Re: [R] Constrained Regression

2010-10-31 Thread Jim Silverton
Hello everyone,
I have 3 variables Y, X1 and X2. Each variables lies between 0 and 1. I want
to do a constrained regression such that a0 and (1-a) 0

for the model:

Y = a*X1 + (1-a)*X2

I tried the help on the constrained regression in R but I concede that it
was not helpful.

Any help is greatly appreciated
-- 
Thanks,
Jim.

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[R] make many barplot into one plot

2010-10-31 Thread Sibylle Stöckli
Dear R users


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

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

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

Thanks
Sibylle


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

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

2010-10-31 Thread Carabiniero

I haven't been able to fully make sense of the conflicting online information
about whether and how to specify nesting structure for a nested, mixed
model.  I'll describe my experiment and hopefully somebody who knows lme4
well can help.

We're measuring the fluorescence intensity of brain slices from frogs that
have undergone various treatments.  We want to use multcomp to look for
differences between treatments, while accounting for the variance introduced
by the random effects of brain and slice.  There are a few measurements per
slice, several slices per brain, and several brains per treatment.  In the
data file, the numbering for slices starts over from 1 for each brain, and
the numbering for brains starts over from 1 for each treatment.  

In other words:  Treatment is a fixed effect, brain is a random effect
nested in treatment, and slice is a random effect nested in brain.

As I understood the documentation, this is the correct specification:

log(Intensity) ~ Treatment + (1|Brain) + (1|Slice)

However, I don't see how lmer understands the correct nesting structure from
that.  How does it know brain isn't crossed with treatment?

Here are two other things I tried, and each gave different results:

log(Intensity) ~ Treatment + (1|Slice/Brain/Treatment)
log(Intensity) ~ Treatment + (1|Brain/Treatment) + (1|Slice/Brain)

I'm not sure why these things give different results, or which one (if any)
is right.  Can anyone help?

Thanks!
-- 
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[R] How to control in order of groups in xyplot

2010-10-31 Thread Jie Liu
Hi guys,

I used the following R code to generate one plot

library(lattice)
xyplot(Y~X1|as.factor(X2)*as.factor(X3), groups = as.factor(X4),
data=mydata)

Both X2 and X3 have three values. X4 has two values. I got 3x3 grids and in
each grid there are two curves about y~x1 for the two X4 values. I am quite
happy with the plot except that I need a different layout of the 3x3 layout.
For example, X2={m=2, m=5, and m=10} and it plots with the order
m=10, m=2, and m=5. Is there any way I can control the order of the
groups in the whole plot? Thanks a lot,

--Jerry

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[R] biglm: how it handles large data set?

2010-10-31 Thread noclue_


I am trying to figure out why 'biglm' can handle large data set... 

According to the R document - biglm creates a linear model object that uses
only p^2  memory for p variables. It can be updated with more data using
update. This allows linear regression on data sets larger than memory.

After reading the source code below, I still could not figure out how
'update'  implements the algorithm...

Thanks for any light shed upon this ... 

 biglm::biglm

function (formula, data, weights = NULL, sandwich = FALSE) 
{
tt - terms(formula)
if (!is.null(weights)) {
if (!inherits(weights, formula)) 
stop(`weights' must be a formula)
w - model.frame(weights, data)[[1]]
}
else w - NULL
mf - model.frame(tt, data)
mm - model.matrix(tt, mf)
qr - bigqr.init(NCOL(mm))
qr - update(qr, mm, model.response(mf), w)
rval - list(call = sys.call(), qr = qr, assign = attr(mm, 
assign), terms = tt, n = NROW(mm), names = colnames(mm), 
weights = weights)
if (sandwich) {
p - ncol(mm)
n - nrow(mm)
xyqr - bigqr.init(p * (p + 1))
xx - matrix(nrow = n, ncol = p * (p + 1))
xx[, 1:p] - mm * model.response(mf)
for (i in 1:p) xx[, p * i + (1:p)] - mm * mm[, i]
xyqr - update(xyqr, xx, rep(0, n), w * w)
rval$sandwich - list(xy = xyqr)
}
rval$df.resid - rval$n - length(qr$D)
class(rval) - biglm
rval
}
environment: namespace:biglm
---
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Re: [R] How to control in order of groups in xyplot

2010-10-31 Thread Rainer Hurling

A working minimal example would have been better, see FAQ.

But I think you are looking for the following:

X2 - factor(c(m=2, m=5, m=10), levels=c(m=2, m=5, m=10))

Here levels are ordered in your way. There might be other solutions for 
this ordering problem.


Hope it helps,
Rainer


On 31.10.2010 06:55 (UTC+1), Jie Liu wrote:

Hi guys,

I used the following R code to generate one plot

library(lattice)
xyplot(Y~X1|as.factor(X2)*as.factor(X3), groups = as.factor(X4),
data=mydata)

Both X2 and X3 have three values. X4 has two values. I got 3x3 grids and in
each grid there are two curves about y~x1 for the two X4 values. I am quite
happy with the plot except that I need a different layout of the 3x3 layout.
For example, X2={m=2, m=5, and m=10} and it plots with the order
m=10, m=2, and m=5. Is there any way I can control the order of the
groups in the whole plot? Thanks a lot,

--Jerry


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Re: [R] for loop

2010-10-31 Thread Matevž Pavlič
Hi Dennis, 

 

Thank you for your extensive explanations. Yes, I guess I did not explain what 
I would like to do.

Basically I would like to conduct a linear regression for each of 15 classes. 
Your answers gave me new

Perspective on how R works. 

 

Thanks again for the help, 

m

 

From: Dennis Murphy [mailto:djmu...@gmail.com] 
Sent: Sunday, October 31, 2010 4:40 AM
To: Matevž Pavlič
Cc: r-help@r-project.org
Subject: Re: [R] for loop

 

Hi:

If your objective is to make 15 plots, one for each level of razred, then you 
don't need to make 15 individual data frames first. The lattice and ggplot2 
packages allow conditioning plots. You haven't mentioned what types of plots 
you're interested in getting, but if it's something simple like a scatterplot 
of y vs. x for each level of razred, it's not that hard to do. Let's fake some 
data:

d - data.frame(razred = rep(LETTERS[1:15], each = 10),
   x = rep(1:10, 15),
   y = rep(2 + 0.5 * 1:10, 15) + rnorm(150))

d has 15 levels of razred with 10 observations at each level. razred is a 
factor, the other variables are either integer or numeric.

Produce scatterplots of y vs. x for each level of razred, using both the 
lattice and ggplot2 packages:

library(lattice)
# each plot adds a new feature - run one plot at a time.
xyplot(y ~ x | razred, data = d, type = c('p', 'r'))
xyplot(y ~ x | razred, data = d, type = c('p', 'r'), layout = c(3, 5))
xyplot(y ~ x | razred, data = d, type = c('p', 'r'), layout = c(3, 5), as.table 
= TRUE)

library(ggplot2)
ggplot(d, aes(x, y)) + geom_point() + geom_smooth(method = 'lm') +
facet_wrap( ~ razred, ncol = 3)
ggplot(d, aes(x, y)) + geom_point() + geom_smooth(method = 'lm', se = FALSE) +
   facet_wrap( ~ razred, ncol = 3)

If instead you want something like a scatterplot matrix for each data subset 
defined by level of razred, then maybe something like this (?):

# add a new variable to the data frame
# splom is the scatterplot matrix function in lattice
d$z1 - rnorm(150)
splom(~ d[, -1] | razred, data = d, layout = c(2, 2, 4))

Just guessing here since you didn't make your objective explicit. 

It's entirely possible that you can conduct a significant part of your data 
analysis without having to split the data into subsets. Several summary 
functions, for example, can compute a number of summary functions by group with 
a one-line call. Here are a couple of examples, one using aggregate() from the 
base package and another using function ddply() from the plyr package:

aggregate(y ~ razred, data = d, FUN = mean)
   razredy
1   A 4.816841
2   B 4.520804
3   C 5.196329
4   D 4.615575
5   E 3.982240
6   F 4.466559
7   G 4.938669
8   H 4.539541
9   I 4.354991
10  J 4.573654
11  K 4.450624
12  L 5.138087
13  M 4.93
14  N 4.879493
15  O 5.087452

library(plyr)
ddply(d, 'razred', summarise, mx = mean(x), my = mean(y), mz1 = mean(z1))
   razred  mx   my mz1
1   A 5.5 4.816841 -0.01745305
2   B 5.5 4.520804  0.24724069
3   C 5.5 5.196329  0.18717750
4   D 5.5 4.615575  0.18885590
5   E 5.5 3.982240 -0.91284339
6   F 5.5 4.466559  0.36479266
7   G 5.5 4.938669 -0.36359562
8   H 5.5 4.539541  0.06061162
9   I 5.5 4.354991  0.05138409
10  J 5.5 4.573654  0.31160018
11  K 5.5 4.450624  0.17458712
12  L 5.5 5.138087 -0.26482357
13  M 5.5 4.93 -0.39194953
14  N 5.5 4.879493  0.33154075
15  O 5.5 5.087452  0.32816931

There are a number of functions and packages that will do this sort of thing 
quite well - I'll mention doBy, data.table, Hmisc and sqldf as excellent 
options, noting that there are other packages and functions in the apply family 
that can perform groupwise processing seamlessly. The point of mentioning this 
is so that you don't automatically think you have to split the data in myriad 
ways before you can process a function. The good folks that designed this 
language, and the many people who have contributed code to the R project, are 
pretty smart, and have devised fairly simple ways to process data, even if it's 
large. 

Of course, it's always possible that splitting is necessary; if you're willing 
to be a little more forthcoming about your analysis goals, you might get a 
better targeted response..

HTH,
Dennis



On Sat, Oct 30, 2010 at 12:00 PM, Matevž Pavlič matevz.pav...@gi-zrmk.si 
wrote:

Just one more thing...
I get a list with 15 data.frames :

List of 15
 $ 1:'data.frame':  7 obs. of  9 variables:
 ..$ vrtina : Factor w/ 6 levels T1A-1,T1A-2,..: 1 1 2 2 5 5 5
 ..$ globina.meritve: num [1:7] 7.6 8.5 10.4 17.4 12.5 15.5 16.5
 ..$ E0 : num [1:7] 4109 2533 491 810 2374 ...
 ..$ Eur1   : num [1:7] 6194 4713 605 1473 NA ...
 ..$ Eur2   : num [1:7] 3665 7216 266 4794 7387 ...
 ..$ Eur3   : num [1:7] 3221 3545 920 3347 6768 ...
 ..$ H  : num [1:7] 8 5.9 5.9 6.9 9.3 

Re: [R] R-help Digest, Vol 92, Issue 31

2010-10-31 Thread Uwe Ligges



On 31.10.2010 05:22, Neyra Peña wrote:

Hi, I'd like to unsubscribe from the list.
Thanks

Neyra




Have you ever read the first lines (those I still cite below) of the 
message you got that already tells you how to unsubscribe?


Uwe Ligges








De: r-help-requ...@r-project.orgr-help-requ...@r-project.org
Para: r-help@r-project.org
Enviado: sáb, octubre 30, 2010 5:30:07 AM
Asunto: R-help Digest, Vol 92, Issue 31

Send R-help mailing list submissions to
 r-help@r-project.org

To subscribe or unsubscribe via the World Wide Web, visit
 https://stat.ethz.ch/mailman/listinfo/r-help



[SNIP]

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[R] Downloading Package Sources

2010-10-31 Thread Santosh Srinivas
Dear Group, any idea how I can download the source code for all packages in
Windows 7?

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Re: [R] return value for grep

2010-10-31 Thread Duncan Murdoch

Ulrich wrote:

Hi,

is it possible to easily change the return value for the grep function  
for cases where there is no match, for example the value 0 or No  
instead of integer (0) )?


It sounds like you might want grepl (which returns a vector of TRUE and 
FALSE values) rather than grep (which returns a vector of indices or 
matches).


Duncan Murdoch

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Re: [R] R VBA

2010-10-31 Thread julien cuisinier
Many thanks for the feedback  the mailing list reference - will do

Obviously I am feeling very stupid now, not sure I missed that =0

sorry for that silly question

Rgds,
Julien


On Oct 30, 2010, at 8:56 PM, RICHARD M. HEIBERGER wrote:

 Look at the macro
 RInterface.PutArrayFromVBA
 documented in the RExcel help file.

 Please send followup questions on RExcel and statconnDCOM to
 the mailing list at

 statconnDCOM, rcom, and RExcel server related issues 
 rco...@mailman.csd.univie.ac.at 
 
 Rich
 On Sat, Oct 30, 2010 at 1:44 PM, julien cuisinier j_cuisin...@hotmail.com 
  wrote:

 Hi List,



 Is there any way to pass on directly VBA variable content into a R  
 variable?


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Re: [R] For loop

2010-10-31 Thread Astabenefica
Thanks for the suggestions.
I add some more detail to clarify:

h=matrix(nrow=1,ncol=22)

In my data h is: 
(0.25  0.25  0  0  0  0  -0.25  -0.25  -0.25  -0.25  -0.5  -0.5  0  0.25  0.25  
0.25  0.25  0.25  0.25  0.25  0.25  0.25)



xx-seq(0,1,0.5)

v=matrix(nrow=20001,ncol=22)
vv=matrix(nrow=20001,ncol=22)
for (y in 1:20001) {
v[y,22]=h[1,22]
}
vv[20001,22]=v[20001,22]
for(k in 21:1) {
 for(j in 20001:2) {
   vv[j-1,k+1]=min(xx[j-1]*v[j-1,k+1],vv[j,k+1])
   v[j,k]=h[1,k]+vv[j-1,k+1]
 }
vv[20001,k]=v[20001,k]
}



The idea of using Rcpp seems to be good. Having never used this package will 
give a look.
Somewhere I read an example like this:


for (i in 1:R) {
  res[i]-f()
  NULL
}

where f() is a function, but I don't how can I trasform my code:
vv[j-1,k+1]=min(xx[j-1]*v[j-1,k+1],vv[j,k+1])
v[j,k]=h[1,k]+vv[j-1,k+1]
in a function








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Re: [R] How to control in order of groups in xyplot

2010-10-31 Thread Peter Ehlers

On 2010-10-31 01:19, Rainer Hurling wrote:

A working minimal example would have been better, see FAQ.

But I think you are looking for the following:

X2- factor(c(m=2, m=5, m=10), levels=c(m=2, m=5, m=10))

Here levels are ordered in your way. There might be other solutions for
this ordering problem.


That would be the best way for this (common) situation since
the ordering is not likely to be wanted in any other way.

For playing with different orderings, lattice provides
the index.cond argument described in the '...' part of
help(xyplot).

  -Peter Ehlers


Hope it helps,
Rainer


On 31.10.2010 06:55 (UTC+1), Jie Liu wrote:

Hi guys,

I used the following R code to generate one plot

library(lattice)
xyplot(Y~X1|as.factor(X2)*as.factor(X3), groups = as.factor(X4),
data=mydata)

Both X2 and X3 have three values. X4 has two values. I got 3x3 grids and in
each grid there are two curves about y~x1 for the two X4 values. I am quite
happy with the plot except that I need a different layout of the 3x3 layout.
For example, X2={m=2, m=5, and m=10} and it plots with the order
m=10, m=2, and m=5. Is there any way I can control the order of the
groups in the whole plot? Thanks a lot,

--Jerry


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Re: [R] Constrained Regression

2010-10-31 Thread Ravi Varadhan
What is the stochastic mechanism that generates the data? In other words, what 
distribution is Y, conditioned on X1 and X2, supposed to be?  

You can also get `a' if you do not wanrt to specify the probability mechanism 
by just doing a least squares fitting, but then making inferences can be a bit 
tricky.

Ravi.



Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University

Ph. (410) 502-2619
email: rvarad...@jhmi.edu


- Original Message -
From: Jim Silverton jim.silver...@gmail.com
Date: Sunday, October 31, 2010 2:45 am
Subject: Re: [R] Constrained Regression
To: r-help@r-project.org


 Hello everyone,
  I have 3 variables Y, X1 and X2. Each variables lies between 0 and 1. 
 I want
  to do a constrained regression such that a0 and (1-a) 0
  
  for the model:
  
  Y = a*X1 + (1-a)*X2
  
  I tried the help on the constrained regression in R but I concede 
 that it
  was not helpful.
  
  Any help is greatly appreciated
  -- 
  Thanks,
  Jim.
  
   [[alternative HTML version deleted]]
  
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  and provide commented, minimal, self-contained, reproducible code.

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[R] Randomly split a sample in two equal subsamples

2010-10-31 Thread Yoan Mihov
Dear all,

I would like to randomly split a sample in two equally large
subsamples. The sample data is stored as a matrix with each row
representing an individual and each column representing some variable
(e.g., name, age, sex, etc.); the first row contains the names of the
variables; the first column contains the individual number (1:n, for n
individuals); the number of individuals is even (so, the overall
number of rows is odd).

I found similar threads (like random subset), but I don't know how
to apply the information from them in my case. Could somebody help me
a little bit? Thanks in advance!

Yoan

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[R] extracting named vector from dataframe

2010-10-31 Thread James Hirschorn
Suppose df is a dataframe with one named row of numeric observations. I want
to coerce df into a named vector.

 

as.vector does not work as I expected: as.vector(df) returns the original
dataframe, while as.vector(df,mode=numeric) returns an unnamed vector of
NAs.

 

This works: 

 

 v - as.numeric(as.matrix(df)); names(v) - names(df);

 

I just wanted check if there was a better/more natural way of doing this?


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Re: [R] Constrained Regression

2010-10-31 Thread David Winsemius


On Oct 31, 2010, at 2:44 AM, Jim Silverton wrote:


Hello everyone,
I have 3 variables Y, X1 and X2. Each variables lies between 0 and  
1. I want

to do a constrained regression such that a0 and (1-a) 0

for the model:

Y = a*X1 + (1-a)*X2



It would not accomplish the constraint that a  0 but you could  
accomplish the other constraint within an lm fit:


X3 - X1-X2
lm(Y ~ X3 + offset(X2) )

Since beta1 is for the model Y ~ 1 + beta1(X1- X2) + 1*X2)
 Y ~ intercept + beta1*X1 + (1 -beta1)*X2

... so beta1 is a.

In the case beta  0 then I suppose a would be assigned 0. This might  
be accomplished within an iterative calculation framework by a large  
penalization for negative values. In a reply (1) to a question by  
Carlos Alzola in 2008 on rhalp, Berwin Turlach offered a solution to a  
similar problem ( sum(coef) == 1 AND coef non-negative). Modifying his  
code to incorporate the above strategy (and choosing two variables for  
which parameter values might be inside the constraint boundaries) we  
get:


library(MASS)   ## to access the Boston data
  designmat - model.matrix(medv~I(age-lstat) +offset(lstat),  
data=Boston)

  Dmat -crossprod(designmat, designmat); dvec - crossprod(designmat,
  Boston$medv)
  Amat - cbind(1, diag(NROW(Dmat)))
  bvec - c(1,rep(0,NROW(Dmat)))
  meq - 1
library(quadprog)
  res - solve.QP(Dmat, dvec, Amat, bvec, meq)

 zapsmall(res$solution)
[1] 0.686547 0.313453

Turlach specifically advised against any interpretation of this  
particular result which was only contructed to demonstrate the  
mathematical mechanics.




I tried the help on the constrained regression in R but I concede  
that it

was not helpful.


I must not have that package installed because I got nothing that  
appeared to be useful with ??constrained regression .



David Winsemius, MD
West Hartford, CT

1) http://finzi.psych.upenn.edu/Rhelp10/2008-March/155990.html

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Re: [R] extracting named vector from dataframe

2010-10-31 Thread baptiste auguie
Hi,

I think you want ?unlist

d = data.frame(x=1, y=2, z=3)
v = unlist(d)
is(v)
[1] numeric vector

HTH,

baptiste

On 31 October 2010 16:54, James Hirschorn james.hirsch...@hotmail.com wrote:
 Suppose df is a dataframe with one named row of numeric observations. I want
 to coerce df into a named vector.



 as.vector does not work as I expected: as.vector(df) returns the original
 dataframe, while as.vector(df,mode=numeric) returns an unnamed vector of
 NAs.



 This works:



 v - as.numeric(as.matrix(df)); names(v) - names(df);



 I just wanted check if there was a better/more natural way of doing this?


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Re: [R] extracting named vector from dataframe

2010-10-31 Thread David Winsemius


On Oct 31, 2010, at 11:54 AM, James Hirschorn wrote:

Suppose df is a dataframe with one named row of numeric  
observations. I want

to coerce df into a named vector.


I don't think you understand the structure of dataframes. They are  
named lists of component columns. The names you are attributing to the  
rows are not attached to the observations but rather are column names.  
So that row is not in any sense a named vector. If you created a  
dataframe with the first column a named vector its names would become  
the rownames.




as.vector does not work as I expected: as.vector(df) returns the  
original
dataframe, while as.vector(df,mode=numeric) returns an unnamed  
vector of

NAs.

This works:


v - as.numeric(as.matrix(df)); names(v) - names(df);


Right. You are now assigning the column names to the elements in a  
row, but in some ways that is an unnatural act, and not something that  
would be expected to work in the general case where a row might be a  
diverse set of types and even different classes. Your as.matrix  
operation coerced all of the values to be of one type.


I just wanted check if there was a better/more natural way of doing  
this?


--

David Winsemius, MD
West Hartford, CT

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Re: [R] Randomly split a sample in two equal subsamples

2010-10-31 Thread Wu Gong

Hi Yoan,

Please try ?sample.

Suppose you have 1:n ids of total observations where n is even, you want to
randomly split it into two subsamples, the following code should work.

n - 20
one.sample - sort(sample(1:n, n/2))
another.sample - (1:n)[-one.sample]


Good luck.

Wu

-
A R learner.
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Re: [R] Need help with lmer model specification syntax for nested mixed model

2010-10-31 Thread Douglas Bates
On Sun, Oct 31, 2010 at 2:35 AM, Carabiniero ja...@troutnut.com wrote:

 I haven't been able to fully make sense of the conflicting online information
 about whether and how to specify nesting structure for a nested, mixed
 model.  I'll describe my experiment and hopefully somebody who knows lme4
 well can help.

 We're measuring the fluorescence intensity of brain slices from frogs that
 have undergone various treatments.  We want to use multcomp to look for
 differences between treatments, while accounting for the variance introduced
 by the random effects of brain and slice.  There are a few measurements per
 slice, several slices per brain, and several brains per treatment.  In the
 data file, the numbering for slices starts over from 1 for each brain, and
 the numbering for brains starts over from 1 for each treatment.

This is what I call implicit nesting in the definition of the
variables.  My general recommendation is to create new variables that
reflect the actual structure of the data, as in

mydata - within(mydata, {
ubrain - factor(Treatment:Brain)
uslice - factor(Treatment:Brain:Slice)
}

then define the model in terms of these factors, ubrain and uslice,
that have the desirable property that each distinct brain has a
distinct label.

 In other words:  Treatment is a fixed effect, brain is a random effect
 nested in treatment, and slice is a random effect nested in brain.

 As I understood the documentation, this is the correct specification:

 log(Intensity) ~ Treatment + (1|Brain) + (1|Slice)

That will work with ubrain and uslice instead of the implicitly nested
Brain and Slice.

 However, I don't see how lmer understands the correct nesting structure from
 that.  How does it know brain isn't crossed with treatment?

lmer can determine the crossed or nested structure from the data
whenever the data reflect the structure.  Implicitly nested factors
don't reflect the structure of the data and rely on external
information to augment the data given.

The computational methods used in lmer don't depend on whether the
grouping factors for the random effects are nested or not.  However
they do require that the grouping factors are well-defined.

 Here are two other things I tried, and each gave different results:

 log(Intensity) ~ Treatment + (1|Slice/Brain/Treatment)
 log(Intensity) ~ Treatment + (1|Brain/Treatment) + (1|Slice/Brain)

 I'm not sure why these things give different results, or which one (if any)
 is right.  Can anyone help?

I have taken the liberty of cc:ing the R-SIG-Mixed-Models mailing list
on this reply and suggest that any follow-ups be on that list.

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Re: [R] Constrained Regression

2010-10-31 Thread Spencer Graves

Have you tried the 'sos' package?


install.packages('sos') # if not already installed
library(sos)
cr - ???'constrained regression' # found 149 matches
summary(cr) # in 69 packages
cr # opens a table in a browser listing all 169 matches with links to 
the help pages



  However, I agree with Ravi Varadhan:  I'd want to understand the 
physical mechanism generating the data.  If each is, for example, a 
proportion, then I'd want to use logistic regression, possible after 
some approximate logistic transformation of X1 and X2 that prevents 
logit(X) from going to +/-Inf.  This is a different model, but it 
achieves the need to avoid predictions of Y going outside the range (0, 1).



  Spencer


On 10/31/2010 9:01 AM, David Winsemius wrote:


On Oct 31, 2010, at 2:44 AM, Jim Silverton wrote:


Hello everyone,
I have 3 variables Y, X1 and X2. Each variables lies between 0 and 1. 
I want

to do a constrained regression such that a0 and (1-a) 0

for the model:

Y = a*X1 + (1-a)*X2



It would not accomplish the constraint that a  0 but you could 
accomplish the other constraint within an lm fit:


X3 - X1-X2
lm(Y ~ X3 + offset(X2) )

Since beta1 is for the model Y ~ 1 + beta1(X1- X2) + 1*X2)
 Y ~ intercept + beta1*X1 + (1 -beta1)*X2

... so beta1 is a.

In the case beta  0 then I suppose a would be assigned 0. This might 
be accomplished within an iterative calculation framework by a large 
penalization for negative values. In a reply (1) to a question by 
Carlos Alzola in 2008 on rhalp, Berwin Turlach offered a solution to a 
similar problem ( sum(coef) == 1 AND coef non-negative). Modifying his 
code to incorporate the above strategy (and choosing two variables for 
which parameter values might be inside the constraint boundaries) we get:


library(MASS)   ## to access the Boston data
  designmat - model.matrix(medv~I(age-lstat) +offset(lstat), 
data=Boston)

  Dmat -crossprod(designmat, designmat); dvec - crossprod(designmat,
  Boston$medv)
  Amat - cbind(1, diag(NROW(Dmat)))
  bvec - c(1,rep(0,NROW(Dmat)))
  meq - 1
library(quadprog)
  res - solve.QP(Dmat, dvec, Amat, bvec, meq)

 zapsmall(res$solution)
[1] 0.686547 0.313453

Turlach specifically advised against any interpretation of this 
particular result which was only contructed to demonstrate the 
mathematical mechanics.




I tried the help on the constrained regression in R but I concede 
that it

was not helpful.


I must not have that package installed because I got nothing that 
appeared to be useful with ??constrained regression .



David Winsemius, MD
West Hartford, CT

1) http://finzi.psych.upenn.edu/Rhelp10/2008-March/155990.html

__
R-help@r-project.org mailing list
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http://www.R-project.org/posting-guide.html

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--
Spencer Graves, PE, PhD
President and Chief Operating Officer
Structure Inspection and Monitoring, Inc.
751 Emerson Ct.
San José, CA 95126
ph:  408-655-4567

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

2010-10-31 Thread govindas


I am sorry, i think the link was broken..! here is the correct one!!!

http://www.4shared.com/file/4zV0g3JR/RF_80-85.html  


[[alternative HTML version deleted]]

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[R] complicated graphic -- persp+map

2010-10-31 Thread claudia tebaldi
Hello

I'm trying to render in 3D what I usually plot by image(), or image.plot()
from the library fields, followed by a map(world,add=TRUE) type of
command. More concretely, I have a field of temperature values, for a given
geographic area, and I would like to plot a 3D surface whose x and y axes
consist of longitude and latitude values and whose height and color-coding
correspond to the temperature values but also -- and here is the problem --
superimpose a map of the region.
persp() can easily do the color-coded surface but I cannot figure out how
(if at all possible) to overlay geographic boundaries, in this specific case
a simple map of the land area outlines.

Thank you in advance for any help

Claudia



-- 
Claudia Tebaldi
Research Scientist, Climate Central
http://www.climatecentral.org
(303) 775 5365 c  (Colorado area code)

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Re: [R] make many barplot into one plot

2010-10-31 Thread Thomas Levine
hierobarp or barNest from {plotrix} may do this more neatly.

2010/10/31 Sibylle Stöckli sibylle.stoec...@gmx.ch

 Dear R users


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

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

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

 Thanks
 Sibylle



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

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

2010-10-31 Thread David Winsemius


On Oct 31, 2010, at 12:54 PM, Spencer Graves wrote:


Have you tried the 'sos' package?


I have, and I am taking this opportunity to load it with my .Rprofile  
to make it more accessible. It works very well. Very clean display. I  
also have constructed a variant of RSiteSearch that I find more useful  
than the current defaults.


rhelpSearch - function(string,
  restrict = c(Rhelp10, Rhelp08, Rhelp02,  
functions ),

   matchesPerPage = 100, ...) {
  RSiteSearch(string=string,
  restrict = restrict,
  matchesPerPage = matchesPerPage, ...)}



install.packages('sos') # if not already installed
library(sos)
cr - ???'constrained regression' # found 149 matches
summary(cr) # in 69 packages
cr # opens a table in a browser listing all 169 matches with links  
to the help pages


 However, I agree with Ravi Varadhan:  I'd want to understand  
the physical mechanism generating the data.  If each is, for  
example, a proportion, then I'd want to use logistic regression,  
possible after some approximate logistic transformation of X1 and X2  
that prevents logit(X) from going to +/-Inf.  This is a different  
model, but it achieves the need to avoid predictions of Y going  
outside the range (0, 1).


No argument. I defer to both of your greater experiences in such  
problems and your interest in educating us less knowledgeable users. I  
also need to amend my suggested strategy in situations where a linear  
model _might_ be appropriate, since I think the inclusion of the  
surrogate variable in the solve.QP setup is very probably creating  
confusion. After reconsideration I think one should keep the two  
approaches separate. These are two approaches to the non-intercept  
versions of the model that yield the same estimate (but only because  
the constraints do not get invoked ):


 lm(medv~I(age-lstat) +offset(lstat) -1, data=Boston)

Call:
lm(formula = medv ~ I(age - lstat) + offset(lstat) - 1, data = Boston)

Coefficients:
I(age - lstat)
0.1163

 library(MASS)   ## to access the Boston data
  designmat - model.matrix(medv~age+lstat-1, data=Boston)
  Dmat -crossprod(designmat, designmat); dvec - crossprod(designmat,
+  Boston$medv)
   Amat - cbind(1, diag(NROW(Dmat)));  bvec -  
c(1,rep(0,NROW(Dmat))); meq - 1

  library(quadprog);
  res - solve.QP(Dmat, dvec, Amat, bvec, meq)
 zapsmall(res$solution)  # zapsmall not really needed in this instance
[1] 0.1163065 0.8836935

--
David.



 Spencer


On 10/31/2010 9:01 AM, David Winsemius wrote:


On Oct 31, 2010, at 2:44 AM, Jim Silverton wrote:


Hello everyone,
I have 3 variables Y, X1 and X2. Each variables lies between 0 and  
1. I want

to do a constrained regression such that a0 and (1-a) 0

for the model:

Y = a*X1 + (1-a)*X2



It would not accomplish the constraint that a  0 but you could  
accomplish the other constraint within an lm fit:


X3 - X1-X2
lm(Y ~ X3 + offset(X2) )

Since beta1 is for the model Y ~ 1 + beta1(X1- X2) + 1*X2)
Y ~ intercept + beta1*X1 + (1 -beta1)*X2

... so beta1 is a.

In the case beta  0 then I suppose a would be assigned 0. This  
might be accomplished within an iterative calculation framework by  
a large penalization for negative values. In a reply (1) to a  
question by Carlos Alzola in 2008 on rhalp, Berwin Turlach offered  
a solution to a similar problem ( sum(coef) == 1 AND coef non- 
negative). Modifying his code to incorporate the above strategy  
(and choosing two variables for which parameter values might be  
inside the constraint boundaries) we get:


library(MASS)   ## to access the Boston data
 designmat - model.matrix(medv~I(age-lstat) +offset(lstat),  
data=Boston)

 Dmat -crossprod(designmat, designmat); dvec - crossprod(designmat,
 Boston$medv)
 Amat - cbind(1, diag(NROW(Dmat)))
 bvec - c(1,rep(0,NROW(Dmat)))
 meq - 1
library(quadprog)
 res - solve.QP(Dmat, dvec, Amat, bvec, meq)

 zapsmall(res$solution)
[1] 0.686547 0.313453

Turlach specifically advised against any interpretation of this  
particular result which was only contructed to demonstrate the  
mathematical mechanics.




I tried the help on the constrained regression in R but I concede  
that it

was not helpful.


I must not have that package installed because I got nothing that  
appeared to be useful with ??constrained regression .



David Winsemius, MD
West Hartford, CT

1) http://finzi.psych.upenn.edu/Rhelp10/2008-March/155990.html

__


David Winsemius, MD
West Hartford, CT

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Re: [R] Randomly split a sample in two equal subsamples

2010-10-31 Thread yoan

Thanks, but I just don't know how to translate that to a dataset with
rows and columns.
Initially, I was thinking about something like that:

# Create some data:
a - c(10,20,15,43,76,41,25,46)
b - factor(c(m, w, m, w, m, w, m, w))
c - c(2,5,8,3,6,1,5,6)
number - c(1:8)
myframe - data.frame(a,b,c, number)

# Randomly sample a subset of numbers:
v1 - sample(number, 4, replace=FALSE)
v2 - number[-v1]

# Use the subset command like this:
firsthalf - subset(myframe, number=v1)

Of course, the last line doesn't work. Is this generally a wrong
approach, or is just my writing wrong?
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Re: [R] Randomly split a sample in two equal subsamples

2010-10-31 Thread Wu Gong

firsthalf - myframe[v1,]

or

firsthalf - subset(myframe, number %in% v1) 


-
A R learner.
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Re: [R] extracting named vector from dataframe

2010-10-31 Thread James Hirschorn
Of course you are right that this would not be appropriate in general, but
what I'm doing--which as Baptiste explained can be done much more nicely
with unlist()---seems reasonable in my context: The dataframe has a computed
statistic for each input, but I need a vector so that I can do operations
such as sort().

-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net] 
Sent: Sunday, October 31, 2010 12:24 PM
To: James Hirschorn
Cc: R-help@r-project.org
Subject: Re: [R] extracting named vector from dataframe


On Oct 31, 2010, at 11:54 AM, James Hirschorn wrote:

 Suppose df is a dataframe with one named row of numeric  
 observations. I want
 to coerce df into a named vector.

I don't think you understand the structure of dataframes. They are  
named lists of component columns. The names you are attributing to the  
rows are not attached to the observations but rather are column names.  
So that row is not in any sense a named vector. If you created a  
dataframe with the first column a named vector its names would become  
the rownames.


 as.vector does not work as I expected: as.vector(df) returns the  
 original
 dataframe, while as.vector(df,mode=numeric) returns an unnamed  
 vector of
 NAs.

 This works:

 v - as.numeric(as.matrix(df)); names(v) - names(df);

Right. You are now assigning the column names to the elements in a  
row, but in some ways that is an unnatural act, and not something that  
would be expected to work in the general case where a row might be a  
diverse set of types and even different classes. Your as.matrix  
operation coerced all of the values to be of one type.

 I just wanted check if there was a better/more natural way of doing  
 this?

--

David Winsemius, MD
West Hartford, CT

__
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Re: [R] Downloading Package Sources

2010-10-31 Thread Uwe Ligges



On 31.10.2010 11:30, Santosh Srinivas wrote:

Dear Group, any idea how I can download the source code for all packages in
Windows 7?


Either apply wget on yourCRANmirror/src/contrib/

or

- choose CRAN mirror
- select CRAN as the only repository (unless yoiu want other packages as 
well)

- ask R to
download.packages(available.packages(type=source)[,1], type=source, 
destdir=.)


Maybe you want to redefine the filters. for available.packages. See its 
help page for details.


Uwe Ligges






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[R] Questions about Probit Analysis

2010-10-31 Thread Lorenzo Isella

Dear All,
I have some questions about probit regressions.
I saw a nice introduction at

http://bit.ly/bU9xL5

and I mainly have two questions.

(1) The first is almost about data manipulation. Consider the following 
snippet


##

mydata - read.csv(url(http://www.ats.ucla.edu/stat/r/dae/binary.csv;))
names(mydata) - c(outcome,x1,x2,x3)

myprobit - glm(mydata$outcome~mydata$x1+mydata$x2+as.factor(mydata$x3), 
family=binomial(link=probit))


print(summary(myprobit))


#Now assume I can make a regression only on x1


myprobit2 - glm(mydata$outcome~mydata$x1, family=binomial(link=probit))

print(summary(myprobit2))

#express in terms of counts

md - t(table(mydata$outcome, mydata$x1))

# create new dataframe


mydatanew - data.frame(as.numeric(row.names(md)))

names(mydatanew) - c(x1)

mydatanew$successes -as.numeric(md[ ,2])

mydatanew$failures -as.numeric(md[ ,1])




where first I carry out a logit regression of the binary outcome (i.e. 
taking only 0/1 as values) on 3 regressors, then I simply regress the 
outcome on the x1 variable.


Finally, I generate the data frame mydatanew (see some of its entries below)

 mydatanew
x1 successes failures
1  220 01
2  300 12
3  340 13
4  360 04
5  380 08
[...]

where for every value of x1 I count the number of 0 and 1 outcomes 
(namely number of failures and number of successes). This is equivalent 
to having a full list of x1 values with an associated 0/1 outcome (I 
have simply counted them) hence it is all the info I need to again 
perform a logit regression of the binary outcome on x1, but the data 
format is now different. How can I actually feed R with mydatanew to 
perform again a logistic regression on x1 only?
(2) This is a bit more conceptual. Let us say that you have a set of 
products A,B,C,D,E,F. Each product has a list of features: x_A for 
product A, x_B for B etc...
Each customer has its own set of parameters (age, sex, income etc..) I 
call x_cust. Finally, the customer is confronted with two products (e.g. 
A and D; combinations may vary, I call each combination of two products 
a scenario) and asked which one he would like to buy. Bottom line: your 
data are in the format



1 x_A x_cust
0 x_D x_cust

meaning that a certain customer chose product A against product D; similarly

1 x_C x_cust
0 x_B x_cust

would mean that the customer choosing between C and B finally selected 
C.  Every customer needs to choose a product in a variety of different 
scenarios.  How would you analyze this kind of data? Is there any way I 
can express, in my probit analysis, the fact that my binary outcome (but 
this product or not) arises always from the comparison of two products 
only (customers are never given a choice between more than two products 
in a given scenario). Or should I simply run my logistic regression on 
my 0/1 outcome without any extra worry (like in the snippet above)?

Many thanks

Lorenzo

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Re: [R] Randomly split a sample in two equal subsamples

2010-10-31 Thread yoan

Thanks, it works!
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Re: [R] extracting named vector from dataframe

2010-10-31 Thread Gabor Grothendieck
On Sun, Oct 31, 2010 at 12:11 PM, baptiste auguie
baptiste.aug...@googlemail.com wrote:
 Hi,

 I think you want ?unlist

 d = data.frame(x=1, y=2, z=3)
 v = unlist(d)
 is(v)
 [1] numeric vector


Here are a few other possibilities too:

   drop(as.matrix(d))

   do.call(c, d)

   sapply(d, identity)

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email: ggrothendieck at gmail.com

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[R] latex on summary.formula from Hmisc doesn't remove N

2010-10-31 Thread Thijs Muizelaar


Hi All,

I'm trying to get a summary table of my datasets, in which I want the 
mean of different groups calculated and presented in a table. I want 
this table to easily be exported to LaTeX. However, I'm not able to 
remove the N in this summary table. It gives the following error:

Error in 1:lt : argument of length 0

What have I tried:
 tmp-summary.formula(traveltime ~ guidanceID + gender, data=event2, 
fun=mean, method='cross', prn=FALSE)


 latex(tmp, file=, title=,cdec=0, prn = FALSE, booktabs=TRUE)
Error in 1:lt : argument of length 0

Without prn = FALSE or with prn = TRUE, the latex function works, 
but gives the unwanted result with the number of values. Using:

 print(tmp, prn=FALSE)
gives the proper table, but this cannot be convert to a LaTeX table.

Am I doing something wrong here? Or is it a bug in Hmisc?

Any ideas?

Thijs

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[R] transfer string to expression

2010-10-31 Thread Yilong Zhang
Dear all:

when I use parse() there is some problems. Below is an example:

b0-1
b1-1
x-1
str2expr-function(x){eval(parse(text=x))}
test1-b0+b1*sqrt(x)
test2-b0+b1
str2expr(test1)
str2expr(test2)

it can work well for test2 but not for test1.

Could you tell me how to fix this problem or is there other more stable
method to transfer an string to expression?

best regards

Elong

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Re: [R] transfer string to expression

2010-10-31 Thread Duncan Murdoch

On 31/10/2010 3:22 PM, Yilong Zhang wrote:

Dear all:

when I use parse() there is some problems. Below is an example:

b0-1
b1-1
x-1
str2expr-function(x){eval(parse(text=x))}
test1-b0+b1*sqrt(x)
test2-b0+b1
str2expr(test1)
str2expr(test2)

it can work well for test2 but not for test1.

Could you tell me how to fix this problem or is there other more stable
method to transfer an string to expression?


You are evaluating those expressions in the evaluation frame of your 
function, where x is the expression.  Just specify the environment in 
which you want to evaluate them, e.g.


str2expr-function(x){eval(parse(text=x), envir=parent.frame() )}

Duncan Murdoch



best regards

Elong

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Re: [R] transfer string to expression

2010-10-31 Thread David Winsemius


On Oct 31, 2010, at 3:22 PM, Yilong Zhang wrote:


Dear all:

when I use parse() there is some problems. Below is an example:

b0-1
b1-1
x-1
str2expr-function(x){eval(parse(text=x))}
test1-b0+b1*sqrt(x)
test2-b0+b1
str2expr(test1)
str2expr(test2)


I don't think the scoping rules are up to the task of keeping the x's  
distinct:


 b0-1
 b1-1
 xx-1
 str2expr-function(x){eval(parse(text=x))}
 test1-b0+b1*sqrt(xx)
 test2-b0+b1
 str2expr(test1)
[1] 2
 str2expr(test2)
[1] 2



it can work well for test2 but not for test1.

Could you tell me how to fix this problem or is there other more  
stable

method to transfer an string to expression?


I generally try to keep the variables distinct so my brain can handle  
the various levels. So I would probably not have used x in that  
manner. I think it made the interpreter attempt to take the sqrt of  
b0+b1*sqrt(x), which gave you the Error in sqrt(x) : Non-numeric  
argument to mathematical function


And apropos is this:
 fortune(If the answer is parse())

If the answer is parse() you should usually rethink the question.
   -- Thomas Lumley
  R-help (February 2005)

--
David Winsemius, MD
West Hartford, CT

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Re: [R] complicated graphic -- persp+map

2010-10-31 Thread Duncan Murdoch

On 31/10/2010 1:29 PM, claudia tebaldi wrote:

Hello

I'm trying to render in 3D what I usually plot by image(), or image.plot()
from the library fields, followed by a map(world,add=TRUE) type of
command. More concretely, I have a field of temperature values, for a given
geographic area, and I would like to plot a 3D surface whose x and y axes
consist of longitude and latitude values and whose height and color-coding
correspond to the temperature values but also -- and here is the problem --
superimpose a map of the region.
persp() can easily do the color-coded surface but I cannot figure out how
(if at all possible) to overlay geographic boundaries, in this specific case
a simple map of the land area outlines.


This is possible in rgl; see example(persp3d).  (This is working in R 
2.11.1, but not in the CRAN build of R 2.12.0 at the moment.  If you 
build it yourself it works there.)


Duncan Murdoch

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Re: [R] transfer string to expression

2010-10-31 Thread Wu Gong

Hi Duncan:

I'm curious about the environment setting. ?eval says: 

If envir is not specified, then the default is parent.frame() (the
environment where the call to eval was made). 

So what's the difference between set envir=parent.frame() or not?

Thank you.

Wu

-
A R learner.
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Re: [R] transfer string to expression

2010-10-31 Thread Duncan Murdoch

On 31/10/2010 4:47 PM, Wu Gong wrote:


Hi Duncan:

I'm curious about the environment setting. ?eval says:

If envir is not specified, then the default is parent.frame() (the
environment where the call to eval was made). 

So what's the difference between set envir=parent.frame() or not?


If you pass parent.frame() as an argument, it will be one level up: 
it's the parent of the function that calls eval.  If you leave it at the 
default, it's the parent of eval, i.e. the function that calls it.


Duncan Murdoch

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

2010-10-31 Thread steven mosher
Ok I downloaded it and showed you how to get your data out. How to read it
into a raster brick,
how to plot the data, how to get the mean rainfall of every day.lots more
you can do.

there is a  bad bit of data in the last time step.

check my blog.

In the future what you should do is write code to emulate your problem. for
example, in your problem you had created a ncdf file with a 3D matrix of
65,69,2192.
You should just do a subset of that, show the code to create a ncdf with
random numbers in it.

creating working code that emulates your problem is key if you want help.

Off list for the rest.

On Sun, Oct 31, 2010 at 10:21 AM, govin...@msu.edu wrote:



 I am sorry, i think the link was broken..! here is the correct one!!!

 http://www.4shared.com/file/4zV0g3JR/RF_80-85.html


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Re: [R] For loop

2010-10-31 Thread Joshua Wiley
Hi,

Using the code below I got almost a 60% speedup.  Obviously this is
largely dependent on there being relatively fewer columns than rows.

HTH,

Josh


#
## Create data
h - structure(c(0.25, 0.25, 0, 0, 0, 0, -0.25, -0.25, -0.25, -0.25,
-0.5, -0.5, 0, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25,
0.25), .Dim = c(1L, 22L), .Dimnames = list(NULL, NULL))
xx - seq(0, 1, 0.5)
v - matrix(nrow=20001,ncol=22)
vv - matrix(nrow=20001,ncol=22)
## Create duplicates for comparison
h.2 - h; xx.2 - xx; v.2 - v; vv.2 - vv

## Old
system.time(for (y in 1:20001) {v[y, 22] - h[1, 22]})
## New
system.time(v.2[, 22] - h.2[1, 22])

## no speedup possible here
vv[20001,22] - v[20001,22]
vv.2[20001,22] - v.2[20001,22]


## Old
system.time(
for(k in 21:1) {
 for(j in 20001:2) {
  vv[j-1, k + 1] - min(xx[j-1] * v[j-1,k+1], vv[j,k+1])
  v[j, k] - h[1, k] + vv[j-1, k+1]
 }
vv[20001, k] - v[20001, k]
})
#   user  system elapsed
# 20.733   0.044  25.869

## New
system.time(for(k in 21:1) {
  tmp - xx.2 * v.2[, k+1]
 for(j in 20001:2) {
  vv.2[j-1, k + 1] - min(tmp[j-1], vv.2[j, k+1])
 }
 v.2[-1, k] - h.2[1, k] + vv.2[-20001, k+1]
 vv.2[20001, k] - v.2[20001, k]
})
#   user  system elapsed
#  9.184   0.012  10.500

## Check that new is identical to old
identical(vv, vv.2)
identical(v, v.2)

On Sun, Oct 31, 2010 at 3:58 AM, Astabenefica astabenef...@hotmail.it wrote:
 Thanks for the suggestions.
 I add some more detail to clarify:

 h=matrix(nrow=1,ncol=22)

 In my data h is:
 (0.25  0.25  0  0  0  0  -0.25  -0.25  -0.25  -0.25  -0.5  -0.5  0  0.25  
 0.25  0.25  0.25  0.25  0.25  0.25  0.25  0.25)



 xx-seq(0,1,0.5)

 v=matrix(nrow=20001,ncol=22)
 vv=matrix(nrow=20001,ncol=22)
 for (y in 1:20001) {
 v[y,22]=h[1,22]
 }
 vv[20001,22]=v[20001,22]
 for(k in 21:1) {
  for(j in 20001:2) {
   vv[j-1,k+1]=min(xx[j-1]*v[j-1,k+1],vv[j,k+1])
   v[j,k]=h[1,k]+vv[j-1,k+1]
  }
 vv[20001,k]=v[20001,k]
 }



 The idea of using Rcpp seems to be good. Having never used this package will 
 give a look.
 Somewhere I read an example like this:


 for (i in 1:R) {
  res[i]-f()
  NULL
 }

 where f() is a function, but I don't how can I trasform my code:
 vv[j-1,k+1]=min(xx[j-1]*v[j-1,k+1],vv[j,k+1])
 v[j,k]=h[1,k]+vv[j-1,k+1]
 in a function








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

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Re: [R] For loop

2010-10-31 Thread Joshua Wiley
Actually, I'm able to gain another second by limiting -/+s in the
subscripts (maybe just an artifact?  I ran it a couple of times to
check, but still).


system.time(for(k in 22:2) {
  tmp - xx.2 * v.2[, k]
 for(j in 2:1) {
  vv.2[j, k] - min(tmp[j], vv.2[j+1, k])
 }
 v.2[-1, k-1] - h.2[1, k-1] + vv.2[-20001, k]
 vv.2[20001, k-1] - v.2[20001, k-1]
})


On Sun, Oct 31, 2010 at 2:39 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
 Hi,

 Using the code below I got almost a 60% speedup.  Obviously this is
 largely dependent on there being relatively fewer columns than rows.

 HTH,

 Josh


 #
 ## Create data
 h - structure(c(0.25, 0.25, 0, 0, 0, 0, -0.25, -0.25, -0.25, -0.25,
 -0.5, -0.5, 0, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25,
 0.25), .Dim = c(1L, 22L), .Dimnames = list(NULL, NULL))
 xx - seq(0, 1, 0.5)
 v - matrix(nrow=20001,ncol=22)
 vv - matrix(nrow=20001,ncol=22)
 ## Create duplicates for comparison
 h.2 - h; xx.2 - xx; v.2 - v; vv.2 - vv

 ## Old
 system.time(for (y in 1:20001) {v[y, 22] - h[1, 22]})
 ## New
 system.time(v.2[, 22] - h.2[1, 22])

 ## no speedup possible here
 vv[20001,22] - v[20001,22]
 vv.2[20001,22] - v.2[20001,22]


 ## Old
 system.time(
 for(k in 21:1) {
  for(j in 20001:2) {
  vv[j-1, k + 1] - min(xx[j-1] * v[j-1,k+1], vv[j,k+1])
  v[j, k] - h[1, k] + vv[j-1, k+1]
  }
 vv[20001, k] - v[20001, k]
 })
 #   user  system elapsed
 # 20.733   0.044  25.869

 ## New
 system.time(for(k in 21:1) {
  tmp - xx.2 * v.2[, k+1]
  for(j in 20001:2) {
  vv.2[j-1, k + 1] - min(tmp[j-1], vv.2[j, k+1])
  }
  v.2[-1, k] - h.2[1, k] + vv.2[-20001, k+1]
  vv.2[20001, k] - v.2[20001, k]
 })
 #   user  system elapsed
 #  9.184   0.012  10.500

 ## Check that new is identical to old
 identical(vv, vv.2)
 identical(v, v.2)

 On Sun, Oct 31, 2010 at 3:58 AM, Astabenefica astabenef...@hotmail.it wrote:
 Thanks for the suggestions.
 I add some more detail to clarify:

 h=matrix(nrow=1,ncol=22)

 In my data h is:
 (0.25  0.25  0  0  0  0  -0.25  -0.25  -0.25  -0.25  -0.5  -0.5  0  0.25  
 0.25  0.25  0.25  0.25  0.25  0.25  0.25  0.25)



 xx-seq(0,1,0.5)

 v=matrix(nrow=20001,ncol=22)
 vv=matrix(nrow=20001,ncol=22)
 for (y in 1:20001) {
 v[y,22]=h[1,22]
 }
 vv[20001,22]=v[20001,22]
 for(k in 21:1) {
  for(j in 20001:2) {
   vv[j-1,k+1]=min(xx[j-1]*v[j-1,k+1],vv[j,k+1])
   v[j,k]=h[1,k]+vv[j-1,k+1]
  }
 vv[20001,k]=v[20001,k]
 }



 The idea of using Rcpp seems to be good. Having never used this package will 
 give a look.
 Somewhere I read an example like this:


 for (i in 1:R) {
  res[i]-f()
  NULL
 }

 where f() is a function, but I don't how can I trasform my code:
 vv[j-1,k+1]=min(xx[j-1]*v[j-1,k+1],vv[j,k+1])
 v[j,k]=h[1,k]+vv[j-1,k+1]
 in a function








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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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[R] parallel for loop

2010-10-31 Thread sachinthaka . abeywardana

Hi all,

Just following on from a previous thread (for loop). Is there a parallel
'for' loop like matlab (parfor maybe?). I know there was a Nvidia GPU
version for blas somewhere. But is there a CPU or a GPU version of the for
loop?

Thanks,
Sachin
p.s. sorry about the corporate notice below: cant get rid of it. Don't have
other mail client access at the office :(

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


Re: [R] parallel for loop

2010-10-31 Thread Jeffrey Spies
Take a look at the parallel computing section of:

http://cran.r-project.org/web/views/HighPerformanceComputing.html

specifically the line concerning the foreach package.

Jeff.

On Sun, Oct 31, 2010 at 6:38 PM,
sachinthaka.abeyward...@allianz.com.au wrote:

 Hi all,

 Just following on from a previous thread (for loop). Is there a parallel
 'for' loop like matlab (parfor maybe?). I know there was a Nvidia GPU
 version for blas somewhere. But is there a CPU or a GPU version of the for
 loop?

 Thanks,
 Sachin
 p.s. sorry about the corporate notice below: cant get rid of it. Don't have
 other mail client access at the office :(

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[R] Rserve error

2010-10-31 Thread Anand Bambhania
Hi all,

I'm trying to run Rserve on windows RGui. It installs successfully but when
I use Rserve() to invoke the service it shows following error:

The program can't start because R.dll is missing from your computer. Try
reinstalling the program to fix this problem. I even tried reinstalling R
but it still shows the same error.

Thanks,

Anand

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[R] how to save this result in a vector

2010-10-31 Thread Changbin Du
HI, Dear R community,

I have the following codes to calculate the commulative coverage. I want to
save the output in a vector, How to do this?

test-seq(10, 342, by=2)

#cover is a vector
cover_per-function (cover) {
for (i in min(cover):max(cover)) {print(100*sum(ifelse(cover = i, 1,
0))/length(cover))}
}

result-cover_per(test)

 result
NULL

Can anyone help me this this?




-- 
Sincerely,
Changbin
--

[[alternative HTML version deleted]]

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Re: [R] how to save this result in a vector

2010-10-31 Thread Joshua Wiley
Hi Changbin,

The yek is that you need to save the results of your for loop rather
than just printing it.  It also may be possible to vectorize your for
loop which might simplify things and speed them up, but I did not look
at that.  Here is one way to save the results, see inline comments for
more details.

Cheers,

Josh


##

test-seq(10, 342, by=2)

#cover is a vector
cover_per - function(cover) {
  ## create a vector to store the results of your for loop
  output - vector(numeric, length(min(cover):max(cover)))
  for (i in min(cover):max(cover)) {
## rather than print()ing the output, assign it to an object
output[i] - 100*sum(ifelse(cover = i, 1, 0))/length(cover)
  }
  ## have the return value from the function be
  ## the object 'output'
  return(output)
}

## here the results will be printed to the screen
## (the results are just the 'output' object)
cover_per(test)
## if you assign it to another object, nothing is printed
## but you can easily print 'result' if you want
result - cover_per(test)

##


On Sun, Oct 31, 2010 at 5:35 PM, Changbin Du changb...@gmail.com wrote:
 HI, Dear R community,

 I have the following codes to calculate the commulative coverage. I want to
 save the output in a vector, How to do this?

 test-seq(10, 342, by=2)

 #cover is a vector
 cover_per-function (cover) {
 for (i in min(cover):max(cover)) {print(100*sum(ifelse(cover = i, 1,
 0))/length(cover))}
 }

 result-cover_per(test)

 result
 NULL

 Can anyone help me this this?




 --
 Sincerely,
 Changbin
 --

        [[alternative HTML version deleted]]

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 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] how to save this result in a vector

2010-10-31 Thread David Winsemius


On Oct 31, 2010, at 8:35 PM, Changbin Du wrote:


HI, Dear R community,

I have the following codes to calculate the commulative coverage.


Not sure exactly what you mean by this. My guess is implemented below.


I want to
save the output in a vector, How to do this?

test-seq(10, 342, by=2)

#cover is a vector
cover_per-function (cover) {
for (i in min(cover):max(cover))


Using either for (i in cover) { ...} or for (i in seq_along(cover) )  
{...} would be more typical.



{print(100*sum(ifelse(cover = i, 1,
0))/length(cover))}
}

result-cover_per(test)


Are you looking for cumsum?

 test-seq(10, 34, by=2)
 100*cumsum(test)/sum(test)
 [1]   3.496503   7.692308  12.587413  18.181818  24.475524   
31.468531  39.160839

 [8]  47.552448  56.643357  66.433566  76.923077  88.111888 100.00

 print(100*cumsum(test)/sum(test), digits=2)
 [1]   3.5   7.7  12.6  18.2  24.5  31.5  39.2  47.6  56.6  66.4   
76.9  88.1 100.0


--

David Winsemius, MD
West Hartford, CT

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Re: [R] how to save this result in a vector

2010-10-31 Thread Joshua Wiley
On Sun, Oct 31, 2010 at 5:44 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
snip
 #cover is a vector
 cover_per - function(cover) {
  ## create a vector to store the results of your for loop
  output - vector(numeric, length(min(cover):max(cover)))
  for (i in min(cover):max(cover)) {
    ## rather than print()ing the output, assign it to an object
    output[i] - 100*sum(ifelse(cover = i, 1, 0))/length(cover)
  }
  ## have the return value from the function be
  ## the object 'output'
  return(output)
 }

I did not catch that i was not necessarily starting at 1 going
sequentially up, so that would have to be done manually (or use cumsum
per David rather than the function you wrote).

cover_per2 - function(cover) {
  output - vector(numeric, length(min(cover):max(cover)))
  j - 1
  for (i in min(cover):max(cover)) {
output[j] - 100*sum(ifelse(cover = i, 1, 0))/length(cover)
j - j + 1
  }
  return(output)
}

Josh

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Re: [R] how to save this result in a vector

2010-10-31 Thread Changbin Du
HI, David, Juan, and Joshua,

Thanks so much for your help!  The following codes works. Appreciated!

test-seq(10, 342, by=2)

#data is a vector
cover_per-function (data) {

output-vector(numeric,length(min(data):max(data)))

for (i in min(data):max(data)) {
  output[i]-(100*sum(ifelse(data = i, 1, 0))/length(data))
  }

return(output)

}


result-cover_per(test)

#***

#data is a vector
cover_per-function (data) {

output-numeric(0)
for (i in min(data):max(data)) {
  x-(100*sum(ifelse(data = i, 1, 0))/length(data))
  output-c(output, x)
   }

return(output)
}


result-cover_per(test)














On Sun, Oct 31, 2010 at 5:46 PM, David Winsemius dwinsem...@comcast.netwrote:


 On Oct 31, 2010, at 8:35 PM, Changbin Du wrote:

  HI, Dear R community,

 I have the following codes to calculate the commulative coverage.


 Not sure exactly what you mean by this. My guess is implemented below.


  I want to
 save the output in a vector, How to do this?

 test-seq(10, 342, by=2)

 #cover is a vector
 cover_per-function (cover) {
 for (i in min(cover):max(cover))


 Using either for (i in cover) { ...} or for (i in seq_along(cover) ) {...}
 would be more typical.


  {print(100*sum(ifelse(cover = i, 1,
 0))/length(cover))}
 }

 result-cover_per(test)


 Are you looking for cumsum?

  test-seq(10, 34, by=2)
  100*cumsum(test)/sum(test)
  [1]   3.496503   7.692308  12.587413  18.181818  24.475524  31.468531
  39.160839
  [8]  47.552448  56.643357  66.433566  76.923077  88.111888 100.00

  print(100*cumsum(test)/sum(test), digits=2)
  [1]   3.5   7.7  12.6  18.2  24.5  31.5  39.2  47.6  56.6  66.4  76.9
  88.1 100.0

 --

 David Winsemius, MD
 West Hartford, CT




-- 
Sincerely,
Changbin
--

Changbin Du
DOE Joint Genome Institute

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Re: [R] how to save this result in a vector

2010-10-31 Thread Changbin Du
Thanks Joshua! Yes, i is not going up sequentially  by 1, as i here is the
raw number of reads for each DNA base. Thanks so much for the great help!


On Sun, Oct 31, 2010 at 6:03 PM, Joshua Wiley jwiley.ps...@gmail.comwrote:

 On Sun, Oct 31, 2010 at 5:44 PM, Joshua Wiley jwiley.ps...@gmail.com
 wrote:
 snip
  #cover is a vector
  cover_per - function(cover) {
   ## create a vector to store the results of your for loop
   output - vector(numeric, length(min(cover):max(cover)))
   for (i in min(cover):max(cover)) {
 ## rather than print()ing the output, assign it to an object
 output[i] - 100*sum(ifelse(cover = i, 1, 0))/length(cover)
   }
   ## have the return value from the function be
   ## the object 'output'
   return(output)
  }

 I did not catch that i was not necessarily starting at 1 going
 sequentially up, so that would have to be done manually (or use cumsum
 per David rather than the function you wrote).

 cover_per2 - function(cover) {
   output - vector(numeric, length(min(cover):max(cover)))
   j - 1
   for (i in min(cover):max(cover)) {
 output[j] - 100*sum(ifelse(cover = i, 1, 0))/length(cover)
j - j + 1
  }
  return(output)
 }

 Josh




-- 
Sincerely,
Changbin
--

Changbin Du
DOE Joint Genome Institute
Bldg 400 Rm 457
2800 Mitchell Dr
Walnut Creet, CA 94598
Phone: 925-927-2856

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Re: [R] Constrained Regression

2010-10-31 Thread Jim Silverton
I thought I would 'add' some meat to the problem I sent.  This is all I know
(1) f = a*X1 + (1-a)*X2
(2) I know n values of f and X1 which happens to be probabilities
(3) I know nothing about X2 except that it also lies in (0,1)
(4) X1 is the probability under the null (fisher's exact test) and X2 is the
alternative. But I have no idea what the alternative to fisher's exact test
can be..
(5) I need to estimate a (which is supposed to be a proportion).
(6) I was thinking about imputing values from (0,1) using  a beta as the
values of X2.

Any help is greatly appreciated.

Jim...

On Sun, Oct 31, 2010 at 1:44 PM, David Winsemius dwinsem...@comcast.netwrote:


 On Oct 31, 2010, at 12:54 PM, Spencer Graves wrote:

  Have you tried the 'sos' package?


 I have, and I am taking this opportunity to load it with my .Rprofile to
 make it more accessible. It works very well. Very clean display. I also have
 constructed a variant of RSiteSearch that I find more useful than the
 current defaults.

 rhelpSearch - function(string,
  restrict = c(Rhelp10, Rhelp08, Rhelp02, functions
 ),
   matchesPerPage = 100, ...) {
  RSiteSearch(string=string,
  restrict = restrict,
  matchesPerPage = matchesPerPage, ...)}



 install.packages('sos') # if not already installed
 library(sos)
 cr - ???'constrained regression' # found 149 matches
 summary(cr) # in 69 packages
 cr # opens a table in a browser listing all 169 matches with links to the
 help pages

 However, I agree with Ravi Varadhan:  I'd want to understand the
 physical mechanism generating the data.  If each is, for example, a
 proportion, then I'd want to use logistic regression, possible after some
 approximate logistic transformation of X1 and X2 that prevents logit(X) from
 going to +/-Inf.  This is a different model, but it achieves the need to
 avoid predictions of Y going outside the range (0, 1).


 No argument. I defer to both of your greater experiences in such problems
 and your interest in educating us less knowledgeable users. I also need to
 amend my suggested strategy in situations where a linear model _might_ be
 appropriate, since I think the inclusion of the surrogate variable in the
 solve.QP setup is very probably creating confusion. After reconsideration I
 think one should keep the two approaches separate. These are two approaches
 to the non-intercept versions of the model that yield the same estimate (but
 only because the constraints do not get invoked ):

  lm(medv~I(age-lstat) +offset(lstat) -1, data=Boston)

 Call:
 lm(formula = medv ~ I(age - lstat) + offset(lstat) - 1, data = Boston)

 Coefficients:
 I(age - lstat)
0.1163


  library(MASS)   ## to access the Boston data
   designmat - model.matrix(medv~age+lstat-1, data=Boston)

   Dmat -crossprod(designmat, designmat); dvec - crossprod(designmat,
 +  Boston$medv)
Amat - cbind(1, diag(NROW(Dmat)));  bvec - c(1,rep(0,NROW(Dmat)));
 meq - 1
   library(quadprog);
   res - solve.QP(Dmat, dvec, Amat, bvec, meq)
  zapsmall(res$solution)  # zapsmall not really needed in this instance
 [1] 0.1163065 0.8836935

 --
 David.



 Spencer


 On 10/31/2010 9:01 AM, David Winsemius wrote:


 On Oct 31, 2010, at 2:44 AM, Jim Silverton wrote:

  Hello everyone,
 I have 3 variables Y, X1 and X2. Each variables lies between 0 and 1. I
 want
 to do a constrained regression such that a0 and (1-a) 0

 for the model:

 Y = a*X1 + (1-a)*X2



 It would not accomplish the constraint that a  0 but you could
 accomplish the other constraint within an lm fit:

 X3 - X1-X2
 lm(Y ~ X3 + offset(X2) )

 Since beta1 is for the model Y ~ 1 + beta1(X1- X2) + 1*X2)
Y ~ intercept + beta1*X1 + (1 -beta1)*X2

 ... so beta1 is a.

 In the case beta  0 then I suppose a would be assigned 0. This might be
 accomplished within an iterative calculation framework by a large
 penalization for negative values. In a reply (1) to a question by Carlos
 Alzola in 2008 on rhalp, Berwin Turlach offered a solution to a similar
 problem ( sum(coef) == 1 AND coef non-negative). Modifying his code to
 incorporate the above strategy (and choosing two variables for which
 parameter values might be inside the constraint boundaries) we get:

 library(MASS)   ## to access the Boston data
  designmat - model.matrix(medv~I(age-lstat) +offset(lstat), data=Boston)
  Dmat -crossprod(designmat, designmat); dvec - crossprod(designmat,
  Boston$medv)
  Amat - cbind(1, diag(NROW(Dmat)))
  bvec - c(1,rep(0,NROW(Dmat)))
  meq - 1
 library(quadprog)
  res - solve.QP(Dmat, dvec, Amat, bvec, meq)

  zapsmall(res$solution)
 [1] 0.686547 0.313453

 Turlach specifically advised against any interpretation of this
 particular result which was only contructed to demonstrate the mathematical
 mechanics.


 I tried the help on the constrained regression in R but I concede that
 it
 was not helpful.


 I must not have that package installed because I got 

[R] Mean and individual growth curve trajectories

2010-10-31 Thread jlwoodard

I'm trying to understand how to plot individual growth curve trajectories,
with the overall mean trajectory superimposed (preferably in a slightly
thicker line, maybe in black) over the individual trajectories.  Using the
sleepstudy data in lme4, here is the code I have so far:

library(lme4)
library(lattice) 
xyplot(Reaction ~ Days, data = sleepstudy, group = Subject, type = 'l')

This plot produces the individual growth curves nicely, but I'd like to be
able to plot the mean for each day (averaged over subjects) on top of this
graph.

Many thanks in advance for any help/suggestions.

John

-- 
View this message in context: 
http://r.789695.n4.nabble.com/Mean-and-individual-growth-curve-trajectories-tp3021672p3021672.html
Sent from the R help mailing list archive at Nabble.com.

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


Re: [R] Constrained Regression

2010-10-31 Thread Spencer Graves


in line


On 10/31/2010 6:26 PM, Jim Silverton wrote:

I thought I would 'add' some meat to the problem I sent.  This is all I know
(1) f = a*X1 + (1-a)*X2


How do you know f = a*X1 + (1-a)*X2?


Why does this relationship make more sense than, e.g., log(f/(1-f)) = 
a*X1 + (1-a)*X2?



(2) I know n values of f and X1 which happens to be probabilities


What do you mean that f and X1 are probabilities?  Where do they come from?


Are the the results of binomial experiments like tosses of a biased coin?



(3) I know nothing about X2 except that it also lies in (0,1)
(4) X1 is the probability under the null (fisher's exact test) and X2 is the
alternative. But I have no idea what the alternative to fisher's exact test
can be..


How do you compute X1?  Do you compute it from data?  If yes, then it's 
more like an estimate of a probability rather than a probability 
itself.  Ditto for X2.



The preferred method for estimating almost anything whenever feasible is 
to write a likelihood function, i.e., the probability of data as a 
function of unknown parameters, then estimate those parameters to 
maximize the likelihood.  Even most nonparametric procedures can be 
expressed this way for an appropriately chosen probability model.



In most situations, I prefer to eliminate constraints by 
transformations, replacing f by ph = log(f/(1-f), X1 and X2 by xi1 = 
log(X1/(1-X1)) and X2 by xi2 = log(X2/(1-X2)), a by alpha = 
log(a/(1-a)), then write a relationship between these unconstrained 
variables and try to estimate alpha by maximum likelihood.



Hope this helps.
Spencer


(5) I need to estimate a (which is supposed to be a proportion).
(6) I was thinking about imputing values from (0,1) using  a beta as the
values of X2.

Any help is greatly appreciated.

Jim...

On Sun, Oct 31, 2010 at 1:44 PM, David Winsemiusdwinsem...@comcast.netwrote:


On Oct 31, 2010, at 12:54 PM, Spencer Graves wrote:

  Have you tried the 'sos' package?
I have, and I am taking this opportunity to load it with my .Rprofile to
make it more accessible. It works very well. Very clean display. I also have
constructed a variant of RSiteSearch that I find more useful than the
current defaults.

rhelpSearch- function(string,
  restrict = c(Rhelp10, Rhelp08, Rhelp02, functions
),
   matchesPerPage = 100, ...) {
  RSiteSearch(string=string,
  restrict = restrict,
  matchesPerPage = matchesPerPage, ...)}




install.packages('sos') # if not already installed
library(sos)
cr- ???'constrained regression' # found 149 matches
summary(cr) # in 69 packages
cr # opens a table in a browser listing all 169 matches with links to the
help pages

 However, I agree with Ravi Varadhan:  I'd want to understand the
physical mechanism generating the data.  If each is, for example, a
proportion, then I'd want to use logistic regression, possible after some
approximate logistic transformation of X1 and X2 that prevents logit(X) from
going to +/-Inf.  This is a different model, but it achieves the need to
avoid predictions of Y going outside the range (0, 1).


No argument. I defer to both of your greater experiences in such problems
and your interest in educating us less knowledgeable users. I also need to
amend my suggested strategy in situations where a linear model _might_ be
appropriate, since I think the inclusion of the surrogate variable in the
solve.QP setup is very probably creating confusion. After reconsideration I
think one should keep the two approaches separate. These are two approaches
to the non-intercept versions of the model that yield the same estimate (but
only because the constraints do not get invoked ):


lm(medv~I(age-lstat) +offset(lstat) -1, data=Boston)

Call:
lm(formula = medv ~ I(age - lstat) + offset(lstat) - 1, data = Boston)

Coefficients:
I(age - lstat)
0.1163



library(MASS)   ## to access the Boston data
  designmat- model.matrix(medv~age+lstat-1, data=Boston)
  Dmat-crossprod(designmat, designmat); dvec- crossprod(designmat,

+  Boston$medv)

   Amat- cbind(1, diag(NROW(Dmat)));  bvec- c(1,rep(0,NROW(Dmat)));

meq- 1

  library(quadprog);
  res- solve.QP(Dmat, dvec, Amat, bvec, meq)
zapsmall(res$solution)  # zapsmall not really needed in this instance

[1] 0.1163065 0.8836935

--
David.



 Spencer


On 10/31/2010 9:01 AM, David Winsemius wrote:


On Oct 31, 2010, at 2:44 AM, Jim Silverton wrote:

  Hello everyone,

I have 3 variables Y, X1 and X2. Each variables lies between 0 and 1. I
want
to do a constrained regression such that a0 and (1-a)0

for the model:

Y = a*X1 + (1-a)*X2



It would not accomplish the constraint that a  0 but you could
accomplish the other constraint within an lm fit:

X3- X1-X2
lm(Y ~ X3 + offset(X2) )

Since beta1 is for the model Y ~ 1 + beta1(X1- X2) + 1*X2)
Y ~ intercept + beta1*X1 + (1 -beta1)*X2

... so beta1 is a.

In the case beta  0 then I suppose a would be assigned 0. This 

Re: [R] Mean and individual growth curve trajectories

2010-10-31 Thread Michael Bibo
jlwoodard john.woodard at wayne.edu writes:

 
 
 I'm trying to understand how to plot individual growth curve trajectories,
 with the overall mean trajectory superimposed (preferably in a slightly
 thicker line, maybe in black) over the individual trajectories.  Using the
 sleepstudy data in lme4, here is the code I have so far:
 
 library(lme4)
 library(lattice) 
 xyplot(Reaction ~ Days, data = sleepstudy, group = Subject, type = 'l')
 
 This plot produces the individual growth curves nicely, but I'd like to be
 able to plot the mean for each day (averaged over subjects) on top of this
 graph.
 
Is this what you want?

xyplot(Reaction ~ Days, data = sleepstudy, group = Subject, type = 'l',
panel=function(...){
  panel.xyplot(...)
  panel.average(...,fun=mean,horizontal=FALSE,col='red',lwd=3)
}
  )

and have you considered:

xyplot(Reaction ~ Days, data = sleepstudy, group = Subject, type = 'l',
panel=function(...){
  panel.xyplot(...)
  panel.loess(...,fun=mean,horizontal=FALSE,col='red',lwd=3)
}
  )

for a smoother curve?



Hope it helps,

Michael Bibo
Queensland Health

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[R] combining plots (curve + Plot functions)

2010-10-31 Thread mmstat
Hello, 
  
What I really want to do is to add a rejection region in the 
form of a long rectangle to a density plot I have drawn. 
I am getting  2 plots.  

How can I add rectangle to first plot?  see code below. 
First section works fine.  It just is not quite what I want. 

# NORMAL DISTRIBUTION PLOT OF RAW DATA WITH UPPER CRITICAL LEVEL - ok 
xcrit=144.1 
# *** single-sample Upper one-tailed hypothesis Test, z statistic *** 
cord.x - c(xcrit,seq(xcrit,200,0.01),200) 
cord.y - c(0,dnorm(seq(xcrit,200,0.01),140,15),0) 
curve(dnorm(x,140,15),xlim=c(80,200),main='Normal PDF',ylab=Probability) 
polygon(cord.x,cord.y,col='orange') 

# NORMAL DISTRIBUTION PLOT OF RAW DATA WITH UPPER CRITICAL LEVEL - 2 plots? 
xcrit=144.1 
# *** single-sample Upper one-tailed hypothesis Test, z statistic *** 
curve(dnorm(x,140,15),xlim=c(80,200),main='Normal PDF',ylab=Probability) 
plot(c(144.1, 200), c(0, .03), type= n) 

Thanks. 
Mary A. Marion 

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Re: [R] Mean and individual growth curve trajectories

2010-10-31 Thread jlwoodard

Thank you so much, Michael.  This solution is just what I was looking for. 
Many thanks!

John
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