = m)

HTH,
Colin 



-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Anastasia
Sent: 27 November 2009 16:01
To: r-help@r-project.org
Subject: [R] Simple Function doesn't work?

Hello,

I am new to R program, therefore, I am sorry if this is a really stupid
question.
I wrote a simple function and for some reason it doesn't work

ReturnsGrid = function(x,y,m){
for (i in 1:m){
   grid[i] - x + (i-1)*(y-x)/m
}
grid
}

xx=ReturnsGrid(0,9,3)

Thanks a lot!

[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Best way to preallocate numeric NA array?

2009-11-27 Thread William Dunlap

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Rob Steele
 Sent: Thursday, November 26, 2009 8:03 AM
 To: r-h...@stat.math.ethz.ch
 Subject: [R] Best way to preallocate numeric NA array?

 These are the ways that occur to me.

 ## This produces a logical vector, which will get converted 
 to a numeric
 ## vector the first time a number is assigned to it.  That seems
 ## wasteful.
 x - rep(NA, n)

 ## This does the conversion ahead of time but it's still creating a
 ## logical vector first, which seems wasteful.
 x - as.numeric(rep(NA, n))

 ## This avoids type conversion but still involves two assignments for
 ## each element in the vector.
 x - numeric(n)
 x[] - NA

 ## This seems reasonable.
 x - rep(as.numeric(NA), n)

 Comments?

I like the rep-based ones because they are more 'functional':
I can pass the output directly into another function.  If speed
is a big concern, I think that
   x - rep.int(NA_real_, n)
may be the fastest but then it is not immediately apparent how
to do the same for integer or fancier classes.  (NA_real_ is
to the numeric class as NA_integer_ is to the integer class.)
Your last one, rep(as.class(NA), n), seems to me to be a
good compromise between readability and speed.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com 

 Thanks,
 Rob

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[R] R 2.10 Line Type Issue...



Morning folks (at least here on the western side of the U.S.)...

This morning I constructed a contour plot of some bivariate  
distributions I'm working with. When I attempted to add a second  
contour to the plot using a dashed line (lty=2), R immediately went  
off to la-la land, requiring a force-quit. I finally tied the problem  
down to this statement:


contour(u,v,ci,levels=c(0.30),add=TRUE,lty=2)

I tried specifying lty=dash as well without any change. The plot  
works with no lty specification (but produces a solid line, of course).


In researching the issue, I learned that the demo, demo(graphics),  
should produce a variety of lines on the first plot. On my system, the  
first plot is completely blank.


My system: PowerMac dual-G5, Leopard 10.5.8, R 2.10 fresh install. I  
attempted re-installing with no change.


Any ideas?

Thanks...

-=d

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Re: [R] problem with dynformula from plm package [RE-POST]

Great! That did the trick, thanks David.

To summarize for the list, to get dynformula to work (example):

R a = 
dynformula(emp~wage+capital,log.form=list(capital=FALSE,TRUE),lag.form=list(emp=2,c(2,3)),diff.form=list(FALSE,capital=TRUE))

it might be necessary to use:

R plm(formula = formula(a), data=EmplUK)

rather than just:

R plm(formula = a, data = EmplUK)

~Owen

2009/11/27 David Winsemius dwinsem...@comcast.net:
 You might also note that the authors hint on the help page that one might
 want to use the formula() operation on the result of dynformula. Following
 that path would have gotten us to a more successful conclusion.

 grun.fe - plm(formula = formula(a),data=EmplUK)
 grun.fe

 Model Formula: log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) +
 lag(log(emp),
    1) + lag(log(emp), 2) + lag(log(wage), 2) + lag(log(wage),
    3) + diff(capital, 2) + diff(capital, 3)

 Coefficients:
  lag(log(emp), 1)  lag(log(emp), 2) lag(log(wage), 2) lag(log(wage), 3)
  diff(capital, 2)
        0.8678675        -0.1936447        -0.1632724         0.3200785
   0.0037612
  diff(capital, 3)
        0.0137866


 --
 David.
 On Nov 27, 2009, at 12:04 PM, Owen Powell wrote:

 Hi David,

 Thank you for the response.

 I forgot to mention that I'd already tried what (I think) you propose
 (adding .form to the end of the lag, log and diff) and I still
 see the same results (posted below). Specifically, I still see no
 lags, logs or diffs in my model.

 Any other ideas?

 ~Owen

 R rm(list = ls())
 R options(prompt= R )
 R library(plm)
 R data(EmplUK, package=plm)
 R EmplUK - plm.data(EmplUK, index = c(firm, year))
 R
 log(emp)~lag(log(emp),1)+lag(log(emp),2)+lag(log(wage),2)+lag(log(wage),3)+diff(capital,2)+diff(capital,3)
 log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(wage),
   2) + lag(log(wage), 3) + diff(capital, 2) + diff(capital,
   3)
 R a =
 dynformula(emp~wage+capital,log.form=list(capital=FALSE,TRUE),lag.form=list(emp=2,c(2,3)),diff.form=list(FALSE,capital=TRUE))
 R grun.fe - plm(formula = a,data=EmplUK,model=within)
 [1] 1031    2
 R summary(grun.fe)
 Oneway (individual) effect Within Model

 Call:
 plm(formula = a, data = EmplUK, model = within)

 Unbalanced Panel: n=140, T=7-9, N=1031

 Residuals :
   Min.  1st Qu.   Median  3rd Qu.     Max.
 -17.1000  -0.3060   0.0137   0.3070  27.3000

 Coefficients :
        Estimate Std. Error t-value  Pr(|t|)
 wage    -0.143626   0.032790 -4.3802 1.186e-05 ***
 capital  0.801495   0.064088 12.5062  2.2e-16 ***
 ---
 Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

 Total Sum of Squares:    5030.6
 Residual Sum of Squares: 4207.8
 F-statistic: 86.9179 on 2 and 889 DF, p-value:  2.22e-16

 2009/11/27 David Winsemius dwinsem...@comcast.net

 On Nov 27, 2009, at 10:25 AM, Owen Powell wrote:

 Hello list,

 I'm following the paper (http://www.jstatsoft.org/v27/i02/paper) on
 how to use plm to run panel regressions, and am having trouble with
 what I believe should be something very basic.

 When I run the command (p.9 in the paper):

 R

 dynformula(emp~wage+capital,log=list(capital=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))


 Perhaps you could have read the help page for the current version of the
 package which says the argument have been modified. Using the current
 arguments:


 dynformula(emp~wage+capital,log.form=list(capital=FALSE,TRUE),lag.form=list(emp=2,c(2,3)),diff.form=list(FALSE,capital=TRUE))

 log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(emp),
   1) + lag(log(emp), 2) + lag(log(wage), 2) + lag(log(wage),
   3) + diff(capital, 2) + diff(capital, 3)

 --
 David Winsemius, MD

 I see:

 emp ~ wage + capital

 rather than the complete model that is given in the paper:

 log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(wage), 2) +
 lag(log(wage), 3) + diff(capital, 2) + diff(capital, 3)

 And indeed, when I try to run a regression using that formula, it
 appears to not contain any lags or logs (output below).

 Any ideas? Thanks in advance,

 ~Owen

 --
 Owen Powell
 http://center.uvt.nl/phd_stud/powell

 R library(plm)
 R data(EmplUK, package=plm)
 R a =

 dynformula(emp~wage+capital,log=list(capital=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))

 snipped


 David Winsemius, MD
 Heritage Laboratories
 West Hartford, CT




 --
 Owen Powell
 http://center.uvt.nl/phd_stud/powell

 David Winsemius, MD
 Heritage Laboratories
 West Hartford, CT





-- 
Owen Powell
http://center.uvt.nl/phd_stud/powell

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] problem with dynformula from plm package [RE-POST]

You might also note that the authors hint on the help page that one  
might want to use the formula() operation on the result of dynformula.  
Following that path would have gotten us to a more successful  
conclusion.


 grun.fe - plm(formula = formula(a),data=EmplUK)
 grun.fe

Model Formula: log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) +  
lag(log(emp),

1) + lag(log(emp), 2) + lag(log(wage), 2) + lag(log(wage),
3) + diff(capital, 2) + diff(capital, 3)

Coefficients:
 lag(log(emp), 1)  lag(log(emp), 2) lag(log(wage), 2) lag(log(wage),  
3)  diff(capital, 2)
0.8678675-0.1936447-0.1632724  
0.3200785 0.0037612

 diff(capital, 3)
0.0137866


--
David.
On Nov 27, 2009, at 12:04 PM, Owen Powell wrote:


Hi David,

Thank you for the response.

I forgot to mention that I'd already tried what (I think) you propose
(adding .form to the end of the lag, log and diff) and I still
see the same results (posted below). Specifically, I still see no
lags, logs or diffs in my model.

Any other ideas?

~Owen

R rm(list = ls())
R options(prompt= R )
R library(plm)
R data(EmplUK, package=plm)
R EmplUK - plm.data(EmplUK, index = c(firm, year))
R log(emp)~lag(log(emp),1)+lag(log(emp),2)+lag(log(wage), 
2)+lag(log(wage),3)+diff(capital,2)+diff(capital,3)

log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(wage),
   2) + lag(log(wage), 3) + diff(capital, 2) + diff(capital,
   3)
R a = dynformula(emp~wage 
+ 
capital 
,log 
.form 
= 
list 
(capital 
= 
FALSE 
,TRUE),lag.form=list(emp=2,c(2,3)),diff.form=list(FALSE,capital=TRUE))

R grun.fe - plm(formula = a,data=EmplUK,model=within)
[1] 10312
R summary(grun.fe)
Oneway (individual) effect Within Model

Call:
plm(formula = a, data = EmplUK, model = within)

Unbalanced Panel: n=140, T=7-9, N=1031

Residuals :
   Min.  1st Qu.   Median  3rd Qu. Max.
-17.1000  -0.3060   0.0137   0.3070  27.3000

Coefficients :
Estimate Std. Error t-value  Pr(|t|)
wage-0.143626   0.032790 -4.3802 1.186e-05 ***
capital  0.801495   0.064088 12.5062  2.2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Total Sum of Squares:5030.6
Residual Sum of Squares: 4207.8
F-statistic: 86.9179 on 2 and 889 DF, p-value:  2.22e-16

2009/11/27 David Winsemius dwinsem...@comcast.net


On Nov 27, 2009, at 10:25 AM, Owen Powell wrote:


Hello list,

I'm following the paper (http://www.jstatsoft.org/v27/i02/paper) on
how to use plm to run panel regressions, and am having trouble  
with

what I believe should be something very basic.

When I run the command (p.9 in the paper):

R
dynformula(emp~wage 
+ 
capital 
,log 
= 
list 
(capital 
=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))




Perhaps you could have read the help page for the current version  
of the package which says the argument have been modified. Using  
the current arguments:


dynformula(emp~wage 
+ 
capital 
,log 
.form 
= 
list 
(capital 
= 
FALSE 
,TRUE 
),lag.form=list(emp=2,c(2,3)),diff.form=list(FALSE,capital=TRUE))


log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(emp),
   1) + lag(log(emp), 2) + lag(log(wage), 2) + lag(log(wage),
   3) + diff(capital, 2) + diff(capital, 3)

--
David Winsemius, MD


I see:

emp ~ wage + capital

rather than the complete model that is given in the paper:

log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(wage), 2) +
lag(log(wage), 3) + diff(capital, 2) + diff(capital, 3)

And indeed, when I try to run a regression using that formula, it
appears to not contain any lags or logs (output below).

Any ideas? Thanks in advance,

~Owen

--
Owen Powell
http://center.uvt.nl/phd_stud/powell

R library(plm)
R data(EmplUK, package=plm)
R a =
dynformula(emp~wage 
+ 
capital 
,log 
= 
list 
(capital 
=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))


snipped


David Winsemius, MD
Heritage Laboratories
West Hartford, CT





--
Owen Powell
http://center.uvt.nl/phd_stud/powell


David Winsemius, MD
Heritage Laboratories
West Hartford, CT

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] Modular inverses

2009-11-27 Thread SJ Robson-Davis

I want to find the inverse of an integer k mod p (prime.) Is there a
function that can do this for me? I know i could simply write (k^(p-2)) %%
p, but i need to do this for large primes (above 100) and this gives the
warning message:
Warning message:
probable complete loss of accuracy in modulus
so this method does not work. Any ideas?

Thanks,

Samuel

--

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Re: [R] problem with dynformula from plm package [RE-POST]



On Nov 27, 2009, at 12:04 PM, Owen Powell wrote:


Hi David,

Thank you for the response.

I forgot to mention that I'd already tried what (I think) you propose
(adding .form to the end of the lag, log and diff) and I still
see the same results (posted below). Specifically, I still see no
lags, logs or diffs in my model.

Any other ideas?

~Owen

R rm(list = ls())
R options(prompt= R )
R library(plm)
R data(EmplUK, package=plm)
R EmplUK - plm.data(EmplUK, index = c(firm, year))
R log(emp)~lag(log(emp),1)+lag(log(emp),2)+lag(log(wage), 
2)+lag(log(wage),3)+diff(capital,2)+diff(capital,3)

log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(wage),
   2) + lag(log(wage), 3) + diff(capital, 2) + diff(capital,
   3)
R a = dynformula(emp~wage 
+ 
capital 
,log 
.form 
= 
list 
(capital 
= 
FALSE 
,TRUE),lag.form=list(emp=2,c(2,3)),diff.form=list(FALSE,capital=TRUE))

R grun.fe - plm(formula = a,data=EmplUK,model=within)
[1] 10312


If I instead use the results of dynformula in plm as copied as a  
string from the output, I get what appears to be more what one might  
expect:


 grun.fe - plm(log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) +  
lag(log(emp),

+ 1) + lag(log(emp), 2) + lag(log(wage), 2) + lag(log(wage),
+ 3) + diff(capital, 2) + diff(capital, 3) ,data=EmplUK)
 grun.fe

Model Formula: log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) +  
lag(log(emp),

1) + lag(log(emp), 2) + lag(log(wage), 2) + lag(log(wage),
3) + diff(capital, 2) + diff(capital, 3)

Coefficients:
 lag(log(emp), 1)  lag(log(emp), 2) lag(log(wage), 2) lag(log(wage),  
3)  diff(capital, 2)
0.8678675-0.1936447-0.1632724  
0.3200785 0.0037612

 diff(capital, 3)
0.0137866

It certainly appears that the function plm is not decoding its  
arguments as do most regression functions. Perhaps you should take up  
your concerns for this non-standard behavior up with the package  
authors?


--
David.


R summary(grun.fe)
Oneway (individual) effect Within Model

Call:
plm(formula = a, data = EmplUK, model = within)

Unbalanced Panel: n=140, T=7-9, N=1031

Residuals :
   Min.  1st Qu.   Median  3rd Qu. Max.
-17.1000  -0.3060   0.0137   0.3070  27.3000

Coefficients :
Estimate Std. Error t-value  Pr(|t|)
wage-0.143626   0.032790 -4.3802 1.186e-05 ***
capital  0.801495   0.064088 12.5062  2.2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Total Sum of Squares:5030.6
Residual Sum of Squares: 4207.8
F-statistic: 86.9179 on 2 and 889 DF, p-value:  2.22e-16

2009/11/27 David Winsemius dwinsem...@comcast.net


On Nov 27, 2009, at 10:25 AM, Owen Powell wrote:


Hello list,

I'm following the paper (http://www.jstatsoft.org/v27/i02/paper) on
how to use plm to run panel regressions, and am having trouble  
with

what I believe should be something very basic.

When I run the command (p.9 in the paper):

R
dynformula(emp~wage 
+ 
capital 
,log 
= 
list 
(capital 
=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))




Perhaps you could have read the help page for the current version  
of the package which says the argument have been modified. Using  
the current arguments:


dynformula(emp~wage 
+ 
capital 
,log 
.form 
= 
list 
(capital 
= 
FALSE 
,TRUE 
),lag.form=list(emp=2,c(2,3)),diff.form=list(FALSE,capital=TRUE))


log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(emp),
   1) + lag(log(emp), 2) + lag(log(wage), 2) + lag(log(wage),
   3) + diff(capital, 2) + diff(capital, 3)

--
David Winsemius, MD


I see:

emp ~ wage + capital

rather than the complete model that is given in the paper:

log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(wage), 2) +
lag(log(wage), 3) + diff(capital, 2) + diff(capital, 3)

And indeed, when I try to run a regression using that formula, it
appears to not contain any lags or logs (output below).

Any ideas? Thanks in advance,

~Owen

--
Owen Powell
http://center.uvt.nl/phd_stud/powell

R library(plm)
R data(EmplUK, package=plm)
R a =
dynformula(emp~wage 
+ 
capital 
,log 
= 
list 
(capital 
=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))


snipped


David Winsemius, MD
Heritage Laboratories
West Hartford, CT





--
Owen Powell
http://center.uvt.nl/phd_stud/powell


David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] problem with dynformula from plm package [RE-POST]

Hi David,

Thank you for the response.

I forgot to mention that I'd already tried what (I think) you propose
(adding .form to the end of the lag, log and diff) and I still
see the same results (posted below). Specifically, I still see no
lags, logs or diffs in my model.

Any other ideas?

~Owen

R rm(list = ls())
R options(prompt= R )
R library(plm)
R data(EmplUK, package=plm)
R EmplUK - plm.data(EmplUK, index = c(firm, year))
R 
log(emp)~lag(log(emp),1)+lag(log(emp),2)+lag(log(wage),2)+lag(log(wage),3)+diff(capital,2)+diff(capital,3)
log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(wage),
2) + lag(log(wage), 3) + diff(capital, 2) + diff(capital,
3)
R a = 
dynformula(emp~wage+capital,log.form=list(capital=FALSE,TRUE),lag.form=list(emp=2,c(2,3)),diff.form=list(FALSE,capital=TRUE))
R grun.fe - plm(formula = a,data=EmplUK,model=within)
[1] 10312
R summary(grun.fe)
Oneway (individual) effect Within Model

Call:
plm(formula = a, data = EmplUK, model = within)

Unbalanced Panel: n=140, T=7-9, N=1031

Residuals :
Min.  1st Qu.   Median  3rd Qu. Max.
-17.1000  -0.3060   0.0137   0.3070  27.3000

Coefficients :
 Estimate Std. Error t-value  Pr(|t|)
wage-0.143626   0.032790 -4.3802 1.186e-05 ***
capital  0.801495   0.064088 12.5062  2.2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Total Sum of Squares:5030.6
Residual Sum of Squares: 4207.8
F-statistic: 86.9179 on 2 and 889 DF, p-value:  2.22e-16

2009/11/27 David Winsemius dwinsem...@comcast.net

 On Nov 27, 2009, at 10:25 AM, Owen Powell wrote:

 Hello list,

 I'm following the paper (http://www.jstatsoft.org/v27/i02/paper) on
 how to use plm to run panel regressions, and am having trouble with
 what I believe should be something very basic.

 When I run the command (p.9 in the paper):

 R
 dynformula(emp~wage+capital,log=list(capital=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))


 Perhaps you could have read the help page for the current version of the 
 package which says the argument have been modified. Using the current 
 arguments:

 dynformula(emp~wage+capital,log.form=list(capital=FALSE,TRUE),lag.form=list(emp=2,c(2,3)),diff.form=list(FALSE,capital=TRUE))

 log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(emp),
    1) + lag(log(emp), 2) + lag(log(wage), 2) + lag(log(wage),
    3) + diff(capital, 2) + diff(capital, 3)

 --
 David Winsemius, MD

 I see:

 emp ~ wage + capital

 rather than the complete model that is given in the paper:

 log(emp) ~ lag(log(emp), 1) + lag(log(emp), 2) + lag(log(wage), 2) +
 lag(log(wage), 3) + diff(capital, 2) + diff(capital, 3)

 And indeed, when I try to run a regression using that formula, it
 appears to not contain any lags or logs (output below).

 Any ideas? Thanks in advance,

 ~Owen

 --
 Owen Powell
 http://center.uvt.nl/phd_stud/powell

 R library(plm)
 R data(EmplUK, package=plm)
 R a =
 dynformula(emp~wage+capital,log=list(capital=FALSE,TRUE),lag=list(emp=2,c(2,3)),diff=list(FALSE,capital=TRUE))

 snipped


 David Winsemius, MD
 Heritage Laboratories
 West Hartford, CT




--
Owen Powell
http://center.uvt.nl/phd_stud/powell

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[R] Replace Values in Matrix

2009-11-27 Thread Romildo Martins

Hello,

how to replace the NA by number zero?

 matrizt
 [,1]   [,2] [,3][,4]
[1,] 1.000NA NA NA
[2,] 0.6717685 0.1453253 NA NA
[3,] 0.3971276 0.1493241 0.14532526 NA
[4,] 0.1493241 0.1453253 0.06795269 0.06217967


Thanks!

Romildo

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Re: [R] Replace Values in Matrix

2009-11-27 Thread baptiste auguie

Hi,

Try this,

matrizt[is.na(matrizt)] - 0

HTH,

baptiste

2009/11/27 Romildo Martins romildo.mart...@gmail.com:
 Hello,

 how to replace the NA by number zero?

 matrizt
                 [,1]           [,2]             [,3]        [,4]
 [1,] 1.000            NA             NA         NA
 [2,] 0.6717685 0.1453253             NA         NA
 [3,] 0.3971276 0.1493241 0.14532526         NA
 [4,] 0.1493241 0.1453253 0.06795269 0.06217967


 Thanks!

 Romildo

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Re: [R] Modular inverses

2009-11-27 Thread William Dunlap

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of SJ Robson-Davis
 Sent: Friday, November 27, 2009 8:52 AM
 To: r-help@r-project.org
 Subject: [R] Modular inverses

 I want to find the inverse of an integer k mod p (prime.) Is there a
 function that can do this for me? I know i could simply write 
 (k^(p-2)) %%
 p, but i need to do this for large primes (above 100) and 
 this gives the
 warning message:
 Warning message:
 probable complete loss of accuracy in modulus
 so this method does not work. Any ideas?

Brute force works:
f-function(k, p) which(((1:(p-1))*k)%%p == 1)
f(10,p=13)
   [1] 4
You can precompute the mapping from k to 1/k mod p
if you will be working with the same p a lot.  Use
sapply() or mapply() if you will be working on a vector
of k and p.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

 Thanks,

 Samuel

 --

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 http://www.R-project.org/posting-guide.html
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[R] using reshape to do ANOVA mixed models

2009-11-27 Thread Or Duek

Hi,
I just started with R and I found that there are many options to rearrange
the data to do mixed models.
I want to use the reshape function. I have 2 between subject variables and
one within.
I was able to change the data structure but still - the result of the aov
functions are calculating everything as a within subject.
the table looks like this:
SerialNobreedtreatmentdistance_1   distance_2
1   c57   dfp 235  3253
etc.

I changed it to look like this:
SerialNo   breed   treatmentexposure   distance
1  c57 dfp   1  235
1  c57dfp2  3253
etc.

Then I do:
dt-aov(distance~(exposure*treatment*breed)+Error(SerialNo/exposure) +
(treatment*breed), dataframe)

what am I doing wrong?
thank you.

[[alternative HTML version deleted]]

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Re: [R] Replace Values in Matrix

2009-11-27 Thread smu


matritz[is.na(matritz)] - 0


On Fri, Nov 27, 2009 at 04:15:45PM -0200, Romildo Martins wrote:
 Hello,
 
 how to replace the NA by number zero?
 
  matrizt
  [,1]   [,2] [,3][,4]
 [1,] 1.000NA NA NA
 [2,] 0.6717685 0.1453253 NA NA
 [3,] 0.3971276 0.1493241 0.14532526 NA
 [4,] 0.1493241 0.1453253 0.06795269 0.06217967
 
 
 Thanks!
 
 Romildo
 
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Graphic Device - View/get all graphics

2009-11-27 Thread MarcioRibeiro


Hello,
After some research, I didn't resolve my doubt yet...
Maybe I didn't explain very well my question... So, I have a bootstrap
simulation, but suppose just a simple part of my function:

teste -function(x){
rw_mean_app-rnorm(x, mean = 10, sd = 2)
rwy_mean_app-rnorm(x, mean = 10, sd = 2)
rw_median_app_ori-rnorm(x, mean = 10, sd = 2)
rw_median_app_mod-rnorm(x, mean = 10, sd = 2)

histogram-par(mfrow=c(1,2),ask=TRUE)
hist(rw_mean_app,main='Method RWOriginal',xlab='Mean',ylab=' ')
hist(rwy_mean_app,main='Method RWY',xlab='Mean',ylab=' ')
par(histogram)

histogram-par(mfrow=c(1,2) ,ask=TRUE)
hist(rw_median_app_ori,main='Method RWOriginal',xlab='Median',ylab=' ')
hist(rw_median_app_mod,main='Method RWModified',xlab='Median',ylab=' ')
par(histogram)

}   #END OF FUNCTION
GRAF-teste(100)

I cannot save rnorm objects in a list cause they are distribution from 1000
simulation... So, I would like to produce the graphics in my function and
find out a way that I could save the graphics... I can save the object using
the LIST procedure, but I dont know how I save an object that is a
graphic...
Listers, I just want some advices in order to make a research about what
functions of R I could save the graphics in my function...
Thanks again,
Marcio




MarcioRibeiro wrote:
 
 Hi Listers,
 I am producing some graphics that the commands are in a FUNCTION...
 The problem is that I end up viewing just last graphic and in my FUNCTION
 there are 4 graphics with the PAR command function... Like those below...
 How do I view/get the other 3 graphics? Any help?
 Thanks in advance...
 
 histogram-par(mfrow=c(1,2))
 hist(rw_mean_app,main='Bootstrap Method RWOriginal',xlab='Mean',ylab=' ')
 hist(rwy_mean_app,main='Bootstrap Method RWY',xlab='Mean',ylab=' ')
 par(histogram)
 
 histogram-par(mfrow=c(1,3))
 hist(rw_median_app_ori,main='Bootstrap Method
 RWOriginal',xlab='Median',ylab=' ')
 hist(rw_median_app_mod,main='Bootstrap Method
 RWModified',xlab='Median',ylab=' ')
 hist(rwy_median_app,main='Bootstrap Method RWY',xlab='Median',ylab=' ')
 par(histogram)
 
 

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[R] Build of XML package failed

2009-11-27 Thread Luis Tito de Morais

Hi list,

It may be a FAQ, but I searched the web and Uni of Newcastle Maths and Stats and
R mailing list archive on this issue but was unable to find a solution. I would
appreciate any pointer to help me solving this.

I am using R version 2.10.0 (2009-10-26) on linux mandriva 2010.0

I tried to install the XML_2.6-0.tar.gz package both with
install.packages('XML', dep=T) from within R and the R CMD INSTALL using a
local tar.gz file.

I am having the following error message (sorry it is partly in french):

Dans le fichier inclus à partir de DocParse.c:13:
Utils.h:175:2: attention : #warning Redefining COPY_TO_USER_STRING to use
encoding from XML parser
DocParse.c: In function notifyError:
DocParse.c:1051: erreur: le format n'est pas une chaîne littérale et pas
d'argument de format

This last error message means:
error: format not a string literal and no format arguments

In the past when having such errors with other packages, I have been able to
solve it with the help of this tip:
http://mario79t.wordpress.com/2009/06/23/warning-format-not-a-string-literal-and-no-format-arguments/

and modifying the faulty source file accordingly.

But in this specific case, I have been unable to find what to modify in the
source file. Line 1051 in the DocParse.c source file only has the command
WARN. I don't know anything about C programming and could not figure out what
to modify in this case.

I would appreciate any help on this issue.

Best regards,

Tito

__
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Re: [R] using reshape to do ANOVA mixed models

2009-11-27 Thread Peter Dalgaard


Or Duek wrote:

Hi,
I just started with R and I found that there are many options to rearrange
the data to do mixed models.
I want to use the reshape function. I have 2 between subject variables and
one within.
I was able to change the data structure but still - the result of the aov
functions are calculating everything as a within subject.
the table looks like this:
SerialNobreedtreatmentdistance_1   distance_2
1   c57   dfp 235  3253
etc.

I changed it to look like this:
SerialNo   breed   treatmentexposure   distance
1  c57 dfp   1  235
1  c57dfp2  3253
etc.

Then I do:
dt-aov(distance~(exposure*treatment*breed)+Error(SerialNo/exposure) +
(treatment*breed), dataframe)

what am I doing wrong?


For a start, not showing us the output Also not showing us enough of 
the data and not convincing us that you are actually using the reshaped 
data frame.



--
   O__   Peter Dalgaard Øster Farimagsgade 5, Entr.B
  c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
 (*) \(*) -- University of Copenhagen   Denmark  Ph:  (+45) 35327918
~~ - (p.dalga...@biostat.ku.dk)  FAX: (+45) 35327907

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[R] Iteration idioms laziness

2009-11-27 Thread Alexander Søndergaard

Hi all,

I'm new to R. Having a functional background, I was wondering what's
the idiomatic way to iterate. It seems that for loops are the default
given there's no tail-call optimization.

I'm curious to know whether there is a way to transform the following
toy snippet into something that doesn't eat up gigabytes of memory
(like it's for loop counterpart) using laziness:

Reduce('+', seq(1,1e6))

Thanks!

Best regards,
A.S.

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Re: [R] Build of XML package failed

2009-11-27 Thread Duncan Temple Lang

Hi Luis.

You can change the two lines

PROBLEM buf
WARN;

to the one line

warning(buf);

That should compile.

If not, please show us the compilation command for DocParse.c, i.e. all the
arguments
to the compiler, just above the error messages.

Luis Tito de Morais wrote:
Hi list,

It may be a FAQ, but I searched the web and Uni of Newcastle Maths and Stats
and
R mailing list archive on this issue but was unable to find a solution. I
would
appreciate any pointer to help me solving this.

I am using R version 2.10.0 (2009-10-26) on linux mandriva 2010.0

I tried to install the XML_2.6-0.tar.gz package both with
install.packages('XML', dep=T) from within R and the R CMD INSTALL using a
local tar.gz file.

I am having the following error message (sorry it is partly in french):

Dans le fichier inclus à partir de DocParse.c:13:
Utils.h:175:2: attention : #warning Redefining COPY_TO_USER_STRING to use
encoding from XML parser
DocParse.c: In function ‘notifyError’:
DocParse.c:1051: erreur: le format n'est pas une chaîne littérale et pas
d'argument de format

This last error message means:
error: format not a string literal and no format arguments

In the past when having such errors with other packages, I have been able to
solve it with the help of this tip:
http://mario79t.wordpress.com/2009/06/23/warning-format-not-a-string-literal-and-no-format-arguments/

and modifying the faulty source file accordingly.

But in this specific case, I have been unable to find what to modify in the
source file. Line 1051 in the DocParse.c source file only has the command
WARN. I don't know anything about C programming and could not figure out
what
to modify in this case.

I would appreciate any help on this issue.

Best regards,

Tito

__
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Re: [R] Export kde object as shapefile

2009-11-27 Thread T.D.Rudolph


On Thu, Nov 26, 2009 at 6:56 PM, David Winsemius dwinsem...@comcast.net
wrote:

The situation that I see (after looking at the documentation for adehabit)
is that for some as yet unarticulated reason, you have decided that the
methods used in prior publications in your domain are not the best and you
are going to invent new ones,

There is nothing particularly cutting edge about wanting to estimate home
range size from a utilization distribution - Worton (1987, 1989) applied
kernel methods to estimate the UD and specifically home range size over 20
years ago.  Based on the literature (Gitzen et al. 2006) and the spatial
distribution of my data, it is reasonable to conclude the plug-in method of
bandwidth selection is the best possible option. 

but you are as yet unable to provided a detailed specification or offer an
implementation of the methods. Furthermore, you are unable even to
manipulate the objects you have thus far created in service toward this
effort.

I never said I was a programmer.  I simply asked if there was a way to
convert a kde object into a spatial object (e.g. SpatialPixels,
SpatialPolygons) for export into a GIS.  Or, alternatively, to generate the
2-D spatial area underneath the 3-D utilisation distribution of a kde
object.
 
I am being somewhat harsh in my assessment because it seems that you really
need is a statistical collaborator with whom you can bang ideas together
and arrive, first at a more detailed rationale and plan, and then a forward
effort using your data and his programming expertise to demonstrate the
superiority of your method. That would seem to be most reasonably a joint
effort with joint authorship as an outcome.
 
Perhaps you are speculating beyond what is warranted under the
circumstances.  I'm sure this is well-intentioned, but perhaps it is not as
appropriate as you would have wished.

Tyler
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Graphic Device - View/get all graphics

2009-11-27 Thread Sarah Goslee

I'm not entirely sure I understand the problem, but one possible solution
would be to use some device like postscript that saves the graphics to a
file. (Or some other device appropriate for your intended use and OS.)

You could still save the data to a list if you wanted - just make it part
of what your function returns.

Sarah

On Fri, Nov 27, 2009 at 12:10 PM, MarcioRibeiro mes...@pop.com.br wrote:

 Hello,
 After some research, I didn't resolve my doubt yet...
 Maybe I didn't explain very well my question... So, I have a bootstrap
 simulation, but suppose just a simple part of my function:

 teste -function(x){
 rw_mean_app-rnorm(x, mean = 10, sd = 2)
 rwy_mean_app-rnorm(x, mean = 10, sd = 2)
 rw_median_app_ori-rnorm(x, mean = 10, sd = 2)
 rw_median_app_mod-rnorm(x, mean = 10, sd = 2)

 histogram-par(mfrow=c(1,2),ask=TRUE)
 hist(rw_mean_app,main='Method RWOriginal',xlab='Mean',ylab=' ')
 hist(rwy_mean_app,main='Method RWY',xlab='Mean',ylab=' ')
 par(histogram)

 histogram-par(mfrow=c(1,2) ,ask=TRUE)
 hist(rw_median_app_ori,main='Method RWOriginal',xlab='Median',ylab=' ')
 hist(rw_median_app_mod,main='Method RWModified',xlab='Median',ylab=' ')
 par(histogram)

 }   #END OF FUNCTION
 GRAF-teste(100)

 I cannot save rnorm objects in a list cause they are distribution from 1000
 simulation... So, I would like to produce the graphics in my function and
 find out a way that I could save the graphics... I can save the object using
 the LIST procedure, but I dont know how I save an object that is a
 graphic...
 Listers, I just want some advices in order to make a research about what
 functions of R I could save the graphics in my function...
 Thanks again,
 Marcio


-- 
Sarah Goslee
http://www.functionaldiversity.org

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[R] my failing understanding ...

2009-11-27 Thread Kjetil Halvorsen

The following I do not understand, but then I did'nt really use
S4 methods ...

 showMethods(plot)
Function: plot (package graphics)
x=ANY
x=lmList.confint
x=merMCMC
(inherited from: x=ANY)

 plot(x=moda0MCMC)
Error in as.double(y) :
  cannot coerce type 'S4' to vector of type 'double'
 class(moda0MCMC)
[1] merMCMC
attr(,package)
[1] lme4


Kjetil

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[R] Orthogonal regression in R

2009-11-27 Thread Georg D. Blind


Dear R-Users!

The most recent entries on orthogonal regression that I have found on 
this list date back to 2005.
Is there, by now, any package available that allows for a multivariate 
orthogonal regression in R?


Thanks and best regards.
Georg

georg.blind at gmx.net

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[R] Questions about use of multinomial for discrimination.

2009-11-27 Thread Ricardo Antunes


Dear All,

I am looking at discriminating among several individuals based on a few 
variable sets (I think some variables do not make sense unless they are 
entered together, so I force them into the models together, hence 
datasets). I have done so with linear discriminant analysis (LDA) using 
MASS::lda,  with acceptable results. However, one of my collaborators 
suggested I use multinomial regression instead. I think his suggestion 
is mainly concerned with the choice of which variables (sets) best 
describe the data. I have used a stepwise approach (using 
klaR::stepclass) using the proportion of correct classifications to 
choose among the sets of variables. However I've been suggested that use 
a method that will give out an AIC instead, that will penalize the use 
of more variables. I have never done multinomial regression, and am 
uncertain about some details. I am looking into using R for this, and 
function multinom from MASS in particular.


In my previous analysis with LDA I have measured the proportion of 
correct classifications using a jackknife procedure (i.e. leaving each 
datum out of the LDA at a time, and using the obtained discriminant 
functions to classify it). I am thinking about doing the same with the 
multinomial regression. I would appreciate any ideas about if this may 
not be good for some reason.


Also, with the LDA I have looked at how much better the discriminant 
functions are compared with random assignment of individual identity. To 
do this I randomly shuffle the categories prior to running the LDA, then 
run the LDA, and measure the proportion of correct classifications using 
the above described jackknife procedure. I run this for many iterations 
and compare the distribution of proportion of correct classifications 
obtained from random assignment, with the one I obtained initially.
Again, I though about repeating this with the multinom. Is this 
unnecessary as another way of looking at this already included in the 
multinom function?


Perhaps this is more of a general statistics question, that one about 
the use of R, but I would appreciate any helpful comments.


Thank you in advance.

Ricardo Antunes

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[R] Debugging the unexpected string constant error in the zp2ssg2.R

2009-11-27 Thread Jason Rupert

I posted the full extent of this question to Nabble, but essentially I'm having 
difficulty determining the route cause of the  unexpected string constant 
error in the zp2ssg2.R (which I'm trying to port over from Octave).


Here is my current Nabble posting:
http://n2.nabble.com/Need-help-solving-the-unexpected-string-constant-error-td4077984.html

Thanks for any insights and feedback.

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[R] How to combine matrix with sapply function?

2009-11-27 Thread Peng Yu

I want an 'apply' function that can give me results by bind the result
from each function call of g. Could somebody let me know how to do it?

g-function(x){
  if(x==1){
rbind(c(1,3,8))
  } else if(x==2) {
rbind(c(4,7,2), c(3,7,3)) #if I use line, sapply() below gives me a list
#rbind(c(4,7,2)) #if I use this line, sapply() below gives me a matrix
  } else if(x==3) {
rbind(c(4,3,1))
  }
}

x=1:3

sapply(x,g)

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Re: [R] [R-es] R en Windows 7 a 64 bits

2009-11-27 Thread Jorge Ivan Velez

Buenas noches César,

Tal y como se estipula en la sección 8.1 de [1], la R 32-bits funciona en
Windows 64-bits aunque con las limitaciones de RAM que todos conocemos (solo
puedes usar hasta 4 GB de RAM en el mejor de los casos). Para instalarlo
sólo descarga el ejecutable usual de CRAN, haz doble click y sigue el
procedimiento de instalación como normalmente lo has hecho.

Hasta donde tengo entendido no existe una versión disponible (gratuita) de R
64-bits, aunque esta puede adquirirse a través de REvolution Computing [2].
Ahora, si por el tipo de trabajo que realizas necesitas gran cantidad de
RAM, entonces las plataformas Linux o Mac son buenas alternativas para ello.

Espero sea de utilidad,

  Jorge Ivan Velez

[1] http://cran.r-project.org/doc/manuals/R-admin.html
 http://cran.r-project.org/doc/manuals/R-admin.html[2]
http://www.revolution-computing.com

http://www.revolution-computing.com/

2009/11/27 Cesar Escalante 

 Buenas noches para cada uno.

 Me alegra saber que superamos los 200 usuarios y tengos expectativas con
 respecto a la I Conferencia de R-hispano en Murcia.

 No soy usuario de Linux por desconocimiento, algo que lamento. Debo
 instalar
 R en un equipo con Windows 7 a 64 bits. ¿Podrían indicarme por favor cómo
 instalarlo? ¿Se hace con el mismo archivo R-2.10.0-win32? ¿Debo cuidar o
 cambiar algo antes o después?

 Gracias de antemano por la amable atención.

 Saludos.

 César Escalante C.

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Re: [R] Export kde object as shapefile



I'm not a programmer either, if by programmer you mean one who took  
a computer science degree and is employed with some title that implies  
a creator of code. I am a user of software toward particular purposes.  
The purposes need to be defined in sufficiently specific manner to  
guide that activity and be used as a basis for successful completion  
of the task.


--
David.

On Nov 27, 2009, at 3:47 PM, T.D.Rudolph wrote:



On Thu, Nov 26, 2009 at 6:56 PM, David Winsemius dwinsem...@comcast.net 


wrote:

The situation that I see (after looking at the documentation for  
adehabit)
is that for some as yet unarticulated reason, you have decided that  
the
methods used in prior publications in your domain are not the best  
and you

are going to invent new ones,

There is nothing particularly cutting edge about wanting to estimate  
home
range size from a utilization distribution - Worton (1987, 1989)  
applied
kernel methods to estimate the UD and specifically home range size  
over 20
years ago.  Based on the literature (Gitzen et al. 2006) and the  
spatial
distribution of my data, it is reasonable to conclude the plug-in  
method of

bandwidth selection is the best possible option.

but you are as yet unable to provided a detailed specification or  
offer an

implementation of the methods. Furthermore, you are unable even to
manipulate the objects you have thus far created in service toward  
this

effort.

I never said I was a programmer.  I simply asked if there was a way to
convert a kde object into a spatial object (e.g. SpatialPixels,
SpatialPolygons) for export into a GIS.  Or, alternatively, to  
generate the

2-D spatial area underneath the 3-D utilisation distribution of a kde
object.

I am being somewhat harsh in my assessment because it seems that  
you really
need is a statistical collaborator with whom you can bang ideas  
together
and arrive, first at a more detailed rationale and plan, and then a  
forward
effort using your data and his programming expertise to demonstrate  
the
superiority of your method. That would seem to be most reasonably a  
joint

effort with joint authorship as an outcome.

Perhaps you are speculating beyond what is warranted under the
circumstances.  I'm sure this is well-intentioned, but perhaps it is  
not as

appropriate as you would have wished.

Tyler
--
View this message in context: 
http://old.nabble.com/Export-kde-object-as-shapefile-tp26532782p26545381.html
Sent from the R help mailing list archive at Nabble.com.

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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] How to combine matrix with sapply function?

2009-11-27 Thread Ben Bolker

Peng Yu pengyu.ut at gmail.com writes:

 
 I want an 'apply' function that can give me results by bind the result
 from each function call of g. Could somebody let me know how to do it?
 
 g-function(x){
  [snip]
 }
 
 x=1:3
 
 sapply(x,g) 

do.call(rbind,sapply(x,g))?

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Re: [R] my failing understanding ...

I get a different result, undoubtedly because I have different  
packages loaded:


 showMethods(plot)

Function plot:
 not a generic function

 sessionInfo()
R version 2.10.0 Patched (2009-10-29 r50258)
x86_64-apple-darwin9.8.0

locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] splines   stats graphics  grDevices utils datasets   
methods   base


other attached packages:
[1] plm_1.2-1sandwich_2.2-1   zoo_1.5-8 
MASS_7.3-3   Formula_0.2-0

[6] kinship_1.1.0-23 lattice_0.17-26  nlme_3.1-96  survival_2.35-7

loaded via a namespace (and not attached):
[1] grid_2.10.0  tools_2.10.0

But I get a much longer and informative result with:

methods(plot)

--
David.

On Nov 27, 2009, at 4:23 PM, Kjetil Halvorsen wrote:


The following I do not understand, but then I did'nt really use
S4 methods ...


showMethods(plot)

Function: plot (package graphics)
x=ANY
x=lmList.confint
x=merMCMC
   (inherited from: x=ANY)


plot(x=moda0MCMC)

Error in as.double(y) :
 cannot coerce type 'S4' to vector of type 'double'

class(moda0MCMC)

[1] merMCMC
attr(,package)
[1] lme4


Kjetil

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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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Re: [R] my failing understanding ...