[R] Help required to remove \\N

[Mohan L]

Dear All,

I have .csv file it looks like this :
rawdata <- read.csv(file='/home/Mohan/Rworks/tmp/VMList_User.txt',sep='\t'
, header=FALSE)

> head(rawdata,n=5)
  TenantDomain Owner  Current State
1\\N  ROOTadmin Running
2\\N  ROOTadmin Stopped
3\\N  ROOTadmin Running
4\\N  ROOTadmin Running
5\\N ROOTadmin Running
20  DEMO   ROOTadmin Stopped
21  DEMOROOT   admin Stopped
22  Demo ROOT   admin Stopped


The first column contain the \\n up to 19 row. I need to replace the
\\N value to  "Blankspace" .  Any help will  really appreciated.

Thanks for your time.

Thanks & Rg
Mohan L

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Re: [R] About 5.1 Arrays

[Stephen Liu]

Hi Richard,

> ## for an array with
> ## dim(a) == c(3,4,2)
> ## a[i,j,k] means select the element in position
> ##i + (j-1)*3 + (k-1)*3*4


My understanding;

e.g.
1)
dim(a) == c(3,4,2)

3 + (4-1)*3 + (2-1)*3*4
3+9+12=24

2)
## dim(a) == c(1,2,1)

1 + (2-1)*3 + (1-1)*3*4
1+3+0=4

3)
## dim(a) == c(2,3,1)

2 + (3-1)*3 + (1-1)*3*4
2+6+0=8

etc.


It is NOT always the product of i*j*k as I thought before.  Thanks for your 
explanation.  



What are the value of 3 and 4?  The values of position and dimension?

e.g.

> a <- sample(24)
> a
 [1] 22 18 17 10 24  1 11 13  9 19 20  8  2 21 23 16  7 14 12 15  4  5  3  6
> dim(a) <- c(3,4,2)
> a
, , 1

 [,1] [,2] [,3] [,4]   <-- positions 1, 2, 3, 4 of the second dimension ?
[1,]   22   10   11   19
[2,]   18   24   13   20
[3,]   17198
^  the first dimension ?

, , 2

 [,1] [,2] [,3] [,4]
[1,]2   16   125
[2,]   217   153
[3,]   23   1446

?  If I'm wrong pls correct me.  TIA


Now I'm going to digest Joshua's advice.


B.R.
Stephen L





From: RICHARD M. HEIBERGER 

Cc: Daniel Nordlund ; r-help@r-project.org
Sent: Sat, November 6, 2010 12:48:35 AM
Subject: Re: [R] About 5.1 Arrays


Continuing with Daniel's example, but with different data values



a <- sample(24)
a
dim(a) <- c(3,4,2)
a
as.vector(a)

## for an array with
## dim(a) == c(3,4,2)
## a[i,j,k] means select the element in position
##i + (j-1)*3 + (k-1)*3*4

index <- function(i,j,k) {
   i + (j-1)*3 + (k-1)*3*4
}

## find the vector position described by row 2, column 1, layer 2
index(2,1,2)## this is the position in the original vector
a[2,1,2]## this is the value in that position with 3D indexing
a[index(2,1,2)] ## this is the same value with 1D vector indexing
a[14]   ## this is the same value with 1D vector indexing

## find the position in row 3, column 4, layer 1
index(3,4,1)## this is the position in the original vector
a[3,4,1]## this is the value in that position with 3D indexing
a[index(3,4,1)] ## this is the same value with 1D vector indexing
a[12]   ## this is the same value with 1D vector indexing


index(1,1,1)## this is the position in the original vector
index(2,1,1)## this is the position in the original vector
index(3,1,1)## this is the position in the original vector
index(1,2,1)## this is the position in the original vector
index(2,2,1)## this is the position in the original vector
index(3,2,1)## this is the position in the original vector
index(1,3,1)## this is the position in the original vector
index(2,3,1)## this is the position in the original vector
index(3,3,1)## this is the position in the original vector
index(1,4,1)## this is the position in the original vector
index(2,4,1)## this is the position in the original vector
index(3,4,1)## this is the position in the original vector
index(1,1,2)## this is the position in the original vector
index(2,1,2)## this is the position in the original vector
index(3,1,2)## this is the position in the original vector
index(1,2,2)## this is the position in the original vector
index(2,2,2)## this is the position in the original vector
index(3,2,2)## this is the position in the original vector
index(1,3,2)## this is the position in the original vector
index(2,3,2)## this is the position in the original vector
index(3,3,2)## this is the position in the original vector
index(1,4,2)## this is the position in the original vector
index(2,4,2)## this is the position in the original vector
index(3,4,2)## this is the position in the original vector



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[R] anova(lme.model)

[Sibylle Stöckli]

Dear R users

Topic: Linear effect model fitting using the nlme package (recomended by 
Pinheiro et al. 2008 for unbalanced data set).

The R help provides much info about the controversy to use the anova(lme.model) 
function to present numerator df and F values. Additionally different p-values 
calculated by lme and anova are reported. However, I come across the same 
problem, and I would very much appreciate some R help to fit an anova function 
to get similar p-values compared to the lme function and additionally to 
provide corresponding F-values. I tried to use contrasts and to deal with the 
‚unbalanced data set’.

Thanks
Sibylle

> Kaltenborn<-read.table("Kaltenborn_YEARS.txt", na.strings="*", header=TRUE)
> 
> 
> library(nlme)

> model5c<-lme(asin(sqrt(PropMortality))~Diversity+ 
> Management+Species+Height+Height*Diversity, data=Kaltenborn, 
> random=~1|Plot/SubPlot, na.action=na.omit, weights=varPower(form=~Diversity), 
> subset=Kaltenborn$ADDspecies!=1, method="ML")

> summary(model5c)
Linear mixed-effects model fit by maximum likelihood
 Data: Kaltenborn 
  Subset: Kaltenborn$ADDspecies != 1 
AIC   BIC   logLik
  -249.3509 -205.4723 137.6755

Random effects:
 Formula: ~1 | Plot
(Intercept)
StdDev:  0.06162279

 Formula: ~1 | SubPlot %in% Plot
(Intercept)   Residual
StdDev:  0.03942785 0.05946185

Variance function:
 Structure: Power of variance covariate
 Formula: ~Diversity 
 Parameter estimates:
power 
0.7302087 
Fixed effects: asin(sqrt(PropMortality)) ~ Diversity + Management + Species +   
   Height + Height * Diversity 
  Value  Std.Error  DF   t-value p-value
(Intercept)   0.5422893 0.05923691 163  9.154585  0.
Diversity-0.0734688 0.02333159  14 -3.148896  0.0071
Managementm+  0.0217734 0.02283375  30  0.953562  0.3479
Managementu  -0.0557160 0.02286694  30 -2.436532  0.0210
SpeciesPab   -0.2058763 0.02763737 163 -7.449198  0.
SpeciesPm 0.0308005 0.02827782 163  1.089210  0.2777
SpeciesQp 0.0968051 0.02689327 163  3.599602  0.0004
Height   -0.0017579 0.00031667 163 -5.551251  0.
Diversity:Height  0.0005122 0.00014443 163  3.546270  0.0005
 Correlation: 
 (Intr) Dvrsty Mngmn+ Mngmnt SpcsPb SpcsPm SpcsQp Height
Diversity-0.867 
Managementm+ -0.173 -0.019  
Managementu  -0.206  0.005  0.499   
SpeciesPab   -0.253  0.085  0.000  0.035
SpeciesPm-0.239  0.058  0.001  0.064  0.521 
SpeciesQp-0.250  0.041 -0.001  0.032  0.502  0.506  
Height   -0.518  0.532 -0.037 -0.004  0.038  0.004  0.033   
Diversity:Height  0.492 -0.581  0.031 -0.008 -0.149 -0.099 -0.069 -0.904

Standardized Within-Group Residuals:
Min  Q1 Med  Q3 Max 
-2.99290873 -0.60522612 -0.05756772  0.62163049  2.80811502 

Number of Observations: 216
Number of Groups: 
 Plot SubPlot %in% Plot 
   1648 

> anova(model5c)
 numDF denDF   F-value p-value
(Intercept)  1   163 244.67887  <.0001
Diversity114   1.53025  0.2364
Management   230   6.01972  0.0063
Species  3   163  51.86699  <.0001
Height   1   163  30.08090  <.0001
Diversity:Height 1   163  12.57603  0.0005
>

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[R] calculate probability

[Jumlong Vongprasert]

 Dear All
I have some problem with calculate probability.
Assume I have data with normal distribution with mean = 5 sd = 2.
I want to approximate probability = 2.4.
I used pnorm(2.4, 5, 2) - pnorm(2.4, 5, 2, lower.tail = FLASE), correct 
or not.

Many Thanks
Jumlong

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Re: [R] Help required to remove \\N

[Henrique Dallazuanna]

Try this:

gsub("N", "BlankSpace", rawdata$Tenant)

On Sat, Nov 6, 2010 at 8:15 AM, Mohan L  wrote:

> Dear All,
>
> I have .csv file it looks like this :
> rawdata <- read.csv(file='/home/Mohan/Rworks/tmp/VMList_User.txt',sep='\t'
> , header=FALSE)
>
> > head(rawdata,n=5)
>  TenantDomain Owner  Current State
> 1\\N  ROOTadmin Running
> 2\\N  ROOTadmin Stopped
> 3\\N  ROOTadmin Running
> 4\\N  ROOTadmin Running
> 5\\N ROOTadmin Running
> 20  DEMO   ROOTadmin Stopped
> 21  DEMOROOT   admin Stopped
> 22  Demo ROOT   admin Stopped
>
>
> The first column contain the \\n up to 19 row. I need to replace the
> \\N value to  "Blankspace" .  Any help will  really appreciated.
>
> Thanks for your time.
>
> Thanks & Rg
> Mohan L
>
> __
> 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.
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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Re: [R] calculate probability

[Ted Harding]

On 06-Nov-10 11:16:28, Jumlong Vongprasert wrote:
> Dear All
> I have some problem with calculate probability.
> Assume I have data with normal distribution with mean = 5 sd = 2.
> I want to approximate probability = 2.4.
> I used pnorm(2.4, 5, 2) - pnorm(2.4, 5, 2, lower.tail = FLASE),
> correct or not.
> Many Thanks
> Jumlong

[A]
Not correct because "FLASE" should be "FALSE".

[B]
Not correct because
a) pnorm(2.4, 5, 2) is the probability that a value from a Normal
distribution with mean 5 will be less than 2.4, and this is less
than 1/2 (since 2.4 is less than the mean).
b) pnorm(2.4, 5, 2, lower.tail = FALSE) is the probability that
such a value will be greater than 2.4, and this is greater than 1/2
(for the same reason.
c) The difference will therefore be (< 1/2) - (> 1/2) which will
be less than 0, so cannot be a probability.

[C]
Not correct because a value sampled from a Normal distribution
(which is a continuous distribution) has probability 0 of being
exactly equal to any given value (e.g. 2.4); so I think your
question does not express what you want to know.

One possibility which could make your question realistic is
that the value "2.4" that you are interested in is a value
sampled from Normal(mean=5, sd=2) **that has been rounded to
1 decimal place** and so could have been any value between
2.35 and 2.45; in that case it makes sense to ask what is the
probability of a value in this range from Normal(mean=5, sd=2).

This would be

  pnorm(2.45, 5, 2) - pnorm(2.35, 5, 2) = 0.008569045

Of course, the degree of rounding may be different -- for
example rounding to *even* values of the first decimal place,
i.e. to values ... , 2.0, 2.2, 2.4, 2.6, 2.8, ...
in which case the event whose probability you want is that
the sampled value is between 2.3 and 2.5, whose probability is

  pnorm(2.5, 5, 2) - pnorm(2.3, 5, 2) = 0.01714178

Since the question you are really interested in cannot be
identified from what you have asked (see examples above),
you should try to make your question clear and definite!

Hoping this helps,
Ted.


E-Mail: (Ted Harding) 
Fax-to-email: +44 (0)870 094 0861
Date: 06-Nov-10   Time: 12:06:17
-- XFMail --

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[R] Extracting elements of a particular slot from S4 object

[Megh Dal]

Hi there, can anyone tell me how to extract to values of a particular slot for 
some S4 object? Let take following example:

> library(fOptions)
> val <-GBSOption(TypeFlag = "c", S = 60, X = 65, Time = 1/4, r = 0.08, b = 
> 0.08, sigma = 0.30)
> val

Title:
 Black Scholes Option Valuation 

Call:
 GBSOption(TypeFlag = "c", S = 60, X = 65, Time = 1/4, r = 0.08, 
 b = 0.08, sigma = 0.3)

Parameters:
  Value:
 TypeFlag c 
 S60
 X65
 Time 0.25  
 r0.08  
 b0.08  
 sigma0.3   

Option Price:
 2.133372 

Description:
 Sat Nov 06 19:25:39 2010 

Here I have tried with following however slapped with some error:


> val@"Option Price"
Error: no slot of name "Option Price" for this object of class "fOPTION"

What is the ideal way to do that?

Thanks,

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Re: [R] Extracting elements of a particular slot from S4 object

[Henrique Dallazuanna]

Try this: v...@price



On Sat, Nov 6, 2010 at 11:41 AM, Megh Dal  wrote:

> Hi there, can anyone tell me how to extract to values of a particular slot
> for some S4 object? Let take following example:
>
> > library(fOptions)
> > val <-GBSOption(TypeFlag = "c", S = 60, X = 65, Time = 1/4, r = 0.08, b =
> 0.08, sigma = 0.30)
> > val
>
> Title:
>  Black Scholes Option Valuation
>
> Call:
>  GBSOption(TypeFlag = "c", S = 60, X = 65, Time = 1/4, r = 0.08,
> b = 0.08, sigma = 0.3)
>
> Parameters:
>  Value:
>  TypeFlag c
>  S60
>  X65
>  Time 0.25
>  r0.08
>  b0.08
>  sigma0.3
>
> Option Price:
>  2.133372
>
> Description:
>  Sat Nov 06 19:25:39 2010
>
> Here I have tried with following however slapped with some error:
>
>
> > val@"Option Price"
> Error: no slot of name "Option Price" for this object of class "fOPTION"
>
> What is the ideal way to do that?
>
> Thanks,
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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[R] Removing NA in ggplot

[Ottar Kvindesland]

Hi list,

I just got stuck with this one:

In Data I have the sets age (numbers 1 to 99 and NA) and gender (M, F and
NA). Then getting some nice plots using

ggplot(data, aes(age[na.exclude(gender)])) +
 geom_histogram( binwidth = 3, aes(y = ..density.. ), fill = "lightblue" )
+
  facet_grid( gender ~ .)

I am trying to get a faceted graph of age distribution excluding the NA data
for gender

Unfortunately I end up with the error message:

Error in data.frame(..., check.names = FALSE) :
arguments imply differing number of rows: 206, 219

How do i Wash out NA's in this situation?


Regards

ottar

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Re: [R] Removing NA in ggplot

[Jeff Newmiller]

Create a subset of your data that excludes the NAs before you feed it to ggplot.

"Ottar Kvindesland"  wrote:

>Hi list,
>
>I just got stuck with this one:
>
>In Data I have the sets age (numbers 1 to 99 and NA) and gender (M, F
>and
>NA). Then getting some nice plots using
>
>ggplot(data, aes(age[na.exclude(gender)])) +
>geom_histogram( binwidth = 3, aes(y = ..density.. ), fill = "lightblue"
>)
>+
>  facet_grid( gender ~ .)
>
>I am trying to get a faceted graph of age distribution excluding the NA
>data
>for gender
>
>Unfortunately I end up with the error message:
>
>Error in data.frame(..., check.names = FALSE) :
>arguments imply differing number of rows: 206, 219
>
>How do i Wash out NA's in this situation?
>
>
>Regards
>
>ottar
>
>   [[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.

---
Jeff NewmillerThe .   .  Go Live...
DCN:Basics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
---
Sent from my phone. Please excuse my brevity.

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Re: [R] About 5.1 Arrays

[Stephen Liu]

Hi Joshua,

Thanks for your advice.

1)
Re your advice:-[quote]
> a3d
, , 1 <--- this is the first position of the third dimension

 [,1] [,2] [,3] [,4]  <--- positions 1, 2, 3, 4 of the second dimension
[1,]147   10
[2,]258   11
[3,]369   12
^  the first dimension

, , 2 <--- the second position of the third dimension
...
[/quote]

Where is the third dimension?


2)
Re your advice:-[quote]
so you can think that in the original vector "a":
1 maps to a[1, 1, 1] in the 3d array
2 maps to a[2, 1, 1].
3 maps to a[3, 1, 1]
4 maps to a[1, 2, 1]
12 maps to a[3, 4, 1]
20 maps to a[2, 3, 2]
24 maps to a[3, 4, 2]
[/quote]

My finding;

# 1 maps to a[1, 1, 1] in the 3d array
> a3d <- array(a, dim = c(1, 1, 1))
> a3d
, , 1

 [,1]
[1,]1

Correct

# 2 maps to a[2, 1, 1].
> a3d <- array(a, dim = c(2, 1, 1))
> a3d
, , 1

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

Correct

# 3 maps to a[3, 1, 1]
> a3d <- array(a, dim = c(3, 1, 1))
> a3d
, , 1

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

Correct

# 4 maps to a[1, 2, 1]
> a3d <- array(a, dim = c(1, 2, 1))
> a3d
, , 1

 [,1] [,2]
[1,]12

Incorrect.  It is "2"


# 12 maps to a[3, 4, 1]
> a3d <- array(a, dim = c(3, 4, 1))
> a3d
, , 1

 [,1] [,2] [,3] [,4]
[1,]147   10
[2,]258   11
[3,]369   12

Correct

# 20 maps to a[2, 3, 2]
> a3d <- array(a, dim = c(2, 3, 2))
> a3d
, , 1

 [,1] [,2] [,3]
[1,]135
[2,]246

, , 2

 [,1] [,2] [,3]
[1,]79   11
[2,]8   10   12

Incorrect.  It is "12"


#  24 maps to a[3, 4, 2]
> a3d <- array(a, dim = c(3, 4, 2))
> a3d
, , 1

 [,1] [,2] [,3] [,4]
[1,]147   10
[2,]258   11
[3,]369   12

, , 2

 [,1] [,2] [,3] [,4]
[1,]   13   16   19   22
[2,]   14   17   20   23
[3,]   15   18   21   24

Correct.

If I'm wrong, pls correct me.  Thanks


B.R.
Stephen




- Original Message 
From: Joshua Wiley 
To: Stephen Liu 
Cc: r-help@r-project.org
Sent: Sat, November 6, 2010 12:48:27 AM
Subject: Re: [R] About 5.1 Arrays

On Fri, Nov 5, 2010 at 9:17 AM, Stephen Liu  wrote:
> Hi Daniel,
>
> Thanks for your detail advice.  I completely understand your explain.
>
> But I can't resolve what does "a" stand for there?

the "a" just represents some vector.  It is the name of the object
that stores your data.  Like you might tell someone to go look in a
book to find some information.

>
> a[1,1,1] is 1 * 1 * 1 = 1
> a[2,1,1] is 2 * 1 * 1 = 2
> a[2,4,2] is 2 * 4 * 2 = 16
> a[3,4,2] is 3 * 4 * 2 = 24

That is the basic idea, but it may not be the most helpful way to
think of it because it depends on the length of the each dimension.
For example

a[1, 2, 1] is not 1 * 2 * 1 = 2
a[1, 1, 2] is not 1 * 1 * 2 = 2

in the little 3d array I show below, it would actually be

a[1, 2, 1] = 4
a[1, 1, 2] = 13

>
> ?
>
>
> B.R.
> Stephen L
>

> - Original Message 
> From: Daniel Nordlund 
> To: r-help@r-project.org
> Sent: Fri, November 5, 2010 11:54:15 PM
> Subject: Re: [R] About 5.1 Arrays
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
>> On Behalf Of Stephen Liu
>> Sent: Friday, November 05, 2010 7:57 AM
>> To: Steve Lianoglou
>> Cc: r-help@r-project.org
>> Subject: Re: [R] About 5.1 Arrays
>>
>> Hi Steve,
>>
>> > It's not clear what you're having problems understanding. By
>> > setting the "dim" attribute of your (1d) vector, you are changing
>> > itsdimenensions.
>>
>> I'm following An Introduction to R to learn R
>>
>> On
>>
>> 5.1 Arrays
>> http://cran.r-project.org/doc/manuals/R-intro.html#Vectors-and-assignment
>>
>>
>> It mentions:-
>> ...
>> For example if the dimension vector for an array, say a, is c(3,4,2) then
>> there
>> are 3 * 4 * 2 = 24 entries in a and the data vector holds them in the
>> order
>> a[1,1,1], a[2,1,1], ..., a[2,4,2], a[3,4,2].
>>
>>
>> I don't understand "on  =24 entries in a and the data vector holds
>> them in
>> the order a[1,1,1], a[2,1,1], ..., a[2,4,2], a[3,4,2]."  the order
>> a[1,1,1],
>> a[2,1,1], ..., a[2,4,2], a[3,4,2]?  What does it mean "the order a[1,1,1],
>> a[2,1,1], ..., a[2,4,2], a[3,4,2]"?

because it is actually stored as a 1 dimensional vector, it is just
telling you the order.  For example, given some vector "a" that
contains the numbers 1 through 24, you could reshape this into a three
dimensional object.  It would be stored like:

# make a vector "a" and an array (built from "a") called a3d
> a <- 1:24
> a3d <- array(a, dim = c(3, 4, 2))
> a
[1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
> a3d
, , 1 <--- this is the first position of the third dimension

 [,1] [,2] [,3] [,4]  <--- positions 1, 2, 3, 4 of the second dimension
[1,]147   10
[2,]258   11
[3,]369   12
^  the first dimension

, , 2 <--- the second position of the third dimension

 [,1] [,2] [,3] [,4]
[1,]   13   16   19   22
[2,]   14   17   20  

Re: [R] anova(lme.model)

[Bert Gunter]

Sounds to me like you should really be seeking help from your local
statistician, not this list. What you request probably cannot be done.

What is wrong with what you get from lme, whose results seem fairly
clear whether the P values are accurate or not?

Cheers,
Bert





On Sat, Nov 6, 2010 at 4:04 AM, "Sibylle Stöckli"
 wrote:
> Dear R users
>
> Topic: Linear effect model fitting using the nlme package (recomended by 
> Pinheiro et al. 2008 for unbalanced data set).
>
> The R help provides much info about the controversy to use the 
> anova(lme.model) function to present numerator df and F values. Additionally 
> different p-values calculated by lme and anova are reported. However, I come 
> across the same problem, and I would very much appreciate some R help to fit 
> an anova function to get similar p-values compared to the lme function and 
> additionally to provide corresponding F-values. I tried to use contrasts and 
> to deal with the ‚unbalanced data set’.
>
> Thanks
> Sibylle
>
>> Kaltenborn<-read.table("Kaltenborn_YEARS.txt", na.strings="*", header=TRUE)
>>
>>
>> library(nlme)
>
>> model5c<-lme(asin(sqrt(PropMortality))~Diversity+ 
>> Management+Species+Height+Height*Diversity, data=Kaltenborn, 
>> random=~1|Plot/SubPlot, na.action=na.omit, 
>> weights=varPower(form=~Diversity), subset=Kaltenborn$ADDspecies!=1, 
>> method="ML")
>
>> summary(model5c)
> Linear mixed-effects model fit by maximum likelihood
>  Data: Kaltenborn
>  Subset: Kaltenborn$ADDspecies != 1
>        AIC       BIC   logLik
>  -249.3509 -205.4723 137.6755
>
> Random effects:
>  Formula: ~1 | Plot
>        (Intercept)
> StdDev:  0.06162279
>
>  Formula: ~1 | SubPlot %in% Plot
>        (Intercept)   Residual
> StdDev:  0.03942785 0.05946185
>
> Variance function:
>  Structure: Power of variance covariate
>  Formula: ~Diversity
>  Parameter estimates:
>    power
> 0.7302087
> Fixed effects: asin(sqrt(PropMortality)) ~ Diversity + Management + Species + 
>      Height + Height * Diversity
>                      Value  Std.Error  DF   t-value p-value
> (Intercept)       0.5422893 0.05923691 163  9.154585  0.
> Diversity        -0.0734688 0.02333159  14 -3.148896  0.0071
> Managementm+      0.0217734 0.02283375  30  0.953562  0.3479
> Managementu      -0.0557160 0.02286694  30 -2.436532  0.0210
> SpeciesPab       -0.2058763 0.02763737 163 -7.449198  0.
> SpeciesPm         0.0308005 0.02827782 163  1.089210  0.2777
> SpeciesQp         0.0968051 0.02689327 163  3.599602  0.0004
> Height           -0.0017579 0.00031667 163 -5.551251  0.
> Diversity:Height  0.0005122 0.00014443 163  3.546270  0.0005
>  Correlation:
>                 (Intr) Dvrsty Mngmn+ Mngmnt SpcsPb SpcsPm SpcsQp Height
> Diversity        -0.867
> Managementm+     -0.173 -0.019
> Managementu      -0.206  0.005  0.499
> SpeciesPab       -0.253  0.085  0.000  0.035
> SpeciesPm        -0.239  0.058  0.001  0.064  0.521
> SpeciesQp        -0.250  0.041 -0.001  0.032  0.502  0.506
> Height           -0.518  0.532 -0.037 -0.004  0.038  0.004  0.033
> Diversity:Height  0.492 -0.581  0.031 -0.008 -0.149 -0.099 -0.069 -0.904
>
> Standardized Within-Group Residuals:
>        Min          Q1         Med          Q3         Max
> -2.99290873 -0.60522612 -0.05756772  0.62163049  2.80811502
>
> Number of Observations: 216
> Number of Groups:
>             Plot SubPlot %in% Plot
>               16                48
>
>> anova(model5c)
>                 numDF denDF   F-value p-value
> (Intercept)          1   163 244.67887  <.0001
> Diversity            1    14   1.53025  0.2364
> Management           2    30   6.01972  0.0063
> Species              3   163  51.86699  <.0001
> Height               1   163  30.08090  <.0001
> Diversity:Height     1   163  12.57603  0.0005
>>
>
> --
>
> __
> 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.
>



-- 
Bert Gunter
Genentech Nonclinical Biostatistics

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[R] Where to get rcom for Linux

[Stephen Liu]

Hi folks,

Debian 600 64-bit

Is rcom for Linux available? 

rcom
rcom: R COM Client Interface and internal COM Server
http://cran.r-project.org/web/packages/rcom/index.html

If YES please advise where to get it.

TIA

B.R.
Stephen L




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Re: [R] Where to get rcom for Linux

[Shige Song]

isn't COM a Windows-only technology?

Shige

On Sat, Nov 6, 2010 at 12:12 PM, Stephen Liu  wrote:
> Hi folks,
>
> Debian 600 64-bit
>
> Is rcom for Linux available?
>
> rcom
> rcom: R COM Client Interface and internal COM Server
> http://cran.r-project.org/web/packages/rcom/index.html
>
> If YES please advise where to get it.
>
> TIA
>
> B.R.
> Stephen L
>
>
>
>
> __
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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.
>

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[R] compare and replace

[Robert Ruser]

Hello R Users,
I'm wondering if there exists any elegant and simple way to do the
following: I have a data.frame X fills in numbers. I have a vector y with
numbers as well. Every value in X that is equal to any values in y should
be replaced by e.g. 1. Any idea?

Robert

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Re: [R] compare and replace

[Ben Bolker]

Robert Ruser  gmail.com> writes:

> 
> Hello R Users,
> I'm wondering if there exists any elegant and simple way to do the
> following: I have a data.frame X fills in numbers. I have a vector y with
> numbers as well. Every value in X that is equal to any values in y should
> be replaced by e.g. 1. Any idea?
> 
> Robert
> 

  If the data frames are completely filled with numbers you
should be able to operate on them as matrices (?as.matrix,
?as.data.frame).

X <- matrix(1:16,nrow=4)
Y <- matrix(rep(2:5,4),nrow=4)

(X[X %in% c(Y)] <- 1)

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Re: [R] compare and replace

[Erik Iverson]

On 11/06/2010 11:36 AM, Robert Ruser wrote:

Hello R Users,
I'm wondering if there exists any elegant and simple way to do the
following: I have a data.frame X fills in numbers. I have a vector y with
numbers as well. Every value in X that is equal to any values in y should
be replaced by e.g. 1. Any idea?


Keep in mind FAQ 7.31 when programming this...

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Re: [R] table with values as dots in increasing sizes

[Michael Friendly]

This is a "tableplot", available on R-Forge at
https://r-forge.r-project.org/projects/tableplot/

install.packages("tableplot", repos="http://R-Forge.R-project.org";)
will install, as long as you are using R 2.12.x; otherwise, you'll
have to download the source package and install from source.


-Michael


On 11/5/2010 4:45 AM, fugelpitch wrote:


I was just thinking of a way to present data and if it is possible in R.

I have a data frame that looks as follows (this is just mockup data).

df
location,"species1","species2","species3","species4","species5"
"loc1",0.44,0.28,0.37,-0.24,0.41
"loc2",0.54,0.62,0.34,0.52,0.71
"loc3",-0.33,0.75,-0.34,0.48,0.61

location is a factor while all the species are numerical vectors.

I would like to present this as a table (or something that looks like a
table) but instead of the numbers I would like to present circles (pch = 19)
that increases in size with increasing number. Is it also possible to make
it change color if the value is negative. (E.g. larger blue circles
represent larger +values while larger red circles represent larger -values)?


Jonas


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Re: [R] compare and replace

[Robert Ruser]

Thank you vary much Ben and Erik.
It's exactly what I want. Below is my a little modified example.

set.seed(12345)
X = sample(c(40:60),40,replace=TRUE)
x = matrix(X,nc=5)
y = c(40,43,55,60)
x[x %in% y] <- -1

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Re: [R] About 5.1 Arrays

[Daniel Nordlund]

> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Stephen Liu
> Sent: Saturday, November 06, 2010 7:38 AM
> To: Joshua Wiley
> Cc: r-help@r-project.org
> Subject: Re: [R] About 5.1 Arrays
> 
> Hi Joshua,
> 
> Thanks for your advice.
> 
> 1)
> Re your advice:-[quote]
> > a3d
> , , 1 <--- this is the first position of the third dimension
> 
>  [,1] [,2] [,3] [,4]  <--- positions 1, 2, 3, 4 of the second
> dimension
> [1,]147   10
> [2,]258   11
> [3,]369   12
> ^  the first dimension
> 
> , , 2 <--- the second position of the third dimension
> ...
> [/quote]
> 
> Where is the third dimension?
> 
> 
> 2)
> Re your advice:-[quote]
> so you can think that in the original vector "a":
> 1 maps to a[1, 1, 1] in the 3d array
> 2 maps to a[2, 1, 1].
> 3 maps to a[3, 1, 1]
> 4 maps to a[1, 2, 1]
> 12 maps to a[3, 4, 1]
> 20 maps to a[2, 3, 2]
> 24 maps to a[3, 4, 2]
> [/quote]
> 
> My finding;
> 
> # 1 maps to a[1, 1, 1] in the 3d array
> > a3d <- array(a, dim = c(1, 1, 1))
> > a3d
> , , 1
> 
>  [,1]
> [1,]1
> 
> Correct
> 
> # 2 maps to a[2, 1, 1].
> > a3d <- array(a, dim = c(2, 1, 1))
> > a3d
> , , 1
> 
>  [,1]
> [1,]1
> [2,]2
> 
> Correct
> 
> # 3 maps to a[3, 1, 1]
> > a3d <- array(a, dim = c(3, 1, 1))
> > a3d
> , , 1
> 
>  [,1]
> [1,]1
> [2,]2
> [3,]3
> 
> Correct
> 
> # 4 maps to a[1, 2, 1]
> > a3d <- array(a, dim = c(1, 2, 1))
> > a3d
> , , 1
> 
>  [,1] [,2]
> [1,]12
> 
> Incorrect.  It is "2"
> 
> 
> # 12 maps to a[3, 4, 1]
> > a3d <- array(a, dim = c(3, 4, 1))
> > a3d
> , , 1
> 
>  [,1] [,2] [,3] [,4]
> [1,]147   10
> [2,]258   11
> [3,]369   12
> 
> Correct
> 
> # 20 maps to a[2, 3, 2]
> > a3d <- array(a, dim = c(2, 3, 2))
> > a3d
> , , 1
> 
>  [,1] [,2] [,3]
> [1,]135
> [2,]246
> 
> , , 2
> 
>  [,1] [,2] [,3]
> [1,]79   11
> [2,]8   10   12
> 
> Incorrect.  It is "12"
> 
> 
> #  24 maps to a[3, 4, 2]
> > a3d <- array(a, dim = c(3, 4, 2))
> > a3d
> , , 1
> 
>  [,1] [,2] [,3] [,4]
> [1,]147   10
> [2,]258   11
> [3,]369   12
> 
> , , 2
> 
>  [,1] [,2] [,3] [,4]
> [1,]   13   16   19   22
> [2,]   14   17   20   23
> [3,]   15   18   21   24
> 
> Correct.
> 
> If I'm wrong, pls correct me.  Thanks
> 
> 
> B.R.
> Stephen
> 

Stephen,

I am correcting you. :-)  You are using dim() incorrectly, and not accessing 
the array correctly.  In all of your examples you should be using dim(3,4,2).  
Then you need to specify the indexes of the array element you want to look at. 
So, to use your example

> a<-1:24
> a3d <- array(a, dim = c(3,4,2))
> a3d
, , 1

 [,1] [,2] [,3] [,4]
[1,]147   10
[2,]258   11
[3,]369   12

, , 2

 [,1] [,2] [,3] [,4]
[1,]   13   16   19   22
[2,]   14   17   20   23
[3,]   15   18   21   24

> 
> # 1 maps to a[1, 1, 1] in the 3d array
> a3d[1, 1, 1]
[1] 1
> 
> # 2 maps to a[2, 1, 1].
> a3d[2, 1, 1]
[1] 2
> 
> # 3 maps to a[3, 1, 1]
> a3d[3, 1, 1]
[1] 3
> 
> # 4 maps to a[1, 2, 1]
> a3d[1, 2, 1]
[1] 4


Hope this is helpful,

Dan

Daniel Nordlund
Bothell, WA USA
 

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Re: [R] anova(lme.model)

[Mike Marchywka]

> Date: Sat, 6 Nov 2010 07:45:26 -0700
> From: gunter.ber...@gene.com
> To: sibylle.stoec...@gmx.ch
> CC: r-help@r-project.org
> Subject: Re: [R] anova(lme.model)
>
> Sounds to me like you should really be seeking help from your local
> statistician, not this list. What you request probably cannot be done.


I'm still bringing my install up to speed so I can't immediately
read the cited R stuff below but it sounds like the OP
mentions a controversy documented in the R packages. Is there
a list for discussing these topics? Offhand that seems legitimate
for a user help list unless you want people to believe that 
" it came out of a computer so it must be right, whatever a P value
is." 


>
> What is wrong with what you get from lme, whose results seem fairly
> clear whether the P values are accurate or not?
>
> Cheers,
> Bert
>
>
>
>
>
> On Sat, Nov 6, 2010 at 4:04 AM, "Sibylle Stöckli"
>  wrote:
> > Dear R users
> >
> > Topic: Linear effect model fitting using the nlme package (recomended by 
> > Pinheiro et al. 2008 for unbalanced data set).
> >
> > The R help provides much info about the controversy to use the 
> > anova(lme.model) function to present numerator df and F values. 
> > Additionally different p-values calculated by lme and anova are reported. 
> > However, I come across the same problem, and I would very much appreciate 
> > some R help to fit an anova function to get similar p-values compared to 
> > the lme function and additionally to provide corresponding F-values. I 
> > tried to use contrasts and to deal with the ‚unbalanced data set’.
> >
> > Thanks
> > Sibylle
> >
> >> Kaltenborn<-read.table("Kaltenborn_YEARS.txt", na.strings="*", header=TRUE)
> >>
> >>
> >> library(nlme)
> >
> >> model5c<-lme(asin(sqrt(PropMortality))~Diversity+ 
> >> Management+Species+Height+Height*Diversity, data=Kaltenborn, 
> >> random=~1|Plot/SubPlot, na.action=na.omit, 
> >> weights=varPower(form=~Diversity), subset=Kaltenborn$ADDspecies!=1, 
> >> method="ML")
> >
> >> summary(model5c)
> > Linear mixed-effects model fit by maximum likelihood
> >  Data: Kaltenborn
> >  Subset: Kaltenborn$ADDspecies != 1
> >AIC   BIC   logLik
> >  -249.3509 -205.4723 137.6755
> >
> > Random effects:
> >  Formula: ~1 | Plot
> >(Intercept)
> > StdDev:  0.06162279
> >
> >  Formula: ~1 | SubPlot %in% Plot
> >(Intercept)   Residual
> > StdDev:  0.03942785 0.05946185
> >
> > Variance function:
> >  Structure: Power of variance covariate
> >  Formula: ~Diversity
> >  Parameter estimates:
> >power
> > 0.7302087
> > Fixed effects: asin(sqrt(PropMortality)) ~ Diversity + Management + Species 
> > +  Height + Height * Diversity
> >  Value  Std.Error  DF   t-value p-value
> > (Intercept)   0.5422893 0.05923691 163  9.154585  0.
> > Diversity-0.0734688 0.02333159  14 -3.148896  0.0071
> > Managementm+  0.0217734 0.02283375  30  0.953562  0.3479
> > Managementu  -0.0557160 0.02286694  30 -2.436532  0.0210
> > SpeciesPab   -0.2058763 0.02763737 163 -7.449198  0.
> > SpeciesPm 0.0308005 0.02827782 163  1.089210  0.2777
> > SpeciesQp 0.0968051 0.02689327 163  3.599602  0.0004
> > Height   -0.0017579 0.00031667 163 -5.551251  0.
> > Diversity:Height  0.0005122 0.00014443 163  3.546270  0.0005
> >  Correlation:
> > (Intr) Dvrsty Mngmn+ Mngmnt SpcsPb SpcsPm SpcsQp Height
> > Diversity-0.867
> > Managementm+ -0.173 -0.019
> > Managementu  -0.206  0.005  0.499
> > SpeciesPab   -0.253  0.085  0.000  0.035
> > SpeciesPm-0.239  0.058  0.001  0.064  0.521
> > SpeciesQp-0.250  0.041 -0.001  0.032  0.502  0.506
> > Height   -0.518  0.532 -0.037 -0.004  0.038  0.004  0.033
> > Diversity:Height  0.492 -0.581  0.031 -0.008 -0.149 -0.099 -0.069 -0.904
> >
> > Standardized Within-Group Residuals:
> >Min  Q1 Med  Q3 Max
> > -2.99290873 -0.60522612 -0.05756772  0.62163049  2.80811502
> >
> > Number of Observations: 216
> > Number of Groups:
> > Plot SubPlot %in% Plot
> >   1648
> >
> >> anova(model5c)
> > numDF denDF   F-value p-value
> > (Intercept)  1   163 244.67887  <.0001
> > Diversity114   1.53025  0.2364
> > Management   230   6.01972  0.0063
> > Species  3   163  51.86699  <.0001
> > Height   1   163  30.08090  <.0001
> > Diversity:Height 1   163  12.57603  0.0005
> >>
> >
>
> --
> Bert Gunter
> Genentech Nonclinical Biostatistics
>






Mike Marchywka | V.P. Technology

415-264-8477
marchy...@phluant.com

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[R] 3-way interaction simple slopes

[Michael Wood]

Can anyone show me how to test for significant simple slopes of a 3-way
interaction, with covariates.

my equation
tmod<-(glm(PCL~ rank.f + gender.f + MONTHS + CEXPOSE.M + bf.m +
MONTHS*CEXPOSE.M*bf.m,
data=mhatv, family=gaussian ,na.action=na.omit))

Thank you
Mike

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[R] Using changing names in loop in R

[Tuatara]

Hello everybody, 

I have usually solved this problem by repeating lines of codes instead of a
loop, but it's such a waste of time, I thought I should really learn how to
do it with loops:

What I want to do:

Say, I have several data files that differ only in a number, e.g. data
points (or vector, or matrix...) Data_1, Data_2, Data_3,... and I want to
manipulate them 

e.g. a simple sum of several data points

>data <- c(NA,n)
>for (i in 1:n){
>data[i] <- Data_i + Data_[i-1]
>  } 

I know that the above code doesn't work, and I don't want to combine the
files into one vector to solve the problem etc. - I would just like to know
who make sure R recognizes the extension "_i". I have the same problem for
say, reading in datafiles that only differ by one digit in the extension,
and I want to (instead of repeating code) combine the process in a loop.

I hope I made myself clear to what my problem is.

Thanks for your help,

//F
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Re: [R] Using changing names in loop in R

[Daniel Nordlund]

> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Tuatara
> Sent: Saturday, November 06, 2010 9:22 AM
> To: r-help@r-project.org
> Subject: [R] Using changing names in loop in R
> 
> 
> Hello everybody,
> 
> I have usually solved this problem by repeating lines of codes instead of
> a
> loop, but it's such a waste of time, I thought I should really learn how
> to
> do it with loops:
> 
> What I want to do:
> 
> Say, I have several data files that differ only in a number, e.g. data
> points (or vector, or matrix...) Data_1, Data_2, Data_3,... and I want to
> manipulate them
> 
> e.g. a simple sum of several data points
> 
> >data <- c(NA,n)
> >for (i in 1:n){
> >data[i] <- Data_i + Data_[i-1]
> >  }
> 
> I know that the above code doesn't work, and I don't want to combine the
> files into one vector to solve the problem etc. - I would just like to
> know
> who make sure R recognizes the extension "_i". I have the same problem for
> say, reading in datafiles that only differ by one digit in the extension,
> and I want to (instead of repeating code) combine the process in a loop.
> 
> I hope I made myself clear to what my problem is.
> 
> Thanks for your help,
> 

This is one of those cases where a commented, self-contained, reproducible 
example would be very helpful in helping you.  You mention you have several 
"data files", but I see no reference to data files in your code.  Did you mean 
data frames?  What is Data_i?  A data frame or something else?

You said you normally do this by repeating lines of code.  Can you show us a 
simple example?  Someone should be able to show you how to optimize it.

Dan

Daniel Nordlund
Bothell, WA USA
 

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[R] Prettier axis labels when using log scales in Lattice

[Marc Paterno]

Hello,

I am trying to alter the way in which lattice functions (specifically xyplot) 
print the axis labels when one uses the 'scales' parameter.
I can obtain the effect I want by using
  scales=list(y=list(log=10, labels=expression(yvalues)))
where yvalues are the values that would have been printed as the y-axis labels 
if the "labels" argument had not been present. To help clarify what I am 
looking for, compare the first of the following plots with the second:

data(iris)
xyplot(Sepal.Length~Sepal.Width, iris, scales=list(y=list(log=10)))

xyplot(Sepal.Length~Sepal.Width, iris, scales=list(y=list(log=10, 
labels=expression(10^0.65,10^0.7,10^0.75,10^0.8,10^0.85,10^0.85,10^0.9

The second is the effect I am trying to achieve. Is there a way to do this 
without explicitly entering the expressions to be printed on the y-axis?

thanks,
Marc Paterno

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Re: [R] Prettier axis labels when using log scales in Lattice

[Gabor Grothendieck]

On Sat, Nov 6, 2010 at 3:02 PM, Marc Paterno  wrote:
> Hello,
>
> I am trying to alter the way in which lattice functions (specifically xyplot) 
> print the axis labels when one uses the 'scales' parameter.
> I can obtain the effect I want by using
>  scales=list(y=list(log=10, labels=expression(yvalues)))
> where yvalues are the values that would have been printed as the y-axis 
> labels if the "labels" argument had not been present. To help clarify what I 
> am looking for, compare the first of the following plots with the second:
>
> data(iris)
> xyplot(Sepal.Length~Sepal.Width, iris, scales=list(y=list(log=10)))
>
> xyplot(Sepal.Length~Sepal.Width, iris, scales=list(y=list(log=10, 
> labels=expression(10^0.65,10^0.7,10^0.75,10^0.8,10^0.85,10^0.85,10^0.9
>
> The second is the effect I am trying to achieve. Is there a way to do this 
> without explicitly entering the expressions to be printed on the y-axis?
>

Try:

library(latticeExtra)
xyplot(Sepal.Length~Sepal.Width, iris, scales=list(y=list(log=10)),
   yscale.components = yscale.components.logpower)


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

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Re: [R] Using changing names in loop in R

[Liviu Andronic]

On Sat, Nov 6, 2010 at 5:22 PM, Tuatara  wrote:
>
> Hello everybody,
>
> I have usually solved this problem by repeating lines of codes instead of a
> loop, but it's such a waste of time, I thought I should really learn how to
> do it with loops:
>

Would the following construct help?
> for(i in 1:10) assign(paste('x', i, sep=''), c(i:10))
> ls()
 [1] "i""pkg"  "tbbt" "x1"   "x10"  "x2"   "x3"   "x4"   "x5"   "x6"
[11] "x7"   "x8"   "x9"
> for(i in 1:10) print(get(paste('x', i, sep='')))
 [1]  1  2  3  4  5  6  7  8  9 10
[1]  2  3  4  5  6  7  8  9 10
[1]  3  4  5  6  7  8  9 10
[1]  4  5  6  7  8  9 10
[1]  5  6  7  8  9 10
[1]  6  7  8  9 10
[1]  7  8  9 10
[1]  8  9 10
[1]  9 10
[1] 10

Read ?assign, ?get, but also this fortune:
> fortune('assign')

The only people who should use the assign function are those who fully
understand why you should never use the assign function.
   -- Greg Snow
  R-help (July 2009)

I haven't yet figured out why I should heed this. Regards
Liviu


> What I want to do:
>
> Say, I have several data files that differ only in a number, e.g. data
> points (or vector, or matrix...) Data_1, Data_2, Data_3,... and I want to
> manipulate them
>
> e.g. a simple sum of several data points
>
>>data <- c(NA,n)
>>for (i in 1:n){
>>data[i] <- Data_i + Data_[i-1]
>>                  }
>
> I know that the above code doesn't work, and I don't want to combine the
> files into one vector to solve the problem etc. - I would just like to know
> who make sure R recognizes the extension "_i". I have the same problem for
> say, reading in datafiles that only differ by one digit in the extension,
> and I want to (instead of repeating code) combine the process in a loop.
>
> I hope I made myself clear to what my problem is.
>
> Thanks for your help,
>
> //F
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/Using-changing-names-in-loop-in-R-tp3030132p3030132.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.
>



-- 
Do you know how to read?
http://www.alienetworks.com/srtest.cfm
http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader
Do you know how to write?
http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail

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Re: [R] anova(lme.model)

[Ben Bolker]

Mike Marchywka  hotmail.com> writes:

> 
> 
> > Date: Sat, 6 Nov 2010 07:45:26 -0700
> > From: gunter.berton  gene.com
> > To: sibylle.stoeckli  gmx.ch
> > CC: r-help  r-project.org
> > Subject: Re: [R] anova(lme.model)
> >
> > Sounds to me like you should really be seeking help from your local
> > statistician, not this list. What you request probably cannot be done.
> 
> I'm still bringing my install up to speed so I can't immediately
> read the cited R stuff below but it sounds like the OP
> mentions a controversy documented in the R packages. Is there
> a list for discussing these topics? Offhand that seems legitimate
> for a user help list unless you want people to believe that 
> " it came out of a computer so it must be right, whatever a P value
> is." 

  It's not documented within the packages themselves, 
it's documented in the r-sig-mixed-models archives, and at 
,
and in the R FAQ

and here


  cheers
   Ben Bolker

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[R] Hashing and environments

[Levy, Roger]

Hi,

I'm trying to write a general-purpose "lexicon" class and associated methods 
for storing and accessing information about large numbers of specific words 
(e.g., their frequencies in different genres).  Crucial to making such a class 
practically useful is to get hashing working correctly so that information 
about specific words can be accessed quickly.  But I've never really understood 
very well how hashing works, so I'm having trouble.

Here is an example of what I've done so far:

***

setClass("Lexicon",representation(e="environment"))
setMethod("initialize","Lexicon",function(.Object,wfreqs) {
.obj...@e <- new.env(hash=T,parent=emptyenv())
assign("wfreqs",wfreqs,envir=.obj...@e)
return(.Object)
})

## function to access word frequencies
wfreq <- function(lexicon,word) {
return(get("wfreqs",envir=lexi...@e)[word])
}

## example of use
my.lexicon <- new("Lexicon",wfreqs=c("the"=2,"person"=1))
wfreq(my.lexicon,"the")

***

However, testing indicates that the way I have set this up does not achieve the 
intended benefits of having the environment hashed:

***

sample.wfreqs <- trunc(runif(1e5,max=100))
names(sample.wfreqs) <- as.character(1:length(sample.wfreqs))
lex <- new("Lexicon",wfreqs=sample.wfreqs)
words.to.lookup <- trunc(runif(100,min=1,max=1e5))
## look up the words directly from the sample.wfreqs vector
system.time({
for(i in words.to.lookup)
sample.wfreqs[as.character(i)]
},gcFirst=TRUE)
## look up the words through the wfreq() function; time approx the same
system.time({
for(i in words.to.lookup)
wfreq(lex,as.character(i))
},gcFirst=TRUE)

***

I'm guessing that the problem is that the indexing of the wfreqs vector in my 
wfreq() function is not happening inside the actual lexicon's environment.  
However, I have not been able to figure out the proper call to get the lookup 
to happen inside the lexicon's environment.  I've tried

wfreq1 <- function(lexicon,word) {
return(eval(wfreqs[word],envir=lexi...@e))
}

which I'd thought should work, but this gives me an error:

> wfreq1(my.lexicon,'the')
Error in eval(wfreqs[word], envir = lexi...@e) : 
  object 'wfreqs' not found

Any advice would be much appreciated!

Best & many thanks in advance,

Roger

--

Roger Levy  Email: rl...@ucsd.edu
Assistant Professor Phone: 858-534-7219
Department of Linguistics   Fax:   858-534-4789
UC San DiegoWeb:   http://ling.ucsd.edu/~rlevy

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[R] SMATR common slopes test

[Eugenio Larios]

Hi All,

I am confused with SMATR's test for common slope. My null hypothesis here is
that all slopes are parallel (common slopes?), right?
So if I get a p value < 0.05 means that we can have confidence to reject it?
That slopes are different?
Or the other way around? it means that we have statistical confidence that
the slopes are parallel?
thanks
-- 
Eugenio Larios
PhD Student
University of Arizona.
Ecology & Evolutionary Biology.
(520) 481-2263
elari...@email.arizona.edu

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[R] saddle points in optim

[Jonathan Phillips]

Hi,
I've been trying to use optim to minimise least squares for a
function, and then get a guess at the error using the hessian matrix
(calculated from numDeriv::hessian, which I read in some other r-help
post was meant to be more accurate than the hessian given in optim).

To get the standard error estimates, I'm calculating
sqrt(diag(solve(x))), hope that's correct.

I've found that using numDeriv's hessian gets me some NaNs for errors,
whereas the one from optim gets me numbers for all parameters.  If I
look for eigenvalues for numDeriv::hessian, I get two negative numbers
(and six positive - I'm fitting to eight parameters), so does this
mean that optim hasn't converged correctly, and has hit a saddle
point?  If so, is there any way I could assist it to find the minimum?

Thanks,
Jon Phillips

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[R] I want to call some R function in C code

[刘力平]

Hi, all:

I want the user to give a function g as the parameter of our function f. In
function f, we use user's function g to compute something.  Since our
function f is implemented in C++, how do I further pass this function to C++
code?

Thank you very much!

best
Liping LIU

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Re: [R] likelyhood maximization problem with polr

[tjb]

blackscorpio wrote:
> 
> Dear community,
> 
> I am currently trying to fit an ordinal logistic regression model with the
> polr function. I often get the same error message :
> 
> "attempt to find suitable starting values failed", for example with :
> ...
> Does anyone have a clue ?
> 

Yes. The code that generates a starting value for the optimization is
flakey. lrm in the Design library appear to be more robust.
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[R] How to generate multivariate uniform distribution random numbers?

[michael]

I wish to generate 100 by 1 vector of x1 and x2 both are uniform distributed
with covariance matrix \Sigma.

Thanks,

Michael

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Re: [R] bugs and misfeatures in polr(MASS).... fixed!

[tjb]

The start value generation code in polr is also known to fail quite
frequently. For example, against the Iris data as recently posted to this
list by blackscorpio (  Sep 6, 2010).

> polr(Species~Sepal_Length+Sepal_Width+Petal_Length+Petal_Width,data=iris)
Error in polr(Species ~ Sepal_Length + Sepal_Width + Petal_Length +
Petal_Width,  : 
  attempt to find suitable starting values failed
In addition: Warning messages:
1: In glm.fit(X, y1, wt, family = binomial(), offset = offset) :
  algorithm did not converge
2: In glm.fit(X, y1, wt, family = binomial(), offset = offset) :
  fitted probabilities numerically 0 or 1 occurred

I suggest that simply setting the coefficients beta to zero and the
cutpoints zeta to sensible values will always produce a feasible starting
point for non-pathological data. Here is my code that does this:

if(missing(start)) {
  # try something that should always work -tjb
  u <- as.integer(table(y))
  u <- (cumsum(u)/sum(u))[1:q]
  zetas <-
 switch(method,
"logistic"= qlogis(u),
"probit"=   qnorm(u),
"cauchit"=  qcauchy(u),
"cloglog"=  -log(-log(u)) )
  s0 <- c(rep(0,pc),zetas[1],log(diff(zetas)))

Using this start code the problem is not manifested. 

> source('fixed-polr.R')
> polr(Species~Sepal_Length+Sepal_Width+Petal_Length+Petal_Width,data=iris)
Call:
polr(formula = Species ~ Sepal_Length + Sepal_Width + Petal_Length + 
Petal_Width, data = iris)

Coefficients:
Sepal_Length  Sepal_Width Petal_Length  Petal_Width 
   -2.466724-6.671515 9.43168918.270058 

Intercepts:
   setosa|versicolor versicolor|virginica 
4.08018942.639320 

Residual Deviance: 11.89857 
AIC: 23.89857 

My change would also likely fix the problem reported by Kevin Coombes on May
6, 2010.
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Re: [R] bugs and misfeatures in polr(MASS).... fixed!

[tjb]

Note that that the enhancements in my original post solve the unresolved
problem of Chaehyung Ahn (22 Mar 2005) whose data I reproduce:

y,x,lx
0,3.2e-02,-1.49485
0,3.2e-02,-1.49485
0,1.0e-01,-1.0
0,1.0e-01,-1.0
0,3.2e-01,-0.49485
0,3.2e-01,-0.49485
1,1.0e+00,0.0
0,1.0e+00,0.0
1,3.2e+00,0.50515
1,3.2e+00,0.50515
0,1.0e+01,1.0
1,1.0e+01,1.0
1,3.2e+01,1.50515
2,3.2e+01,1.50515
2,1.0e+02,2.0
1,1.0e+02,2.0
2,3.2e+02,2.50515
1,3.2e+02,2.50515
2,1.0e+03,3.0
2,1.0e+03,3.0

Using the MASS version we get

> ahn$y<-as.factor(ahn$y)
> summary(polr(y~lx,data=ahn))

Re-fitting to get Hessian

Error in optim(s0, fmin, gmin, method = "BFGS", hessian = Hess, ...) : 
  initial value in 'vmmin' is not finite

Whereas,

> source('fixed-polr.R')
> summary(polr(y~lx,data=ahn))

Re-fitting to get Hessian

Call:
polr(formula = y ~ lx, data = ahn)

Coefficients:
   Value Std. Error t value
lx 2.421 0.8146   2.971

Intercepts:
Value  Std. Error t value
0|1 0.5865 0.8118 0.7224 
1|2 4.8966 1.7422 2.8106 

Residual Deviance: 20.43286 
AIC: 26.43286 

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Re: [R] Hashing and environments

[William Dunlap]

I would make make an environemnt called wfreqsEnv
whose entry names are your words and whose entry
values are the information about the words.  I find
it convenient to use [[ to make it appear to be
a list (instead of using exists(), assign(), and get()).
E.g., the following enters the 100,000 words from a
list of 17,576 and records their id numbers and the
number of times each is found in the sample.

> wfreqsEnv <- new.env(hash=TRUE, parent = emptyenv())
> words <- do.call("paste", c(list(sep=""), expand.grid(LETTERS,
letters, letters)))
# length(words) == 17576
> set.seed(1)
> samp <- sample(seq_along(words), size=10, replace=TRUE)
> system.time(for(i in samp) {
+word <- words[i]
+if (is.null(wfreqsEnv[[word]])) { # new entry
+wfreqsEnv[[word]] <- list(Count=1, EntryNo=i)
+} else { # update existing entry
+wfreqsEnv[[word]]$Count <- wfreqsEnv[[word]]$Count + 1
+}
+})
   user  system elapsed 
   2.280.002.14 
(The time, in seconds, is from an ancient Windows laptop, c. 2002.)

Here is a small check that we are getting what we expect:
> words[14736]
[1] "Tuv"
> wfreqsEnv[["Tuv"]]
$Count
[1] 8

$EntryNo
[1] 14736

> sum(samp==14736)
[1] 8

If we do this with a non-hashed environment we get the same
answers but the elapsed time is now 34.81 seconds instead of
2.14.  If you make wfreqEnv be a list instead of an environment
then that time is 74.12 seconds (and the answers are the same).

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

> -Original Message-
> From: r-help-boun...@r-project.org 
> [mailto:r-help-boun...@r-project.org] On Behalf Of Levy, Roger
> Sent: Saturday, November 06, 2010 1:39 PM
> To: r-help@r-project.org
> Subject: [R] Hashing and environments
> 
> Hi,
> 
> I'm trying to write a general-purpose "lexicon" class and 
> associated methods for storing and accessing information 
> about large numbers of specific words (e.g., their 
> frequencies in different genres).  Crucial to making such a 
> class practically useful is to get hashing working correctly 
> so that information about specific words can be accessed 
> quickly.  But I've never really understood very well how 
> hashing works, so I'm having trouble.
> 
> Here is an example of what I've done so far:
> 
> ***
> 
> setClass("Lexicon",representation(e="environment"))
> setMethod("initialize","Lexicon",function(.Object,wfreqs) {
>   .obj...@e <- new.env(hash=T,parent=emptyenv())
>   assign("wfreqs",wfreqs,envir=.obj...@e)
>   return(.Object)
>   })
> 
> ## function to access word frequencies
> wfreq <- function(lexicon,word) {
>   return(get("wfreqs",envir=lexi...@e)[word])
> }
> 
> ## example of use
> my.lexicon <- new("Lexicon",wfreqs=c("the"=2,"person"=1))
> wfreq(my.lexicon,"the")
> 
> ***
> 
> However, testing indicates that the way I have set this up 
> does not achieve the intended benefits of having the 
> environment hashed:
> 
> ***
> 
> sample.wfreqs <- trunc(runif(1e5,max=100))
> names(sample.wfreqs) <- as.character(1:length(sample.wfreqs))
> lex <- new("Lexicon",wfreqs=sample.wfreqs)
> words.to.lookup <- trunc(runif(100,min=1,max=1e5))
> ## look up the words directly from the sample.wfreqs vector
> system.time({
>   for(i in words.to.lookup)
>   sample.wfreqs[as.character(i)]
>   },gcFirst=TRUE)
> ## look up the words through the wfreq() function; time 
> approx the same
> system.time({
>   for(i in words.to.lookup)
>   wfreq(lex,as.character(i))
>   },gcFirst=TRUE)
> 
> ***
> 
> I'm guessing that the problem is that the indexing of the 
> wfreqs vector in my wfreq() function is not happening inside 
> the actual lexicon's environment.  However, I have not been 
> able to figure out the proper call to get the lookup to 
> happen inside the lexicon's environment.  I've tried
> 
> wfreq1 <- function(lexicon,word) {
>   return(eval(wfreqs[word],envir=lexi...@e))
> }
> 
> which I'd thought should work, but this gives me an error:
> 
> > wfreq1(my.lexicon,'the')
> Error in eval(wfreqs[word], envir = lexi...@e) : 
>   object 'wfreqs' not found
> 
> Any advice would be much appreciated!
> 
> Best & many thanks in advance,
> 
> Roger
> 
> --
> 
> Roger Levy  Email: rl...@ucsd.edu
> Assistant Professor Phone: 858-534-7219
> Department of Linguistics   Fax:   858-534-4789
> UC San DiegoWeb:   http://ling.ucsd.edu/~rlevy
> 
> __
> 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] Using changing names in loop in R

[Joshua Wiley]

Hi,

If you have data that is similar enough to warrant only changing the
extension (i.e., 1, 2, etc.) and that you (at least at times) wish to
perform operations on together, it is time to start thinking of a more
flexible framework.  Fortunately, such a framework already exists in
lists.  Lists let you keep diverse data structures (i.e., you do not
have to combine everything into a simple vector).  Lists make it
tremendously easy to do many tasks (including the two you mentioned).
For example, suppose I want to read in 10 files and then do some
manipulations:

## initialize an empty list
dat <- vector(mode = "list", length = 10)

for(i in 1:10) {
  dat[[i]] <- read.table(paste(myfilename, i, sep = ''), header = TRUE, etc.)
}

# presumably these are data frames at this point

Now everything is nicely stored, dat[[1]] is even intuitively similar
to Data_1, Data_2, etc.  At this point, suppose I want to create some
new stuff:

dat[[11]] <- matrix(1, ncol = 1, nrow = 5) # add in a matrix

for(i in 1:n) {
  dat[[i]] <- dat[[i]] + dat[[i - 1]]
}

dat[[12]] <- 5 # just a little vector

Another great advantage of lists is that it is possible to name their
elements.  This can make things more meaningful or aid the memory.
However, even when an element is named, it can still be accessed by
its index

# name the first element 'price'
names(dat)[[1]] <- "price"

now I could access it as any of these:

dat$price
dat[["price"]]
dat[[1]]

If that weren't enough, you can easily use functions on every element
of a list with constructs such as lapply(), no for loop required!

lapply(X = dat, FUN = mean, na.rm = TRUE)

It is possible to not use lists and still do what you are after, but
frankly it is messier, more prone to error, and less effective in many
cases.  It is generally a very nice feature of the assignment operator
that it is aware of its environment and does not go assigning or
overwriting things where you do not expect.  You're left to the wolves
and your own wits with assign().

HTH,

Josh

On Sat, Nov 6, 2010 at 9:22 AM, Tuatara  wrote:
>
> Hello everybody,
>
> I have usually solved this problem by repeating lines of codes instead of a
> loop, but it's such a waste of time, I thought I should really learn how to
> do it with loops:
>
> What I want to do:
>
> Say, I have several data files that differ only in a number, e.g. data
> points (or vector, or matrix...) Data_1, Data_2, Data_3,... and I want to
> manipulate them
>
> e.g. a simple sum of several data points
>
>>data <- c(NA,n)
>>for (i in 1:n){
>>data[i] <- Data_i + Data_[i-1]
>>                  }
>
> I know that the above code doesn't work, and I don't want to combine the
> files into one vector to solve the problem etc. - I would just like to know
> who make sure R recognizes the extension "_i". I have the same problem for
> say, reading in datafiles that only differ by one digit in the extension,
> and I want to (instead of repeating code) combine the process in a loop.
>
> I hope I made myself clear to what my problem is.
>
> Thanks for your help,
>
> //F
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/Using-changing-names-in-loop-in-R-tp3030132p3030132.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.
>



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

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Re: [R] SMATR common slopes test

[Kevin Middleton]

Eugenio -

> I am confused with SMATR's test for common slope. My null hypothesis here is
> that all slopes are parallel (common slopes?), right?
> So if I get a p value < 0.05 means that we can have confidence to reject it?
> That slopes are different?
> Or the other way around? it means that we have statistical confidence that
> the slopes are parallel?

Try this:

set.seed(5)
n <- 20
x <- rnorm(n)

y1 <- 2 * x + rnorm(n)
y2 <- 2 * x + rnorm(n)
y3 <- 4 * x + rnorm(n)

# Slopes approximately equal
slope.com(x = c(x, x), y = c(y1, y2), groups = rep(c(1,2), each = n))

#$p
#[1] 0.4498037

# Slopes of 2 and 4
slope.com(x = c(x, x), y = c(y1, y3), groups = rep(c(1,2), each = n))

#$p
#[1] 0.0003850332


Cheers,
Kevin

-
Kevin M. Middleton
Department of Biology
California State University San Bernardino

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Re: [R] How to generate multivariate uniform distribution random

[Ted Harding]

On 06-Nov-10 21:54:41, michael wrote:
> I wish to generate 100 by 1 vector of x1 and x2 both are uniform
> distributed with covariance matrix \Sigma.
> 
> Thanks,
> Michael

First, some comments.

1. I don't think you mean a "100 by 1 vector of x1 and x2" since
   you have two variables. "100 by 2" perhaps?

2. Given the range over which x1 is to be uniformly distributed
   (say of length A) and the range for x2 (say of length B),
   then by virtue of the uniform distribution the variance of x1
   will be (A^2)/12 and the variance of x2 will be (B^2)/12,
   so you don't have a choice about these. So your only "free
   parameter" is the covariance (or the correlation) between
   x1 and x2.

That said, it is a curious little problem. Here is one simple
solution (though it may have properties that you do not want).
Using X and Y for x1 and x2 (for simplicity), and using ranges
(-1/2,1/2) for the ranges of X and Y (you could rescale later):

1. Generate X uniformly distributed on (-1/2,1/2).
2. With probability p (0 < p < 1) let Y=X.
   Otherwise, let Y be independently uniform on (-1/2,1/2).

This can be implemented by the following code:

  n  <- 100
  p  <- 0.6# (say)
  X  <- runif(n)-1/2
  Z  <- runif(n)-1/2
  U  <- rbinom(n,1,p)  # =1 where Y=X, = 0 when Y independent of X
  Y  <- X*U + Z*(1-U)
  XY <- cbind(X,Y) # your n by 2 matrix of bivariate (X,Y)

Now the marginal distributions of X and Y are both uniform.
var(X) = 1/12 and var(Y) = 1/12. Since the means are 0,
cov(X,Y) = Exp(X*Y) = p*Exp(X^2) = p/12
cor(X,Y) = cov(X,Y)/sqrt(var(X)*var(Y)) = (p/12)/(1/12) = p.

So all you need to do to get a desired correlation rho between
X and Y is to set p = rho.

The one thing you may not like about this is the diagonal line
you will get for the cases where X=Y:

  plot(X,Y)

Test:

  n <- 10
  p  <- 0.6 # (say)
  X  <- runif(n)-1/2
  Z  <- runif(n)-1/2
  U  <- rbinom(n,1,p)
  Y  <- X*U + Z*(1-U)
  var(X)
  # [1] 0.08340525  # theory: var = 1/12 = 0.0833
>   var(Y)
[1] 0.08318914  # theory: var = 1/12 = 0.0833
>   cov(X,Y)
[1] 0.04953733  # theory: cov = p/12 = 0.6/12 = 0.05 
>   cor(X,Y)
[1] 0.5947063   # theory: cor = p= 0.6

It would be interesting to see a solution which did not involve
having cases with X=Y!

Hoping this helps,
Ted.


E-Mail: (Ted Harding) 
Fax-to-email: +44 (0)870 094 0861
Date: 06-Nov-10   Time: 23:07:26
-- XFMail --

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

Re: [R] Hashing and environments

[Kjetil Halvorsen]

some of this can be automated using the CRAN package
hash.

Kjetil

On Sat, Nov 6, 2010 at 10:43 PM, William Dunlap  wrote:
> I would make make an environemnt called wfreqsEnv
> whose entry names are your words and whose entry
> values are the information about the words.  I find
> it convenient to use [[ to make it appear to be
> a list (instead of using exists(), assign(), and get()).
> E.g., the following enters the 100,000 words from a
> list of 17,576 and records their id numbers and the
> number of times each is found in the sample.
>
>> wfreqsEnv <- new.env(hash=TRUE, parent = emptyenv())
>> words <- do.call("paste", c(list(sep=""), expand.grid(LETTERS,
> letters, letters)))
> # length(words) == 17576
>> set.seed(1)
>> samp <- sample(seq_along(words), size=10, replace=TRUE)
>> system.time(for(i in samp) {
> +    word <- words[i]
> +    if (is.null(wfreqsEnv[[word]])) { # new entry
> +        wfreqsEnv[[word]] <- list(Count=1, EntryNo=i)
> +    } else { # update existing entry
> +        wfreqsEnv[[word]]$Count <- wfreqsEnv[[word]]$Count + 1
> +    }
> +})
>   user  system elapsed
>   2.28    0.00    2.14
> (The time, in seconds, is from an ancient Windows laptop, c. 2002.)
>
> Here is a small check that we are getting what we expect:
>> words[14736]
> [1] "Tuv"
>> wfreqsEnv[["Tuv"]]
> $Count
> [1] 8
>
> $EntryNo
> [1] 14736
>
>> sum(samp==14736)
> [1] 8
>
> If we do this with a non-hashed environment we get the same
> answers but the elapsed time is now 34.81 seconds instead of
> 2.14.  If you make wfreqEnv be a list instead of an environment
> then that time is 74.12 seconds (and the answers are the same).
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] On Behalf Of Levy, Roger
>> Sent: Saturday, November 06, 2010 1:39 PM
>> To: r-help@r-project.org
>> Subject: [R] Hashing and environments
>>
>> Hi,
>>
>> I'm trying to write a general-purpose "lexicon" class and
>> associated methods for storing and accessing information
>> about large numbers of specific words (e.g., their
>> frequencies in different genres).  Crucial to making such a
>> class practically useful is to get hashing working correctly
>> so that information about specific words can be accessed
>> quickly.  But I've never really understood very well how
>> hashing works, so I'm having trouble.
>>
>> Here is an example of what I've done so far:
>>
>> ***
>>
>> setClass("Lexicon",representation(e="environment"))
>> setMethod("initialize","Lexicon",function(.Object,wfreqs) {
>>       .obj...@e <- new.env(hash=T,parent=emptyenv())
>>       assign("wfreqs",wfreqs,envir=.obj...@e)
>>       return(.Object)
>>       })
>>
>> ## function to access word frequencies
>> wfreq <- function(lexicon,word) {
>>       return(get("wfreqs",envir=lexi...@e)[word])
>> }
>>
>> ## example of use
>> my.lexicon <- new("Lexicon",wfreqs=c("the"=2,"person"=1))
>> wfreq(my.lexicon,"the")
>>
>> ***
>>
>> However, testing indicates that the way I have set this up
>> does not achieve the intended benefits of having the
>> environment hashed:
>>
>> ***
>>
>> sample.wfreqs <- trunc(runif(1e5,max=100))
>> names(sample.wfreqs) <- as.character(1:length(sample.wfreqs))
>> lex <- new("Lexicon",wfreqs=sample.wfreqs)
>> words.to.lookup <- trunc(runif(100,min=1,max=1e5))
>> ## look up the words directly from the sample.wfreqs vector
>> system.time({
>>       for(i in words.to.lookup)
>>               sample.wfreqs[as.character(i)]
>>       },gcFirst=TRUE)
>> ## look up the words through the wfreq() function; time
>> approx the same
>> system.time({
>>       for(i in words.to.lookup)
>>               wfreq(lex,as.character(i))
>>       },gcFirst=TRUE)
>>
>> ***
>>
>> I'm guessing that the problem is that the indexing of the
>> wfreqs vector in my wfreq() function is not happening inside
>> the actual lexicon's environment.  However, I have not been
>> able to figure out the proper call to get the lookup to
>> happen inside the lexicon's environment.  I've tried
>>
>> wfreq1 <- function(lexicon,word) {
>>       return(eval(wfreqs[word],envir=lexi...@e))
>> }
>>
>> which I'd thought should work, but this gives me an error:
>>
>> > wfreq1(my.lexicon,'the')
>> Error in eval(wfreqs[word], envir = lexi...@e) :
>>   object 'wfreqs' not found
>>
>> Any advice would be much appreciated!
>>
>> Best & many thanks in advance,
>>
>> Roger
>>
>> --
>>
>> Roger Levy                      Email: rl...@ucsd.edu
>> Assistant Professor             Phone: 858-534-7219
>> Department of Linguistics       Fax:   858-534-4789
>> UC San Diego                    Web:   http://ling.ucsd.edu/~rlevy
>>
>> __
>> 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 c

Re: [R] SMATR common slopes test

[Eugenio Larios]

great thanks a lot!

On Sat, Nov 6, 2010 at 3:54 PM, Kevin Middleton  wrote:

>
> Eugenio -
>
> > I am confused with SMATR's test for common slope. My null hypothesis here
> is
> > that all slopes are parallel (common slopes?), right?
> > So if I get a p value < 0.05 means that we can have confidence to reject
> it?
> > That slopes are different?
> > Or the other way around? it means that we have statistical confidence
> that
> > the slopes are parallel?
>
> Try this:
>
> set.seed(5)
> n <- 20
> x <- rnorm(n)
>
> y1 <- 2 * x + rnorm(n)
> y2 <- 2 * x + rnorm(n)
> y3 <- 4 * x + rnorm(n)
>
> # Slopes approximately equal
> slope.com(x = c(x, x), y = c(y1, y2), groups = rep(c(1,2), each = n))
>
> #$p
> #[1] 0.4498037
>
> # Slopes of 2 and 4
> slope.com(x = c(x, x), y = c(y1, y3), groups = rep(c(1,2), each = n))
>
> #$p
> #[1] 0.0003850332
>
>
> Cheers,
> Kevin
>
> -
> Kevin M. Middleton
> Department of Biology
> California State University San Bernardino
>
>


-- 
Eugenio Larios
PhD Student
University of Arizona.
Ecology & Evolutionary Biology.
(520) 481-2263
elari...@email.arizona.edu

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Re: [R] How to generate multivariate uniform distribution random

[michael]

Ted,

  Thanks for your help, it is right on the money!

for your comments:
1. Yes I mean 100 by 2, each variable x1, x2 is 100 by 1.
2. The correlation is the only free parameter.

Michael



On Sat, Nov 6, 2010 at 7:07 PM, Ted Harding wrote:

> On 06-Nov-10 21:54:41, michael wrote:
> > I wish to generate 100 by 1 vector of x1 and x2 both are uniform
> > distributed with covariance matrix \Sigma.
> >
> > Thanks,
> > Michael
>
> First, some comments.
>
> 1. I don't think you mean a "100 by 1 vector of x1 and x2" since
>   you have two variables. "100 by 2" perhaps?
>
> 2. Given the range over which x1 is to be uniformly distributed
>   (say of length A) and the range for x2 (say of length B),
>   then by virtue of the uniform distribution the variance of x1
>   will be (A^2)/12 and the variance of x2 will be (B^2)/12,
>   so you don't have a choice about these. So your only "free
>   parameter" is the covariance (or the correlation) between
>   x1 and x2.
>
> That said, it is a curious little problem. Here is one simple
> solution (though it may have properties that you do not want).
> Using X and Y for x1 and x2 (for simplicity), and using ranges
> (-1/2,1/2) for the ranges of X and Y (you could rescale later):
>
> 1. Generate X uniformly distributed on (-1/2,1/2).
> 2. With probability p (0 < p < 1) let Y=X.
>   Otherwise, let Y be independently uniform on (-1/2,1/2).
>
> This can be implemented by the following code:
>
>  n  <- 100
>  p  <- 0.6# (say)
>  X  <- runif(n)-1/2
>  Z  <- runif(n)-1/2
>  U  <- rbinom(n,1,p)  # =1 where Y=X, = 0 when Y independent of X
>  Y  <- X*U + Z*(1-U)
>  XY <- cbind(X,Y) # your n by 2 matrix of bivariate (X,Y)
>
> Now the marginal distributions of X and Y are both uniform.
> var(X) = 1/12 and var(Y) = 1/12. Since the means are 0,
> cov(X,Y) = Exp(X*Y) = p*Exp(X^2) = p/12
> cor(X,Y) = cov(X,Y)/sqrt(var(X)*var(Y)) = (p/12)/(1/12) = p.
>
> So all you need to do to get a desired correlation rho between
> X and Y is to set p = rho.
>
> The one thing you may not like about this is the diagonal line
> you will get for the cases where X=Y:
>
>  plot(X,Y)
>
> Test:
>
>  n <- 10
>  p  <- 0.6 # (say)
>  X  <- runif(n)-1/2
>  Z  <- runif(n)-1/2
>  U  <- rbinom(n,1,p)
>  Y  <- X*U + Z*(1-U)
>  var(X)
>  # [1] 0.08340525  # theory: var = 1/12 = 0.0833
> >   var(Y)
> [1] 0.08318914  # theory: var = 1/12 = 0.0833
> >   cov(X,Y)
> [1] 0.04953733  # theory: cov = p/12 = 0.6/12 = 0.05
> >   cor(X,Y)
> [1] 0.5947063   # theory: cor = p= 0.6
>
> It would be interesting to see a solution which did not involve
> having cases with X=Y!
>
> Hoping this helps,
> Ted.
>
> 
> E-Mail: (Ted Harding) 
> Fax-to-email: +44 (0)870 094 0861
> Date: 06-Nov-10   Time: 23:07:26
> -- XFMail --
>

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Re: [R] Removing NA in ggplot

[Ottar Kvindesland]

OK, any reason why ggplot2 does not allow filtering of NA?


ottar

On 6 November 2010 15:23, Jeff Newmiller  wrote:

> Create a subset of your data that excludes the NAs before you feed it to
> ggplot.
>
> "Ottar Kvindesland"  wrote:
>
> >Hi list,
> >
> >I just got stuck with this one:
> >
> >In Data I have the sets age (numbers 1 to 99 and NA) and gender (M, F
> >and
> >NA). Then getting some nice plots using
> >
> >ggplot(data, aes(age[na.exclude(gender)])) +
> >geom_histogram( binwidth = 3, aes(y = ..density.. ), fill = "lightblue"
> >)
> >+
> >  facet_grid( gender ~ .)
> >
> >I am trying to get a faceted graph of age distribution excluding the NA
> >data
> >for gender
> >
> >Unfortunately I end up with the error message:
> >
> >Error in data.frame(..., check.names = FALSE) :
> >arguments imply differing number of rows: 206, 219
> >
> >How do i Wash out NA's in this situation?
> >
> >
> >Regards
> >
> >ottar
> >
> >   [[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.
>
> ---
> Jeff NewmillerThe .   .  Go Live...
> DCN:Basics: ##.#.   ##.#.  Live
> Go...
>  Live:   OO#.. Dead: OO#..  Playing
> Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
> /Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
> ---
> Sent from my phone. Please excuse my brevity.
>

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Re: [R] Removing NA in ggplot

[Joshua Wiley]

On Sat, Nov 6, 2010 at 4:43 PM, Ottar Kvindesland
 wrote:
> OK, any reason why ggplot2 does not allow filtering of NA?

It is not so much that ggplot2 does not allow the filtering of NA
values, it is that you need to use data from the dataset you
specified.  By subsetting in aes() rather than in data, ggplot2 has
differing datasets that it is being told to work with, so it returns
an error (I'm sure that is a simplification, but the general point).

Do your exclusion in the data argument.  I imagine something like
this, but untested since I have nothing to test it on.

ggplot(data[na.exclude(gender), ], aes(age)) +
geom_histogram( binwidth = 3, aes(y = ..density.. ), fill = "lightblue" ) +
facet_grid( gender ~ .)

HTH,

Josh

>
>
> ottar
>
> On 6 November 2010 15:23, Jeff Newmiller  wrote:
>
>> Create a subset of your data that excludes the NAs before you feed it to
>> ggplot.
>>
>> "Ottar Kvindesland"  wrote:
>>
>> >Hi list,
>> >
>> >I just got stuck with this one:
>> >
>> >In Data I have the sets age (numbers 1 to 99 and NA) and gender (M, F
>> >and
>> >NA). Then getting some nice plots using
>> >
>> >ggplot(data, aes(age[na.exclude(gender)])) +
>> >geom_histogram( binwidth = 3, aes(y = ..density.. ), fill = "lightblue"
>> >)
>> >+
>> >  facet_grid( gender ~ .)
>> >
>> >I am trying to get a faceted graph of age distribution excluding the NA
>> >data
>> >for gender
>> >
>> >Unfortunately I end up with the error message:
>> >
>> >Error in data.frame(..., check.names = FALSE) :
>> >arguments imply differing number of rows: 206, 219
>> >
>> >How do i Wash out NA's in this situation?
>> >
>> >
>> >Regards
>> >
>> >ottar
>> >
>> >       [[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.
>>
>> ---
>> Jeff Newmiller                        The     .       .  Go Live...
>> DCN:        Basics: ##.#.       ##.#.  Live
>> Go...
>>                                      Live:   OO#.. Dead: OO#..  Playing
>> Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
>> /Software/Embedded Controllers)               .OO#.       .OO#.  rocks...1k
>> ---
>> Sent from my phone. Please excuse my brevity.
>>
>
>        [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>



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

__
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Re: [R] Using changing names in loop in R

[Tuatara]

A more detailed example:

Say I would like to read in data files that are set-up identically and have
identical (but somewhat) different text names (see below):

data_1 <- read.csv("data1.txt")
data_2 <- read.csv("data2.txt")
data_3 <- read.csv("data3.txt")

How do I automate this process?

(I assume the way I make R understand that the data file extension is to be
read as a number rather than a string is the same for things like applying
functions to matrices with different extensions, e.g. data_i, i = 1,2,3)
-- 
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Re: [R] How to generate multivariate uniform distribution random

[G. Jay Kerns]

Dear Michael,

On Sat, Nov 6, 2010 at 7:27 PM, michael  wrote:
> Ted,
>
>      Thanks for your help, it is right on the money!
>
> for your comments:
> 1. Yes I mean 100 by 2, each variable x1, x2 is 100 by 1.
> 2. The correlation is the only free parameter.
>
> Michael
>
>

I like Ted's solution.  If all you are looking for is unif(0,1), you
could use the Probability Integral Transform;  something like this:

set.seed(1)

library(MASS)
S <- matrix(c(1, 0.9, 0.9, 1), nrow = 2)
X <- mvrnorm(100, mu = c(0,0), Sigma = S)
Y <- pnorm(X)

var(Y)
cor(Y)

You could also use copulas, but those depend on contributed packages
(and you can read more about them on the CRAN Task View for
probability distributions).

Hope this helps,
Jay


__
G. Jay Kerns, Ph.D.
Youngstown State University
http://people.ysu.edu/~gkerns/

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Re: [R] Using changing names in loop in R

[Joshua Wiley]

On Sat, Nov 6, 2010 at 3:13 PM, Tuatara  wrote:
>
> A more detailed example:
>
> Say I would like to read in data files that are set-up identically and have
> identical (but somewhat) different text names (see below):
>
> data_1 <- read.csv("data1.txt")
> data_2 <- read.csv("data2.txt")
> data_3 <- read.csv("data3.txt")
>
> How do I automate this process?

Well, the (nearly) verbatim automated duplicate would be:

for(i in 1:3) {
  assign(x = paste("data", i, sep = "_"), value =
read.csv(paste("data", i, ".txt", sep = '')))
}

but the preferred way would be:

dat <- lapply(1:3, function(x) {read.csv(paste("data", x, ".txt", sep = ''))})

which would read in and store all three files in one convenient list.

Josh

>
> (I assume the way I make R understand that the data file extension is to be
> read as a number rather than a string is the same for things like applying
> functions to matrices with different extensions, e.g. data_i, i = 1,2,3)
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/Using-changing-names-in-loop-in-R-tp3030132p3030412.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.
>



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

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

Re: [R] How to generate multivariate uniform distribution random

[michael]

Jay,

  Yes I'm looking for unif(0,1) and your method works just fine. I
suppose your method should work for dimensions greater than 2, am I right?

Michael

On Sat, Nov 6, 2010 at 8:05 PM, G. Jay Kerns  wrote:

> Dear Michael,
>
> On Sat, Nov 6, 2010 at 7:27 PM, michael  wrote:
> > Ted,
> >
> >  Thanks for your help, it is right on the money!
> >
> > for your comments:
> > 1. Yes I mean 100 by 2, each variable x1, x2 is 100 by 1.
> > 2. The correlation is the only free parameter.
> >
> > Michael
> >
> >
>
> I like Ted's solution.  If all you are looking for is unif(0,1), you
> could use the Probability Integral Transform;  something like this:
>
> set.seed(1)
>
> library(MASS)
> S <- matrix(c(1, 0.9, 0.9, 1), nrow = 2)
> X <- mvrnorm(100, mu = c(0,0), Sigma = S)
> Y <- pnorm(X)
>
> var(Y)
> cor(Y)
>
> You could also use copulas, but those depend on contributed packages
> (and you can read more about them on the CRAN Task View for
> probability distributions).
>
> Hope this helps,
> Jay
>
>
> __
> G. Jay Kerns, Ph.D.
> Youngstown State University
> http://people.ysu.edu/~gkerns/
>

[[alternative HTML version deleted]]

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

Re: [R] How to generate multivariate uniform distribution random

[G. Jay Kerns]

On Sat, Nov 6, 2010 at 8:22 PM, michael  wrote:
> Jay,
>
>   Yes I'm looking for unif(0,1) and your method works just fine. I
> suppose your method should work for dimensions greater than 2, am I right?
>
> Michael
>

Yes, but it gets that much more tricky to specify the covariance
matrix.  Two ways around this are to suppose that Sigma has a
simplified correlation structure, or again, to use copulas.

Jay

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Re: [R] Where to get rcom for Linux

[Stephen Liu]

- Original Message 

From: Shige Song 
To: Stephen Liu 
Cc: r-help@r-project.org
Sent: Sun, November 7, 2010 12:17:42 AM
Subject: Re: [R] Where to get rcom for Linux

> isn't COM a Windows-only technology?

Hi Shige

Thanks.

I see.  I was surprised for unable to find it after having turned over the 
whole 
Internet World.  Which Windows version Win7/Vista/Win Server 2008 will be more 
suitable running RBloomberg?

B.R.
Stephen L


On Sat, Nov 6, 2010 at 12:12 PM, Stephen Liu  wrote:
> Hi folks,
>
> Debian 600 64-bit
>
> Is rcom for Linux available?
>
> rcom
> rcom: R COM Client Interface and internal COM Server
> http://cran.r-project.org/web/packages/rcom/index.html
>
> If YES please advise where to get it.
>
> TIA
>
> B.R.
> Stephen L
>
>
>
>
> __
> 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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[R] postscript window size

[threshold]

Dear R users, simple figure: 

postscript(file="~/Desktop/figure.ps", horizontal=T, width=20, height=10)
par(mfcol=c(2,5))
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
##-
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
dev.off()

Does not resize when saved on Desktop. I could ignore 'width' and 'height',
and the same result. Particular figures plot(rnorm()) are too rectangular,
when I want to have them more like squares (stretched horizontally).  

Working on Ubuntu and need to paste 'figure.ps' into latex (Kile). 
Thanks a lot, robert

-- 
View this message in context: 
http://r.789695.n4.nabble.com/postscript-window-size-tp3030514p3030514.html
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] postscript window size

[Joshua Wiley]

Hi Robert,

You need to add paper = "special" to the postscript() call.

postscript(file="~/Desktop/figure.ps", horizontal=TRUE,
   width=20, height=10, paper = "special")
plot()
...
plot()
dev.off()

Otherwise it is reset because you are specifying a size outside of
what can fit on the default paper size.

Once that is changed, it should work (tested both ways on Debian :))

Cheers,

Josh

On Sat, Nov 6, 2010 at 6:23 PM, threshold  wrote:
>
> Dear R users, simple figure:
>
> postscript(file="~/Desktop/figure.ps", horizontal=T, width=20, height=10)
> par(mfcol=c(2,5))
> plot(rnorm(100), type='l')
> plot(rnorm(100), type='l')
> plot(rnorm(100), type='l')
> plot(rnorm(100), type='l')
> plot(rnorm(100), type='l')
> ##-
> plot(rnorm(100), type='l')
> plot(rnorm(100), type='l')
> plot(rnorm(100), type='l')
> plot(rnorm(100), type='l')
> plot(rnorm(100), type='l')
> dev.off()
>
> Does not resize when saved on Desktop. I could ignore 'width' and 'height',
> and the same result. Particular figures plot(rnorm()) are too rectangular,
> when I want to have them more like squares (stretched horizontally).
>
> Working on Ubuntu and need to paste 'figure.ps' into latex (Kile).
> Thanks a lot, robert
>
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/postscript-window-size-tp3030514p3030514.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.
>



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

__
R-help@r-project.org mailing list
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[R] can't load nlme on windoze 7

[Mike Marchywka]

Hi,

I've got a problem that sounds a lot like this,

http://r.789695.n4.nabble.com/Re-R-R-2-12-0-hangs-while-loading-RGtk2-on-FreeBSD-td3005929.html

under windoze 7.

but it seems to hang with this stack trace,

#0  0x77830190 in ntdll!LdrFindResource_U ()

   from /cygdrive/c/Windows/system32/ntdll.dll




building goes as follows,

$ ./R CMD INSTALL --no-test-load nlme_3.1-97.tar.gz
* installing to library 'C:/pfs/R/R-2.11.1/library'
* installing *source* package 'nlme' ...
** libs
  making DLL ...
gcc -I"C:/pfs/R/R-2.11.1/include" -O3 -Wall  -std=gnu99 -c corStruct.c -
o corStruct.o
gcc -I"C:/pfs/R/R-2.11.1/include" -O3 -Wall  -std=gnu99 -c gnls.c -o gnl
s.o
gcc -I"C:/pfs/R/R-2.11.1/include" -O3 -Wall  -std=gnu99 -c init.c -o ini
t.o
gcc -I"C:/pfs/R/R-2.11.1/include" -O3 -Wall  -std=gnu99 -c matrix.c -o m
atrix.o
gcc -I"C:/pfs/R/R-2.11.1/include" -O3 -Wall  -std=gnu99 -c nlOptimizer.c
 -o nlOptimizer.o
gcc -I"C:/pfs/R/R-2.11.1/include" -O3 -Wall  -std=gnu99 -c nlme.c -o nlm
e.o
gcc -I"C:/pfs/R/R-2.11.1/include" -O3 -Wall  -std=gnu99 -c nlmefit.c -o
nlmefit.o
gcc -I"C:/pfs/R/R-2.11.1/include" -O3 -Wall  -std=gnu99 -c nls.c -o nls.
o
gcc -I"C:/pfs/R/R-2.11.1/include" -O3 -Wall  -std=gnu99 -c pdMat.c -o pd
Mat.o
gcc -shared -s -static-libgcc -o nlme.dll tmp.def corStruct.o gnls.o init.o matr
ix.o nlOptimizer.o nlme.o nlmefit.o nls.o pdMat.o -LC:/pfs/R/R-2.11.1/bin -lR
installing to C:/pfs/R/R-2.11.1/library/nlme/libs
  ... done
** R
** data
**  moving datasets to lazyload DB
** inst
** preparing package for lazy loading
** help
*** installing help indices
** building package indices ...

* DONE (nlme)


$ gcc --version
gcc (GCC) 4.3.4 20090804 (release) 1
Copyright (C) 2008 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.


$ gdb
GNU gdb 6.8.0.20080328-cvs (cygwin-special)
Copyright (C) 2008 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later 
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  Type "show copying"
and "show warranty" for details.
This GDB was configured as "i686-pc-cygwin".
(gdb) target exec R.exe
(gdb) run
Starting program: /cygdrive/c/pfs/R/R-2.11.1/bin/R.exe
[New thread 20844.0x5368]
Error: dll starting at 0x7742 not found.
Error: dll starting at 0x769c not found.
Error: dll starting at 0x7742 not found.
Error: dll starting at 0x7754 not found.
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
(no debugging symbols found)
Error: dll starting at 0x4a0b not found.

R version 2.11.1 (2010-05-31)
Copyright (C) 2010 The R Foundation for Statistical Computing
ISBN 3-900051-07-0

R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.

R is a collaborative project with many contributors.
Type 'contributors()' for more information and
'citation()' on how to cite R or R packages in publications.

Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.

> library(nlme)
[New thread 20844.0x5154]
[Switching to thread 20844.0x5154]
Quit
(gdb) bt
#0  0x77830190 in ntdll!LdrFindResource_U ()
   from /cygdrive/c/Windows/system32/ntdll.dll
(gdb)






Mike Marchywka | V.P. Technology

415-264-8477
marchy...@phluant.com

Online Advertising and Analytics for Mobile
http://www.phluant.com


  
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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] 3-way interaction simple slopes

[David Winsemius]

On Nov 6, 2010, at 9:06 AM, Michael Wood wrote:

Can anyone show me how to test for significant simple slopes of a 3- 
way

interaction, with covariates.



You might start by defining what you mean by "simple slopes" when  
discussing a model with a three way interaction. You might also  
include a description of the variables involved.




my equation
tmod<-(glm(PCL~ rank.f + gender.f + MONTHS + CEXPOSE.M + bf.m +
MONTHS*CEXPOSE.M*bf.m,
data=mhatv, family=gaussian ,na.action=na.omit))

Thank you
Mike



David Winsemius, MD
West Hartford, CT

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] About 5.1 Arrays

[Stephen Liu]

Hi Daniel,


> I am correcting you. :-)  You are using dim() incorrectly, and not  accessing 
>the array correctly.  
>
> In all of your examples you should be  using dim(3,4,2).  Then you need to 
>specify the indexes 
>
> of the array  element you want to look at. So, to use your example

Thanks for your correction.  So the index to be used in my example should be 
(3,4,2) only.


But still I'm not very clear re your advice on follows

> a<-1:24
> a3d <- array(a, dim = c(3,4,2))
> a3d
, , 1

 [,1] [,2] [,3] [,4]
[1,]147   10
[2,]258   11
[3,]369   12

, , 2

 [,1] [,2] [,3] [,4]
[1,]   13   16   19   22
[2,]   14   17   20   23
[3,]   15   18   21   24

> 
> # 1 maps to a[1, 1, 1] in the 3d array
> a3d[1, 1, 1]
[1] 1
> 
> # 2 maps to a[2, 1, 1].
> a3d[2, 1, 1]
[1] 2
> 
> # 3 maps to a[3, 1, 1]
> a3d[3, 1, 1]
[1] 3
> 
> # 4 maps to a[1, 2, 1]
> a3d[1, 2, 1]
[1] 4


What does it mean;

[1] 1
[1] 2
[1] 3
[1] 4

as mentioned ?

Anyway I'll move/continue on the manual to see what will happen.

B.R.
Stephen L






- Original Message 
From: Daniel Nordlund 
To: r-help@r-project.org
Sent: Sun, November 7, 2010 2:08:04 AM
Subject: Re: [R] About 5.1 Arrays

> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Stephen Liu
> Sent: Saturday, November 06, 2010 7:38 AM
> To: Joshua Wiley
> Cc: r-help@r-project.org
> Subject: Re: [R] About 5.1 Arrays
> 
> Hi Joshua,
> 
> Thanks for your advice.
> 
> 1)
> Re your advice:-[quote]
> > a3d
> , , 1 <--- this is the first position of the third dimension
> 
>  [,1] [,2] [,3] [,4]  <--- positions 1, 2, 3, 4 of the second
> dimension
> [1,]147   10
> [2,]258   11
> [3,]369   12
> ^  the first dimension
> 
> , , 2 <--- the second position of the third dimension
> ...
> [/quote]
> 
> Where is the third dimension?
> 
> 
> 2)
> Re your advice:-[quote]
> so you can think that in the original vector "a":
> 1 maps to a[1, 1, 1] in the 3d array
> 2 maps to a[2, 1, 1].
> 3 maps to a[3, 1, 1]
> 4 maps to a[1, 2, 1]
> 12 maps to a[3, 4, 1]
> 20 maps to a[2, 3, 2]
> 24 maps to a[3, 4, 2]
> [/quote]
> 
> My finding;
> 
> # 1 maps to a[1, 1, 1] in the 3d array
> > a3d <- array(a, dim = c(1, 1, 1))
> > a3d
> , , 1
> 
>  [,1]
> [1,]1
> 
> Correct
> 
> # 2 maps to a[2, 1, 1].
> > a3d <- array(a, dim = c(2, 1, 1))
> > a3d
> , , 1
> 
>  [,1]
> [1,]1
> [2,]2
> 
> Correct
> 
> # 3 maps to a[3, 1, 1]
> > a3d <- array(a, dim = c(3, 1, 1))
> > a3d
> , , 1
> 
>  [,1]
> [1,]1
> [2,]2
> [3,]3
> 
> Correct
> 
> # 4 maps to a[1, 2, 1]
> > a3d <- array(a, dim = c(1, 2, 1))
> > a3d
> , , 1
> 
>  [,1] [,2]
> [1,]12
> 
> Incorrect.  It is "2"
> 
> 
> # 12 maps to a[3, 4, 1]
> > a3d <- array(a, dim = c(3, 4, 1))
> > a3d
> , , 1
> 
>  [,1] [,2] [,3] [,4]
> [1,]147   10
> [2,]258   11
> [3,]369   12
> 
> Correct
> 
> # 20 maps to a[2, 3, 2]
> > a3d <- array(a, dim = c(2, 3, 2))
> > a3d
> , , 1
> 
>  [,1] [,2] [,3]
> [1,]135
> [2,]246
> 
> , , 2
> 
>  [,1] [,2] [,3]
> [1,]79   11
> [2,]8   10   12
> 
> Incorrect.  It is "12"
> 
> 
> #  24 maps to a[3, 4, 2]
> > a3d <- array(a, dim = c(3, 4, 2))
> > a3d
> , , 1
> 
>  [,1] [,2] [,3] [,4]
> [1,]147   10
> [2,]258   11
> [3,]369   12
> 
> , , 2
> 
>  [,1] [,2] [,3] [,4]
> [1,]   13   16   19   22
> [2,]   14   17   20   23
> [3,]   15   18   21   24
> 
> Correct.
> 
> If I'm wrong, pls correct me.  Thanks
> 
> 
> B.R.
> Stephen
> 

Stephen,

I am correcting you. :-)  You are using dim() incorrectly, and not accessing 
the 
array correctly.  In all of your examples you should be using dim(3,4,2).  Then 
you need to specify the indexes of the array element you want to look at. So, 
to 
use your example

> a<-1:24
> a3d <- array(a, dim = c(3,4,2))
> a3d
, , 1

 [,1] [,2] [,3] [,4]
[1,]147   10
[2,]258   11
[3,]369   12

, , 2

 [,1] [,2] [,3] [,4]
[1,]   13   16   19   22
[2,]   14   17   20   23
[3,]   15   18   21   24

> 
> # 1 maps to a[1, 1, 1] in the 3d array
> a3d[1, 1, 1]
[1] 1
> 
> # 2 maps to a[2, 1, 1].
> a3d[2, 1, 1]
[1] 2
> 
> # 3 maps to a[3, 1, 1]
> a3d[3, 1, 1]
[1] 3
> 
> # 4 maps to a[1, 2, 1]
> a3d[1, 2, 1]
[1] 4


Hope this is helpful,

Dan

Daniel Nordlund
Bothell, WA USA


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Re: [R] Hashing and environments

[Levy, Roger]

Wow, that is perfect: the hash package is exactly what I needed.  Thank you!

Roger

On Nov 6, 2010, at 4:09 PM, Kjetil Halvorsen wrote:

> some of this can be automated using the CRAN package
> hash.
> 
> Kjetil
> 
> On Sat, Nov 6, 2010 at 10:43 PM, William Dunlap  wrote:
>> I would make make an environemnt called wfreqsEnv
>> whose entry names are your words and whose entry
>> values are the information about the words.  I find
>> it convenient to use [[ to make it appear to be
>> a list (instead of using exists(), assign(), and get()).
>> E.g., the following enters the 100,000 words from a
>> list of 17,576 and records their id numbers and the
>> number of times each is found in the sample.
>> 
>>> wfreqsEnv <- new.env(hash=TRUE, parent = emptyenv())
>>> words <- do.call("paste", c(list(sep=""), expand.grid(LETTERS,
>> letters, letters)))
>> # length(words) == 17576
>>> set.seed(1)
>>> samp <- sample(seq_along(words), size=10, replace=TRUE)
>>> system.time(for(i in samp) {
>> +word <- words[i]
>> +if (is.null(wfreqsEnv[[word]])) { # new entry
>> +wfreqsEnv[[word]] <- list(Count=1, EntryNo=i)
>> +} else { # update existing entry
>> +wfreqsEnv[[word]]$Count <- wfreqsEnv[[word]]$Count + 1
>> +}
>> +})
>>   user  system elapsed
>>   2.280.002.14
>> (The time, in seconds, is from an ancient Windows laptop, c. 2002.)
>> 
>> Here is a small check that we are getting what we expect:
>>> words[14736]
>> [1] "Tuv"
>>> wfreqsEnv[["Tuv"]]
>> $Count
>> [1] 8
>> 
>> $EntryNo
>> [1] 14736
>> 
>>> sum(samp==14736)
>> [1] 8
>> 
>> If we do this with a non-hashed environment we get the same
>> answers but the elapsed time is now 34.81 seconds instead of
>> 2.14.  If you make wfreqEnv be a list instead of an environment
>> then that time is 74.12 seconds (and the answers are the same).
>> 
>> Bill Dunlap
>> Spotfire, TIBCO Software
>> wdunlap tibco.com
>> 
>>> -Original Message-
>>> From: r-help-boun...@r-project.org
>>> [mailto:r-help-boun...@r-project.org] On Behalf Of Levy, Roger
>>> Sent: Saturday, November 06, 2010 1:39 PM
>>> To: r-help@r-project.org
>>> Subject: [R] Hashing and environments
>>> 
>>> Hi,
>>> 
>>> I'm trying to write a general-purpose "lexicon" class and
>>> associated methods for storing and accessing information
>>> about large numbers of specific words (e.g., their
>>> frequencies in different genres).  Crucial to making such a
>>> class practically useful is to get hashing working correctly
>>> so that information about specific words can be accessed
>>> quickly.  But I've never really understood very well how
>>> hashing works, so I'm having trouble.
>>> 
>>> Here is an example of what I've done so far:
>>> 
>>> ***
>>> 
>>> setClass("Lexicon",representation(e="environment"))
>>> setMethod("initialize","Lexicon",function(.Object,wfreqs) {
>>>   .obj...@e <- new.env(hash=T,parent=emptyenv())
>>>   assign("wfreqs",wfreqs,envir=.obj...@e)
>>>   return(.Object)
>>>   })
>>> 
>>> ## function to access word frequencies
>>> wfreq <- function(lexicon,word) {
>>>   return(get("wfreqs",envir=lexi...@e)[word])
>>> }
>>> 
>>> ## example of use
>>> my.lexicon <- new("Lexicon",wfreqs=c("the"=2,"person"=1))
>>> wfreq(my.lexicon,"the")
>>> 
>>> ***
>>> 
>>> However, testing indicates that the way I have set this up
>>> does not achieve the intended benefits of having the
>>> environment hashed:
>>> 
>>> ***
>>> 
>>> sample.wfreqs <- trunc(runif(1e5,max=100))
>>> names(sample.wfreqs) <- as.character(1:length(sample.wfreqs))
>>> lex <- new("Lexicon",wfreqs=sample.wfreqs)
>>> words.to.lookup <- trunc(runif(100,min=1,max=1e5))
>>> ## look up the words directly from the sample.wfreqs vector
>>> system.time({
>>>   for(i in words.to.lookup)
>>>   sample.wfreqs[as.character(i)]
>>>   },gcFirst=TRUE)
>>> ## look up the words through the wfreq() function; time
>>> approx the same
>>> system.time({
>>>   for(i in words.to.lookup)
>>>   wfreq(lex,as.character(i))
>>>   },gcFirst=TRUE)
>>> 
>>> ***
>>> 
>>> I'm guessing that the problem is that the indexing of the
>>> wfreqs vector in my wfreq() function is not happening inside
>>> the actual lexicon's environment.  However, I have not been
>>> able to figure out the proper call to get the lookup to
>>> happen inside the lexicon's environment.  I've tried
>>> 
>>> wfreq1 <- function(lexicon,word) {
>>>   return(eval(wfreqs[word],envir=lexi...@e))
>>> }
>>> 
>>> which I'd thought should work, but this gives me an error:
>>> 
 wfreq1(my.lexicon,'the')
>>> Error in eval(wfreqs[word], envir = lexi...@e) :
>>>   object 'wfreqs' not found
>>> 
>>> Any advice would be much appreciated!
>>> 
>>> Best & many thanks in advance,
>>> 
>>> Roger
>>> 
>>> --
>>> 
>>> Roger Levy  Email: rl...@ucsd.edu
>>> Assistant Professor Phone: 858-534-7219
>>> Department of Linguistics   Fax:   858-534-4789
>>> UC San DiegoWeb

[R] is this matrix symmetric

[Jun Shen]

Hi,

I have this symmetric matrix, at least I think so.

 col1  col2  col3
[1,] 0.20 0.05 0.06
[2,] 0.05 0.10 0.03
[3,] 0.06 0.03 0.08

or

structure(c(0.2, 0.05, 0.06, 0.05, 0.1, 0.03, 0.06, 0.03, 0.08
), .Dim = c(3L, 3L), .Dimnames = list(NULL, c("var1", "var2",
"var3")))

But isSymmetric() doesn't agree. Any comment? I am on R 2.10.1 Thanks.

Jun

[[alternative HTML version deleted]]

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[R] creating a scale (factor) based on a continuous variable nested within levels of factor

[hind lazrak]

Hello R-helpers


I hope that my subject line is not detering anyone from helping me out:)
I have been stuck of a few hours now, and I don't seem to pinpoint
where the problem is.


I have a data.frame which is structured as follow:
str(hDatPretty)
'data.frame': 1665 obs. of  8 variables:
$ time    : num  0 1.02 2.05 3.07 4.09 ...
$ hr      : num  62.4 63.6 64.6 65.5 66.2 ...
$ emg     : num  3.3 3.42 3.52 3.57 3.6 ...
$ respRate: num  50.4 50.6 50.7 50.8 50.9 ...
$ scr     : num  1.7 1.72 1.73 1.74 1.75 ...
$ skinTemp: num  28.1 28.2 28.2 28.2 28.2 ...
$ rating  : num  4 4 4 4 4 4 4 4 4 4 ...
$ songId  : Factor w/ 37 levels "1","2","3","4",..: 1 1 1 1 1 1 1 1 1 1 ...

It consists of ratings ($rating) given by people (here the id variable
is not indicated as this is a subset with only one person) for each of
the 37 songs ($songId) they listen to.
While they are listening we measure physiological responses (emg,
hr,...) every second over a period of 45 seconds.
Here's a quick peek at the data
head(hDatPretty)

time   hr  emg respRate  scr skinTemp rating songId
1.1 0.00 62.42135 3.300562 50.40538 1.703105 28.14489  4  1
1.2 1.022727 63.59057 3.424884 50.59292 1.718110 28.16189  4  1
1.3 2.045455 64.59840 3.515219 50.73523 1.730594 28.17836  4  1
1.4 3.068182 65.47707 3.573151 50.83909 1.740594 28.19422  4  1
1.5 4.090909 66.22192 3.597183 50.90466 1.748086 28.20948  4  1
1.6 5.113636 66.89209 3.588530 50.91911 1.753385 28.22414  4  1

So, every study participant gives one rating (from -10 to 10) for each song
If we tab the data this is what we have (for the first 10 songs)
table(hDatPretty$songId, hDatPretty$rating)


    -10 -9 -7 -3  0  1  3  4  5  7  8  9 10
 1    0  0  0  0  0  0  0 45  0  0  0  0  0  # song 1 gets a score of 4
 2    0  0  0  0  0  0 45  0  0  0  0  0  0  # song 2 gets a score of 3
 3    0  0 45  0  0  0  0  0  0  0  0  0  0  #.
 4    0 45  0  0  0  0  0  0  0  0  0  0  0
 5    0  0  0  0  0  0  0  0  0 45  0  0  0
 6    0  0  0  0  0  0  0  0  0  0  0  0 45
 7    0  0  0  0  0  0  0  0  0  0 45  0  0  #song 7 gets a score of 8
 8    0  0  0 45  0  0  0  0  0  0  0  0  0
 9    0  0  0  0  0  0  0 45  0  0  0  0  0
 10   0  0  0  0  0 45  0  0  0  0  0  0  0

What I would like to do is to create another scale ( a factor) based
on the ratings with the following levels
-10;-4 == dislike where -4 is included
-4;4 == neutral where -4 is excluded
4;10 == like  where 4 is excluded

My code to obtain this new variable

liking <- numeric(length(hDatPretty$rating))
liking[hDatPretty$rating <= -4] <- 'dislike'
liking[hDatPretty$rating > -4 & hDatPretty$rating <= 4] <- 'neutral'
liking[hDatPretty$rating > 4] <- 'like'

hDatPretty['liking']<- factor(liking)

The problem that I have is that for some reasons it does assign
different values to the same rating for some songs but not all (?)
See for example

  dislike like neutral
1        0    8      37   ## Here is one problem where the song #
1gets two 'liking' scores while the rating is constant
2        0    0      45
3       45    0       0
4       45    0       0
5        0   45       0
6        0   45       0
7        0   45       0
8        0    0      45
9        0   10      35  ## here is a similar problem

Could you PLEASE help me with the proper code to obtain my 'liking'
variable for each of the song based on the rating each song gets?

Many thanks.


Hind
p.s.: I have also tried the cut() in the code as follow...unsuccesfully

hDatPretty$liking <- by(hDatPretty$rating, hDatPretty$songId,
   function (z) { cut(hDatPretty$z, c(-10, -4,4,10),
   labels=c('dislike', 'neutral', 'like'))})

Error in cut.default(hDatPretty$z, c(-10, -4, 4, 10), labels = c("dislike",  :
 'x' must be numeric

again thank you.

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[R] When using ACF, receive error: no applicable method for 'ACF' applied to an object of class "c('double', 'numeric')"

[evt]

I am guessing this is a very simple question, but this is only my second day
with R so it is all still a bit imposing.

I am trying to run an autocorrelation.

I imported a CSV file, which has one column labeled "logistic".

I ran the command:
ACF(data$logistic,maxLag=10)

However, I received the error:
Error in UseMethod("ACF") : 
  no applicable method for 'ACF' applied to an object of class "c('double',
'numeric')"

I am thinking perhaps I need to change the data type, but I really don't
know how (or if that is even the problem).

Your suggestions are very much appreciated!
-- 
View this message in context: 
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