date:20101029

try this:

> x
 [1]  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
> y <- cut(x, breaks=c(-Inf,6,18, Inf), labels=c('a','b','c'))
> levels(y) <- c('night','day','night')
> y
 [1] night night night night night night night day   day   day   day
day   day   day   day   day   day   day
[19] day   night night night night night night
Levels: night day
>


On Fri, Oct 29, 2010 at 8:56 PM, will phillips  wrote:
>
> Hello
>
> I have what is probably a very simple grouping question however, given my
> limited exposure to R, I have not found a solution yet despite my research
> efforts and wild attempts at what I thought "might" produce some sort of
> result.
>
> I have a very simple list of integers that range between 1 and 24.  These
> correspond to hours of the day.
>
> I am trying to create a grouping of Day and Night with
> Day = 6 to 17.99
> Night = 1 to 5.59  and  18 to 24
>
> Using the Cut command I can create the segments but I have not found a
> "combine" type of command to merger the two "night" segments.  No luck with
> if/else either.
>
> Any help would be greatly appreciated
>
> Thank you
>
> Will
>
>
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/grouping-question-tp3019922p3019922.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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[R] grouping question

2010-10-29 Thread will phillips


Hello

I have what is probably a very simple grouping question however, given my
limited exposure to R, I have not found a solution yet despite my research
efforts and wild attempts at what I thought "might" produce some sort of
result.

I have a very simple list of integers that range between 1 and 24.  These
correspond to hours of the day.

I am trying to create a grouping of Day and Night with 
Day = 6 to 17.99
Night = 1 to 5.59  and  18 to 24

Using the Cut command I can create the segments but I have not found a
"combine" type of command to merger the two "night" segments.  No luck with
if/else either.

Any help would be greatly appreciated

Thank you

Will


-- 
View this message in context: 
http://r.789695.n4.nabble.com/grouping-question-tp3019922p3019922.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] plot pdf

2010-10-29 Thread Joshua Wiley

Hi Mary,

I am not sure off hand why savePlot() is not doing what you expect,
but hopefully my inline comments answer your other questions and at
least get you the graphs you want.

Cheers,

Josh

On Fri, Oct 29, 2010 at 3:18 PM,   wrote:
>
>
> I want to plot the unstadardized version of a normal plot.  Can you explain 
> why that is not working?
> Dev.set(1)
> xcrit=-1.645
> cord.x <- c(-3,seq(-3,xcrit,0.01),xcrit)
> cord.y <- c(0,dnorm(seq(-3,xcrit,0.01)),0)   # what does final 0 do here?

You are creating *coordinates* to draw a polygon.  That is, (x, y).
The 0s just match with -3 and xcrit and serve to define the perimeter
of the polygon.  Please read the documentation for ?polygon for more
details.

> curve(dnorm(x,0,1),xlim=c(-3,3),main='Normal PDF')

This makes your nice curve based on the density function for the
normal distribution with a range from -3 to +3.  This part works fine
for me.

> polygon(cord.x,cord.y,col='orange')

This does the coloring, the coordinates define the polygon, which is
closed by connecting the first and last points.  This also seems to
work.  I end up with the lower region of the graph filled orange.

> savePlot(filename="c:\\ssl_z.emf",type="emf")  #unable to create 
> metafile?  why?

hard for me to say right now (I'm on a linux box).  It might help
though if you reported sessionInfo()

> x=seq(80,200,1)
> plot(x,dnorm(x,140,15),type="l",ylab="Probability")

This seems fine to me also.

> dev.set(4)

> xcrit=135.9
> cord.x <- c(80,seq(80,xcrit,.1),xcrit)
> cord.y <- c(0,dnorm(seq(80,xcrit,.1)),0)  #final 0 needs to be 
> changed but to what?

Why does the final zero need to be changed?  What do you want it to
be?  I suspect you do need to change dnorm() though.  You are creating
y coordinates based on a normal distribution with mean 0 and sd = 1,
but then you plot a curve for a mean of 140 and sd = 15.  I would
adjust your code to:

cord.y <- c(0, dnorm(seq(80, xcrit, .1), 140, 15), 0)

which gives the results I *think* you are after.

> curve(dnorm(x,140,15),xlim=c(80,200),main='Normal PDF',ylab="Probability")
> polygon(cord.x,cord.y,col='orange')

Using the adjustment I showed above, I now get a graph that is filled
orange from 80 to xcrit (under the normal curve).

> savePlot(filename="c:\\ssl_z.emf",type="emf") #unable to create 
> metafile?  why?
>
>
> Sincerely,
>
> Mary A. Marion
>
>
>        [[alternative HTML version deleted]]

I manually removed spaces in my reply.  In the future it nice if you
could send plain text emails (since the html ones are scrubbed).

>
>
> __
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> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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Re: [R] vertical list sum

2010-10-29 Thread Jorge Ivan Velez

Hi Duncan,

If vectors of unequal length is the problem, one way to go, using your
example, would be:

# your example
x <- list(c(9, 5, 7, 2, 14, 4, 4, 3), c(3, 6, 25, 2, 14, 3, 3, 4),
c(28, 4, 14, 3, 14, 2, 4, 5), 28)
x

# maximum number of components
k <- max(sapply(x, length))
k

# expanding each list of x to have k elements
lapply(x, function(l) l[1:k])

# average as requested by Greg
colMeans(do.call(rbind, lapply(x, function(l) l[1:k])), na.rm = TRUE)

HTH,
Jorge


On Fri, Oct 29, 2010 at 7:37 PM, Duncan Mackay <> wrote:

> Hi Jorge
>
> I tried your methods for all (which work for complete rows) and then I
> remove the first value of $y and repeated; both fail because of the  unequal
> numbers
> The problem is when there are unequal numbers in the rows and trying to
> make a matrix of them.
>
> I was trying some things with Greg's vaules.
> x <- list()
> x[[1]] <- c(9,5,7,2, 14,  4,  4,  3)
> x[[2]] <- c(3,  6, 25,  2, 14,  3,  3 , 4)
> x[[3]] <- c(28,  4 ,14,  3, 14,  2  ,4 , 5)
> x[4] <- list(28 , 4 ,14 , 3, 14,  2 , 4  )
>
> x.av <- list()
> for(j in seq_along(1:max(sapply(x,length))) ) x.av[j] <-
> mean(sapply(x,"[",j),na.rm=T)
> unlist(x.av) # if you want a vector
>
> which Greg may have used first
>
> Regards
>
> Duncan
>
> Duncan Mackay
> Department of Agronomy and Soil Science
> University of New England
> ARMIDALE NSW 2351
> Email home: mac...@northnet.com.au
>
>
> At 08:53 30/10/2010, you wrote:
>
>> Hi Greg,
>>
>> Here are two ways of doing it:
>>
>> > mylist <- list(x = rpois(10, 10), y = rpois(10, 20), z = rpois(10, 5))
>> > mylist
>> $x
>>  [1]  3 13 14 16 10  7  3  5 12 14
>>
>> $y
>>  [1] 17 16 26 13 23 24 16 28 23 12
>>
>> $z
>>  [1]  2  6  5  5  5  1  9 11  6  4
>>
>> >
>> > colMeans(do.call(rbind, mylist), na.rm = TRUE)
>>  [1]  7.33 11.67 15.00 11.33 12.67 10.67  9.33
>> 14.67 13.67
>> [10] 10.00
>> >
>> > Reduce("+", mylist)/length(mylist)
>>  [1]  7.33 11.67 15.00 11.33 12.67 10.67  9.33
>> 14.67 13.67
>> [10] 10.00
>>
>>
>> HTH,
>> Jorge
>>
>>
>> On Fri, Oct 29, 2010 at 6:46 PM, Gregory Ryslik <> wrote:
>>
>> > Hi Everyone,
>> >
>> > I have a list of vectors like this (in this case it's 3 vectors but
>> assume
>> > the vector count and the length of each vector is not known):
>> >
>> > [[1]]
>> > [1]  9  5  7  2 14  4  4  3
>> >
>> > [[2]]
>> > [1]  3  6 25  2 14  3  3  4
>> >
>> > [[3]]
>> > [1] 28  4 14  3 14  2  4  5
>> >
>> > What I want to do is take the average vertically. Thus I want to do
>> 9+3+28
>> > /3, 5+6+4 /3, etc... and then have it return a vector. I'm assuming that
>> if
>> > I can sum it, I can count it to so summing this would be just as
>> helpful.
>> >
>> > I understand I can first go through each element of the list, get a
>> vector,
>> > cbind into  matrix and sum across but I was hoping to avoid that... I
>> tried
>> > getting it to work with mapply but am having difficulties...
>> >
>> > Thanks!
>> >
>> > Kind regards,
>> > Greg
>> > __
>> > R-help@r-project.org mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> > http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>> >
>>
>>[[alternative HTML version deleted]]
>>
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] R and Matlab

2010-10-29 Thread Henrik Bengtsson

Hi.

On Fri, Oct 29, 2010 at 3:31 AM, Claudia Beleites  wrote:
> Dear Henrik,
>
> sorry for bothering you with a report hastily pasted together and not
> particularly nice for you as I used my toy data flu from a non-standard
> package. I should have better used e.g. the iris.
>
> I'm aware that writeMat doesn't deal with S4 objects. In fact, if I'd
> overlook the error message, there's the 2nd chance to see that the file size
> is 0B.

Yes, it is an unfortunate side effect that if there is an error while
writing the MAT file, it gives a corrupt *.mat file.  I'll put it on
the "to do in the future" list to write to a temporary file (e.g.
*.mat.tmp) which is only renamed to *.mat when writeMat() returns
successfully.  I use that "trick" elsewhere and it has saved us a few
times.

> In fact the attempt to save flu directly was a classical "autopilot" error,
> that's why I tried to save the x afterwards.
>
> So the problem here was the unnamed storing of x.
>
>> I intentionally do not try to "infer" the name "x" from
>> writeMat("flu.mat", x), basically because I think using substitute()
>> should be avoided as far as possible, but also because it is unclear
>> what the name should be in cases such as writeMat("flu.mat", 1:10).
>
> I was just going to suggest a patch that assigns the names of type Vnumber
> to the unnamed objects - but when I wanted to get the source I realized your
> version with the warning is already out.
>
> I think, however, you may forgot a nchar?: any (nchar (names) == 0)

Thanks for spotting that; a merge of (nchar (names) == 0) and (names
== "") incorrectly became (names == 0). Corrected in the next release.

>
> So here's my suggestion for l. 775-777 of writeMat.R:
>
>  if (is.null(names) || any (nchar (names) == 0L)) {
>    names [nchar (names) == 0L] <- paste ("V", which (nchar (names) == 0L),
> sep = "")
>    names (args) <- names
>    warning("All objects written have to be named, e.g. use writeMat(...,
> x=a, y=y) and not writeMat(..., x=a, y): ", deparse(sys.call()), "\nDummy
> names have been assigned.");
>  }

I did think about that, however, it may introduce other ambiguities.
For instance, consider

writeMat("foo.mat", V2=1, 2);

Then there will be two "V2" names.  The analogue to read.table() or
data.frame() is to add ".1" etc when there is a clash, e.g. "V2" and
"V2.1".  However, "V2.1" is not a valid name in Matlab.  What should
then be done?  Of course, you can try to make sure you generate valid
Matlab names.

On a related matter, today you can do writeMat("foo.mat", V2=1,
"V2.1"=2) and there is no warning/error given by writeMat() and it
reads correctly by readMat().  However, in Matlab you get

>> load('foo.mat')
??? Error using ==> load
Invalid field name: 'V2.1'.

If anyone knows a regular expression for testing the validity of names
such that they are valid Matlab variable/field names, please let me
know and I can add additional sanity checks in writeMat().

Also, as your initial example indicates that it could be surprising
that writeMat("foo.mat", x) would become writeMat("foo.mat", V1=x) and
not writeMat("foo.mat", x=x).

After further investigation, I actually think that although Matlab
indeed can read non-named objects using data=load('foo.mat') I don't
think they are accessible.  So I was wrong.  Because of this, I have
bumped up the warning to be an error, preventing non-named objects to
be written.  Will be the case in the next release of the package.

I will postpone adding any bells and whistles trying to make
writeMat() smart such as adding names.  As soon as you do that you
introduce other issues and expectations and have to worry about
backward compatibilities if it turns out to be a bad idea.  My
strategy for now is to have writeMat() assert that only valid MAT
files are written, and give errors otherwise.

/Henrik

>
>
> After all, e.g. data.frame () will also rather create dummy names for
> unnamed columns. And, I think, a warning should make the user aware that
> he's doing something that _may_ not work out as intendet. But here I think
> it is _most likely_ not working as intended.
>
>
>> MISCELLANEOUS:
>> Note that writeMat() cannot write compressed MAT files.  It is
>> documented in help("readMat"), and will be so in help("writeMat") in
>> the next release.  Package Rcompression, loaded or not, has no effect
>> on writeMat().  It is only readMat() that can read them, if
>> Rcompression is installed.  You do not have to load it
>> explicitly/yourself - if readMat() detects a compress MAT file, it
>> will automatically try to load it;
>
> OK, good to know.
>
> Thanks a lot for your explanation in spite of my bad report.
>
> Claudia
>
>
> --
> Claudia Beleites
> Dipartimento dei Materiali e delle Risorse Naturali
> Università degli Studi di Trieste
> Via Alfonso Valerio 6/a
> I-34127 Trieste
>
> phone: +39 0 40 5 58-37 68
> email: cbelei...@units.it
>
> __
> R-help@r-project.org mailing list
> https://s

Re: [R] vertical list sum

2010-10-29 Thread Gregory Ryslik

Hi,

Ah! Thanks for your help! I have even seen Reduce before but for some reason 
just didn't make the connection. This helps a ton! Thanks again.

Kind regards,
Greg
On Oct 29, 2010, at 6:53 PM, Jorge Ivan Velez wrote:

> Hi Greg,
> 
> Here are two ways of doing it:
> 
> > mylist <- list(x = rpois(10, 10), y = rpois(10, 20), z = rpois(10, 5))
> > mylist
> $x
>  [1]  3 13 14 16 10  7  3  5 12 14
> 
> $y
>  [1] 17 16 26 13 23 24 16 28 23 12
> 
> $z
>  [1]  2  6  5  5  5  1  9 11  6  4
> 
> > 
> > colMeans(do.call(rbind, mylist), na.rm = TRUE)
>  [1]  7.33 11.67 15.00 11.33 12.67 10.67  9.33 
> 14.67 13.67
> [10] 10.00
> > 
> > Reduce("+", mylist)/length(mylist)
>  [1]  7.33 11.67 15.00 11.33 12.67 10.67  9.33 
> 14.67 13.67
> [10] 10.00
> 
> 
> HTH,
> Jorge
> 
> 
> On Fri, Oct 29, 2010 at 6:46 PM, Gregory Ryslik <> wrote:
> Hi Everyone,
> 
> I have a list of vectors like this (in this case it's 3 vectors but assume 
> the vector count and the length of each vector is not known):
> 
> [[1]]
> [1]  9  5  7  2 14  4  4  3
> 
> [[2]]
> [1]  3  6 25  2 14  3  3  4
> 
> [[3]]
> [1] 28  4 14  3 14  2  4  5
> 
> What I want to do is take the average vertically. Thus I want to do 9+3+28 
> /3, 5+6+4 /3, etc... and then have it return a vector. I'm assuming that if I 
> can sum it, I can count it to so summing this would be just as helpful.
> 
> I understand I can first go through each element of the list, get a vector, 
> cbind into  matrix and sum across but I was hoping to avoid that... I tried 
> getting it to work with mapply but am having difficulties...
> 
> Thanks!
> 
> Kind regards,
> Greg
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 


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Re: [R] date calculation

2010-10-29 Thread Phil Spector


Tao -
   The documentation for the difftime function says:

Function ‘difftime’ calculates a difference of two date/time
 objects and returns an object of class ‘"difftime"’ with an
 attribute indicating the units.

So that answers your question.

If you want it to be an integer, you're certainly free to make it one:

as.integer(difftime(strptime("24NOV2004", format="%d%b%Y"), 

+ strptime("13MAY2004",format="%d%b%Y"), units="days"))
[1] 195

- Phil Spector
 Statistical Computing Facility
 Department of Statistics
 UC Berkeley
 spec...@stat.berkeley.edu



On Fri, 29 Oct 2010, Shi, Tao wrote:


Hi list,

Could someone explain to me why the following result is not a integer?



difftime(strptime("24NOV2004", format="%d%b%Y"), strptime("13MAY2004",
format="%d%b%Y"), units="days")

Time difference of 195.0417 days

I'm using R2.12.0 on WinXP.

Thanks!

...Tao

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[R] date calculation

2010-10-29 Thread Shi, Tao

Hi list,

Could someone explain to me why the following result is not a integer?


> difftime(strptime("24NOV2004", format="%d%b%Y"), strptime("13MAY2004", 
>format="%d%b%Y"), units="days")
Time difference of 195.0417 days

I'm using R2.12.0 on WinXP.

Thanks!

...Tao

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Re: [R] How to scan df from a specific word?

On Fri, Oct 29, 2010 at 6:34 PM, M.Ribeiro  wrote:
>
> Hi R-helpers,
>
> I need to read some file with different lines (I don't know the number of
> lines to skip) and I would like to find a way to start reading the
> data.frame from the word "source".
>
> ex:
>
> djhsafk
> asdfhkjash
> shdfjkash
> asfhjkash         #those lines contain numbers and words, I want to skip
> then but they have different sizes
> asdfhjkash
> asdfhjksa
>
> source
> tret 2
> res 3
>

Here is a one line solution but it does make use of the external
utility, gawk.  If you using Linux you probably have it on your system
already.  You can also get gawk for Windows or if you download
Duncan's Rtools distribution its included there too -- gawk.exe is
just a single file so just make sure you put it somewhere on your
PATH.

> read.table(pipe('gawk "/Analysis of Variance/ {exit}; /Source/ {i++}; i" 
> myfile.dat'), header = TRUE, fill = TRUE)
   Source Model termsGamma Component Comp.SE X. C
1Residual  8383  8367   NANA  NA NA
2 at(type,1).Nfam6262  10.1131   10.11311.81  0 P
3 at(type,2).Nfam6262  28.1153   28.11532.16  0 P
4rep.iblk   768   768  63.2919   63.2919   10.94  0 P
5  at(type,1).Nfemale4444  29.9049   29.90492.93  0 P
6   at(type,1).Nclone  2689  2689 109.5600  109.5600   12.66  0 P
7  at(type,2).Nfemale4444  14.0305   14.03051.68  0 P
8Variance 0 0 479.0400  479.0400   36.23  0 P
9Variance 0 0 490.5800  490.5800   17.51  0 P
10   Variance 0 0 469.9320  469.9320   36.51  0 P
11   Variance 0 0 544.6540  544.6540   17.86  0 P

-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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Re: [R] vertical list sum

2010-10-29 Thread Duncan Mackay


Aplogies to the original poster - I got names wrong in a cut and paste


Hi Jorge

I tried your methods for all (which work for complete rows) and then I 
remove the first value of $y and repeated; both fail because of 
the  unequal numbers
The problem is when there are unequal numbers in the rows and trying to 
make a matrix of them.


I was trying some things with Greg's vaules.
x <- list()
x[[1]] <- c(9,5,7,2, 14,  4,  4,  3)
x[[2]] <- c(3,  6, 25,  2, 14,  3,  3 , 4)
x[[3]] <- c(28,  4 ,14,  3, 14,  2  ,4 , 5)
x[4] <- list(28 , 4 ,14 , 3, 14,  2 , 4  )

x.av <- list()
for(j in seq_along(1:max(sapply(x,length))) ) x.av[j] <- 
mean(sapply(x,"[",j),na.rm=T)

unlist(x.av) # if you want a vector

which Greg may have used first

Regards

Duncan
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Re: [R] vertical list sum

2010-10-29 Thread Duncan Mackay


Hi Jorge

I tried your methods for all (which work for complete rows) and then I 
remove the first value of $y and repeated; both fail because of 
the  unequal numbers
The problem is when there are unequal numbers in the rows and trying to 
make a matrix of them.


I was trying some things with Greg's vaules.
x <- list()
x[[1]] <- c(9,5,7,2, 14,  4,  4,  3)
x[[2]] <- c(3,  6, 25,  2, 14,  3,  3 , 4)
x[[3]] <- c(28,  4 ,14,  3, 14,  2  ,4 , 5)
x[4] <- list(28 , 4 ,14 , 3, 14,  2 , 4  )

x.av <- list()
for(j in seq_along(1:max(sapply(x,length))) ) x.av[j] <- 
mean(sapply(x,"[",j),na.rm=T)

unlist(x.av) # if you want a vector

which Greg may have used first

Regards

Duncan

Duncan Mackay
Department of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email home: mac...@northnet.com.au

At 08:53 30/10/2010, you wrote:

Hi Greg,

Here are two ways of doing it:

> mylist <- list(x = rpois(10, 10), y = rpois(10, 20), z = rpois(10, 5))
> mylist
$x
 [1]  3 13 14 16 10  7  3  5 12 14

$y
 [1] 17 16 26 13 23 24 16 28 23 12

$z
 [1]  2  6  5  5  5  1  9 11  6  4

>
> colMeans(do.call(rbind, mylist), na.rm = TRUE)
 [1]  7.33 11.67 15.00 11.33 12.67 10.67  9.33
14.67 13.67
[10] 10.00
>
> Reduce("+", mylist)/length(mylist)
 [1]  7.33 11.67 15.00 11.33 12.67 10.67  9.33
14.67 13.67
[10] 10.00


HTH,
Jorge


On Fri, Oct 29, 2010 at 6:46 PM, Gregory Ryslik <> wrote:

> Hi Everyone,
>
> I have a list of vectors like this (in this case it's 3 vectors but assume
> the vector count and the length of each vector is not known):
>
> [[1]]
> [1]  9  5  7  2 14  4  4  3
>
> [[2]]
> [1]  3  6 25  2 14  3  3  4
>
> [[3]]
> [1] 28  4 14  3 14  2  4  5
>
> What I want to do is take the average vertically. Thus I want to do 9+3+28
> /3, 5+6+4 /3, etc... and then have it return a vector. I'm assuming that if
> I can sum it, I can count it to so summing this would be just as helpful.
>
> I understand I can first go through each element of the list, get a vector,
> cbind into  matrix and sum across but I was hoping to avoid that... I tried
> getting it to work with mapply but am having difficulties...
>
> Thanks!
>
> Kind regards,
> Greg
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] How to scan df from a specific word?

2010-10-29 Thread Peter Langfelder

Sorry, this isn't really an R solution, but here it goes anyway. You
can isolate the block from Source to the first following blank line by
the following unix/linux/cygwin command, assuming inFile is your input
file and outFile is the output file:

cat inFile | grep -A 100 Source | grep -m 1 -B 100 ^$ > outFile

(in this command the number of lines in the Source block is limited to
about 100-2=98, you can safely increase the number if necessary)

You will probably still have the problem of successive spaces in the
block you are interested in. This can also be handled on the command
line, for example by adding

| sed 's/   */\t/g'

(note there are 3 spaces between the first / and the *) between the $
and > characters in the command above. The last bit will replace all
occurrences of 2 or more spaces by a tab, so you can read the file as
tab-separated.

Equivalently, you could do the line filtering in R as Phil suggested.

Peter



On Fri, Oct 29, 2010 at 3:39 PM, M.Ribeiro  wrote:
>
> Sorry, the explanation wasn't very good...just to explain better.
>
> I am writing a loop to read and process every time a different file in the
> same script.
> And what I want to load into a variable each time is a data.frame that is
> bellow the word source in all of my files.
>
> So I would like to recognize the word Source in the text fileand read
> the table bellow source until the next blank line (the file has more written
> stuff bellow the data frame that I want to read too)
>
> Here is an example of the file. I want the df to read from source until the
> blank line right above the words "Analysis of Variance
>
> Notice:     37 singularities detected in design matrix.
>   1 LogL=-2664.01     S2=  1.       8367 df    :   2 components
> constrained
>   2 LogL=-2269.45     S2=  1.       8367 df
>   3 LogL=-1698.47     S2=  1.       8367 df
>   4 LogL=-1252.72     S2=  1.       8367 df
>   5 LogL=-1013.52     S2=  1.       8367 df
>   6 LogL=-957.409     S2=  1.       8367 df
>   7 LogL=-944.252     S2=  1.       8367 df
>   8 LogL=-939.976     S2=  1.       8367 df
>   9 LogL=-938.908     S2=  1.       8367 df
>  10 LogL=-938.798     S2=  1.       8367 df
>  11 LogL=-938.795     S2=  1.       8367 df
>  12 LogL=-938.795     S2=  1.       8367 df
>
>  Source                Model  terms     Gamma     Component    Comp/SE   % C
>  Residual               8383   8367
>  at(type,1).Nfam          62     62   10.1131       10.1131       1.81   0 P
>  at(type,2).Nfam          62     62   28.1153       28.1153       2.16   0 P
>  rep.iblk                768    768   63.2919       63.2919      10.94   0 P
>  at(type,1).Nfemale       44     44   29.9049       29.9049       2.93   0 P
>  at(type,1).Nclone      2689   2689   109.560       109.560      12.66   0 P
>  at(type,2).Nfemale       44     44   14.0305       14.0305       1.68   0 P
>  Variance                  0      0   479.040       479.040      36.23   0 P
>  Variance                  0      0   490.580       490.580      17.51   0 P
>  Variance                  0      0   469.932       469.932      36.51   0 P
>  Variance                  0      0   544.654       544.654      17.86   0 P
>
>  Analysis of Variance              NumDF              F_inc
>  27 mu                                1            5860.84
>
>  12 culture                           1               0.07
>  10 type                              1              29.59
>  28 culture.rep                       6              14.06
>  30 culture.rep.type                  7               2.17
>  36 at(type,1).Nfam                      62 effects fitted
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/How-to-scan-df-from-a-specific-word-tp3019841p3019846.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] How to scan df from a specific word?

2010-10-29 Thread Phil Spector


It might be easier to preprocess the file before passing it
to R, but you an do what you want using a connection:


f = file('s.dat','r')
while(1){

+   txt = readLines(f,1)
+   if(length(grep('Source',txt)))break;
+ }

while(1){

+   more = readLines(f,1)
+   if(nchar(more) == 0)break;
+   txt = paste(txt,more,sep='\n')
+ }

try = read.table(textConnection(txt),header=TRUE,fill=TRUE)
try

   Source Model termsGamma Component Comp.SE X. C
1Residual  8383  8367   NANA  NA NA 
2 at(type,1).Nfam6262  10.1131   10.11311.81  0 P

3 at(type,2).Nfam6262  28.1153   28.11532.16  0 P
4rep.iblk   768   768  63.2919   63.2919   10.94  0 P
5  at(type,1).Nfemale4444  29.9049   29.90492.93  0 P
6   at(type,1).Nclone  2689  2689 109.5600  109.5600   12.66  0 P
7  at(type,2).Nfemale4444  14.0305   14.03051.68  0 P
8Variance 0 0 479.0400  479.0400   36.23  0 P
9Variance 0 0 490.5800  490.5800   17.51  0 P
10   Variance 0 0 469.9320  469.9320   36.51  0 P
11   Variance 0 0 544.6540  544.6540   17.86  0 P


The first loop reads up until the word "Source" appears in the 
line, and the second loop builds a text representation of what

you want so that you can read it using textConnection.

Hope this helps.
- Phil Spector
 Statistical Computing Facility
 Department of Statistics
 UC Berkeley
 spec...@stat.berkeley.edu



On Fri, 29 Oct 2010, M.Ribeiro wrote:



Sorry, the explanation wasn't very good...just to explain better.

I am writing a loop to read and process every time a different file in the
same script.
And what I want to load into a variable each time is a data.frame that is
bellow the word source in all of my files.

So I would like to recognize the word Source in the text fileand read
the table bellow source until the next blank line (the file has more written
stuff bellow the data frame that I want to read too)

Here is an example of the file. I want the df to read from source until the
blank line right above the words "Analysis of Variance

Notice: 37 singularities detected in design matrix.
  1 LogL=-2664.01 S2=  1.   8367 df:   2 components
constrained
  2 LogL=-2269.45 S2=  1.   8367 df
  3 LogL=-1698.47 S2=  1.   8367 df
  4 LogL=-1252.72 S2=  1.   8367 df
  5 LogL=-1013.52 S2=  1.   8367 df
  6 LogL=-957.409 S2=  1.   8367 df
  7 LogL=-944.252 S2=  1.   8367 df
  8 LogL=-939.976 S2=  1.   8367 df
  9 LogL=-938.908 S2=  1.   8367 df
 10 LogL=-938.798 S2=  1.   8367 df
 11 LogL=-938.795 S2=  1.   8367 df
 12 LogL=-938.795 S2=  1.   8367 df

SourceModel  terms Gamma ComponentComp/SE   % C
Residual   8383   8367
at(type,1).Nfam  62 62   10.1131   10.1131   1.81   0 P
at(type,2).Nfam  62 62   28.1153   28.1153   2.16   0 P
rep.iblk768768   63.2919   63.2919  10.94   0 P
at(type,1).Nfemale   44 44   29.9049   29.9049   2.93   0 P
at(type,1).Nclone  2689   2689   109.560   109.560  12.66   0 P
at(type,2).Nfemale   44 44   14.0305   14.0305   1.68   0 P
Variance  0  0   479.040   479.040  36.23   0 P
Variance  0  0   490.580   490.580  17.51   0 P
Variance  0  0   469.932   469.932  36.51   0 P
Variance  0  0   544.654   544.654  17.86   0 P

Analysis of Variance  NumDF  F_inc
 27 mu15860.84

 12 culture   1   0.07
 10 type  1  29.59
 28 culture.rep   6  14.06
 30 culture.rep.type  7   2.17
 36 at(type,1).Nfam  62 effects fitted
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Re: [R] vertical list sum

2010-10-29 Thread Jorge Ivan Velez

Hi Greg,

Here are two ways of doing it:

> mylist <- list(x = rpois(10, 10), y = rpois(10, 20), z = rpois(10, 5))
> mylist
$x
 [1]  3 13 14 16 10  7  3  5 12 14

$y
 [1] 17 16 26 13 23 24 16 28 23 12

$z
 [1]  2  6  5  5  5  1  9 11  6  4

>
> colMeans(do.call(rbind, mylist), na.rm = TRUE)
 [1]  7.33 11.67 15.00 11.33 12.67 10.67  9.33
14.67 13.67
[10] 10.00
>
> Reduce("+", mylist)/length(mylist)
 [1]  7.33 11.67 15.00 11.33 12.67 10.67  9.33
14.67 13.67
[10] 10.00


HTH,
Jorge


On Fri, Oct 29, 2010 at 6:46 PM, Gregory Ryslik <> wrote:

> Hi Everyone,
>
> I have a list of vectors like this (in this case it's 3 vectors but assume
> the vector count and the length of each vector is not known):
>
> [[1]]
> [1]  9  5  7  2 14  4  4  3
>
> [[2]]
> [1]  3  6 25  2 14  3  3  4
>
> [[3]]
> [1] 28  4 14  3 14  2  4  5
>
> What I want to do is take the average vertically. Thus I want to do 9+3+28
> /3, 5+6+4 /3, etc... and then have it return a vector. I'm assuming that if
> I can sum it, I can count it to so summing this would be just as helpful.
>
> I understand I can first go through each element of the list, get a vector,
> cbind into  matrix and sum across but I was hoping to avoid that... I tried
> getting it to work with mapply but am having difficulties...
>
> Thanks!
>
> Kind regards,
> Greg
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

[[alternative HTML version deleted]]

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[R] Differenciate numbers from reference for rows


So, I am having a tricky reference file to extract information from.

The format of the file is

x   1 + 4 * 3 + 5 + 6 + 11 * 0.5

So, the elements that are not being multiplied (1, 5 and 6) and the elements
before the multiplication sign (4 and 11) means actually the reference for
the row in a matrix where I need to extract the element from.

The numbers after the multiplication sign are regular numbers 
Ex:

> x<-matrix(20:35)
> x
  [,1]
 [1,]   20
 [2,]   21
 [3,]   22
 [4,]   23
 [5,]   24
 [6,]   25
 [7,]   26
 [8,]   27
 [9,]   28
[10,]   29
[11,]   30
[12,]   31
[13,]   32
[14,]   33
[15,]   34
[16,]   35

I would like to read the rows 1,4,5,6 and 11 and sum then. However the
numbers in the elements row 4 and 11 are multiplied by 3 and 0.5

So it would be
20 + 23 * 3 + 24 + 25 + 30 * 0.5.

And I have this format in different files so I can't do all by hand.
Can anybody help me with a script that can differentiate this?
Thanks
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[R] vertical list sum

2010-10-29 Thread Gregory Ryslik

Hi Everyone,

I have a list of vectors like this (in this case it's 3 vectors but assume the 
vector count and the length of each vector is not known):

[[1]]
[1]  9  5  7  2 14  4  4  3

[[2]]
[1]  3  6 25  2 14  3  3  4

[[3]]
[1] 28  4 14  3 14  2  4  5

What I want to do is take the average vertically. Thus I want to do 9+3+28 /3, 
5+6+4 /3, etc... and then have it return a vector. I'm assuming that if I can 
sum it, I can count it to so summing this would be just as helpful.

I understand I can first go through each element of the list, get a vector, 
cbind into  matrix and sum across but I was hoping to avoid that... I tried 
getting it to work with mapply but am having difficulties...

Thanks!

Kind regards,
Greg
__
R-help@r-project.org mailing list
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Re: [R] How to scan df from a specific word?


Sorry, the explanation wasn't very good...just to explain better.

I am writing a loop to read and process every time a different file in the
same script.
And what I want to load into a variable each time is a data.frame that is
bellow the word source in all of my files.

So I would like to recognize the word Source in the text fileand read
the table bellow source until the next blank line (the file has more written
stuff bellow the data frame that I want to read too)

Here is an example of the file. I want the df to read from source until the
blank line right above the words "Analysis of Variance

Notice: 37 singularities detected in design matrix.
   1 LogL=-2664.01 S2=  1.   8367 df:   2 components
constrained
   2 LogL=-2269.45 S2=  1.   8367 df
   3 LogL=-1698.47 S2=  1.   8367 df
   4 LogL=-1252.72 S2=  1.   8367 df
   5 LogL=-1013.52 S2=  1.   8367 df
   6 LogL=-957.409 S2=  1.   8367 df
   7 LogL=-944.252 S2=  1.   8367 df
   8 LogL=-939.976 S2=  1.   8367 df
   9 LogL=-938.908 S2=  1.   8367 df
  10 LogL=-938.798 S2=  1.   8367 df
  11 LogL=-938.795 S2=  1.   8367 df
  12 LogL=-938.795 S2=  1.   8367 df

 SourceModel  terms Gamma ComponentComp/SE   % C
 Residual   8383   8367
 at(type,1).Nfam  62 62   10.1131   10.1131   1.81   0 P  
 at(type,2).Nfam  62 62   28.1153   28.1153   2.16   0 P  
 rep.iblk768768   63.2919   63.2919  10.94   0 P  
 at(type,1).Nfemale   44 44   29.9049   29.9049   2.93   0 P  
 at(type,1).Nclone  2689   2689   109.560   109.560  12.66   0 P  
 at(type,2).Nfemale   44 44   14.0305   14.0305   1.68   0 P  
 Variance  0  0   479.040   479.040  36.23   0 P  
 Variance  0  0   490.580   490.580  17.51   0 P  
 Variance  0  0   469.932   469.932  36.51   0 P  
 Variance  0  0   544.654   544.654  17.86   0 P  

 Analysis of Variance  NumDF  F_inc  
  27 mu15860.84

  12 culture   1   0.07
  10 type  1  29.59
  28 culture.rep   6  14.06
  30 culture.rep.type  7   2.17
  36 at(type,1).Nfam  62 effects fitted 
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Re: [R] How to scan df from a specific word?


Sorry, the explanation wasn't very good...just to explain better.

I am writing a loop to read and process different files in the same script.
And what I want to load into a variable is a data.frame that is above the
word source in all of my files.

So I would like to recognize the word Source in the text fileand read
the table bellow source until the next blank line (the file has more written
stuff bellow the data frame that I want to read too)

Here is an example of the file. I want the df to read from source until the
blank line right above the words "Analysis of Variance

Notice: 37 singularities detected in design matrix.
   1 LogL=-2664.01 S2=  1.   8367 df:   2 components
constrained
   2 LogL=-2269.45 S2=  1.   8367 df 
   3 LogL=-1698.47 S2=  1.   8367 df 
   4 LogL=-1252.72 S2=  1.   8367 df 
   5 LogL=-1013.52 S2=  1.   8367 df 
   6 LogL=-957.409 S2=  1.   8367 df 
   7 LogL=-944.252 S2=  1.   8367 df 
   8 LogL=-939.976 S2=  1.   8367 df 
   9 LogL=-938.908 S2=  1.   8367 df 
  10 LogL=-938.798 S2=  1.   8367 df 
  11 LogL=-938.795 S2=  1.   8367 df 
  12 LogL=-938.795 S2=  1.   8367 df 

 SourceModel  terms Gamma ComponentComp/SE   % C
 Residual   8383   8367
 at(type,1).Nfam  62 62   10.1131   10.1131   1.81   0 P   
 at(type,2).Nfam  62 62   28.1153   28.1153   2.16   0 P   
 rep.iblk768768   63.2919   63.2919  10.94   0 P   
 at(type,1).Nfemale   44 44   29.9049   29.9049   2.93   0 P   
 at(type,1).Nclone  2689   2689   109.560   109.560  12.66   0 P   
 at(type,2).Nfemale   44 44   14.0305   14.0305   1.68   0 P   
 Variance  0  0   479.040   479.040  36.23   0 P   
 Variance  0  0   490.580   490.580  17.51   0 P   
 Variance  0  0   469.932   469.932  36.51   0 P   
 Variance  0  0   544.654   544.654  17.86   0 P   

 Analysis of Variance  NumDF  F_inc  
  27 mu15860.84 

  12 culture   1   0.07 
  10 type  1  29.59 
  28 culture.rep   6  14.06 
  30 culture.rep.type  7   2.17 
  36 at(type,1).Nfam  62 effects fitted
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[R] How to scan df from a specific word?


Hi R-helpers,

I need to read some file with different lines (I don't know the number of
lines to skip) and I would like to find a way to start reading the
data.frame from the word "source".

ex:

djhsafk
asdfhkjash
shdfjkash
asfhjkash #those lines contain numbers and words, I want to skip
then but they have different sizes
asdfhjkash
asdfhjksa

source
tret 2 
res 3

Can anybody help?
Thanks
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[R] plot pdf

2010-10-29 Thread mmstat



I want to plot the unstadardized version of a normal plot.Â  Can you explain 
why that is not working? 



Dev.set(1) 



xcrit=-1.645 



cord.x <- c(-3,seq(-3,xcrit,0.01),xcrit) 



cord.y <- c(0,dnorm(seq(-3,xcrit,0.01)),0)Â Â Â Â Â Â Â Â Â Â Â  # what does 
final 0 do here? 



curve(dnorm(x,0,1),xlim=c(-3,3),main='Normal PDF') 



polygon(cord.x,cord.y,col='orange') 



savePlot(filename="c:\\ssl_z.emf",type="emf") Â Â Â Â Â Â Â Â Â #unable to 
create metafile?Â  why?Â  



Â  

x=seq(80,200,1) 



plot(x,dnorm(x,140,15),type="l",ylab="Probability") 





Â  

dev.set(4) 



xcrit=135.9 



cord.x <- c(80,seq(80,xcrit,.1),xcrit) 



cord.y <- c(0,dnorm(seq(80,xcrit,.1)),0)Â Â Â Â Â Â Â Â Â Â Â Â Â Â #final 0 
needs to be changed but to what? 

curve(dnorm(x,140,15),xlim=c(80,200),main='Normal PDF',ylab="Probability") 



polygon(cord.x,cord.y,col='orange') 



savePlot(filename="c:\\ssl_z.emf",type="emf")Â Â Â Â Â Â Â Â  #unable to create 
metafile?Â  why?Â  



Â  



Â  

Sincerely, 

Mary A. Marion 





[[alternative HTML version deleted]]

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Re: [R] how to debug (mtrace) a function defined inside a function?

2010-10-29 Thread Duncan Murdoch


On 29/10/2010 5:19 PM, Andre Zege wrote:


Hi, everyone. I am using a fair amount of closures in my code. Problem i am
experiencing is i cannot figure out how to mtrace functions defined within a
function. There must be some way to name such function for mtrace to see it
and let me step into it. For example, say i have code


mymodel<-function(){
  data<-numeric(0)
  build<-function(){
data<<-1
  }

  m<-list()
  m$build<-build
  m
}


How do I mtrace build function defined inside mymodel function so that i can
step into build?


I don't use mtrace, but you can use setBreakpoint() to set a breakpoint 
by line number, and presumably you could set the action to whatever will 
trigger mtrace.  It will modify the source of mymodel so that every time 
it creates a new build(), it creates it with a breakpoint within.


See ?setBreakpoint and ?trace.

Duncan Murdoch

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Re: [R] R version 2-12.0 - running as 32 or as 64 bit?

Actually, Berwin  - you are right. I also did get 2 shortcuts. I just
thought that one of them was old - from my previous R installations.
And indeed the 4th line tells what version is running.
Great!
Thank you!
Dimitri



On Fri, Oct 29, 2010 at 5:02 PM, Berwin A Turlach
 wrote:
> G'day Dimitri,
>
> On Fri, 29 Oct 2010 16:45:00 -0400
> Dimitri Liakhovitski  wrote:
>
>> Question: I installed R verison 2-12.0 on my Windows 7 (64 bit) PC.
>> When I was installing it, it did not ask me anything about 32 vs. 64
>> bit. So, if I run R now - is it running as a 32-bit or a 64-bit?
>
> Well, when I did the same, I got two shortcuts installed on my desktop,
> one named R 2.12.0 and the other named R x64 2.12.0.  If I had any
> doubts which version of R these shortcuts would start, then the fourth
> line of the start-up message would put them to rest.  Missing that
> message, you can always issue the command
>
>> .Machine$sizeof.pointer
>
> if the answer is 4, you are running 32bit, if the answer is 8, then you
> are running 64 bit.
>
> HTH.
>
> Cheers,
>
>        Berwin
>
> == Full address 
> Berwin A Turlach                      Tel.: +61 (8) 6488 3338 (secr)
> School of Maths and Stats (M019)            +61 (8) 6488 3383 (self)
> The University of Western Australia   FAX : +61 (8) 6488 1028
> 35 Stirling Highway
> Crawley WA 6009                e-mail: ber...@maths.uwa.edu.au
> Australia                        http://www.maths.uwa.edu.au/~berwin
>



-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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[R] how to debug (mtrace) a function defined inside a function?

2010-10-29 Thread Andre Zege


Hi, everyone. I am using a fair amount of closures in my code. Problem i am
experiencing is i cannot figure out how to mtrace functions defined within a
function. There must be some way to name such function for mtrace to see it
and let me step into it. For example, say i have code


mymodel<-function(){
 data<-numeric(0)
 build<-function(){
   data<<-1
 }
 
 m<-list()
 m$build<-build
 m
}


How do I mtrace build function defined inside mymodel function so that i can
step into build?
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Re: [R] SARIMA simulation using time series history

2010-10-29 Thread Knut Erik Vedahl

Thank you for your suggestion, Juan. However, I have tried to specify  
the innovations and this doesn't seem to solve the problem.


Are there really no opportunities to simulate from a (S)ARIMA model  
using past observations in R, besides to make the whole script from  
the ground?



Knut


Siterer Juan Pablo Calle :


Hi Knut,

The arima.sim function, in the stats package, has an argument  
innov="YourSerie" for the innovations, you should try something like:


arima.sim(list(order=c(p,d,q), ar=c(ar coefficients), ma=c(ma  
coefficients), seasonal=list(order=c(P,D,Q), period=NA)), n=N,  
innov=YourSerie)


I'm not sure, but maybe this help you.

Salute.

Juan.

On 2010-10-28, at 8:08 PM, Knut Erik Vedahl wrote:


Hi,

I'm currently working with a SARIMA model from which I want to make  
simulations. As I understand, neither sarima.Sim nor the functions  
in the gsarima package use historic realizations of the time series  
to simulate future values. However, I want to use historic values  
as input and simulate future values based on the history. Anyone  
who know whether such a function is available in the libaries or  
anywhere else?



Thanks for any help!

Knut

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Re: [R] Memory use in R

2010-10-29 Thread Duncan Murdoch


On 29/10/2010 12:46 PM, DM2010 wrote:


Dear R Users

I have two questions about how R makes use of memory on a Windows computer.

On my machine certain R jobs seem to stop with messages such as...
"Error: cannot allocate vector of size 215.0 Mb"
...when, according to Windows Task Manager, there are still hundreds of
megabytes of physical memory available. These jobs usually stop when the
peak commit charge is around 2.6GB (in Windows-speak the "commit charge" is
all the memory allocated by system, drivers, applications etc.).

Machine characteristics

Processor: Pentium-D
4GB of installed memory (Task Manager reports 3584MB = 4GB - 512MB total
memory)
Windows booted with /3GB switch
R started with --max-mem-size=3071M
memory.limit() as reported by R: 3071

My questions are:
1. When it starts, does R ask the system to allocate memory for it in a
single chunk?


No, it can handle fragmented allocations.


2. Can R allocate memory either side of the system-preallocated memory in
the middle of address space?


I think so.  However, when it asks for 215 Mb, it needs to get that 
allocation in a single block of contiguous addresses, and it won't move 
things around to make space.  When all you've got is 3 GB to play in and 
you've allocated 2.6 GB, there's a good chance that the remaining space 
is spread out in small pieces and no available unused address ranges are 
big enough.


If you run R in a 64 bit OS you'll find it handles this situation 
better, because the OS can assign the physical memory to whatever 
address it likes, and you'll have lots of big open ranges.  On the other 
hand, with bigger pointers everything takes a bit more space, so you 
might run out sooner.


Duncan Murdoch


Best wishes

David Max


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Re: [R] R version 2-12.0 - running as 32 or as 64 bit?

2010-10-29 Thread Duncan Murdoch


On 29/10/2010 4:45 PM, Dimitri Liakhovitski wrote:

Question: I installed R verison 2-12.0 on my Windows 7 (64 bit) PC.
When I was installing it, it did not ask me anything about 32 vs. 64 bit.
So, if I run R now - is it running as a 32-bit or a 64-bit?


If it didn't ask you about it, then it probably thinks your machine 
can't run 64 bits.  Any ideas why it might think that?  Can you give 
details about what Windows says about itself in Help | About Windows 
(accessible from Explorer)?


Duncan Murdoch

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[R] NetWorkSpace from REvolution; Distributed Computing setup questions

2010-10-29 Thread Timothy Murphy

***Summary:***

I'm setting up a cluster using netWorkSpace, and I'm having issues
with the sleigh initialization. My R function to initialize the sleigh
succeeds and the sleigh appears to be ready, but I get apparently
conflicting information from "status(s)", "rankCount(s)", and "s"; and
basic sleigh functions cause the sleigh to hang indefinitely.

Also, the log file contains an error that indicates that the script is
trying to find a file in a nonexistent directory:
"/usr/local/lib/R/site-library/nws/bin/RNWSSleighWorker.sh: 37:
/Library/Frameworks/R.framework/Resources/bin/R: not found" (see
section 4).

I've spent quite a bit of time trying to debug this, and I've gathered
here all the information that I think may be pertinent to solving the
problem. The following is therefore a bit lengthy, but I think
complete (as far as I'm able to tell from the existing documentation).
It's organized into sections roughly by the topic tested.

So if you're familiar with the workings of netWorkSpaces, I would be
very grateful if you would take a look at my diagnostics below and
tell me if you can identify the problem.

***Details:***

***Section 1***
Currently my setup is:

MASTER:
MacBook Pro running OS X 10.6.4 R-2.11.1, Python 2.6.1, and NWSserver-2.0.0.

WORKER:
Optiplex GX620 running Ubuntu 10.10 (64bit), R-2.11.1, Python 2.6.6,
NWSserver-2.0.0, and NWS-2.0.0.3 (client)
R and Python are in the PATH on both machines; I can start them from
the worker's command line by typing "R" or "python".
The client is able to find the "RNWSSleighWorker.sh" file.

(Note 1: I put the server software on the client because I was getting
an message saying: "No nws server found" each time I tried to install
the client software. I don't know if this is needed)
(Note 2: I plan to set up many more machines if I can get this working)
(Note 3: Originally I was trying this on Windows machines with Cygwin,
but I encountered the same error and figured I could at least rule out
a possible cause by setting it up on a linux machine. Ultimately I
would like to get this working in Windows/Cygwin.)

***Section 2***
The function I used to start the sleigh is:

s=sleigh(
+   nwsHost="172.30.xx.xx",
+   nwsPort=8765,
+   launch=sshcmd,
+   nodeList=c("10.85.xxx.xxx"),
+   scriptExec=envcmd,
+   scriptDir="/usr/local/lib/R/site-library/nws/bin",
+   scriptName="RNWSSleighWorker.sh",
+   workingDir='~/tmp/',
+   logDir='~/tmp/',
+   outfile="outfileTest",
+   user="tj")

This function returns the message below and then clear command prompt:

Executing command:
'/Library/Frameworks/R.framework/Resources/library/nws/bin/SleighWorkerWrapper.sh'
'ssh' '-f' '-x' '-l' 'tj' '10.85.101.109' 'env'
'RSleighName=10.85.101.109'
'RSleighNwsName=sleigh_ride_0450__nwssNGG4LF'
'RSleighUserNwsName=sleigh_user_0452__nwssNGG4LF' 'RSleighID=1'
'RSleighWorkerCount=1'
'RSleighScriptDir=/usr/local/lib/R/site-library/nws/bin'
'RSleighNwsHost=172.30.34.71' 'RSleighNwsPort=8765'
'RSleighWorkingDir=~/tmp/'
'RProg=/Library/Frameworks/R.framework/Resources/bin/R'
'RSleighWorkerOut=sleigh_ride_0450__nwssNGG4LF_0001.txt'
'RSleighLogDir=~/tmp/'
'/usr/local/lib/R/site-library/nws/bin/RNWSSleighWorker.sh'

If I type the name of the sleigh "s" as below, I get information that
makes it look like the sleigh is ready to receive commands:

> s
NWS Sleigh Object
NWS Host:   172.30.xx.xx:8765
Workspace Name: sleigh_ride_0446__nwssNGG4LF
1 Worker Nodes: 10.85.xxx.xxx

Likewise, if I send a simple ssh command to the worker I get a response:
> system('ssh t...@10.85.101.109 date')
Fri Oct 29 15:15:10 EDT 2010

I can also communicate values between the two machines using the NWS
web server and the nwsStore() and nwsFetch() functions.

However, if I check the status of the sleigh using "status(s)" or
"rankCount(s)", I get less encouraging information:

> status(s)
$numWorkers
[1] 0
$closed
[1] 0

> rankCount(s6)
[1] 0

***Section 3***
I can access the NWS server through "localhost:8766" and see that
sleighs are being created. There are two entries: a sleigh_ride and a
sleigh_user; but the worker count in the sleigh_ride is also zero.

If I execute either of the following test sleigh functions, the sleigh
will hang indefinitely (though the R terminal will not hang, since I
used blocking=false):
eachWorker(s5, Sys.info, eo=list(blocking=FALSE))
eachWorker(s, function() library(nws), eo=list(blocking=FALSE))

***Section 4***
***CRUX OF THE ISSUE (probably):***
Finally, three files get created in the "~/tmp" directory that I
specified as the logDir and workingDir, named: "outfileTest",
"RSleighSentinelLog_1000_1" , and
"sleigh_ride_0450__nwssNGG4LF_0001.txt". All three contain exactly the
same information:

"/usr/local/lib/R/site-library/nws/bin/RNWSSleighWorker.sh: 37:
/Library/Frameworks/R.framework/Resources/bin/R: not found"

Whats puzzling is the "Library/Frameworks/R.framework/Resources/bin/R"
part. That looks like an OS X-style path rather tha

Re: [R] One-class SVM

2010-10-29 Thread Steve Lianoglou

Hi,

On Fri, Oct 29, 2010 at 4:24 PM, ANJAN PURKAYASTHA
 wrote:
> Does any R package support one-class SVM?
> I'm trying to develop an application to detect anomalies in genome
> sequencing.

I'm pretty sure kernlab supports this:
http://cran.r-project.org/web/packages/kernlab/index.html

And:
http://www.jstatsoft.org/v11/i09/paper

-steve

-- 
Steve Lianoglou
Graduate Student: Computational Systems Biology
 | Memorial Sloan-Kettering Cancer Center
 | Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact

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Re: [R] R version 2-12.0 - running as 32 or as 64 bit?

2010-10-29 Thread Berwin A Turlach

G'day Dimitri,

On Fri, 29 Oct 2010 16:45:00 -0400
Dimitri Liakhovitski  wrote:

> Question: I installed R verison 2-12.0 on my Windows 7 (64 bit) PC.
> When I was installing it, it did not ask me anything about 32 vs. 64
> bit. So, if I run R now - is it running as a 32-bit or a 64-bit?

Well, when I did the same, I got two shortcuts installed on my desktop,
one named R 2.12.0 and the other named R x64 2.12.0.  If I had any
doubts which version of R these shortcuts would start, then the fourth
line of the start-up message would put them to rest.  Missing that
message, you can always issue the command

> .Machine$sizeof.pointer

if the answer is 4, you are running 32bit, if the answer is 8, then you
are running 64 bit.

HTH.

Cheers,

Berwin

== Full address 
Berwin A Turlach  Tel.: +61 (8) 6488 3338 (secr)
School of Maths and Stats (M019)+61 (8) 6488 3383 (self)
The University of Western Australia   FAX : +61 (8) 6488 1028
35 Stirling Highway   
Crawley WA 6009e-mail: ber...@maths.uwa.edu.au
Australiahttp://www.maths.uwa.edu.au/~berwin

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[R] R version 2-12.0 - running as 32 or as 64 bit?

Question: I installed R verison 2-12.0 on my Windows 7 (64 bit) PC.
When I was installing it, it did not ask me anything about 32 vs. 64 bit.
So, if I run R now - is it running as a 32-bit or a 64-bit?

thank you!

-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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Re: [R] help pages do not open

2010-10-29 Thread Joshua Wiley

Hi Dimitri,

The help pages should be loaded from a local server (which is
corroborated by the 127...ip), so the internet connection should not
matter.

I would check whether IE is the default application for .html
extensions, and you could also try running R as an admin (though I do
not really think that should matter here).

Cheers,

Josh

On Fri, Oct 29, 2010 at 1:20 PM, Dimitri Liakhovitski
 wrote:
> I tried one more thing - I uninstalled R again, and this time
> installed it under:
> C:\Users\Myname\R\R-2.12.0
> And again - am gettting the same error:
> Error in shell.exec(url) :
>  access to 'http://127.0.0.1:30111/library/stats/html/lm.html' denied
>
> Any help is greatly appreciated!
> Dimitri
>
> On Fri, Oct 29, 2010 at 9:22 AM, Dimitri Liakhovitski
>  wrote:
>> I have just installed R-2.12
>> I have Windows 7, 64-bit verison.
>> I currently have IE as my default browser. The internet connection is very 
>> good.
>>
>> Whenever I try to run a help command (?lm, for example), I get this error:
>> Error in shell.exec(url) : access to
>> 'http://127.0.0.1:20271/library/stats/html/lm.html' denied
>>
>> I first got this message when Google Chrome was my default browser.
>> For some reason, Google Chrome stopped opening web pages. At the same
>> time R11 stopped showing help pages. Then, I uninstalled Google Chrome
>> and defined IE as my default browser. Then, I uninstalled R11 and
>> installed R12 instead. But I am getting the same error.
>>
>> Any advice?
>>
>>
>> --
>> Dimitri Liakhovitski
>> Ninah Consulting
>> www.ninah.com
>>
>
>
>
> --
> Dimitri Liakhovitski
> Ninah Consulting
> www.ninah.com
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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Re: [R] help pages do not open

Problem solved. It turned out I had to reset my IE as my default
browser for all extentions (like .html). Now, R help is working!
Dimitri

On Fri, Oct 29, 2010 at 4:20 PM, Dimitri Liakhovitski
 wrote:
> I tried one more thing - I uninstalled R again, and this time
> installed it under:
> C:\Users\Myname\R\R-2.12.0
> And again - am gettting the same error:
> Error in shell.exec(url) :
>  access to 'http://127.0.0.1:30111/library/stats/html/lm.html' denied
>
> Any help is greatly appreciated!
> Dimitri
>
> On Fri, Oct 29, 2010 at 9:22 AM, Dimitri Liakhovitski
>  wrote:
>> I have just installed R-2.12
>> I have Windows 7, 64-bit verison.
>> I currently have IE as my default browser. The internet connection is very 
>> good.
>>
>> Whenever I try to run a help command (?lm, for example), I get this error:
>> Error in shell.exec(url) : access to
>> 'http://127.0.0.1:20271/library/stats/html/lm.html' denied
>>
>> I first got this message when Google Chrome was my default browser.
>> For some reason, Google Chrome stopped opening web pages. At the same
>> time R11 stopped showing help pages. Then, I uninstalled Google Chrome
>> and defined IE as my default browser. Then, I uninstalled R11 and
>> installed R12 instead. But I am getting the same error.
>>
>> Any advice?
>>
>>
>> --
>> Dimitri Liakhovitski
>> Ninah Consulting
>> www.ninah.com
>>
>
>
>
> --
> Dimitri Liakhovitski
> Ninah Consulting
> www.ninah.com
>



-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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[R] One-class SVM

2010-10-29 Thread ANJAN PURKAYASTHA

Does any R package support one-class SVM?
I'm trying to develop an application to detect anomalies in genome
sequencing.
Thanks in advance.
Anjan

-- 
===
anjan purkayastha, phd.
research associate
fas center for systems biology,
harvard university
52 oxford street
cambridge ma 02138
phone-703.740.6939
===

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Re: [R] help pages do not open

I tried one more thing - I uninstalled R again, and this time
installed it under:
C:\Users\Myname\R\R-2.12.0
And again - am gettting the same error:
Error in shell.exec(url) :
  access to 'http://127.0.0.1:30111/library/stats/html/lm.html' denied

Any help is greatly appreciated!
Dimitri

On Fri, Oct 29, 2010 at 9:22 AM, Dimitri Liakhovitski
 wrote:
> I have just installed R-2.12
> I have Windows 7, 64-bit verison.
> I currently have IE as my default browser. The internet connection is very 
> good.
>
> Whenever I try to run a help command (?lm, for example), I get this error:
> Error in shell.exec(url) : access to
> 'http://127.0.0.1:20271/library/stats/html/lm.html' denied
>
> I first got this message when Google Chrome was my default browser.
> For some reason, Google Chrome stopped opening web pages. At the same
> time R11 stopped showing help pages. Then, I uninstalled Google Chrome
> and defined IE as my default browser. Then, I uninstalled R11 and
> installed R12 instead. But I am getting the same error.
>
> Any advice?
>
>
> --
> Dimitri Liakhovitski
> Ninah Consulting
> www.ninah.com
>



-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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[R] Memory use in R

2010-10-29 Thread DM2010


Dear R Users

I have two questions about how R makes use of memory on a Windows computer.

On my machine certain R jobs seem to stop with messages such as...
"Error: cannot allocate vector of size 215.0 Mb"
...when, according to Windows Task Manager, there are still hundreds of
megabytes of physical memory available. These jobs usually stop when the
peak commit charge is around 2.6GB (in Windows-speak the "commit charge" is
all the memory allocated by system, drivers, applications etc.).

Machine characteristics

Processor: Pentium-D
4GB of installed memory (Task Manager reports 3584MB = 4GB - 512MB total
memory)
Windows booted with /3GB switch
R started with --max-mem-size=3071M
memory.limit() as reported by R: 3071

My questions are:
1. When it starts, does R ask the system to allocate memory for it in a
single chunk?
2. Can R allocate memory either side of the system-preallocated memory in
the middle of address space?

Best wishes

David Max
-- 
View this message in context: 
http://r.789695.n4.nabble.com/Memory-use-in-R-tp3019447p3019447.html
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[R] Tukey post hoc comparison (glht?) after 3factorial mixed model (lmer)

2010-10-29 Thread Anna Radtke

Hello, dear R-community.

This is a question about TukeyHSD between factor combinations of a Three-Way
ANOVA, which is - since it is a multi measure ANOVA - not a simple ANOVA but
a Generalised Linear Mixed Model (GLMM), calculated with "lmer".

> growth <-
groupedData(length~meas|box_id,outer=~spec*comp*water,data=all.spec)
> model <- lmer(length~spec*comp*water+(meas|box_id),data=growth)
> summary(model)

This works fine. But now, I would like to calculate Tukey HSD among certain
factor combinations. Formerly (before version 1.0), Tukey contrasts
including all levels of interactions could be calculated with the multcomp
package more or less automatically like this:

> summary (glht (model, linfct = mcp (spec*comp*water = "Tukey")))

Now, Mr Hothorn et al. have changed the function and it is "suggestet to the
users that they write out, manually, the set of contrasts they want". I
think that is not a bad idea, but I got problems with the syntax in my case.
And I would really grateful if anybody could help me. So far I tried:

> K <- cbind(0,diag(length(fixef(model))-1))
> rownames(K) <- names(fixef(model))[-1]
> model_glht <- glht(model,linfct=K)
> summary(model_glht)  # but actually this is not what I want

I'm rather looking for a Tukey output like this

specSt:compControl - specSt:compRoot p that they are equal < 0.05
specSt:compControl - specSt:compShoot
specSt:compControl - specSt:compFull
specSt:compRoot - specSt:compShoot
specSt:compRoot - specSt:compFull
specSt:compShoot - specSt:compFull

and that for each of the 3 studied species, i.e. compare all-pairwise
combinations of the levels of the second factor within each level of the
first factor. We are not interested in the third factor, so data can be
averaged over that.


THANKS A LOT FOR ANY ADVICE, Sincerely

Anna Radtke



-> model output:

Linear mixed model fit by REML
Formula: length ~ spec * comp * water + (meas | box_id)
   Data: growth
  AIC  BIC logLik deviance REMLdev
 4800 4909  -2372 49884744
Random effects:
 Groups   NameVariance Std.Dev. Corr
 box_id   (Intercept)  0.0   0.000
  meas222484.2 471.682NaN
 Residual   9439.4  97.157
Number of obs: 360, groups: box_id, 120

Fixed effects:
Estimate Std. Error t value
(Intercept) -353.554 65.777  -5.375
specSt  -103.111 93.023  -1.108
specSv-4.485 93.023  -0.048
comproot -41.686 93.023  -0.448
compshoot241.847 93.023   2.600
compxfull316.849 93.023   3.406
watermoist   -48.620 93.023  -0.523
specSt:comproot   97.168131.554   0.739
specSv:comproot   54.849131.554   0.417
specSt:compshoot  10.416131.554   0.079
specSv:compshoot   6.584131.554   0.050
specSt:compxfull  90.949131.554   0.691
specSv:compxfull  43.284131.554   0.329
specSt:watermoist  9.222131.554   0.070
specSv:watermoist104.229131.554   0.792
comproot:watermoist   15.931131.554   0.121
compshoot:watermoist  28.999131.554   0.220
compxfull:watermoist  39.363131.554   0.299
specSt:comproot:watermoist45.513186.046   0.245
specSv:comproot:watermoist   -24.350186.046  -0.131
specSt:compshoot:watermoist   46.027186.046   0.247
specSv:compshoot:watermoist1.232186.046   0.007
specSt:compxfull:watermoist   23.749186.046   0.128
specSv:compxfull:watermoist  -57.109186.046  -0.307
"meas" "spec" "comp" "water" "box_id" "sprouts" "leaves" "length" "long.sprout"
"81" 1 "Sf" "xfull" "moist" 81 6.8 13.6 150.4 43.8
"82" 1 "Sf" "root" "moist" 82 6.5 18.5 104.25 25.75
"83" 1 "Sf" "control" "moist" 83 9.25 29 146 27
"84" 1 "Sf" "control" "moist" 84 6.8 31.4 163.8 46.6
"85" 1 "Sf" "shoot" "moist" 85 4.8 19.8 127.6 32.8
"86" 1 "Sf" "control" "moist" 86 8.2 30.6 144.4 31.6
"87" 1 "Sf" "root" "moist" 87 7.8 32.8 174 38.8
"88" 1 "Sf" "shoot" "moist" 88 5 8.5 54 22.25
"89" 1 "Sf" "xfull" "moist" 89 2.25 7 54.25 27
"90" 1 "Sf" "control" "moist" 90 5.6 21.2 103 30
"91" 1 "Sf" "shoot" "moist" 91 4.25 18 84.25 29.25
"92" 1 "Sf" "root" "moist" 92 5.8 26.4 133.2 33
"93" 1 "Sf" "shoot" "moist" 93 11.25 35.25 215 44.25
"94" 1 "Sf" "root" "moist" 94 6 17. 90.6667 19.
"95" 1 "Sf" "root" "moist" 95 4.7 5.3 33.6667 9.3
"96" 1 "Sf" "shoot" "moist" 96 6.5 11.5 88.25 25.5
"97" 1 "Sf" "xfull" "moist" 97 9.8 13 184.4 40.2
"98" 1 "Sf" "xfull" "moist" 98 5.8 15.2 156.6 41.6
"99" 1 "Sf" "control" "moist" 99 7.5 26.5 140.75 36.5
"100" 1 "Sf" "xfull" "moist" 100 4.2 10.4 91 41.2
"101" 1 "Sf" "control" "awater-logged" 101 9 35.8 222.2 43.6
"102" 1 "Sf" "shoot" "awater-logged" 102 10.8 40.6 245.4 47.6
"103" 1 "Sf" "xfull" "awater-logged" 103 4.6 10.6 134.6 47
"104" 1 "Sf" "shoot" "awate

Re: [R] doubt in climate variability analysis in R! - code included!

2010-10-29 Thread govindas



the following code was used  

library(akima)
library(clim.pact)
nc.1 <- "RF_80-05.nc"
nc.rf.in <- open.ncdf(nc.1)

x1 <- retrieve.nc(nc.1, v.nam="Rainfall",l.scale=FALSE,  x.rng=c(70, 80), 
y.rng=c(10, 13.5))

#dimension is checked for the subset. (lon, lat, time) is changed as (time, 
lat, lon)
>dim(x1$dat)
#[1] 2192    8   20

My question is - how can i convert this array into a dataframe so that i have 
"lat", "lon", "precipitation values" in 3 different columns (note, I will have 
it for just a single day). So, my expected dataframe will have rainfall values 
for each given pair of "lon" and "lat". 

Or is there any other better way to do my spatial variogram analysis for a 
single day given the above dataset?

here is the link for the dataset.
HTTP://WWW.4SHARED.COM/FILE/4ZV0G3JR/RF_80-85.HTML

-- 
Regards,
Mahalakshmi
Graduate Student
#20, Department of Geography
Michigan State University
East Lansing, MI 48824 Quoting govin...@msu.edu:

>
>
> Hello all,
>
> I  am trying to use "clim.pact" package for my work, but since this 
> is  the  beginning for me to use gridded datasets in "R", I am having 
> some   trouble.
>
> I want to do seasonal analyses like   trends, anomalies, variograms, 
> EOF and probably kriging too to   downscale my 1 degree gridded data 
> to 0.5.  So, as a first step, I   compiled my entire dataset (with 25 
> yeears of daily dataset which were   present as 25 files) into a 
> single netcdf file.
>
> Then, I downloaded clim.pact to do further analysis, which works but  
> seems  to change dataset's original dimensions' order for  
> "retrieve.nc"   function (i.e. original lon, lat, time order was 
> changed  to time,  lat,  lon after using this function to get a 
> subset). I am not  sure as  to why  this happened and not able to get 
> any plots such as box  plot  (showing  trend in "lon", "lat", 
> "time"), variogram (or variance),   correlation  analysis done 
> because of this conversion problem.
>
> Further, basic "R"  functions seem to work well with objects such as  
>  dataframe, matrix ..etc  with time in a separate column, and the 
> data   values (precipitation, or  temperature) in a separate coulmn 
> with   corresponding station values  (lon/lat). So, now I have very 
> little idea   about what I have to do. Can anyone suggest me a better 
> (probably  more refined way) way than what I am currently doing to 
> analyze these  data?
>
>  
>
> --
> Regards,
> Mahalakshmi
> Graduate Student
> #20, Department of Geography
> Michigan State University
> East Lansing, MI 48824
[[alternative HTML version deleted]]

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

Re: [R] Plotting 2 Lines on the Same Chart

On Fri, Oct 29, 2010 at 2:30 PM, Jason Kwok  wrote:
> Thanks.  1 more question.
>
> When I use
>
> Plot(series1)
> lines(series2)
>
> The graph will use the y axis scaling for series 1 so some of series 2 is
> cut off.  How do I control the y axis scaling?
>

Plot them together as a multivariate series as in the examples already
pointed to.

-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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Re: [R] Plotting 2 Lines on the Same Chart

2010-10-29 Thread Jason Kwok

Thanks.  1 more question.

When I use

Plot(series1)
lines(series2)

The graph will use the y axis scaling for series 1 so some of series 2 is
cut off.  How do I control the y axis scaling?

Thanks,

Jason

On Fri, Oct 29, 2010 at 2:10 PM, Gabor Grothendieck  wrote:

> On Fri, Oct 29, 2010 at 1:41 PM, Jason Kwok  wrote:
> > How do I plot two time series plots on the same chart?
> >
>
> Try this:
>
> example(plot.ts)
> example(ts.plot)
>
> library(zoo)
> example(plot.zoo)
> library(lattice)
> example(xyplot.zoo)
>
> --
> Statistics & Software Consulting
> GKX Group, GKX Associates Inc.
> tel: 1-877-GKX-GROUP
> email: ggrothendieck at gmail.com
>

[[alternative HTML version deleted]]

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

Re: [R] Plotting 2 Lines on the Same Chart

On Fri, Oct 29, 2010 at 1:41 PM, Jason Kwok  wrote:
> How do I plot two time series plots on the same chart?
>

Try this:

example(plot.ts)
example(ts.plot)

library(zoo)
example(plot.zoo)
library(lattice)
example(xyplot.zoo)

-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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Re: [R] Dickey Fuller Test

2010-10-29 Thread Cuckovic Paik


Thanks a lot for your answers,  Dennis and Bernhard!

-- 
View this message in context: 
http://r.789695.n4.nabble.com/Dickey-Fuller-Test-tp3018408p3019544.html
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[R] Plotting 2 Lines on the Same Chart

2010-10-29 Thread Jason Kwok

How do I plot two time series plots on the same chart?

Thanks,

Jason

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Re: [R] doubt in climate variability analysis in R!

2010-10-29 Thread Daniel Nordlund

> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of govin...@msu.edu
> Sent: Friday, October 29, 2010 8:33 AM
> To: r-help@r-project.org
> Subject: [R] doubt in climate variability analysis in R!
> 
> 
> 
> Hello all,
> 
> I  am trying to use "clim.pact" package for my work, but since this is
> the  beginning for me to use gridded datasets in "R", I am having some
> trouble.
> 
> I want to do seasonal analyses like   trends, anomalies, variograms, EOF
> and probably kriging too to   downscale my 1 degree gridded data to 0.5.
> So, as a first step, I   compiled my entire dataset (with 25 yeears of
> daily dataset which were   present as 25 files) into a single netcdf file.
> 
> Then, I downloaded clim.pact to do further analysis, which works but
> seems  to change dataset's original dimensions' order for  "retrieve.nc"
> function (i.e. original lon, lat, time order was changed  to time,  lat,
> lon after using this function to get a subset). I am not  sure as  to why
> this happened and not able to get any plots such as box  plot  (showing
> trend in "lon", "lat", "time"), variogram (or variance),   correlation
> analysis done because of this conversion problem.
> 
> Further, basic "R"  functions seem to work well with objects such as
> dataframe, matrix ..etc  with time in a separate column, and the data
> values (precipitation, or  temperature) in a separate coulmn with
> corresponding station values  (lon/lat). So, now I have very little idea
> about what I have to do. Can anyone suggest me a better (probably  more
> refined way) way than what I am currently doing to analyze these  data?
> 
> 
> 
> --
> Regards,
> Mahalakshmi
> Graduate Student
> #20, Department of Geography
> Michigan State University
> East Lansing, MI 48824
>   [[alternative HTML version deleted]]

You should read the posing guide referenced at the bottom of every email.  It 
will tell you how to ask questions and what information you need to provide in 
order to get useful help.  At a minimum, we need a self-contained reproducible 
example that demonstrates the problem you are having.  Provide say 10 
observations and the code you used to subset the data, and maybe someone can 
help you with a solution.

Dan

Daniel Nordlund
Bothell, WA USA

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Re: [R] Repeated Measures MANOVA

2010-10-29 Thread John Fox

Dear Chris,

See the examples in ?Anova in the car package (but it works with a model fit
by lm).

I hope this helps,
 John


John Fox
Senator William McMaster 
  Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
web: socserv.mcmaster.ca/jfox


> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
> Behalf Of Christopher Kurby
> Sent: October-29-10 12:24 PM
> To: r-help@r-project.org
> Subject: [R] Repeated Measures MANOVA
> 
> Hello all,
> 
> Is there an r function that exists that will perform repeated measures
> MANOVAs? For example, let's say I have 3 DVs, one between-subjects IV, and
> one within-subjects IV. Based on the documentation for the manova command,
a
> function like that below is not appropriate because it cannot take Error
> arguments.
> 
> manova(cbind(DV1,DV2,DV3) ~ BetweenSubjectsIV * WithinSubjectsIV +
> Error(SubjNum/WithinSubjectsIV), data=dat)
> 
> Do you know of any functions that would allow me to perform such an
analysis?
> I realize that I could turn the DVs into levels of a new IV, and then
perform
> a standard repeated measures ANOVA. But, I do not feel that would be
> appropriate in this case.
> 
> Thanks so much for your time,
> Chris
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

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Re: [R] scatterplot3d; scaling point symbols to depth of graph

2010-10-29 Thread Uwe Ligges


I had a look:

I fixed it 5 months ago and forgot to make a new release.
The version on R-forge contains the fix already.

New release on its way to CRAN.

Best wishes;
Uwe


On 28.10.2010 11:28, John Coulthard wrote:


Hi

I'm trying to scale the point symbols on a 3d plot so that the ones at the 
front are larger than the ones at the back.  I'm trying to give the image some 
perspective.

Given this code...

library(scatterplot3d)
data=array(c(0,5,9), c(3,3))
scatterplot3d(data, pch=19, cex.symbols=10-data[,2], 
color=c("red","blue","black"));

data



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


[1,]000


[2,]555


[3,]999



which gives a vector for  cex.symbols as

10-data[,2]


[1] 10  5  1


  I would expect the largest point to be the red one at the origin but
the image I get has the black symbol at co-ords 9,9,9 as the largest and red at 
0,0,0 the smallest.

Then if I do...

data=array(c(0,9,5), c(3,3))
data

  [,1] [,2] [,3]
[1,]000
[2,]999
[3,]555

scatterplot3d(data, pch=19, cex.symbols=10-data[,2], 
color=c("red","blue","black"));
10-data[,2]

[1] 10  1  5

I'd expect the position and size of the points to be the same but the colour of 
blue and black to be exchanged.  But the size of the points also changes such 
that the red point at 0,0,0 is the medium size and the black at 5,5,5 is the 
smallest.

So is it possible to get the points described by each row in data to be scaled 
by the values in data[,2]?

Many thanks

John




sessionInfo()

R version 2.11.1 (2010-05-31)
i386-redhat-linux-gnu

locale:
  [1] LC_CTYPE=en_US.utf8   LC_NUMERIC=C
  [3] LC_TIME=en_US.utf8LC_COLLATE=en_US.utf8
  [5] LC_MONETARY=C LC_MESSAGES=en_US.utf8
  [7] LC_PAPER=en_US.utf8   LC_NAME=C
  [9] LC_ADDRESS=C  LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C

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

other attached packages:
[1] scatterplot3d_0.3-30

loaded via a namespace (and not attached):
[1] tools_2.11.1


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

[R] Repeated Measures MANOVA

2010-10-29 Thread Christopher Kurby

Hello all,

Is there an r function that exists that will perform repeated measures MANOVAs? 
For example, let's say I have 3 DVs, one between-subjects IV, and one 
within-subjects IV. Based on the documentation for the manova command, a 
function like that below is not appropriate because it cannot take Error 
arguments.

manova(cbind(DV1,DV2,DV3) ~ BetweenSubjectsIV * WithinSubjectsIV + 
Error(SubjNum/WithinSubjectsIV), data=dat)

Do you know of any functions that would allow me to perform such an analysis? I 
realize that I could turn the DVs into levels of a new IV, and then perform a 
standard repeated measures ANOVA. But, I do not feel that would be appropriate 
in this case.

Thanks so much for your time,
Chris
__
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Re: [R] Clustering



On Oct 29, 2010, at 12:08 PM, David Winsemius wrote:



On Oct 29, 2010, at 11:37 AM, dpender wrote:


Apologies for being vague,

The structure of the output is as follows:


Still no code?



$ cluster1  : Named num [1:131] 3.05 2.71 3.26 2.91 2.88 3.11 3.21  
-1 2.97

3.39 ...
..- attr(*, "names")= chr [1:131] "6667" "6668" "6669" "6670" ...

With 613 clusters.  What I require is abstracting the first and  
last value

of

- attr(*, "names")= chr [1:131] "6667" "6668" "6669" "6670"


Those values are in an attribute:


Corrections:


? attribute


?attributes


? attr

Your specific request may (perhaps) be addressed by something like:

attrnames <- attr(objname["cluster1"], "names")
   ^  ^   should be doubled square- 
brackets

attrnames[c(1, length(attrnames)]

 ^  missing right-paren

Might work:
attrnames <- attr(clusobj[["cluster1"]], "names")
attrnames[c(1, length(attrnames))]
--

David Winsemius, MD
West Hartford, CT

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[R] Tukey post hoc comparison (glht?) after 3factorial mixed model (lmer)

2010-10-29 Thread Anna Radtke

Hello, dear R-community.

This is a question about TukeyHSD between factor combinations of a Three-Way
ANOVA, which is - since it is a multi measure ANOVA - not a simple ANOVA but
a Generalised Linear Mixed Model (GLMM), calculated with "lmer".

> growth <-
groupedData(length~meas|box_id,outer=~spec*comp*water,data=all.spec)
> model <- lmer(length~spec*comp*water+(meas|box_id),data=growth)
> summary(model)

This works fine. But now, I would like to calculate Tukey HSD among certain
factor combinations. Formerly (before version 1.0), Tukey contrasts
including all levels of interactions could be calculated with the multcomp
package more or less automatically like this:

> summary (glht (model, linfct = mcp (spec*comp*water = "Tukey")))

Now, Mr Hothorn et al. have changed the function and it is "suggestet to the
users that they write out, manually, the set of contrasts they want". I
think that is not a bad idea, but I got problems with the syntax in my case.
And I would really grateful if anybody could help me. So far I tried:

> K <- cbind(0,diag(length(fixef(model))-1))
> rownames(K) <- names(fixef(model))[-1]
> model_glht <- glht(model,linfct=K)
> summary(model_glht)  # but actually this is not what I want

I'm rather looking for a Tukey output like this

specSt:compControl - specSt:compRoot p that they are equal < 0.05
specSt:compControl - specSt:compShoot
specSt:compControl - specSt:compFull
specSt:compRoot - specSt:compShoot
specSt:compRoot - specSt:compFull
specSt:compShoot - specSt:compFull

and that for each of the 3 studied species, i.e. compare all-pairwise
combinations of the levels of the second factor within each level of the
first factor. We are not interested in the third factor, so data can be
averaged over that.


THANKS A LOT FOR ANY ADVICE, Sincerely

Anna Radtke



-> model output:

Linear mixed model fit by REML
Formula: length ~ spec * comp * water + (meas | box_id)
   Data: growth
  AIC  BIC logLik deviance REMLdev
 4800 4909  -2372 49884744
Random effects:
 Groups   NameVariance Std.Dev. Corr
 box_id   (Intercept)  0.0   0.000
  meas222484.2 471.682NaN
 Residual   9439.4  97.157
Number of obs: 360, groups: box_id, 120

Fixed effects:
Estimate Std. Error t value
(Intercept) -353.554 65.777  -5.375
specSt  -103.111 93.023  -1.108
specSv-4.485 93.023  -0.048
comproot -41.686 93.023  -0.448
compshoot241.847 93.023   2.600
compxfull316.849 93.023   3.406
watermoist   -48.620 93.023  -0.523
specSt:comproot   97.168131.554   0.739
specSv:comproot   54.849131.554   0.417
specSt:compshoot  10.416131.554   0.079
specSv:compshoot   6.584131.554   0.050
specSt:compxfull  90.949131.554   0.691
specSv:compxfull  43.284131.554   0.329
specSt:watermoist  9.222131.554   0.070
specSv:watermoist104.229131.554   0.792
comproot:watermoist   15.931131.554   0.121
compshoot:watermoist  28.999131.554   0.220
compxfull:watermoist  39.363131.554   0.299
specSt:comproot:watermoist45.513186.046   0.245
specSv:comproot:watermoist   -24.350186.046  -0.131
specSt:compshoot:watermoist   46.027186.046   0.247
specSv:compshoot:watermoist1.232186.046   0.007
specSt:compxfull:watermoist   23.749186.046   0.128
specSv:compxfull:watermoist  -57.109186.046  -0.307
"meas" "spec" "comp" "water" "box_id" "sprouts" "leaves" "length" "long.sprout"
"81" 1 "Sf" "xfull" "moist" 81 6.8 13.6 150.4 43.8
"82" 1 "Sf" "root" "moist" 82 6.5 18.5 104.25 25.75
"83" 1 "Sf" "control" "moist" 83 9.25 29 146 27
"84" 1 "Sf" "control" "moist" 84 6.8 31.4 163.8 46.6
"85" 1 "Sf" "shoot" "moist" 85 4.8 19.8 127.6 32.8
"86" 1 "Sf" "control" "moist" 86 8.2 30.6 144.4 31.6
"87" 1 "Sf" "root" "moist" 87 7.8 32.8 174 38.8
"88" 1 "Sf" "shoot" "moist" 88 5 8.5 54 22.25
"89" 1 "Sf" "xfull" "moist" 89 2.25 7 54.25 27
"90" 1 "Sf" "control" "moist" 90 5.6 21.2 103 30
"91" 1 "Sf" "shoot" "moist" 91 4.25 18 84.25 29.25
"92" 1 "Sf" "root" "moist" 92 5.8 26.4 133.2 33
"93" 1 "Sf" "shoot" "moist" 93 11.25 35.25 215 44.25
"94" 1 "Sf" "root" "moist" 94 6 17. 90.6667 19.
"95" 1 "Sf" "root" "moist" 95 4.7 5.3 33.6667 9.3
"96" 1 "Sf" "shoot" "moist" 96 6.5 11.5 88.25 25.5
"97" 1 "Sf" "xfull" "moist" 97 9.8 13 184.4 40.2
"98" 1 "Sf" "xfull" "moist" 98 5.8 15.2 156.6 41.6
"99" 1 "Sf" "control" "moist" 99 7.5 26.5 140.75 36.5
"100" 1 "Sf" "xfull" "moist" 100 4.2 10.4 91 41.2
"101" 1 "Sf" "control" "awater-logged" 101 9 35.8 222.2 43.6
"102" 1 "Sf" "shoot" "awater-logged" 102 10.8 40.6 245.4 47.6
"103" 1 "Sf" "xfull" "awater-logged" 103 4.6 10.6 134.6 47
"104" 1 "Sf" "shoot" "awate

Re: [R] Clustering



On Oct 29, 2010, at 11:37 AM, dpender wrote:




Apologies for being vague,

The structure of the output is as follows:


Still no code?



$ cluster1  : Named num [1:131] 3.05 2.71 3.26 2.91 2.88 3.11 3.21  
-1 2.97

3.39 ...
 ..- attr(*, "names")= chr [1:131] "6667" "6668" "6669" "6670" ...

With 613 clusters.  What I require is abstracting the first and last  
value

of

- attr(*, "names")= chr [1:131] "6667" "6668" "6669" "6670"


Those values are in an attribute:

? attribute
? attr

Your specific request may (perhaps) be addressed by something like:

attrnames <- attr(objname["cluster1"], "names")
attrnames[c(1, length(attrnames)]

I don't really think this is optimal. For one thing it won't  
generalize to the rest of the clusters. Generally when data are put  
into an attribute, the programmer also provides an extraction  
function. Since Nabble posters are prone to not retaining thread  
context, the name of your function and package are probably elsewhere  
further up the thread but not available as I read this on a mail- 
client. Perhaps if you read more of the documentation? (Just a guess.)




This will give a start and end point of the cluster and allow for the
spacing to be determined.


Those would be character values and, if you want to do calculations,  
would obviously need to be coerced to numeric.




Thanks,

Doug
--


David Winsemius, MD
West Hartford, CT

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Re: [R] draw path diagram using dot

2010-10-29 Thread eshi


Dear John,

Thank you very much for your help! I appreciate it. 

eshi
-- 
View this message in context: 
http://r.789695.n4.nabble.com/draw-path-diagram-using-dot-tp3017987p3019359.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Cox Proportional Models and Haplotypes

2010-10-29 Thread Brad McNeney

You could do a weighted cox ph model, with possible haplotype
configurations for each subject weighted by their posterior
probabilities given genotype data.

Are your markers SNPs?  If so you can use a utility function from the
hapassoc package to get started. For example, if your data is in a
dataframe dat, with nsnp SNPs in the last nsnps columns, you could
create an augmented data frame (augmented by pseudo-individuals for
each subject with ambiguous phase) with

library(hapassoc)
ph<-pre.hapassoc(dat,nsnps)
augdat<-cbind(ph$nonHaploDM,ph$haploDM)
wts<-ph$wt

and then use coxph with augdat as the data frme and  wts as the
weights. See ?pre.hapassoc for details on accepted formats for the SNP
data.

Brad
--
Brad McNeney
Statistics and Actuarial Science
Simon Fraser University

On Tue, Oct 26, 2010 at 12:01 PM, David Winsemius
 wrote:
>
> On Oct 26, 2010, at 8:57 AM, sr500 wrote:
>
>>
>> Hello,
>>
>> I was wondering if anyone knew of a function that fits haplotype data into
>> a
>> cox proportional hazard model. I have computed my Haplotype frequencies
>> using the haplo.stats package. I have also been using the haplo.glm
>> function
>> but this is a linear regression and is not quite what I am looking for…
>>
>
> I think you need to describe your data situation more completely and ideally
> would include a small extract of your data to allow testing and
> illustration. Cox proportional hazards models are typically used to analyze
> time to event data and you have not alluded to any "events".
>
> --
> David.
>
>> Thank you very much,
>> SR
>>
>> --
>> View this message in context:
>> http://r.789695.n4.nabble.com/Cox-Proportional-Models-and-Haplotypes-tp3013989p3013989.html
>> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] Please help me about Monte Carlo Permutation

2010-10-29 Thread Chitra

Hi dear Mi³ego dnia,
I am sorry I got lost here.
May be it it sound if I write you about what I am going to do.

"To test such null hypotheses we used the Monte Carlo test, i.e. we chose
randomly (5000 times) a value for the
current dependent characteristic (richness or actual species pool) in the
interval 0 to the maximum possible
value of the dependent variable for each observed value of the current
independent variable. The maximum observed
value was either the calculated size of the regional pool, or the measured
size of the actual pool. Each time,
we calculated the correlation coefficient r between the independent and
dependent variables in order to achieve
the empirical distribution of r for the null hypothesis conditions. The
empirical probability of cases with a correlation
between the two studied variables positive and stronger than that observed
in the real data, served as an estimate
of the significance level for rejecting the null hypotheses. Thus, by saying
that there exists a significant
relationship, we mean that the relationship is significantly stronger than
expected from our null model".

I hope I can get all necessary R code for analyses.
Chitra
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Re: [R] Clustering

2010-10-29 Thread dpender


 
Apologies for being vague,

The structure of the output is as follows:

$ cluster1  : Named num [1:131] 3.05 2.71 3.26 2.91 2.88 3.11 3.21 -1 2.97
3.39 ...
  ..- attr(*, "names")= chr [1:131] "6667" "6668" "6669" "6670" ...

With 613 clusters.  What I require is abstracting the first and last value
of 

- attr(*, "names")= chr [1:131] "6667" "6668" "6669" "6670"

This will give a start and end point of the cluster and allow for the
spacing to be determined.

Thanks,

Doug
-- 
View this message in context: 
http://r.789695.n4.nabble.com/Clustering-tp3017056p3019323.html
Sent from the R help mailing list archive at Nabble.com.

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[R] I got lot in the permutation

2010-10-29 Thread cbbaniya

Thanks for your help.
May be I can not make clear by my writing. The statement which I am going to do 
is like this

"To test such null hypotheses we used the Monte Carlo test, i.e. we chose 
randomly (5000 times) a value for the
current dependent characteristic (richness or actual species pool) in the 
interval 0 to the maximum possible
value of the dependent variable for each observed value of the current 
independent variable. The maximum observed
value was either the calculated size of the regional pool, or the measured size 
of the actual pool. Each time,
we calculated the correlation coefficient r between the independent and 
dependent variables in order to achieve
the empirical distribution of r for the null hypothesis conditions. The 
empirical probability of cases with a correlation
between the two studied variables positive and stronger than that observed in 
the real data, served as an estimate
of the significance level for rejecting the null hypotheses. Thus, by saying 
that there exists a significant
relationship, we mean that the relationship is significantly stronger than 
expected from our null model".


I hope you will figure out and provide me all code for analyses.
Thanks.
Chitra

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[R] doubt in climate variability analysis in R!

2010-10-29 Thread govindas



Hello all,

I  am trying to use "clim.pact" package for my work, but since this is  the  
beginning for me to use gridded datasets in "R", I am having some   trouble. 

I want to do seasonal analyses like   trends, anomalies, variograms, EOF and 
probably kriging too to   downscale my 1 degree gridded data to 0.5.  So, as a 
first step, I   compiled my entire dataset (with 25 yeears of daily dataset 
which were   present as 25 files) into a single netcdf file. 

Then, I downloaded clim.pact to do further analysis, which works but  seems  to 
change dataset's original dimensions' order for  "retrieve.nc"   function (i.e. 
original lon, lat, time order was changed  to time,  lat,  lon after using this 
function to get a subset). I am not  sure as  to why  this happened and not 
able to get any plots such as box  plot  (showing  trend in "lon", "lat", 
"time"), variogram (or variance),   correlation  analysis done because of this 
conversion problem. 

Further, basic "R"  functions seem to work well with objects such as   
dataframe, matrix ..etc  with time in a separate column, and the data   values 
(precipitation, or  temperature) in a separate coulmn with   corresponding 
station values  (lon/lat). So, now I have very little idea   about what I have 
to do. Can anyone suggest me a better (probably  more refined way) way than 
what I am currently doing to analyze these  data? 

  

-- 
Regards,
Mahalakshmi
Graduate Student
#20, Department of Geography
Michigan State University
East Lansing, MI 48824
[[alternative HTML version deleted]]

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[R] Contract programming position at Merck (NJ, USA)

2010-10-29 Thread Liaw, Andy

Job: Scientific programmer at Merck, Biostatistics, Rahway, NJ, USA
[Job Description] 
This position works closely with statisticians to process and analyze
ultrasound, MRI, and radiotelemetry longitudinal studies using a series
of programs developed in R and Mathworks/Matlab.  This position provides
support for the analysis of several pre-clinical and clinical functional
MRI studies by preprocessing and processing data using the software FSL.
Qualified candidates must have a proficiency and experience with
statistical software and technical computing packages including Matlab,
R, SAS, and S-Plus as well as familiarity with medical image concepts
(e.g., functional MRI) and an understanding of analysis tools for fMRI
(FSL, SPM).

This is contract position for an ongoing need in Biometrics Research.
It is a term contract position (1 year) with the possibility to extend
up to 2 years in length based on continued business need and available
budget. 

If you are interested, please contact:
amy_gilles...@merck.com

Notice:  This e-mail message, together with any attachme...{{dropped:14}}

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Re: [R] Reading multiple .csv-files and assigning them to variable names

On Fri, Oct 29, 2010 at 5:16 AM, Sarah Moens  wrote:
> Hi all,
>
> I've been trying to find a solution for the problem of reading
> multiple files and storing them in a variable that contains the names
> by which I want to call the datasets later on.
>
> For example (5 filenames):
>
> - The filenames are stored in one variable:
> filenames = paste(paste('name', '_', 1:5, sep = ''), '.csv', sep = '')
>
> - Subsequently I have a variable just containing the meaningful names
> for the dataset
> meaningfulnames = c('name1','name2'...,'name5')
>
> - I want to link each of these names to the data that is read
>
> for (i in 1:5)
> {
>     meaningfulnames[i] = read.csv(filenames[i], header = TRUE, sep = ',')
> }
>
>
> I need to read in quite a lot of datafiles. I have a code doing this
> one at a time, but since the number of datafiles I need to read will
> increase in the future, I want to make sure I have a more flexible
> solution for this.
>

Try this:

   filenames <- sprintf("%s_%d.csv", "name", 1:5)
   L <- sapply(filenames, read.csv, simplify = FALSE)

L will be a list with the data frames as components and the file names
as the component names.  If you wish to change the names from the
filenames to some other names you can do this:

   names(L) <- vector.of.other.names

If the data frames all have the same number of columns and the same
names in the same order then you could also put them into a single
data frame like this:

   do.call("rbind", L)

or if you want to retain the knowledge of which each came from:

   library(lattice)
   do.call("make.groups", L)

-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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Re: [R] Printing data.frame data: alternatives to print?

2010-10-29 Thread Matthew Pettis

I think that'll work... thanks!

matt

On Fri, Oct 29, 2010 at 9:45 AM, jim holtman  wrote:
> Is this what you want:
>
>> df
>       f1      f2
> 1   Maj I Minor A
> 2   Maj I Minor A
> 3   Maj I Minor A
> 4  Maj II Minor A
> 5  Maj II Minor B
> 6  Maj II Minor B
> 7 Maj III Minor B
> 8 Maj III Minor C
> 9 Maj III Minor C
>> df[!duplicated(df),]
>       f1      f2
> 1   Maj I Minor A
> 4  Maj II Minor A
> 5  Maj II Minor B
> 7 Maj III Minor B
> 8 Maj III Minor C
>>
>
>
> On Fri, Oct 29, 2010 at 9:53 AM, Matthew Pettis
>  wrote:
>> Hi,
>>
>> I have a data frame with two factors (well, more, but 2 for simple
>> consideration), and I want to display the different combinations of
>> the them that actually occur in the data.  In reality, there are too
>> many of them to do to do a 'table' call and have one col vertical and
>> one col horizontal (I don't want any of the factors listed
>> horizontally).  Before I try to write a function to do this for me, I
>> was wondering if there were alternate printing styles for data that
>> already exist, and if someone could direct me to them?  Inclded is a
>> sample code and 2 possibilities (others welcome for consideration) of
>> how I want to display some data.
>>
>> Thanks,
>> Matt
>>
>> -
>>
>> df <- data.frame(
>>    f1=rep(c("Maj I", "Maj II", "Maj III"), each=3),
>>    f2=c("Minor A", "Minor A", "Minor A", "Minor A", "Minor B", "Minor
>> B", "Minor B", "Minor C", "Minor C")
>> )
>>
>> -
>>
>> What I want printed is something like:
>>
>> ---
>>      f1         f2
>> Maj I       Minor A
>>
>> Maj II      Minor A
>>            Minor B
>>
>> Maj III     Minor B
>>            Minor C
>> ---
>>
>> or
>>
>> ---
>>      f1         f2
>> Maj I       Minor A
>>
>> Maj II      Minor A
>> Maj II      Minor B
>>
>> Maj III     Minor B
>> Maj III     Minor C
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem that you are trying to solve?
>



-- 
Seven Deadly Sins (Gandhi):
  - Wealth without work             - Politics without principle
  - Pleasure without conscience     - Commerce without morality
  - Science without humanity        - Worship without sacrifice
  - Knowledge without character

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Re: [R] strftime vs strptime ??



On Oct 29, 2010, at 6:55 AM, skan wrote:



Hello

Could anyone explain me the difference between strftime vs strptime,  
please

?

I've read the help but it's a little bit cionfusing for me.


You should focus on the "Value" section of the help page. They return  
vectors of different classes.


--

David Winsemius, MD
West Hartford, CT

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Re: [R] Printing data.frame data: alternatives to print?

Is this what you want:

> df
   f1  f2
1   Maj I Minor A
2   Maj I Minor A
3   Maj I Minor A
4  Maj II Minor A
5  Maj II Minor B
6  Maj II Minor B
7 Maj III Minor B
8 Maj III Minor C
9 Maj III Minor C
> df[!duplicated(df),]
   f1  f2
1   Maj I Minor A
4  Maj II Minor A
5  Maj II Minor B
7 Maj III Minor B
8 Maj III Minor C
>


On Fri, Oct 29, 2010 at 9:53 AM, Matthew Pettis
 wrote:
> Hi,
>
> I have a data frame with two factors (well, more, but 2 for simple
> consideration), and I want to display the different combinations of
> the them that actually occur in the data.  In reality, there are too
> many of them to do to do a 'table' call and have one col vertical and
> one col horizontal (I don't want any of the factors listed
> horizontally).  Before I try to write a function to do this for me, I
> was wondering if there were alternate printing styles for data that
> already exist, and if someone could direct me to them?  Inclded is a
> sample code and 2 possibilities (others welcome for consideration) of
> how I want to display some data.
>
> Thanks,
> Matt
>
> -
>
> df <- data.frame(
>    f1=rep(c("Maj I", "Maj II", "Maj III"), each=3),
>    f2=c("Minor A", "Minor A", "Minor A", "Minor A", "Minor B", "Minor
> B", "Minor B", "Minor C", "Minor C")
> )
>
> -
>
> What I want printed is something like:
>
> ---
>      f1         f2
> Maj I       Minor A
>
> Maj II      Minor A
>            Minor B
>
> Maj III     Minor B
>            Minor C
> ---
>
> or
>
> ---
>      f1         f2
> Maj I       Minor A
>
> Maj II      Minor A
> Maj II      Minor B
>
> Maj III     Minor B
> Maj III     Minor C
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
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+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] strftime vs strptime ??

2010-10-29 Thread Prof Brian Ripley

strptime() takes a character vector and makes a date-time object. 
That is input.


strftime() takes a date-time object and makes an character vector.
That is output, and it is normally called via format() or print().

Let's see what the help says (slightly edited to refer just to these 
two):


Date-time Conversion Functions to and from Character

(One is 'to' and one is 'from.)

 Functions to convert between character representations and objects
 of classes ‘"POSIXlt"’ and ‘"POSIXct"’ representing calendar dates
 and times.

 ‘strftime’ converts objects from the class ‘"POSIXlt"’ to
  character vectors.

 ‘strptime’ converts character vectors to class ‘"POSIXlt"’



On Fri, 29 Oct 2010, skan wrote:



Hello

Could anyone explain me the difference between strftime vs strptime, please
?

I've read the help but it's a little bit cionfusing for me.


cheers
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--
Brian D. Ripley,  rip...@stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
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Re: [R] Checking existance of a directory

2010-10-29 Thread Prof Brian Ripley


On Fri, 29 Oct 2010, Ron Michael wrote:

Hi all, I am wondering is there any way to check whether some 
Directory exists or not, given the parent path of that directory? 
After searching for a while I found that there is a function 
dir.create() to create some directory. However I need to know 
whether such directory already exists or not, otherwise I will 
create that.


Well, as others have also pointed out, you don't need to do that (and 
I created dir.create() deliberately so you don't).


But see ?file_test and op = "-d" if you actually need to in some other 
application.  (This is linked from ?file.exists which is linked from 
?dir.create.)




?
Thanks,


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

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Re: [R] Clustering



On Oct 29, 2010, at 5:14 AM, dpender wrote:




That's helpful but the reason I'm using clusters in evd is that I  
need to

specify a time condition to ensure independence.


I believe this is the first we heard about any particular function or  
package.




I therefore have an output


We would need to know the code that produced that output and either  
the data to which that code was applied or the structure of the  
output. (Use the str() function)



in the form Cluster[[i]][j-k]  where i is the
cluster number and j-k is the range of values above the threshold  
taking

account of the time condition.


Unless you find someone who uses that package in the manner you have,  
you will need to explain in much greater detail than you have so far.




From this I can get durations easily enough but the spacing is  
proving quite

difficult.


There are quite possibly methods using rle() or possibly something  
like rollapply() from the zoo package, but you need to provide a  
specific and richer test case.




The data is for ocean waves and therefore it may be possible that  
the wave
height drops below the threshold for a short period but should still  
be

considered part of the same event, hence the time conditon.

Hope this clarifies the problem.


It clarifies it to the extent that it show how much more you will need  
to further clarify.


--

David Winsemius, MD
West Hartford, CT

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Re: [R] Reading multiple .csv-files and assigning them to variable names

Read them into a list; much easier to handle:

myList <- lapply(filenames, read.csv)


On Fri, Oct 29, 2010 at 5:16 AM, Sarah Moens  wrote:
> Hi all,
>
> I've been trying to find a solution for the problem of reading
> multiple files and storing them in a variable that contains the names
> by which I want to call the datasets later on.
>
> For example (5 filenames):
>
> - The filenames are stored in one variable:
> filenames = paste(paste('name', '_', 1:5, sep = ''), '.csv', sep = '')
>
> - Subsequently I have a variable just containing the meaningful names
> for the dataset
> meaningfulnames = c('name1','name2'...,'name5')
>
> - I want to link each of these names to the data that is read
>
> for (i in 1:5)
> {
>     meaningfulnames[i] = read.csv(filenames[i], header = TRUE, sep = ',')
> }
>
>
> I need to read in quite a lot of datafiles. I have a code doing this
> one at a time, but since the number of datafiles I need to read will
> increase in the future, I want to make sure I have a more flexible
> solution for this.
>
> Thanks a lot for your help. I have tried to look in the help pages and
> also came across dbfread, but I can't seem to find something I can use
> or understand at this point.
>
>
> Sarah
>
> __
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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.
>



-- 
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Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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[R] ksvm problem

2010-10-29 Thread Neeti


Hi to all!

When I use the example from kernlab::ksvm this works fine.. Give me the
result…

> filter <-
> ksvm(type~.,data=spamtrain,kernel="rbfdot",kpar=list(sigma=0.05),C=5,cross=3)

But as soon as I change the type data as follows

> type_train<-spamtrain[,ncol(spamtrain)]
> filter <-
> ksvm(type_train,data=spamtrain,kernel="rbfdot",kpar=list(sigma=0.05),C=5,cross=3)
Error: evaluation nested too deeply: infinite recursion /
options(expressions=)?

This gives me error. I don’t know why?
I have also tried changing the expression values but it still doesn’t work…

> .Options$expressions
[1] 5000
> options(expressions=1)
> .Options$expressions
[1] 1
> filter <-
> ksvm(type_train,data=spamtrain,kernel="rbfdot",kpar=list(sigma=0.05),C=5,cross=3)
Error: evaluation nested too deeply: infinite recursion /
options(expressions=)?

Why am I getting this error? And how to resolve it? 

thank you ..


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[R] Reading multiple .csv-files and assigning them to variable names

2010-10-29 Thread Sarah Moens

Hi all,

I've been trying to find a solution for the problem of reading
multiple files and storing them in a variable that contains the names
by which I want to call the datasets later on.

For example (5 filenames):

- The filenames are stored in one variable:
filenames = paste(paste('name', '_', 1:5, sep = ''), '.csv', sep = '')

- Subsequently I have a variable just containing the meaningful names
for the dataset
meaningfulnames = c('name1','name2'...,'name5')

- I want to link each of these names to the data that is read

for (i in 1:5)
{
 meaningfulnames[i] = read.csv(filenames[i], header = TRUE, sep = ',')
}


I need to read in quite a lot of datafiles. I have a code doing this
one at a time, but since the number of datafiles I need to read will
increase in the future, I want to make sure I have a more flexible
solution for this.

Thanks a lot for your help. I have tried to look in the help pages and
also came across dbfread, but I can't seem to find something I can use
or understand at this point.


Sarah

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Re: [R] quiry on paste() function

2010-10-29 Thread Jan van der Laan


Perhaps, the following construct does what you need:

paste(c("a", "b", "c"), c(",", ":", ""), sep="",collapse="")

Regards,
Jan



On 29-10-2010 10:49, Ron Michael wrote:

Hi all, I want to club different objects (character type) to a single one and using 
paste() function that can be done happily. However the problem with paste function is the 
separator field is unique for all underlying objects. If I put separator as "-" 
then this will come in between all underlying objects which are to be clubbed.
  
Therefore I am wondering whether there is any mechanism to put different separators for different places. Trivially this can be done by applying paste() function repeatedly. However I feel there must be some single mechanism for doing that.
  
Here is one example, where I apply paste() function twice to incorporate 2 different separators:

paste(paste("a", "b", sep=","), "c", sep=":")

[1] "a,b:c"

Can the same thing be done using paste() function once? something like:
paste("a", "b", "c", sep=c(",", ":")) # this put 1st separator everywhere which 
is not intended
  
Thanks,



[[alternative HTML version deleted]]



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Re: [R] Printing data.frame data: alternatives to print?

2010-10-29 Thread Jan van der Laan


Matt,

Below are three (of the probably many more) possible ways of doing this:

aggregate(1:nrow(df), df, length)

ftable(1 ~ f1 + f2, data=df)

library(plyr)
ddply(df, .(f1,f2), nrow)

Regards,
Jan



On 29-10-2010 15:53, Matthew Pettis wrote:

Hi,

I have a data frame with two factors (well, more, but 2 for simple
consideration), and I want to display the different combinations of
the them that actually occur in the data.  In reality, there are too
many of them to do to do a 'table' call and have one col vertical and
one col horizontal (I don't want any of the factors listed
horizontally).  Before I try to write a function to do this for me, I
was wondering if there were alternate printing styles for data that
already exist, and if someone could direct me to them?  Inclded is a
sample code and 2 possibilities (others welcome for consideration) of
how I want to display some data.

Thanks,
Matt

-

df<- data.frame(
 f1=rep(c("Maj I", "Maj II", "Maj III"), each=3),
 f2=c("Minor A", "Minor A", "Minor A", "Minor A", "Minor B", "Minor
B", "Minor B", "Minor C", "Minor C")
)

-

What I want printed is something like:

---
   f1 f2
Maj I   Minor A

Maj II  Minor A
 Minor B

Maj III Minor B
 Minor C
---

or

---
   f1 f2
Maj I   Minor A

Maj II  Minor A
Maj II  Minor B

Maj III Minor B
Maj III Minor C

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Re: [R] Simulating data, loop

2010-10-29 Thread Ivan Calandra


Hi Sarah,

There is one thing you need to think about: how do you choose which 
values should not be removed if you have more than 20 and which should 
be if you have less than 20. In my code, I've just done it with 
sample(), which might not be what you need.


Here is what I have:
if (length(which(ynew==0))>20) {
   y <- y[c(which(ynew==1), 
sample(which(ynew==0),20-length(which(ynew==1]

} else {
   y <- y[sample(which(ynew==1),20)]
}

Does it make what you're looking for?

HTH,
Ivan


Le 10/29/2010 14:46, Sarah a écrit :

Hello,

I would like to run a script in which a loop is included. Since I'm new to
R, I cannot manage the following problem. I really hope someone could help
me out.

Data in the variable Y should be removed from the simulated data set with
probability 0.50 if the variable X has a value below zero, and with
probability 0.10 if X has a value above zero (see script).
However, the total number of removed values from Y should be 20 when X<  0,
and 4 when X>0. Whenever the total number of removed values is less than 20
(when X<0), R should remove more values from Y at random (until a total of
20 values has been removed). When R has removed more than 20 values from Y,
R should restore some removed values such that a maximum of 20 values has
been removed from variable Y.

How can I tell R to put some removed values back in the data set, or to
remove more values until a maximum number of removed values has been
reached?

y<- rnorm(40,1,3)
x<- 1+2*y1+ rnorm(40,0,5)
#Remove values in Y dependent on X:
ynew<- rep(NA,40)
for (j in 1:40)
{
if (x[j]<  0){ynew[j]<- rbinom(1,1,0.50)}
if (x[j]>  0){ynew[j]<- rbinom(1,1,0.10)}
}


--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de

**
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[R] Printing data.frame data: alternatives to print?

2010-10-29 Thread Matthew Pettis

Hi,

I have a data frame with two factors (well, more, but 2 for simple
consideration), and I want to display the different combinations of
the them that actually occur in the data.  In reality, there are too
many of them to do to do a 'table' call and have one col vertical and
one col horizontal (I don't want any of the factors listed
horizontally).  Before I try to write a function to do this for me, I
was wondering if there were alternate printing styles for data that
already exist, and if someone could direct me to them?  Inclded is a
sample code and 2 possibilities (others welcome for consideration) of
how I want to display some data.

Thanks,
Matt

-

df <- data.frame(
f1=rep(c("Maj I", "Maj II", "Maj III"), each=3),
f2=c("Minor A", "Minor A", "Minor A", "Minor A", "Minor B", "Minor
B", "Minor B", "Minor C", "Minor C")
)

-

What I want printed is something like:

---
  f1 f2
Maj I   Minor A

Maj II  Minor A
Minor B

Maj III Minor B
Minor C
---

or

---
  f1 f2
Maj I   Minor A

Maj II  Minor A
Maj II  Minor B

Maj III Minor B
Maj III Minor C

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[R] help pages do not open

I have just installed R 12.
I have Windows 7, 64-bit verison.
I currently have IE as my default browser. The internet connection is very good.

Whenever I try to run a help command (?lm, for example), I get this error:
Error in shell.exec(url) : access to
'http://127.0.0.1:20271/library/stats/html/lm.html' denied

I first got this message when Google Chrome was my default browser.
For some reason, Google Chrome stopped opening web pages. At the same
time R11 stopped showing help pages. Then, I uninstalled Google Chrome
and defined IE as my default browser. Then, I uninstalled R11 and
installed R12 instead. But I am getting the same error.

Any advice?


-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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[R] Simulating data, loop

2010-10-29 Thread Sarah


Hello,

I would like to run a script in which a loop is included. Since I'm new to
R, I cannot manage the following problem. I really hope someone could help
me out.

Data in the variable Y should be removed from the simulated data set with
probability 0.50 if the variable X has a value below zero, and with
probability 0.10 if X has a value above zero (see script).
However, the total number of removed values from Y should be 20 when X < 0,
and 4 when X>0. Whenever the total number of removed values is less than 20
(when X<0), R should remove more values from Y at random (until a total of
20 values has been removed). When R has removed more than 20 values from Y,
R should restore some removed values such that a maximum of 20 values has
been removed from variable Y. 

How can I tell R to put some removed values back in the data set, or to
remove more values until a maximum number of removed values has been
reached?

y <- rnorm(40,1,3)
x <- 1+2*y1+ rnorm(40,0,5)
#Remove values in Y dependent on X:
ynew <- rep(NA,40)
for (j in 1:40)
{
if (x[j] < 0){ynew[j] <- rbinom(1,1,0.50)}
if (x[j] > 0){ynew[j] <- rbinom(1,1,0.10)}
}
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Re: [R] relsurv package

2010-10-29 Thread Laurence Lauvier


Hello,
I have a question about relsurv package particularly rsadd function:
Rsadd(Surv(time,cens)~sex+ratetable(age=age*365.24,sex=sex,year=year),data=,=ratetable=,int=5,method=”max.lik”).
In the tutorial, it is indicated that "the age and year must be given in the
date format, i.e. in number of days since 01.01.1960". Nethertheless, in
Pohar’s article,
http://ibmi.mf.uni-lj.si/ibmi/biostat-center/predtiski/CMPB_Pohar_Stare_relsurv.pdf,
there is no indication about that. What is the true way to use this
function.
Thanks for your help,

Laurence

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[R] make many barplot into one plot

2010-10-29 Thread Sibylle Stöckli

Dear R users


I would like to group my barplot graph (see example on the R help link). The 
proposed R code, adding individual bars to the plot, looks really overwhelming. 
My specific dataset just consists of five groups and three different levels 
within each groups (the individual bars). The .txt file is read as matrix 
(horizontal: group, vertical: levels).

The R trellis barchart (function group="") is an easy function, but 
unfortunately the upper plot part look much different from other graphs. I 
would therefore prefer barplot to stansdardize my plots within the manuscript.

It would be very  helpful for me to know if anyone else has worked on the 
barplot group function.

Thanks
Sibylle


http://onertipaday.blogspot.com/2007/05/make-many-barplot-into-one-plot.html

R code from the link
## I have 4 tables like this:satu <- array(c(5,15,20,68,29,54,84,119), 
dim=c(2,4), dimnames=list(c("Negative", "Positive"), c("Black", "Brown", "Red", 
"Blond")))dua <- array(c(50,105,30,8,29,25,84,9), dim=c(2,4), 
dimnames=list(c("Negative", "Positive"), c("Black", "Brown", "Red", 
"Blond")))tiga <- array(c(9,16,26,68,12,4,84,12), dim=c(2,4), 
dimnames=list(c("Negative", "Positive"), c("Black", "Brown", "Red", 
"Blond")))empat <- array(c(25,13,50,78,19,34,84,101), dim=c(2,4), 
dimnames=list(c("Negative", "Positive"), c("Black", "Brown", "Red", "Blond")))# 
rbind() the tables togetherTAB <- rbind(satu, dua, tiga, empat)# Do the barplot 
and save the bar midpointsmp <- barplot(TAB, beside = TRUE, axisnames = FALSE)# 
Add the individual bar labelsmtext(1, at = mp, text = c("N", "P"),line = 0, cex 
= 0.5)# Get the midpoints of each sequential pair of bars# within each of the 
four groupsat <- t(sapply(seq(1, nrow(TAB), by = 2),function(x) 
colMeans(mp[c(x, x+1), ])))# Add the group labels !
 for each pairmtext(1, at = at, text = rep(c("satu", "dua", "tiga", "empat"), 
4),line = 1, cex = 0.75)# Add the color labels for each groupmtext(1, at = 
colMeans(mp), text = c("Black", "Brown", "Red", "Blond"), line = 2)
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Re: [R] check RAM usage

?memory.size

use before/after a sequence of commands to get an idea of the memory usage.

On Fri, Oct 29, 2010 at 5:21 AM, Joel  wrote:
>
> Hi
>
> Is there any way to check an certain command or procedure's RAM usage?
>
> Im after something similar to system.time(bla) that gives me the time the
> command took to preform but for RAM usage.
>
> Hope you understand what i mean.
>
> Best regards
> Joel
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-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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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] quiry on paste() function