date:20110602



On Jun 2, 2011, at 2:18 AM, teriri wrote:


http://r.789695.n4.nabble.com/file/n3567636/ecdfs.jpg ecdfs.jpg
http://r.789695.n4.nabble.com/file/n3567636/ecdf_curve.gif  
ecdf_curve.gif


Hello,

I have generated a plot of two empirical CDFs (attachment 1). As a  
result,

they are stepwise when plotted. The following code was used:
plot(ecdf(mut), do.points=FALSE, verticals=TRUE, xlim=range(mut,  
non),

col=red)
plot(ecdf(non), do.points=FALSE, verticals=TRUE, add=TRUE,  
col=blue)


But what I need instead are smooth curves, similar to ones that are
generated from a theoretical cdf (attachment 2).

I have looked at so many threads; one suggestion to someone else was  
to use
library(fitdistplot) and look for distributions that may fit (e.g.  
weibull).
But I could really use guidance before spending additional time on  
this.


Two threads up in the answer to the question based on mean and std  
you might find Ellison's answer useful. You and ter...@gmail.com in  
the same class?


--

David Winsemius, MD
West Hartford, CT

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[R] predict with eha package

2011-06-02 Thread Mike Harwood

Hello list, and thank you in advance.

I'm unable to generate predicted values when specifying newdata using
phreg and aftreg models
in the eha package, but I do not have the same problem with a
proportional hazards model from coxph.  Without the newdata argument
the predicted values are returned, but with
newdata=model.dataframe coxph is fine but both aftreg and phreg
models return an  Error in predict.coxph(f.ph.eha,
newdata = mort, type = lp) :  Data is not the same size as it was in
the original fit message.  Since I ultimately want a parametric model
and the real
data is left truncated and right censored, I think the aftreg function
in the eha package is what I must use.  Following is my sample code,
without the output.

#~ All models generated successfully -
f.ph - coxph(Surv(enter, exit, event) ~ ses, data = mort)
f.ph.eha - phreg(Surv(enter, exit, event) ~ ses, data = mort)
f.aft - aftreg(Surv(enter, exit, event) ~ ses, data = mort)

#~ All fits generated successfully ---
f.ph.fit - predict(f.ph, type='lp')
f.ph.eha.fit - predict(f.ph.eha, type='lp')
f.aft.fit - predict(f.aft, type='lp')

#~ First fit generated successfully, others output
error
f.ph.fit - predict(f.ph, newdata=mort, type='lp')
f.ph.eha.fit - predict(f.ph.eha, newdata=mort, type='lp')
f.aft.fit - predict(f.aft, newdata=mort, type='lp')


Mike

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Re: [R] lattice panel fine control

2011-06-02 Thread maxbre

thank you so much for the very detailed indications which turned out to be a
real help in ponting me to the right direction; 

referring back to my previous questions there is something still open:

2- I'm in trouble with the point labels because I would like to rotate them
by an angle of 90 degrees (and I did not find mention of anything like
angle or rot to accomplish this task)
panel.text(x, y, lab = mydata$name, cex = 0.6, pos=3, offset=0.5, here
something to rotate labels)

3- I was referring to the error bar of each point (standard error of
points); I think this could be accomplished by arrows but the following
line is giving me an error
panel.arrows(x=tv.avg-tv.erst, y=ped.avg-ped.erst, x1=tv.avg+tv.erst,
y1=ped.avg+ped.erst, angle=90, code=3)

thanks again for your great help in bootstrapping me to the lattice features

maxbre

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[R] ARM package for R 2.10.1

2011-06-02 Thread jmdpulido

Dear all...

I am looking for the zip file of an old version of the ARM package
compatible for R 2.10.1 version. 

When I try to charge the ARM package I get the following message package
'arm' was built under R version 2.13.0 .

I can not update R to 2.13.0 as I always get this error  the setup files
are corrupted please obtain a new copy of the program.

If you have an old version of the zip file of ARM and LME4 packages that I
can use with R 2.10.1, please, send it to jmdpul...@yahoo.es or send me the
file where I can find them.

Thanks for the attention

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[R] Line histogram for a matrix

2011-06-02 Thread Sakti

Hi guys!

I'm new to R, but I was wondering if one could plot many histograms into a
single graph each having a different color. To make things clear:

Suppose you have a matrix of 100 rows and 10 columns. I'm interested in
plotting the histogram for each row, but it should not appear as bars but
rather as lines connecting the points of the frequencies. Now, I want to do
this for the 100 rows and make all histogram lines appear with different
colors into the same graph.

The graph should look something like:

http://robjhyndman.com/Rfiles/animation/frmale191.jpg
http://robjhyndman.com/Rfiles/animation/frmale191.jpg  

but instead of those values should be more like histogram shape.

How can you do it???

Thanks!!! 

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[R] What type of bootstrapping is used in package vars?

2011-06-02 Thread jessezeng

Hi, 

I used the package vars for my project but when the impulse response plots
were produced, some of the levels are out of the confidence band. I was
wondering what the problem is and also, does anyone know what type of
bootstrapping the package is using: is it parametric, case resampling,
gaussian, or something else? 

Thanks a lot, 

Jesse

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[R] multicore: collect slow on large objects?

2011-06-02 Thread Trond Arild Ydersbond

I use package multicore, and it works very well. There is, however, one thing I 
wonder if I don't do correctly, here is one example:  I read ca 5.5 mill 
records into a dataframe, using read.dta through a one-line function rdam1. It 
runs nicely in parallel with other activitites.

  p1 -  parallel(rdam1()) ;

Then I recover the data frame with collect:

 am1 - (collect(p1))[[1]]

All goes well, but the collecting step takes quite some time, and as the data 
frame is already in memory, should this be necessary? I experience the same 
with all parallelized steps resulting in large objects.  Is there a way to 
avoid this, for instance, accessing the object through a pointer?

Trond Ydersbond




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[R] Counting occurrences in a moving window

2011-06-02 Thread mdvaan

Hi list, based on the following data.frame I would like to create a variable
that indicates the number of occurrences of A in the 3 years prior to the
current year:

DF = data.frame(read.table(textConnection(  A  B
8025  1995
8026  1995
8029  1995
8026  1996
8025  1997
8026  1997
8025  1997
8027  1997
8026  1999
8027  1999
8028  1995
8029  1998
8025  1997
8027  1997
8026  1999
8027  1999
8028  1995
8029  1998),head=TRUE,stringsAsFactors=FALSE))

becomes:

AB  C
8025  1995  0  
8026  1995  0
8029  1995  0
8026  1996  1
8025  1997  1
8026  1997  2
8025  1997  1
8027  1997  0
8026  1999  2
8027  1999  2
8028  1995  0
8029  1998  1
8025  1997  1
8027  1997  0
8026  1999  2
8027  1999  2 
8028  1995  0
8029  2000  1

So 8026 in 1997 = 2 because 8026 can be found in 1995 and 1996 which are
both within the appropriate window (1996 - 1994).

Any ideas? I looked at the rollapply vignette, but couldn't figure out how
to apply it to my data.

Thanks a lot!




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[R] Help in a project

2011-06-02 Thread Tarun Manchanda

To whom it may concern,
I am a new user to R and I need help in my if and for statements as I need a 
place marker.

I would like to convert the following table:
Observation | Variable 1
|-
1 | 1
2 | A 
3 | 631 

Into :
Observation| Variable 1 | Variable 1 Flag
|--|--
1 |  1 | NA
2 | NA | A
3 | 631 | NA


please note |  represents another column.

this is for a large data set and I cannot manually go through this:
my attempt to this was :
l=0 # placeholder
m-c() #newvariable 1
v-c() #newvariable flag
for (z in tablename1$Variable_1)
(if (is.numeric(z) ) m[z:l] l=l+1 else v[z:1] l=l+1

Can you please guide me as to how to approach this problem?
Any help would be greatly appreciated.
Tarun Manchanda


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[R] Using SQLDF to pick values based on word count

2011-06-02 Thread Abraham Mathew

I have a data frame in R with the following  values.

cars
autocar
cars info
what is that
donna drive
car
telephone
i need car...

I want to select all values which contain 'car', values with three
words, and those keywords with car that contain three words.

The first part is done with :
sqldf(SELECT Keyword FROM dat WHERE Keyword like '%car%')

However, I'm not sure how to pick those values with words
and those values with three words AND 'car'

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Re: [R] shading in overlap between two ranges

2011-06-02 Thread Dennis Murphy

Hi:

Here's one approach using geom_ribbon() in ggplot2 - the 'overlap' is
the change in color where the two ribbons intersect. Using your
example data with the same names and the 'one.month' variable removed,

library(ggplot2)
ggplot() +
  geom_ribbon(data = target, aes(x = i.value, ymin = X25, ymax = X75,
 fill = 'Target'), alpha = 0.4) +
  geom_ribbon(data = observed, aes(x = i.value, ymin = X25, ymax = X75,
 fill = 'Observed'), alpha = 0.4) +
  scale_fill_manual(, c('Target' = 'blue', 'Observed' = 'orange')) +
  opts(legend.position = c(0.88, 0.85),
   legend.background = theme_rect(colour = 'transparent'),
   legend.text = theme_text(size = 12)) +
  labs(x = 'Month', y = 'Value')

There is a separate geom_ribbon() for each of target and observed. A
factor variable for fill color is generated on the fly with colors
specified in scale_fill_manual(). The opts() reposition the legend
inside the graphics region (the values represent proportions of the
total graphics area in each direction), make the legend background
transparent and slightly increase the size of the legend labels
(default size = 10 in theme_text).
Alpha transparency is used so that the overlap creates a blend of the
two colors; without it, one overwrites the other.

HTH,
Dennis


On Thu, Jun 2, 2011 at 8:04 AM, Graves, Gregory ggra...@sfwmd.gov wrote:
 I have 2 datafiles 'target' and 'observed' as shown below (I will gladly
 email these 2 small files to whomever).  X25. And X75. Indicate the
 value of 25th and 75th-percentile of the target ('what should be') and
 the observed ('what is').  The i.value is simply the month.

 target
        X        i.value    X25.     X75.
 1  one.month       1 10.845225 17.87237
 2  one.month       2 12.235813 19.74490
 3  one.month       3 14.611749 23.44810
 4  one.month       4 17.529332 28.09647
 5  one.month       5 19.458738 30.56936
 6  one.month       6 15.264505 28.29333
 7  one.month       7 12.370369 23.35455
 8  one.month       8 12.471224 21.82794
 9  one.month       9  9.716685 17.28762
 10 one.month      10  6.470568 12.49830
 11 one.month      11  6.180560 14.24961
 12 one.month      12  9.673738 15.79208

 observed
     X         i.value   X25.     X75.
 1  one.month       1 19.81000 27.63500
 2  one.month       2 23.64062 30.09125
 3  one.month       3 26.04865 35.99104
 4  one.month       4 32.02625 41.50958
 5  one.month       5 34.74479 47.75958
 6  one.month       6 37.48885 46.56448
 7  one.month       7 30.06740 40.10146
 8  one.month       8 26.14917 39.49458
 9  one.month       9 14.12521 32.39406
 10 one.month      10 11.04125 23.55479
 11 one.month      11 13.14917 23.56833
 12 one.month      12 17.17938 27.02458

 The following plots 4 lines on one graph.  The area between the two red
 lines represents the target 'zone', and the area between the two black
 lines is the observed 'zone'.

 with(target, plot(X25.~i.value,ylim=c(0,55),type='l',col='red'))
 par(new=T)
 with(target, plot(X75.~i.value,ylim=c(0,55),type='l',col='red'))
 par(new=T)
 with(observed, plot(X25.~i.value,ylim=c(0,55),type='l'))
 par(new=T)
 with(observed, plot(X75.~i.value,ylim=c(0,55),type='l'))
 par(new=F)

 Ideally, the target and the observed should overlap in every month -
 they don't.  The desire is to visually accentuate the amount of overlap
 by shading in the area where these two zones overlap.  How would you
 do that?  Note, that in some of these characterizations, the overlap
 wanders in and out [I already have routines that calculate the percent
 of overlap, but I have been requested to find a way to shade the
 overlap.]

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Re: [R] an efficient way to calculate correlation matrix

2011-06-02 Thread Dennis Murphy

?cor

Example:

 dd - data.frame(x1 = rnorm(40), x2 = rnorm(40), x3 = runif(40, 0, 10))
'data.frame':   40 obs. of  3 variables:
 $ x1: num  -0.5585 1.3831 -1.7862 0.0572 0.2825 ...
 $ x2: num  -0.5247 -0.8636 -0.0749 0.2399 -0.1592 ...
 $ x3: num  7.698 5.259 0.918 3.251 5.169 ...
 cor(dd)
   x1  x2  x3
x1  1.000 -0.23268659 -0.02915700
x2 -0.2326866  1. -0.07073142
x3 -0.0291570 -0.07073142  1.

It will also run on a matrix of numeric variables. Any factor or
character variables in the set of variables shipped to cor() will
cause an error; for example,

 head(Oats, 3)
Grouped Data: yield ~ nitro | Block
  Block Variety nitro yield
1 I Victory   0.0   111
2 I Victory   0.2   130
3 I Victory   0.4   157
 cor(Oats)
Error in cor(Oats) : 'x' must be numeric
 cor(Oats[, 3:4])
  nitro yield
nitro 1.000 0.6130266
yield 0.6130266 1.000

HTH,
Dennis

On Thu, Jun 2, 2011 at 8:48 AM, Bill Hyman billhym...@yahoo.com wrote:
 Dear all,

 I have a problem. I have m variables each of which has n observations. I want 
 to
 calculate pairwise correlation among the m variables and store the values in 
 a m
 x m matrix. It is extremely slow to use nested 'for' loops if m and n are 
 large.
 Is there any efficient alternative to do this? Many thanks for your
 suggestions!!

 Bill

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Re: [R] Latin Hypercube Sampling with a condition

2011-06-02 Thread Rob Carnell

Duarte Viana viana.sptd at gmail.com writes:

 
 Thanks Rob and Ravi for the replies.
 
 Let me try to explain my problem. I am trying to make a kind of
 sensitivity analysis where I have 5 parameters (the margins of the
 Latin hypercube), 3 of them are proportions that should sum to one. My
 idea is to obtain uniform combinations of the 3 proportion-parameters
 with the other two parameters. The uniformity should be maintained in
 order to guarantee that each parameter (out of 5) have its own range
 of values equally represented (for model output analyses).
 
 Theoretically the 3 proportion-parameters might be regarded as one in
 which the configuration of the proportions that sum to one vary. I
 think I can visualize it like a set of permutations, more or less like
 in the example below:
 
 0.1 - 0.1 - 0.8
 0.1 - 0.2 - 0.7
 0.1 - 0.3 - 0.6
 .
 .
 .
 0.1 - 0.1 - 0.8
 0.2 - 0.1 - 0.7
 0.3 - 0.1 - 0.6
 .
 .
 .
 0.8 - 0.1 - 0.1
 0.7 - 0.2 - 0.1
 0.6 - 0.3 - 0.1
 .
 .
 .
 and so on, until all possible combinations are represented (and doing
 it with more values) and then combined with the other two parameters
 as to form a Latin hypercube.
 
 The solutions given in the thread sent by Ravi work fine for random
 generation of the 3 proportion-parameters, but it is hard to make a
 Latin hypercube out of that with two more parameters.
 
 Cheers,
 
 Duarte
 
 

Duarte,

The commmon practice in your situation is draw the K parameters together as a 
uniform Latin hypercube on 0-1 and then transform the margins of the hypercube 
to the desired distributions.

Easy Example
Parameter 1: normal(1, 2)
Parameter 2: normal(3, 4)
Parameter 3: uniform(5, 10)

require(lhs)
N - 1000
x - randomLHS(N, 3)
y - x
y[,1] - qnorm(x[,1], 1, 2)
y[,2] - qnorm(x[,2], 3, 4)
y[,3] - qunif(x[,3], 5, 10)

par(mfrow=c(2,2))
apply(x, 2, hist)

par(mfrow=c(2,2))
apply(y, 2, hist)

The transformed distributions maintain their Latin properties, but are in 
the form of new distributions.

In your case, you'd like the first three columns to be transformed into a 
correlated set that sums to one.  Still follow the pattern...

x - randomLHS(N, 5)
y - x
y[,1] - x[,1]/rowSums(x[,1:3])
y[,2] - x[,2]/rowSums(x[,1:3])
y[,3] - x[,3]/rowSums(x[,1:3])
y[,4] - x[,4]
y[,5] - x[,5]

par(mfrow=c(2,3))
apply(x, 2, hist)

par(mfrow=c(2,3))
apply(y, 2, hist)

all.equal(rowSums(y[,1:3]), rep(1, nrow(y)))

The uniform properties are gone as you can see here...

par(mfrow=c(1,1))
pairs(x)
paris(y, col=red)

But, the Latin properties of the first three margins are maintained as in 
this smaller example...

N - 10
x - randomLHS(N, 5)
y - x
y[,1] - x[,1]/rowSums(x[,1:3])
y[,2] - x[,2]/rowSums(x[,1:3])
y[,3] - x[,3]/rowSums(x[,1:3])
y[,4] - x[,4]
y[,5] - x[,5]

pairs(x)
pairs(y, col=red)

You could also look into a dirichlet type transform as I posted here
http://tolstoy.newcastle.edu.au/R/e5/help/08/11/8420.html

Rob

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Re: [R] Zero-inflated regression models: predicting no 0s

2011-06-02 Thread geojs

Thanks for the quick reply, 

I understand that the predict(zip1A, type = response) command is computing
the fitted_means and these are different than the probabilities
predict(zip1A, type = prob). Although, according to Martin (2005), the
highest probabilities do not simply lead to the true count estimates: to
get the true estimate of relative mean abundance from the ZIP one must
multiply the estimated relative mean number of individuals at a site by the
probability that the relative mean number of individuals at a site is
generated through a Poisson distribution.

I initially thought that the predicted mean and the observed count could be
compared to estimate the fit of the model, but now I am not sure what to
think with Martin (2005) statement. 

Thank you for your help, 

JM

Martin, T.G. et al. (2005) Zero tolerance ecology: improving ecological
inference by modelling the source of zero observations, Ecology Letters,
Volume 8, Issue 11, pages 1235–1246.


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[R] generically forward low-level graphic parametres in function

2011-06-02 Thread eldor ado

Hello,

following problem:

i have a written my own function  to draw some sophisticated graphic,
which after manipulating data somewhere contains a plot comand
function - function(x) {
...
plot(xy)
}

to tidy up my final graphs, it would be very handy to be able forward
all low-level graphic parameters set in function to plot

i.e.
function(x, cex=0.5) calls plot (xy, cex=0.5)

is there a way to do this?

thanks a lot,
lukas kohl
department of chemical ecology
university of vienna

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Re: [R] Help in a project

2011-06-02 Thread K. Elo


Hello,

three commands might do the job (NOTE: df=your data frame, 
obser=Observation, var1=Variable 1 [TYPE: string], var1flag=Variable 1 
flag [TYPE: string])


1. df$var1flag-NA
2. df$var1flag[ is.na(as.numeric(df$var1)) ]-df$var1[ 
is.na(as.numeric(df$var1)) ]

3. df$var1-as.numeric(df$var1)

{Please note: Var1 must be type string at the beginning. Otherwise this 
would not work (you cannot have an A in a numeric column...)]


After this, 'df' gives:

 df
  obser var1 var1flag
1 1 1000 NA
2 2 NAA
3 3  631 NA

Is this what You are looking for?

HTH,
Kimmo

02.06.2011 18:28, Tarun Manchanda wrote:

To whom it may concern,
I am a new user to R and I need help in my if and for statements as I need a 
place marker.

I would like to convert the following table:
Observation | Variable 1
|-
1 | 1
2 | A
3 | 631

Into :
Observation| Variable 1 | Variable 1 Flag
|--|--
1 |  1 | NA
2 | NA | A
3 | 631 | NA


please note |  represents another column.


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[R] confusion matrix error

2011-06-02 Thread romzero

Hi,
this is my R-script
I need to make a confusion matrix
but the last row return me an error

require(mixOmics)
require(SDMTools)

file - C:\\data.txt
d - read.table(file, header=T, row.names = NULL)

X - as.matrix(d[,2:11])
Y - as.factor(d[,1])

i - 1 
samp - sample(1:3, nrow(X), replace = TRUE) # Creation of a list of the
same size as X

test - which(samp == i) # Search which column in samp has a value of 1
train - setdiff(1:nrow(X), test) # Keeping the column that are not in test

#PLS-DA
plsda.train - plsda(X[train, ], Y[train], ncomp = 10)
test.predict - predict(plsda.train, X[test, ], method = class.dist)

confusion.matrix(test.predict, Y[-train])

Error in confusion.matrix(test.predict, Y[-train]) : 
  this requires the same number of observed  predicted values

I have no idea why it don't work

Thank you for help

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Re: [R] ARM package for R 2.10.1

2011-06-02 Thread Jonathan Daily

I would recommend trying to fix the installation of R 2.13.0 rather
than trying to obtain old packages. Try downloading the installer
again from a different mirror.

On Thu, Jun 2, 2011 at 10:33 AM, jmdpulido jmdpul...@yahoo.es wrote:
 Dear all...

 I am looking for the zip file of an old version of the ARM package
 compatible for R 2.10.1 version.

 When I try to charge the ARM package I get the following message package
 'arm' was built under R version 2.13.0 .

 I can not update R to 2.13.0 as I always get this error  the setup files
 are corrupted please obtain a new copy of the program.

 If you have an old version of the zip file of ARM and LME4 packages that I
 can use with R 2.10.1, please, send it to jmdpul...@yahoo.es or send me the
 file where I can find them.

 Thanks for the attention

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/ARM-package-for-R-2-10-1-tp3568473p3568473.html
 Sent from the R help mailing list archive at Nabble.com.

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




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===
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Technician
===
#!/usr/bin/env outside
# It's great, trust me.

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Re: [R] generically forward low-level graphic parametres in function

2011-06-02 Thread Jonathan Daily

try using `...`

function - function(x, ...) {
#do stuff
plot(xy, ...)
}

On Thu, Jun 2, 2011 at 1:26 PM, eldor ado rat.c...@gmail.com wrote:
 Hello,

 following problem:

 i have a written my own function  to draw some sophisticated graphic,
 which after manipulating data somewhere contains a plot comand
 function - function(x) {
 ...
 plot(xy)
 }

 to tidy up my final graphs, it would be very handy to be able forward
 all low-level graphic parameters set in function to plot

 i.e.
 function(x, cex=0.5) calls plot (xy, cex=0.5)

 is there a way to do this?

 thanks a lot,
 lukas kohl
 department of chemical ecology
 university of vienna

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

Re: [R] Problem with package development

Thanks Prof. Ripley and Duncan for your pointers. Noting down your
points I have modified my way of building package and have done
following so far:

1. In my C: drive I create one working folder naming R_PackageBuild
2. In R console I have written following codes:
 setwd(c:/R_packageBuild)
 package.skeleton(trial1,namespace = TRUE, code_files = f:/trial.r)
3. then I opened cmd and wrote following:
cd C:\R_PackageBuild
Rcmd build –binary trial1

This process halted with following error:
Error: unexpected symbol in “tools:::.test_load_package(‘trial1′,….)”
Execution halted
ERROR: loading failed

What I have missed in this process? Can you please help me how to
solve this issue?

Thanks,

PS: I am sorry I missplet 'Program Files'. Thanks Prof. Ripley for this pointer.

On Wed, Jun 1, 2011 at 11:41 AM, Prof Brian Ripley
rip...@stats.ox.ac.uk wrote:
 On Wed, 1 Jun 2011, Nipesh Bajaj wrote:

 I have been struggling for last one hour but not yet any through.

 However again I recreate the package.skeleton and run R CMD check trial3

 Here are the errors:

 warning in dir.create(pkgoutdir, mode = 0755):
 cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
 reason .
 Error in printLog(Log, , text, \n): object 'Log' not found
 Execution haulted

 Why I am getting this error? what is that Log. I will really
 appreciate if somebody please help me to figure out.

 R CMD check writes a (in your case) trial3.Rcheck directory, and in there in
 file 00check.log a copy of the log.  If it cannot create trial3.Rcheck it
 cannot write the log.

 I would be surprised that even on Windows Vista the message was literally

 reason .

 but if it was, blame Microsoft for their error messages.
 But

 cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',

 is clear enough.  You need to run 'R CMD check' in your user area.
 In case you did this because that is where you though 'R' was, it is not the
 correct R.exe.   You may need to add

 c:\Program Files\R\R-2.13.0\bin\i386

 (assuming 32-bit R) to your path.

 However, your use of e.g. 'Program files' suggests you are not accurately
 transmitting the messages you got.



 Thanks,

 On Wed, Jun 1, 2011 at 1:20 AM, Nipesh Bajaj bajaj141...@gmail.com
 wrote:

 Actually partly I followed. Here is the more details what I have done so
 far:

 1. Edit the help file skeletons in 'man', possibly combining help
 files for multiple functions.
 I have modified with following:
 \name{fn1}
 \alias{fn1}

 \title{
 A function.
 }

 \description{
 A function.
 }

 \usage{
 A function.
 }

 \arguments{
 A function.
 }

 \value{
 A function.
 }

 \author{
 \bold{Me}
 \cr
 \email{m...@me.com}
 }

 2. Edit the exports in 'NAMESPACE', and add necessary imports.
 Actually I really do not know what I would do here. In the
 corresponding file, only exportPattern(^[[:alpha:]]+) is there.
 Therefore I put that unaltered.

 3. Put any C/C++/Fortran code in 'src'.
 I do not have any such code

 4. If you have compiled code, add a useDynLib() directive to 'NAMESPACE'.
 Again I do not know what to do, so ingored this step.

 5. Run R CMD build to build the package tarball.
 * Run R CMD check to check the package tarball.

 I did not follow this step exactly. What I done is, put 'trial3'
 folder in R/R-2.13.0bin folder (after above modification), from the
 R-working folder. Then just run R CMD INSTALL trial3. However
 previously with this job, I could create package effectively. After
 updating R to the current version my problem starts.

 Those are not sufficient?

 Thanks,

 On Wed, Jun 1, 2011 at 1:09 AM, Duncan Murdoch murdoch.dun...@gmail.com
 wrote:

 On 11-05-31 3:36 PM, Nipesh Bajaj wrote:

 Dear all, I am having a strage problem while I was trying to build a
 package. Here is my package skeleton:

 fn1- Vectorize(function(x,y,z) {
                        return(x + y +z)
                }, vectorize.args = c(x), SIMPLIFY = TRUE)
 package.skeleton(trial3,namespace = TRUE)

 Did you follow the instructions that package.skeleton printed?

 Duncan Murdoch


 However when I tun R CMD INSTALL trial3 in CMD, the execution
 stopped with following message:

 *** installing help indices
 ** building package indices...
 ** testing if install package can be loaded
 Error: unexpected symbol in tools:::test_load_package(..
 Execution haulted
 ERROR: loading failed.

 I am using R 2.13.0 in Vista with latest Rtools installed. Can
 somebody guide me where I have done wrong?

 Thanks,

 __
 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-help@r-project.org mailing list
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 PLEASE do read the posting guide

Re: [R] ARM package for R 2.10.1

2011-06-02 Thread Prof Brian Ripley


On Thu, 2 Jun 2011, Jonathan Daily wrote:


I would recommend trying to fix the installation of R 2.13.0 rather
than trying to obtain old packages. Try downloading the installer
again from a different mirror.


He has already been given help (3x) to that effect.  People who fail 
to acknowledege any help are asking not to be helped again.


But the R 2.10 Windows binary tree is still up at
http://cran.r-project.org/bin/windows/contrib/2.10/
and simply using the menus will install the correct versions.



On Thu, Jun 2, 2011 at 10:33 AM, jmdpulido jmdpul...@yahoo.es wrote:

Dear all...

I am looking for the zip file of an old version of the ARM package
compatible for R 2.10.1 version.

When I try to charge the ARM package I get the following message package
'arm' was built under R version 2.13.0 .

I can not update R to 2.13.0 as I always get this error  the setup files
are corrupted please obtain a new copy of the program.

If you have an old version of the zip file of ARM and LME4 packages that I
can use with R 2.10.1, please, send it to jmdpul...@yahoo.es or send me the
file where I can find them.

Thanks for the attention

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===
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--
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Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Latin Hypercube Sampling with a condition

2011-06-02 Thread Duarte Viana

Thanks again Rob for your help.

In terms of parameter comparison there won't be a problem. However, if
one wants to assume a particular distribution (and not the one given
by the imposed condition), for example an uniform distribution to
obtain all the possible combinations (all the multidimensional space
uniformly filled), then a limitation exists. Perhaps it is not
possible to fulfill the three criteria - sum to one, maintain the
uniform distribution and maintain the latin hypercube property. Well,
I will try to do it the way you proposed.

Cheers,

Duarte






On Thu, Jun 2, 2011 at 7:06 PM, Rob Carnell carne...@battelle.org wrote:
 Duarte Viana viana.sptd at gmail.com writes:


 Thanks Rob and Ravi for the replies.

 Let me try to explain my problem. I am trying to make a kind of
 sensitivity analysis where I have 5 parameters (the margins of the
 Latin hypercube), 3 of them are proportions that should sum to one. My
 idea is to obtain uniform combinations of the 3 proportion-parameters
 with the other two parameters. The uniformity should be maintained in
 order to guarantee that each parameter (out of 5) have its own range
 of values equally represented (for model output analyses).

 Theoretically the 3 proportion-parameters might be regarded as one in
 which the configuration of the proportions that sum to one vary. I
 think I can visualize it like a set of permutations, more or less like
 in the example below:

 0.1 - 0.1 - 0.8
 0.1 - 0.2 - 0.7
 0.1 - 0.3 - 0.6
 .
 .
 .
 0.1 - 0.1 - 0.8
 0.2 - 0.1 - 0.7
 0.3 - 0.1 - 0.6
 .
 .
 .
 0.8 - 0.1 - 0.1
 0.7 - 0.2 - 0.1
 0.6 - 0.3 - 0.1
 .
 .
 .
 and so on, until all possible combinations are represented (and doing
 it with more values) and then combined with the other two parameters
 as to form a Latin hypercube.

 The solutions given in the thread sent by Ravi work fine for random
 generation of the 3 proportion-parameters, but it is hard to make a
 Latin hypercube out of that with two more parameters.

 Cheers,

 Duarte



 Duarte,

 The commmon practice in your situation is draw the K parameters together as a
 uniform Latin hypercube on 0-1 and then transform the margins of the hypercube
 to the desired distributions.

 Easy Example
 Parameter 1: normal(1, 2)
 Parameter 2: normal(3, 4)
 Parameter 3: uniform(5, 10)

 require(lhs)
 N - 1000
 x - randomLHS(N, 3)
 y - x
 y[,1] - qnorm(x[,1], 1, 2)
 y[,2] - qnorm(x[,2], 3, 4)
 y[,3] - qunif(x[,3], 5, 10)

 par(mfrow=c(2,2))
 apply(x, 2, hist)

 par(mfrow=c(2,2))
 apply(y, 2, hist)

 The transformed distributions maintain their Latin properties, but are in
 the form of new distributions.

 In your case, you'd like the first three columns to be transformed into a
 correlated set that sums to one.  Still follow the pattern...

 x - randomLHS(N, 5)
 y - x
 y[,1] - x[,1]/rowSums(x[,1:3])
 y[,2] - x[,2]/rowSums(x[,1:3])
 y[,3] - x[,3]/rowSums(x[,1:3])
 y[,4] - x[,4]
 y[,5] - x[,5]

 par(mfrow=c(2,3))
 apply(x, 2, hist)

 par(mfrow=c(2,3))
 apply(y, 2, hist)

 all.equal(rowSums(y[,1:3]), rep(1, nrow(y)))

 The uniform properties are gone as you can see here...

 par(mfrow=c(1,1))
 pairs(x)
 paris(y, col=red)

 But, the Latin properties of the first three margins are maintained as in
 this smaller example...

 N - 10
 x - randomLHS(N, 5)
 y - x
 y[,1] - x[,1]/rowSums(x[,1:3])
 y[,2] - x[,2]/rowSums(x[,1:3])
 y[,3] - x[,3]/rowSums(x[,1:3])
 y[,4] - x[,4]
 y[,5] - x[,5]

 pairs(x)
 pairs(y, col=red)

 You could also look into a dirichlet type transform as I posted here
 http://tolstoy.newcastle.edu.au/R/e5/help/08/11/8420.html

 Rob

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Re: [R] Problem with package development

2011-06-02 Thread Duncan Murdoch


On 02/06/2011 2:03 PM, Nipesh Bajaj wrote:

Thanks Prof. Ripley and Duncan for your pointers. Noting down your
points I have modified my way of building package and have done
following so far:

1. In my C: drive I create one working folder naming R_PackageBuild
2. In R console I have written following codes:
  setwd(c:/R_packageBuild)
  package.skeleton(trial1,namespace = TRUE, code_files = f:/trial.r)
3. then I opened cmd and wrote following:
cd C:\R_PackageBuild
Rcmd build –binary trial1

This process halted with following error:
Error: unexpected symbol in “tools:::.test_load_package(‘trial1′,….)”
Execution halted
ERROR: loading failed

What I have missed in this process? Can you please help me how to
solve this issue?


You haven't done the manual changes required between steps 2 and 3.  
package.skeleton() creates the skeleton of a package; you run it once as 
you are starting development, the do a lot of manual updates, described 
on the ?package.skeleton help page, and in the  ‘Read-and-delete-me’ 
file.  Once those are done, step 3 should succeed.


Duncan Murdoch


Thanks,

PS: I am sorry I missplet 'Program Files'. Thanks Prof. Ripley for this pointer.

On Wed, Jun 1, 2011 at 11:41 AM, Prof Brian Ripley
rip...@stats.ox.ac.uk  wrote:
  On Wed, 1 Jun 2011, Nipesh Bajaj wrote:

  I have been struggling for last one hour but not yet any through.

  However again I recreate the package.skeleton and run R CMD check trial3

  Here are the errors:

  warning in dir.create(pkgoutdir, mode = 0755):
  cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
  reason .
  Error in printLog(Log, , text, \n): object 'Log' not found
  Execution haulted

  Why I am getting this error? what is that Log. I will really
  appreciate if somebody please help me to figure out.

  R CMD check writes a (in your case) trial3.Rcheck directory, and in there in
  file 00check.log a copy of the log.  If it cannot create trial3.Rcheck it
  cannot write the log.

  I would be surprised that even on Windows Vista the message was literally

  reason .

  but if it was, blame Microsoft for their error messages.
  But

  cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',

  is clear enough.  You need to run 'R CMD check' in your user area.
  In case you did this because that is where you though 'R' was, it is not the
  correct R.exe.   You may need to add

  c:\Program Files\R\R-2.13.0\bin\i386

  (assuming 32-bit R) to your path.

  However, your use of e.g. 'Program files' suggests you are not accurately
  transmitting the messages you got.



  Thanks,

  On Wed, Jun 1, 2011 at 1:20 AM, Nipesh Bajajbajaj141...@gmail.com
  wrote:

  Actually partly I followed. Here is the more details what I have done so
  far:

  1. Edit the help file skeletons in 'man', possibly combining help
  files for multiple functions.
  I have modified with following:
  \name{fn1}
  \alias{fn1}

  \title{
  A function.
  }

  \description{
  A function.
  }

  \usage{
  A function.
  }

  \arguments{
  A function.
  }

  \value{
  A function.
  }

  \author{
  \bold{Me}
  \cr
  \email{m...@me.com}
  }

  2. Edit the exports in 'NAMESPACE', and add necessary imports.
  Actually I really do not know what I would do here. In the
  corresponding file, only exportPattern(^[[:alpha:]]+) is there.
  Therefore I put that unaltered.

  3. Put any C/C++/Fortran code in 'src'.
  I do not have any such code

  4. If you have compiled code, add a useDynLib() directive to 'NAMESPACE'.
  Again I do not know what to do, so ingored this step.

  5. Run R CMD build to build the package tarball.
  * Run R CMD check to check the package tarball.

  I did not follow this step exactly. What I done is, put 'trial3'
  folder in R/R-2.13.0bin folder (after above modification), from the
  R-working folder. Then just run R CMD INSTALL trial3. However
  previously with this job, I could create package effectively. After
  updating R to the current version my problem starts.

  Those are not sufficient?

  Thanks,

  On Wed, Jun 1, 2011 at 1:09 AM, Duncan Murdochmurdoch.dun...@gmail.com
  wrote:

  On 11-05-31 3:36 PM, Nipesh Bajaj wrote:

  Dear all, I am having a strage problem while I was trying to build a
  package. Here is my package skeleton:

  fn1- Vectorize(function(x,y,z) {
  return(x + y +z)
  }, vectorize.args = c(x), SIMPLIFY = TRUE)
  package.skeleton(trial3,namespace = TRUE)

  Did you follow the instructions that package.skeleton printed?

  Duncan Murdoch


  However when I tun R CMD INSTALL trial3 in CMD, the execution
  stopped with following message:

  *** installing help indices
  ** building package indices...
  ** testing if install package can be loaded
  Error: unexpected symbol in tools:::test_load_package(..
  Execution haulted
  ERROR: loading failed.

  I am using R 2.13.0 in Vista with latest Rtools installed. Can
  somebody guide

Re: [R] Problem with package development

What else I need to do? In the Read-and-delete-me file following steps
are asked to perform:
* Edit the help file skeletons in 'man', possibly combining help files
for multiple functions.
* Edit the exports in 'NAMESPACE', and add necessary imports.
* Put any C/C++/Fortran code in 'src'.
* If you have compiled code, add a useDynLib() directive to 'NAMESPACE'.
* Run R CMD build to build the package tarball.
* Run R CMD check to check the package tarball.

I editied the help page for fn1() function (as I already communicated
in previous mail) as follows:
\name{fn1}
\alias{fn1}

\title{
A function.
}

\description{
A function.
}

\usage{
A function.
}

\arguments{
A function.
}

\value{
A function.
}

\author{
\bold{Me}
\cr
\email{m...@me.com}
}

And regarding th Namespace file, this time I put
package.skeleton(trial1,namespace = FALSE, code_files =
f:/trial.r)


I do not have any C/C++ code so I ignored 3rd step.

then Read-and-delete-me file asking me to build the package, so in
cmd, I run following:
cd C:\R_PackageBuild
Rcmd build –binary trial1

What I am missing in this entire process? Do you please point me?

Thanks,


On Thu, Jun 2, 2011 at 11:40 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
 On 02/06/2011 2:03 PM, Nipesh Bajaj wrote:

 Thanks Prof. Ripley and Duncan for your pointers. Noting down your
 points I have modified my way of building package and have done
 following so far:

 1. In my C: drive I create one working folder naming R_PackageBuild
 2. In R console I have written following codes:
   setwd(c:/R_packageBuild)
   package.skeleton(trial1,namespace = TRUE, code_files = f:/trial.r)
 3. then I opened cmd and wrote following:
 cd C:\R_PackageBuild
 Rcmd build –binary trial1

 This process halted with following error:
 Error: unexpected symbol in “tools:::.test_load_package(‘trial1′,….)”
 Execution halted
 ERROR: loading failed

 What I have missed in this process? Can you please help me how to
 solve this issue?

 You haven't done the manual changes required between steps 2 and 3.
  package.skeleton() creates the skeleton of a package; you run it once as
 you are starting development, the do a lot of manual updates, described on
 the ?package.skeleton help page, and in the  ‘Read-and-delete-me’ file.
  Once those are done, step 3 should succeed.

 Duncan Murdoch

 Thanks,

 PS: I am sorry I missplet 'Program Files'. Thanks Prof. Ripley for this
 pointer.

 On Wed, Jun 1, 2011 at 11:41 AM, Prof Brian Ripley
 rip...@stats.ox.ac.uk  wrote:
   On Wed, 1 Jun 2011, Nipesh Bajaj wrote:
 
   I have been struggling for last one hour but not yet any through.
 
   However again I recreate the package.skeleton and run R CMD check
  trial3
 
   Here are the errors:
 
   warning in dir.create(pkgoutdir, mode = 0755):
   cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
   reason .
   Error in printLog(Log, , text, \n): object 'Log' not found
   Execution haulted
 
   Why I am getting this error? what is that Log. I will really
   appreciate if somebody please help me to figure out.
 
   R CMD check writes a (in your case) trial3.Rcheck directory, and in
  there in
   file 00check.log a copy of the log.  If it cannot create trial3.Rcheck
  it
   cannot write the log.
 
   I would be surprised that even on Windows Vista the message was
  literally
 
   reason .
 
   but if it was, blame Microsoft for their error messages.
   But
 
   cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
 
   is clear enough.  You need to run 'R CMD check' in your user area.
   In case you did this because that is where you though 'R' was, it is
  not the
   correct R.exe.   You may need to add
 
   c:\Program Files\R\R-2.13.0\bin\i386
 
   (assuming 32-bit R) to your path.
 
   However, your use of e.g. 'Program files' suggests you are not
  accurately
   transmitting the messages you got.
 
 
 
   Thanks,
 
   On Wed, Jun 1, 2011 at 1:20 AM, Nipesh Bajajbajaj141...@gmail.com
   wrote:
 
   Actually partly I followed. Here is the more details what I have done
  so
   far:
 
   1. Edit the help file skeletons in 'man', possibly combining help
   files for multiple functions.
   I have modified with following:
   \name{fn1}
   \alias{fn1}
 
   \title{
   A function.
   }
 
   \description{
   A function.
   }
 
   \usage{
   A function.
   }
 
   \arguments{
   A function.
   }
 
   \value{
   A function.
   }
 
   \author{
   \bold{Me}
   \cr
   \email{m...@me.com}
   }
 
   2. Edit the exports in 'NAMESPACE', and add necessary imports.
   Actually I really do not know what I would do here. In the
   corresponding file, only exportPattern(^[[:alpha:]]+) is there.
   Therefore I put that unaltered.
 
   3. Put any C/C++/Fortran code in 'src'.
   I do not have any such code
 
   4. If you have compiled code, add a useDynLib() directive to
  'NAMESPACE'.
   Again I do not know what to do, so ingored this step.
 
   5. Run R CMD build to build the package

Re: [R] Line histogram for a matrix



On Jun 2, 2011, at 11:04 AM, Sakti wrote:


Hi guys!

I'm new to R, but I was wondering if one could plot many histograms  
into a

single graph each having a different color. To make things clear:

Suppose you have a matrix of 100 rows and 10 columns. I'm interested  
in
plotting the histogram for each row, but it should not appear as  
bars but
rather as lines connecting the points of the frequencies. Now, I  
want to do
this for the 100 rows and make all histogram lines appear with  
different

colors into the same graph.


You should look at the matplot function.



The graph should look something like:

http://robjhyndman.com/Rfiles/animation/frmale191.jpg
http://robjhyndman.com/Rfiles/animation/frmale191.jpg

but instead of those values should be more like histogram shape.

How can you do it???

Thanks!!!




David Winsemius, MD
West Hartford, CT

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[R] plotmath: paste string and expression [from a vector of expressions]

Dear all,

I have a vector of expressions and would like to paste some string to it 
before using it in a plot:

vars - vector(expression, 2)
vars[1] - expression(alpha)
vars[2] - expression(beta)
plot(0, 0, main=substitute(bold(Foo ~~ VAR), list(VAR=vars[2]) ))

Although I tried hard, I just can't figure out how to solve this. The title 
should be Foo theta, where theta is the greek letter. I tried some 
constructions with bquote but that wasn't successful... I also looked in the 
mailing list but couldn't find anything helpful [I am sure I overlooked 
something].

Cheers,

Marius

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Re: [R] Problem with Snowball RWeka

2011-06-02 Thread flobede

Greetings to all,
I have a similar issue with Snowball.
I am runing R version 2.12.1 (2010-12-16) on windows 7

Here is my script : 

library(tm)

custom.xml  -  system.file(texts,  custom.xml,  package  =  tm)
print(readLines(custom.xml),  quote  =  FALSE)

myXMLReader - readXML(
  spec = list(
Language = list(node, /document/language),
DateTimeStamp = list(node, /document/date),
Origin = list(node, /document/source),
Description = list(node, /document/subject),
Type = list(node, /document/country),
  Heading = list(node, /document/title),
Content = list(node, /document/contenu),
Author = list(node, /document/author)),
doc = PlainTextDocument())

mySource  -  function(x,  encoding  =  UTF-8)
  XMLSource(x,  function(tree)  XML::xmlRoot(tree)$children,  myXMLReader, 
encoding)

corpusmf  -  Corpus(mySource(custom.xml))
meta(corpusmf[[1]])
meta(corpusmf[[2]])

corpusmf - tm_map(corpusmf, stripWhitespace)
corpusmf - tm_map(corpusmf, removeNumbers)
corpusmf - tm_map(corpusmf, removePunctuation)
corpusmf - tm_map(corpusmf,stemDocument)

matrix - TermDocumentMatrix(corpusmf,control=list(weighting =weightBin ))
print(matrix)
 
-
stemDocument returns an error message :
Stemmer 'porter' unknown!
Stemmer 'english' unknown!
Stemmer 'porter' unknown!
Stemmer 'english' unknown!

I tried to invoke library(Snowball) before, but it's the same.
I found a clue on Weka website
http://weka.wikispaces.com/The+snowball+stemmers+don%27t+work,+what+am+I+doing+wrong%3F
but I don't understand what I should do with this archives
I would be grateful if someone could help on this;
Kind regards,

--
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[R] complex search between dataframes

2011-06-02 Thread Filippo Beleggia

Hi!

I am very new to R, I hope someone can help me.

I have two dataframes:

data1-data.frame(from=c(1,12,16,40,55,81,101),to=c(10,13,23,45,67,99,123))
data2-data.frame(name=c(1,2,3,4,5,6,7,8,9),position=c(2,14,20,50,150,2000,2001,2002,85))


I want to know which of the entries in position of data2 are included between 
any from and the corresponding to of data1.

So in this case I would need to somehow be able to extract 2,20 and 85, 
corrisponding to the names 1,3 and 9.

Thank you very much!
Filippo

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[R] Matrix Question

2011-06-02 Thread Ben Ganzfried

Hi,

First of all, I would like to introduce myself as I will probably have many
questions over the next few weeks and want to thank you guys in advance for
your help.  I'm a cancer researcher and I need to learn R to complete a few
projects.  I have an introductory background in Python.

My questions at the moment are based on the following sample input file:
*Sample_Input_File*
 characteristics_ch1.3  Stage: T1N0  Stage: T2N1  Stage: T0N0  Stage:
T1N0  Stage:
T0N3

characteristics_ch1.3 is a column header in  the input excel file.

T's represent stage and N's represent degree of disease spreading.

I want to create output that looks like this:
*Sample_Output_File*
T N
1 0
2 1
0 0
1 0
0 3

As it currently stands, my code is the following:

rm(list=ls())
source(../../functions.R)

uncurated - read.csv(../uncurated/Sample_Input_File_full_pdata.csv,as.is
=TRUE,row.names=1)

##initial creation of curated dataframe
curated -
initialCuratedDF(rownames(uncurated),template.filename=Sample_Template_File.csv)

##
##start the mappings
##


##title - alt_sample_name
curated$alt_sample_name - uncurated$title

#T
tmp - uncurated$characteristics_ch1.3
tmp - *??*
curated$T - tmp

#N
tmp - uncurated$characteristics_ch1.3
tmp - *??*
curated$N - tmp

write.table(curated, row.names=FALSE,
file=../curated/Sample_Output_File_curated_pdata.txt,sep=\t)

My question is the following:

What code gets me the desired output (replacing the *??*'s above)?  I
want to: a) Find the integer value one element to the right of T; and b)
find the integer value one element to the right of N.  I've read the
regular expression tutorial for R, but could only figure out how to grab an
integer value if it is the only integer value in the row (ie more than one
integer value makes this basic regular expression unsuccessful).

Thank you very much for any help you can provide.

Sincerely,

Ben Ganzfried

[[alternative HTML version deleted]]

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

Re: [R] lattice panel fine control

2011-06-02 Thread Bert Gunter

You did not read the help files carefully enough.

The Help for panel.text tells you that it is the same function as
ltext. ltext lists a bunch of parameters, srt among them, and refers
you to the corresponding base R graphics function, which is text().

?text then refers you to par for this and other miscellaneous parameters.

?par then tells you that srt gives the string rotation in degrees, the
answer to your question.

So, yes, it's a bit rough going; but careful attention to the docs
DOES get you there.

-- Bert

On Thu, Jun 2, 2011 at 7:12 AM, maxbre mbres...@arpa.veneto.it wrote:
 thank you so much for the very detailed indications which turned out to be a
 real help in ponting me to the right direction;

 referring back to my previous questions there is something still open:

 2- I'm in trouble with the point labels because I would like to rotate them
 by an angle of 90 degrees (and I did not find mention of anything like
 angle or rot to accomplish this task)
 panel.text(x, y, lab = mydata$name, cex = 0.6, pos=3, offset=0.5, here
 something to rotate labels)

 3- I was referring to the error bar of each point (standard error of
 points); I think this could be accomplished by arrows but the following
 line is giving me an error
 panel.arrows(x=tv.avg-tv.erst, y=ped.avg-ped.erst, x1=tv.avg+tv.erst,
 y1=ped.avg+ped.erst, angle=90, code=3)

 thanks again for your great help in bootstrapping me to the lattice features

 maxbre

 --
 View this message in context: 
 http://r.789695.n4.nabble.com/lattice-panel-fine-control-tp3566347p3568424.html
 Sent from the R help mailing list archive at Nabble.com.

 __
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-- 
Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions.

-- Maimonides (1135-1204)

Bert Gunter
Genentech Nonclinical Biostatistics

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[R] Error message while mapping probes with hgu133plus2.db

2011-06-02 Thread Vickie S


Hi,
eset is an expressionset based on hgu133plus2 platform. 

Using hgu133plus2.db package, I want to map probes to UniGene cluster IDs. 
It results in an error message.

xx=hgu133plus2UNIGENE
uniaf1=xx[[as.character(featureNames(eset))]]
Error in .checkKeys(value, Lkeys(x), x@ifnotfound) : 
  value for AFFX-r2-Hs18SrRNA-3_s_at not found

Could you please help me figure out what creates the error ?

Many thanks

Vickie


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

Re: [R] Problem with package development

I have run R CMD check trial1 and saw an error. This says that:

* checking pdf version of manual without hyperrefs or index  ... ERROR
Re-running with no redirection of stdout/stderr.
Hmm... looks like a package
Error in texi2dvi(Rd2.tex, pdf = (out_ext == pdf), quiet = FALSE, :
pdflatex is not available
Error in running tools:: texi2dvi

Does this information hwlp you to suggest something? Please let me
know what else I can provide.

Thanks,

On Fri, Jun 3, 2011 at 12:00 AM, Nipesh Bajaj bajaj141...@gmail.com wrote:
 What else I need to do? In the Read-and-delete-me file following steps
 are asked to perform:
 * Edit the help file skeletons in 'man', possibly combining help files
 for multiple functions.
 * Edit the exports in 'NAMESPACE', and add necessary imports.
 * Put any C/C++/Fortran code in 'src'.
 * If you have compiled code, add a useDynLib() directive to 'NAMESPACE'.
 * Run R CMD build to build the package tarball.
 * Run R CMD check to check the package tarball.

 I editied the help page for fn1() function (as I already communicated
 in previous mail) as follows:
 \name{fn1}
 \alias{fn1}

 \title{
 A function.
 }

 \description{
 A function.
 }

 \usage{
 A function.
 }

 \arguments{
 A function.
 }

 \value{
 A function.
 }

 \author{
 \bold{Me}
 \cr
 \email{m...@me.com}
 }

 And regarding th Namespace file, this time I put
 package.skeleton(trial1,namespace = FALSE, code_files =
 f:/trial.r)
 

 I do not have any C/C++ code so I ignored 3rd step.

 then Read-and-delete-me file asking me to build the package, so in
 cmd, I run following:
 cd C:\R_PackageBuild
 Rcmd build –binary trial1

 What I am missing in this entire process? Do you please point me?

 Thanks,


 On Thu, Jun 2, 2011 at 11:40 PM, Duncan Murdoch
 murdoch.dun...@gmail.com wrote:
 On 02/06/2011 2:03 PM, Nipesh Bajaj wrote:

 Thanks Prof. Ripley and Duncan for your pointers. Noting down your
 points I have modified my way of building package and have done
 following so far:

 1. In my C: drive I create one working folder naming R_PackageBuild
 2. In R console I have written following codes:
   setwd(c:/R_packageBuild)
   package.skeleton(trial1,namespace = TRUE, code_files = f:/trial.r)
 3. then I opened cmd and wrote following:
 cd C:\R_PackageBuild
 Rcmd build –binary trial1

 This process halted with following error:
 Error: unexpected symbol in “tools:::.test_load_package(‘trial1′,….)”
 Execution halted
 ERROR: loading failed

 What I have missed in this process? Can you please help me how to
 solve this issue?

 You haven't done the manual changes required between steps 2 and 3.
  package.skeleton() creates the skeleton of a package; you run it once as
 you are starting development, the do a lot of manual updates, described on
 the ?package.skeleton help page, and in the  ‘Read-and-delete-me’ file.
  Once those are done, step 3 should succeed.

 Duncan Murdoch

 Thanks,

 PS: I am sorry I missplet 'Program Files'. Thanks Prof. Ripley for this
 pointer.

 On Wed, Jun 1, 2011 at 11:41 AM, Prof Brian Ripley
 rip...@stats.ox.ac.uk  wrote:
   On Wed, 1 Jun 2011, Nipesh Bajaj wrote:
 
   I have been struggling for last one hour but not yet any through.
 
   However again I recreate the package.skeleton and run R CMD check
  trial3
 
   Here are the errors:
 
   warning in dir.create(pkgoutdir, mode = 0755):
   cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
   reason .
   Error in printLog(Log, , text, \n): object 'Log' not found
   Execution haulted
 
   Why I am getting this error? what is that Log. I will really
   appreciate if somebody please help me to figure out.
 
   R CMD check writes a (in your case) trial3.Rcheck directory, and in
  there in
   file 00check.log a copy of the log.  If it cannot create trial3.Rcheck
  it
   cannot write the log.
 
   I would be surprised that even on Windows Vista the message was
  literally
 
   reason .
 
   but if it was, blame Microsoft for their error messages.
   But
 
   cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
 
   is clear enough.  You need to run 'R CMD check' in your user area.
   In case you did this because that is where you though 'R' was, it is
  not the
   correct R.exe.   You may need to add
 
   c:\Program Files\R\R-2.13.0\bin\i386
 
   (assuming 32-bit R) to your path.
 
   However, your use of e.g. 'Program files' suggests you are not
  accurately
   transmitting the messages you got.
 
 
 
   Thanks,
 
   On Wed, Jun 1, 2011 at 1:20 AM, Nipesh Bajajbajaj141...@gmail.com
   wrote:
 
   Actually partly I followed. Here is the more details what I have done
  so
   far:
 
   1. Edit the help file skeletons in 'man', possibly combining help
   files for multiple functions.
   I have modified with following:
   \name{fn1}
   \alias{fn1}
 
   \title{
   A function.
   }
 
   \description{
   A function.
   }
 
   \usage{
   A function.
   }
 
   \arguments{
   A function.
   }
 
   \value{
   A

Re: [R] Regex Question: return digits after particular letters



On Jun 2, 2011, at 2:54 PM, Ben Ganzfried wrote:


Hi,

First of all, I would like to introduce myself as I will probably  
have many
questions over the next few weeks and want to thank you guys in  
advance for
your help.  I'm a cancer researcher and I need to learn R to  
complete a few

projects.  I have an introductory background in Python.

My questions at the moment are based on the following sample input  
file:

*Sample_Input_File*
characteristics_ch1.3  Stage: T1N0  Stage: T2N1  Stage: T0N0  Stage:
T1N0  Stage:
T0N3



I haven't quite figured out what your structure really is, and for  
that you should learn to post the output of dput()  on the R object...  
but see if this helps:


 stg - c('Stage: T1N0',  'Stage: T2N1', 'Stage: T0N0', 'Stage:  
T1N0', 'Stage: T0N3')

 Tstg - sub(.*T(\\d)N., \\1, stg)
 Tstg
#[1] 1 2 0 1 0
 Nstg - sub(.*T\\dN(\\d), \\1, stg)
 Nstg
#[1] 0 1 0 0 3



characteristics_ch1.3 is a column header in  the input excel file.

T's represent stage and N's represent degree of disease spreading.

I want to create output that looks like this:
*Sample_Output_File*
T N
1 0
2 1
0 0
1 0
0 3

As it currently stands, my code is the following:






# rm(list=ls())


AND PLEASE DONT POST THAT CODE WITHOUT A COMMENT.

I noticed it this time, but it is very aggravating to accidentally  
wide out hours of work while trying to offer help.



source(../../functions.R)

uncurated - read.csv(../uncurated/ 
Sample_Input_File_full_pdata.csv,as.is

=TRUE,row.names=1)

##initial creation of curated dataframe
curated -
initialCuratedDF 
(rownames(uncurated),template.filename=Sample_Template_File.csv)


##
##start the mappings
##


##title - alt_sample_name
curated$alt_sample_name - uncurated$title

#T
tmp - uncurated$characteristics_ch1.3
tmp - *??*
curated$T - tmp


So here Tstg is tmp


#N
tmp - uncurated$characteristics_ch1.3
tmp - *??*
curated$N - tmp

And Nstg is tmp


write.table(curated, row.names=FALSE,
file=../curated/Sample_Output_File_curated_pdata.txt,sep=\t)

My question is the following:

What code gets me the desired output (replacing the *??*'s  
above)?  I
want to: a) Find the integer value one element to the right of T;  
and b)

find the integer value one element to the right of N.  I've read the
regular expression tutorial for R, but could only figure out how to  
grab an
integer value if it is the only integer value in the row (ie more  
than one

integer value makes this basic regular expression unsuccessful).


Just surround it with a pattern and use the ()  , \\n mechanism


Thank you very much for any help you can provide.

Sincerely,

Ben Ganzfried

[[alternative HTML version deleted]]



David Winsemius, MD
West Hartford, CT

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Re: [R] re-write plot function for ggplot

2011-06-02 Thread Hadley Wickham

Doesn't deal with what problems?
Hadley

On Thursday, June 2, 2011, rmje robinmje...@gmail.com wrote:
 I have been browsing the pages about ggplot and it really doesn't deal with
 such problems as far as I can see.



 --
 View this message in context: 
 http://r.789695.n4.nabble.com/re-write-plot-function-for-ggplot-tp3565868p3568025.html
 Sent from the R help mailing list archive at Nabble.com.

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-- 
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/

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[R] Paid R Help

2011-06-02 Thread Hess, Michael

Hello R people,

I am looking to pay someone to help write some R code.

Inputs:
Study identifier: ID Number for the study, each ID number is for one study
only each block set should only be used for that study. This will require
that you store the results from the blocks someplace on the file system.
Trait #1: dichotomous rural / urban
Trait #2: dichotomous sick / healthy
Assignment Ratio: a number between 0 and 1, usually .5 but, for this study it
would be .67. Indicating the % of participants to be randomized to the
intervention arm. 1-the Assignment Ratio will be the number of participants to
be randomized to the control arm. So for example, for each 6 person block,
4 would be assigned to intervention and 2 to control.
Four blocks will be in action at any given time one each for each of the
following combinations FOR EACH STUDY ID.
Rural + sick Rural + healthy urban + sick urban + healthy
The status of where we are in each block for each study will need to be stored
somewhere.

Returned value
Study Arm: dichotomous Intervention / Control

There might be some code that helps do this in R already.
http://rss.acs.unt.edu/Rdoc/library/blockrand/html/blockrand.package.html

http://docs.google.com/viewer?a=vq=cache:BnSasjcO0xQJ:personality-project.org/revelle/syllabi/205/block.randomization.pdf+r+block+randomizationhl=engl=uspid=blsrcid=ADGEESj9HQShWGzlfTJZOQQdCyohIcLV8HptDj8JZqsXbzDIZLUM6J3BDe6dpTsw95JcG6QDeiPn

The purpose of block randomization is to insure equal distribution the two
traits (in this case urban / rural and sick/healthy) across the two study arms.

This will be called as each person enrolls in the study. So you will need to
store the block data some place per study id.

I am happy to pay someone to work on this problem for me. Please contact me
off list, if you are willing and able.

Thanks,
Michael Hess
University of Michigan

**
Electronic Mail is not secure, may not be read every day, and should not be
used for urgent or sensitive issues

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[R] Plot cdf, pdf

2011-06-02 Thread Alaios

Dear all,
do you know any easy way based on a vector input how to plot easily cdf and pdf.

I would like to thank you in advance for your help

Regards
Alex

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[R] Adding a line to a beside=TRUE barplot

2011-06-02 Thread Galen Moore

Greetings  

 

Grateful for any help on this one:

 

In the following demo code, I am trying to get the points in the line to
appear over the same x-axis labels as are used by the paired Bars.  It
appears, however, that R/lattice ignores the x-axis points used by the bars
and plots the x points for the line at ½ points.

 

Can you help me tweak this code so that the nth bump in the line appears
over the same nth pair of bars?  Im open to any options besides
lattice/barplot. 

 

library(lattice)

aa - abs(rnorm(c(1:10)))*5

bb - abs(rnorm(c(1:10)))*5

cc - abs(rnorm(c(1:10)))*5

 

dd - as.matrix(cbind(aa, bb))

 

barplot(t(dd), beside=TRUE, ylim=c(0,10))

lines(cc)

 

Many thanks,

 

Galen


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Re: [R] plotmath: paste string and expression [from a vector of expressions]

2011-06-02 Thread Uwe Ligges




On 02.06.2011 20:43, Marius Hofert wrote:

Dear all,

I have a vector of expressions and would like to paste some string to it 
before using it in a plot:

vars- vector(expression, 2)
vars[1]- expression(alpha)
vars[2]- expression(beta)
plot(0, 0, main=substitute(bold(Foo ~~ VAR), list(VAR=vars[2]) ))

Although I tried hard, I just can't figure out how to solve this. The title should be 
Footheta, wheretheta  is the greek letter. I tried some constructions 
with bquote but that wasn't successful... I also looked in the mailing list but couldn't find 
anything helpful [I am sure I overlooked something].



plot(0, 0, main=expression(Foo ~~ theta))

Uwe Ligges




Cheers,

Marius

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[R] Removal of elements from nomograms

2011-06-02 Thread Rob James

The rms package includes the nomogram function, which generates a list 
object that can be passed to plot for graphical production of nomograms.


I would like to remove  the linear predictor line in the graph, which 
means (I suspect) removing it from the nomogram output object.  I've 
looked at the nomogram output object, but it is not clear to me if or 
how it might be edited to remove the linear predictor content. Similarly 
, I do not see how to coax  nomogram()  into not producing this output 
in the first place.


As ever, thanks in advance are likely due to Frank Harrell, without whom 
many things would be much more difficult.


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

Re: [R] plotmath: paste string and expression [from a vector of expressions]

Dear Uwe,

thanks for your help. Actually, I first thought about writing your solution in 
the email in order to make clear that it is not the solution I'm looking for 
:-) My goal is to work with the vector vars of expressions. The example is 
only a minimal example and for that your solution is perfectly fine, but my 
original problem is more complicated and there it makes sense to work with a 
vector of expressions. Do you know a solution to that? I tried many things... 
the obvious plot(0, 0, main=substitute(bold(Foo ~~ VAR), list(VAR=vars[2]) )) 
did not work...

Cheers,

Marius

On 2011-06-02, at 22:14 , Uwe Ligges wrote:

 
 
 On 02.06.2011 20:43, Marius Hofert wrote:
 Dear all,
 
 I have a vector of expressions and would like to paste some string to it 
 before using it in a plot:
 
 vars- vector(expression, 2)
 vars[1]- expression(alpha)
 vars[2]- expression(beta)
 plot(0, 0, main=substitute(bold(Foo ~~ VAR), list(VAR=vars[2]) ))
 
 Although I tried hard, I just can't figure out how to solve this. The title 
 should be Footheta, wheretheta  is the greek letter. I tried some 
 constructions with bquote but that wasn't successful... I also looked in the 
 mailing list but couldn't find anything helpful [I am sure I overlooked 
 something].
 
 
 plot(0, 0, main=expression(Foo ~~ theta))
 
 Uwe Ligges
 
 
 
 Cheers,
 
 Marius
 
 __
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Re: [R] Regex Question: return digits after particular letters

2011-06-02 Thread Ben Ganzfried

Thank you very much for your help.  It saved me a lot of time and it worked
perfectly.  I have a quick follow-up as I'm not sure I understand yet why
the code works and where it comes from.

For example, in: Tstg - sub(.*T(\\d)N., \\1, tmp)

*How exactly does the substitution operation work?

*On a high-level, I get that we are taking the values in the vector tmp, and
replacing each tmp value with the integer immediately after the T.  But
more lower-level, how does .*T(\\d)N., \\1 actually get us there?  I'll
undoubtedly face similar but different situations many times in the future
and I want to make sure that I know how to solve them.

Thanks again--I really appreciate your kindness.

Ben Ganzfried

On Thu, Jun 2, 2011 at 3:33 PM, David Winsemius dwinsem...@comcast.netwrote:


 On Jun 2, 2011, at 2:54 PM, Ben Ganzfried wrote:

  Hi,

 First of all, I would like to introduce myself as I will probably have
 many
 questions over the next few weeks and want to thank you guys in advance
 for
 your help.  I'm a cancer researcher and I need to learn R to complete a
 few
 projects.  I have an introductory background in Python.

 My questions at the moment are based on the following sample input file:
 *Sample_Input_File*
 characteristics_ch1.3  Stage: T1N0  Stage: T2N1  Stage: T0N0  Stage:
 T1N0  Stage:
 T0N3


 I haven't quite figured out what your structure really is, and for that you
 should learn to post the output of dput()  on the R object... but see if
 this helps:

  stg - c('Stage: T1N0',  'Stage: T2N1', 'Stage: T0N0', 'Stage: T1N0',
 'Stage: T0N3')
  Tstg - sub(.*T(\\d)N., \\1, stg)
  Tstg
 #[1] 1 2 0 1 0
  Nstg - sub(.*T\\dN(\\d), \\1, stg)
  Nstg
 #[1] 0 1 0 0 3


  characteristics_ch1.3 is a column header in  the input excel file.

 T's represent stage and N's represent degree of disease spreading.

 I want to create output that looks like this:
 *Sample_Output_File*
 T N
 1 0
 2 1
 0 0
 1 0
 0 3

 As it currently stands, my code is the following:




  # rm(list=ls())

 
 AND PLEASE DONT POST THAT CODE WITHOUT A COMMENT.

 I noticed it this time, but it is very aggravating to accidentally wide out
 hours of work while trying to offer help.

  source(../../functions.R)

 uncurated - read.csv(../uncurated/Sample_Input_File_full_pdata.csv,
 as.is
 =TRUE,row.names=1)

 ##initial creation of curated dataframe
 curated -

 initialCuratedDF(rownames(uncurated),template.filename=Sample_Template_File.csv)

 ##
 ##start the mappings
 ##


 ##title - alt_sample_name
 curated$alt_sample_name - uncurated$title

 #T
 tmp - uncurated$characteristics_ch1.3
 tmp - *??*
 curated$T - tmp


 So here Tstg is tmp


 #N
 tmp - uncurated$characteristics_ch1.3
 tmp - *??*
 curated$N - tmp

 And Nstg is tmp

  write.table(curated, row.names=FALSE,
 file=../curated/Sample_Output_File_curated_pdata.txt,sep=\t)

 My question is the following:

 What code gets me the desired output (replacing the *??*'s above)?  I
 want to: a) Find the integer value one element to the right of T; and b)
 find the integer value one element to the right of N.  I've read the
 regular expression tutorial for R, but could only figure out how to grab
 an
 integer value if it is the only integer value in the row (ie more than one
 integer value makes this basic regular expression unsuccessful).


 Just surround it with a pattern and use the ()  , \\n mechanism


 Thank you very much for any help you can provide.

 Sincerely,

 Ben Ganzfried

[[alternative HTML version deleted]]



 David Winsemius, MD
 West Hartford, CT



[[alternative HTML version deleted]]

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

Re: [R] plotmath: paste string and expression [from a vector of expressions]

2011-06-02 Thread Dennis Murphy

Hi:

This seems to work:

vars2 - c(quote(alpha), quote(beta))   # returns a list of mode call
plot(0, 0, main = bquote(bold('Foo '~.(vars2[[2]]

Expressions are only evaluated once, which means that inner
expressions are not evaluated. You need a call object rather than an
expression inside of bquote().

HTH,
Dennis

On Thu, Jun 2, 2011 at 11:43 AM, Marius Hofert m_hof...@web.de wrote:
 Dear all,

 I have a vector of expressions and would like to paste some string to it 
 before using it in a plot:

 vars - vector(expression, 2)
 vars[1] - expression(alpha)
 vars[2] - expression(beta)
 plot(0, 0, main=substitute(bold(Foo ~~ VAR), list(VAR=vars[2]) ))

 Although I tried hard, I just can't figure out how to solve this. The title 
 should be Foo theta, where theta is the greek letter. I tried some 
 constructions with bquote but that wasn't successful... I also looked in the 
 mailing list but couldn't find anything helpful [I am sure I overlooked 
 something].

 Cheers,

 Marius

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


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

Re: [R] Regex Question: return digits after particular letters


On Jun 2, 2011, at 4:21 PM, Ben Ganzfried wrote:

 Thank you very much for your help.  It saved me a lot of time and it  
 worked perfectly.  I have a quick follow-up as I'm not sure I  
 understand yet why the code works and where it comes from.

 For example, in: Tstg - sub(.*T(\\d)N., \\1, tmp)

 How exactly does the substitution operation work?

 On a high-level, I get that we are taking the values in the vector  
 tmp, and replacing each tmp value with the integer immediately after  
 the T.  But more lower-level, how does .*T(\\d)N., \\1  
 actually get us there?  I'll undoubtedly face similar but different  
 situations many times in the future and I want to make sure that I  
 know how to solve them.

The parentheses in the first pattern encloses the portion that can be  
referred to with \\1 in the second argument. Since I only enclosed  
the \\d (which is a single digit), that's what got substituted for  
the entire matched pattern. The initial part of the pattern was  
dotstar == .* which will match anything (or nothing)  before the  
T, but since it wasn't in the parens, it gets dropped.

It's actually all described on the regex page, but it helps to work  
through some examples to get the hang of it.

-- 
David.


 Thanks again--I really appreciate your kindness.

 Ben Ganzfried

 On Thu, Jun 2, 2011 at 3:33 PM, David Winsemius dwinsem...@comcast.net 
  wrote:

 On Jun 2, 2011, at 2:54 PM, Ben Ganzfried wrote:

 Hi,

 First of all, I would like to introduce myself as I will probably  
 have many
 questions over the next few weeks and want to thank you guys in  
 advance for
 your help.  I'm a cancer researcher and I need to learn R to  
 complete a few
 projects.  I have an introductory background in Python.

 My questions at the moment are based on the following sample input  
 file:
 *Sample_Input_File*
 characteristics_ch1.3  Stage: T1N0  Stage: T2N1  Stage: T0N0  Stage:
 T1N0  Stage:
 T0N3


 I haven't quite figured out what your structure really is, and for  
 that you should learn to post the output of dput()  on the R  
 object... but see if this helps:

  stg - c('Stage: T1N0',  'Stage: T2N1', 'Stage: T0N0', 'Stage:  
 T1N0', 'Stage: T0N3')
  Tstg - sub(.*T(\\d)N., \\1, stg)
  Tstg
 #[1] 1 2 0 1 0
  Nstg - sub(.*T\\dN(\\d), \\1, stg)
  Nstg
 #[1] 0 1 0 0 3


 characteristics_ch1.3 is a column header in  the input excel file.

 T's represent stage and N's represent degree of disease spreading.

 I want to create output that looks like this:
 *Sample_Output_File*
 T N
 1 0
 2 1
 0 0
 1 0
 0 3

 As it currently stands, my code is the following:




 # rm(list=ls())
 
 AND PLEASE DONT POST THAT CODE WITHOUT A COMMENT.

 I noticed it this time, but it is very aggravating to accidentally  
 wide out hours of work while trying to offer help.

 source(../../functions.R)

 uncurated - read.csv(../uncurated/ 
 Sample_Input_File_full_pdata.csv,as.is
 =TRUE,row.names=1)

 ##initial creation of curated dataframe
 curated -
 initialCuratedDF 
 (rownames(uncurated),template.filename=Sample_Template_File.csv)

 ##
 ##start the mappings
 ##


 ##title - alt_sample_name
 curated$alt_sample_name - uncurated$title

 #T
 tmp - uncurated$characteristics_ch1.3
 tmp - *??*
 curated$T - tmp

 So here Tstg is tmp

 #N
 tmp - uncurated$characteristics_ch1.3
 tmp - *??*
 curated$N - tmp
 And Nstg is tmp

 write.table(curated, row.names=FALSE,
 file=../curated/Sample_Output_File_curated_pdata.txt,sep=\t)

 My question is the following:

 What code gets me the desired output (replacing the *??*'s  
 above)?  I
 want to: a) Find the integer value one element to the right of T;  
 and b)
 find the integer value one element to the right of N.  I've read the
 regular expression tutorial for R, but could only figure out how to  
 grab an
 integer value if it is the only integer value in the row (ie more  
 than one
 integer value makes this basic regular expression unsuccessful).

 Just surround it with a pattern and use the ()  , \\n mechanism

 Thank you very much for any help you can provide.

 Sincerely,

 Ben Ganzfried

[[alternative HTML version deleted]]


 David Winsemius, MD
 West Hartford, CT



David Winsemius, MD
West Hartford, CT


[[alternative HTML version deleted]]

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[R] print overlaid plots to a pdf file.. - Help please...

2011-06-02 Thread Santosh

Dear Rxperts!
Below are relevant lines of code, I am having problems with.. I am not sure
if it is a bug in R or OS related or something else...

When viewing the pdf file, I notice overlaid plots in a panel of the plot
are shifted to the right. The Y-axes and X-labels are not exactly on top of
each other.

Would highly appreciate your suggestions..
Thanks,
Santosh


p1a -
bwplot(Y~X,notch=T,data=d3,subset=TYPE==3,ylab='',scales=list(relation='free',alternating=1),do.out=F)
p2a -
stripplot(Y~X,notch=T,data=d3,subset=TYPE==3,as.table=T,ylab='',col='gray50',cex=0.5,jitter.data=T,scales=list(relation='free',alternating=1))
p1b -
bwplot(Y~X,notch=T,data=d3,subset=TYPE==4,ylab='',scales=list(relation='free',alternating=1),do.out=F)
p2b -
stripplot(Y~X,notch=T,data=d3,subset=TYPE==4,as.table=T,ylab='',col='gray50',cex=0.5,jitter.data=T,scales=list(relation='free',alternating=1))
p1c -
bwplot(Y~X,notch=T,data=d3,subset=TYPE==5,ylab='',scales=list(relation='free',alternating=1),do.out=F)
p2c -
stripplot(Y~X,notch=T,data=d3,subset=TYPE==5,as.table=T,col='gray50',cex=0.5,jitter.data=T,scales=list(relation='free',alternating=1))

pdf(paste(file.pdf,sep=''),h=4,w=11)
print(p2a,split=c(1,1,2,1),more=T);print(p1a,split=c(1,1,2,1),more=T)
print(p2b,split=c(2,1,2,1),more=T);print(p1b,split=c(2,1,2,1),more=T)
dev.off()



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Re: [R] plotmath: paste string and expression [from a vector of expressions]



On Jun 2, 2011, at 4:19 PM, Marius Hofert wrote:


Dear Uwe,

thanks for your help. Actually, I first thought about writing your  
solution in the email in order to make clear that it is not the  
solution I'm looking for :-) My goal is to work with the vector  
vars of expressions. The example is only a minimal example and for  
that your solution is perfectly fine, but my original problem is  
more complicated and there it makes sense to work with a vector of  
expressions. Do you know a solution to that? I tried many things...  
the obvious plot(0, 0, main=substitute(bold(Foo ~~ VAR),  
list(VAR=vars[2]) )) did not work...


 vars - vector(expression, 2)
 vars[[1]] - quote(alpha)
 vars[[2]] - quote(beta)
 plot(0, 0, main= bquote( paste(Foo , .(vars[[2]] )) ) )

--
DAvid.


Cheers,

Marius

On 2011-06-02, at 22:14 , Uwe Ligges wrote:




On 02.06.2011 20:43, Marius Hofert wrote:

Dear all,

I have a vector of expressions and would like to paste some  
string to it before using it in a plot:


vars- vector(expression, 2)
vars[1]- expression(alpha)
vars[2]- expression(beta)
plot(0, 0, main=substitute(bold(Foo ~~ VAR), list(VAR=vars[2]) ))

Although I tried hard, I just can't figure out how to solve this.  
The title should be Footheta, wheretheta  is the greek  
letter. I tried some constructions with bquote but that wasn't  
successful... I also looked in the mailing list but couldn't find  
anything helpful [I am sure I overlooked something].



plot(0, 0, main=expression(Foo ~~ theta))

Uwe Ligges




Cheers,

Marius

__
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.


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


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] Paid R Help

2011-06-02 Thread Marc Schwartz

On Jun 2, 2011, at 2:41 PM, Hess, Michael wrote:

Hello R people,

I am looking to pay someone to help write some R code.

Inputs:
Study identifier: ID Number for the study, each ID number is for one study
only each block set should only be used for that study. This will require
that you store the results from the blocks someplace on the file system.
Trait #1: dichotomous rural / urban
Trait #2: dichotomous sick / healthy
Assignment Ratio: a number between 0 and 1, usually .5 but, for this study
it would be .67. Indicating the % of participants to be randomized to the
intervention arm. 1-the Assignment Ratio will be the number of participants
to be randomized to the control arm. So for example, for each 6 person
block, 4 would be assigned to intervention and 2 to control.
Four blocks will be in action at any given time one each for each of the
following combinations FOR EACH STUDY ID.
Rural + sick Rural + healthy urban + sick urban + healthy
The status of where we are in each block for each study will need to be
stored somewhere.

Returned value
Study Arm: dichotomous Intervention / Control

There might be some code that helps do this in R already.
http://rss.acs.unt.edu/Rdoc/library/blockrand/html/blockrand.package.html

The purpose of block randomization is to insure equal distribution the two
traits (in this case urban / rural and sick/healthy) across the two study
arms.

This will be called as each person enrolls in the study. So you will need to
store the block data some place per study id.

I am happy to pay someone to work on this problem for me. Please contact me
off list, if you are willing and able.

Thanks,
Michael Hess
University of Michigan

Michael,

The means to do this is pretty straightforward using the blockrand() function
in the package of the same name, as you reference above.

You essentially have a stratified randomization, based upon two dichotomous
factors. You are then doing a 2:1 randomization to the two arms in the study,
within each of the four strata.

Using blockrand():

# Set the RNG seed so that you can reproduce the sequence again in the future
set.seed(1)

# Generate 18 randomizations in one of the four strata, such that there will be
# 3 blocks, each of size 6, with 4 Intervention and 2 Control subjects in each
block
# The actual block size used will be num.levels * block.sizes (6 * 1)

Blocks1 - blockrand(18, num.levels = 6, levels = rep(c(I, C), c(4, 2)),
stratum = Rural + Sick, block.sizes = 1)

Blocks1
id stratum block.id block.size treatment
1 1 Rural + Sick1 6 I
2 2 Rural + Sick1 6 C
3 3 Rural + Sick1 6 I
4 4 Rural + Sick1 6 I
5 5 Rural + Sick1 6 I
6 6 Rural + Sick1 6 C
7 7 Rural + Sick2 6 I
8 8 Rural + Sick2 6 I
9 9 Rural + Sick2 6 C
10 10 Rural + Sick2 6 C
11 11 Rural + Sick2 6 I
12 12 Rural + Sick2 6 I
13 13 Rural + Sick3 6 I
14 14 Rural + Sick3 6 I
15 15 Rural + Sick3 6 C
16 16 Rural + Sick3 6 I
17 17 Rural + Sick3 6 C
18 18 Rural + Sick3 6 I

It is then up to whatever data management system you are using to properly use
the blocks in each stratum correctly, based upon the two characteristics that
you are using for stratification.

Just modify the 'stratum' value for each of the other 3 strata and be sure to
change the RNG seed before each run so that each subsequent strata has a
different sequence. You DO want to keep track of the RNG seed used before each
run of blockrand(), so that you can reproduce the sequencing again in the
future, if required.

HTH,

Marc Schwartz

Re: [R] Removal of elements from nomograms

2011-06-02 Thread Frank Harrell

The documentation includes this: nomogram(fit, ..., lp=FALSE)
Frank


Rob James-2 wrote:
 
 The rms package includes the nomogram function, which generates a list 
 object that can be passed to plot for graphical production of nomograms.
 
 I would like to remove  the linear predictor line in the graph, which 
 means (I suspect) removing it from the nomogram output object.  I've 
 looked at the nomogram output object, but it is not clear to me if or 
 how it might be edited to remove the linear predictor content. Similarly 
 , I do not see how to coax  nomogram()  into not producing this output 
 in the first place.
 
 As ever, thanks in advance are likely due to Frank Harrell, without whom 
 many things would be much more difficult.
 
 __
 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.
 


-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context: 
http://r.789695.n4.nabble.com/Removal-of-elements-from-nomograms-tp3569301p3569364.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Removal of elements from nomograms



On Jun 2, 2011, at 4:19 PM, Rob James wrote:

The rms package includes the nomogram function, which generates a  
list object that can be passed to plot for graphical production of  
nomograms.


I would like to remove  the linear predictor line in the graph,  
which means (I suspect) removing it from the nomogram output  
object.  I've looked at the nomogram output object, but it is not  
clear to me if or how it might be edited to remove the linear  
predictor content. Similarly , I do not see how to coax  nomogram()   
into not producing this output in the first place.


Setting the lp element to NULL would remove it from the object,  
but ... why? The plot and print functions depend on it, so your goals  
need a better elaboration.


If you want to suppress plotting of the lp then looking at the code  
it appears that the plot.nomogram function uses the info attributes  
and a bit of experimentation shows that this surgery allows plotting  
of a nomogram without the Linear Predictor line:


 attr(nom, info)$lp - FALSE
 plot(nom, xfrac=.45)

At least with the first example in the help(nomogram) page it does not  
seem to do violence to the result. And looking again at the help page,  
I see it is the fourth argument, so you could have just done it in the  
nomogram call with nomogram( . , lp=FALSE).




As ever, thanks in advance are likely due to Frank Harrell, without  
whom many things would be much more difficult.


But he cannot sit at our shoulders and read the help page to us it  
would seem.


--

David Winsemius, MD
West Hartford, CT

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Re: [R] shading in overlap between two ranges



(1) For crying out loud don't muck about with par(new=TRUE)
like that.  Use points() and lines() (and other plot functions) to
add graphical constructs to existing plots.  (And use TRUE
not T --- it's a lot safer.)

(2) In general for shading in regions between two lines
on a plot, use polygon().

cheers,

Rolf Turner

On 03/06/11 03:04, Graves, Gregory wrote:

I have 2 datafiles 'target' and 'observed' as shown below (I will gladly
email these 2 small files to whomever).  X25. And X75. Indicate the
value of 25th and 75th-percentile of the target ('what should be') and
the observed ('what is').  The i.value is simply the month.


target

Xi.valueX25. X75.
1  one.month   1 10.845225 17.87237
2  one.month   2 12.235813 19.74490
3  one.month   3 14.611749 23.44810
4  one.month   4 17.529332 28.09647
5  one.month   5 19.458738 30.56936
6  one.month   6 15.264505 28.29333
7  one.month   7 12.370369 23.35455
8  one.month   8 12.471224 21.82794
9  one.month   9  9.716685 17.28762
10 one.month  10  6.470568 12.49830
11 one.month  11  6.180560 14.24961
12 one.month  12  9.673738 15.79208


observed

  X i.value   X25. X75.
1  one.month   1 19.81000 27.63500
2  one.month   2 23.64062 30.09125
3  one.month   3 26.04865 35.99104
4  one.month   4 32.02625 41.50958
5  one.month   5 34.74479 47.75958
6  one.month   6 37.48885 46.56448
7  one.month   7 30.06740 40.10146
8  one.month   8 26.14917 39.49458
9  one.month   9 14.12521 32.39406
10 one.month  10 11.04125 23.55479
11 one.month  11 13.14917 23.56833
12 one.month  12 17.17938 27.02458

The following plots 4 lines on one graph.  The area between the two red
lines represents the target 'zone', and the area between the two black
lines is the observed 'zone'.

with(target, plot(X25.~i.value,ylim=c(0,55),type='l',col='red'))
par(new=T)
with(target, plot(X75.~i.value,ylim=c(0,55),type='l',col='red'))
par(new=T)
with(observed, plot(X25.~i.value,ylim=c(0,55),type='l'))
par(new=T)
with(observed, plot(X75.~i.value,ylim=c(0,55),type='l'))
par(new=F)

Ideally, the target and the observed should overlap in every month -
they don't.  The desire is to visually accentuate the amount of overlap
by shading in the area where these two zones overlap.  How would you
do that?  Note, that in some of these characterizations, the overlap
wanders in and out [I already have routines that calculate the percent
of overlap, but I have been requested to find a way to shade the
overlap.]

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[R] Use line break at scrip but avoid line break on graphics

2011-06-02 Thread Walmes Zeviani

Hello list,

I have plots with long strings in main=, ylab= or xlab=. So, in I my script
I use break long lines to avoid lines hiden on my monitor and in sweave
document pages. I use graphics like this

plot(1, main=
  )

but I would like a plot result like this

plot(1, main= )

I remember once I saw a meta character like \n that avoid this breack line

plot(1, main=\(?)
  )

Does someone know that?

Thanks.
Walmes

==
Walmes Marques Zeviani
LEG (Laboratório de Estatística e Geoinformação, 25.450418 S, 49.231759 W)
Departamento de Estatística - Universidade Federal do Paraná
fone: (+55) 41 3361 3573
VoIP: (3361 3600) 1053 1173
e-mail: wal...@ufpr.br
twitter: @walmeszeviani
homepage: http://www.leg.ufpr.br/~walmes
linux user number: 531218
==

[[alternative HTML version deleted]]

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Re: [R] plotmath: paste string and expression [from a vector of expressions]

Dear Dennis, Dear Uwe, Dear David,

many thanks for helping. Dennis and David, your solutions seemed perfectly 
fine, but when I applied it to my original problem, it did not show a title. 
Below is a (longer) minimal example (the first part is from the help page of 
bbmle). Is this a bug in bbmle? Hmmm...

library(bbmle)

x - 0:10
y - c(26, 17, 13, 12, 20, 5, 9, 8, 5, 4, 8)
d - data.frame(x,y)

## in general it is best practice to use the `data' argument,
##  but variables can also be drawn from the global environment
LL - function(ymax=15, xhalf=6)
-sum(stats::dpois(y, lambda=ymax/(1+x/xhalf), log=TRUE))

## uses default parameters of LL
(fit - mle2(LL))
ml - mle2(LL, fixed=list(xhalf=6))
mlp - profile(ml)
 
vars - c(quote(theta), quote(beta))
plot(mlp, main=bquote(bold(Foo~.(vars[[2]]

Cheers,

Marius

On 2011-06-02, at 22:23 , Dennis Murphy wrote:

 Hi:
 
 This seems to work:
 
 vars2 - c(quote(alpha), quote(beta))   # returns a list of mode call
 plot(0, 0, main = bquote(bold('Foo '~.(vars2[[2]]
 
 Expressions are only evaluated once, which means that inner
 expressions are not evaluated. You need a call object rather than an
 expression inside of bquote().
 
 HTH,
 Dennis
 
 On Thu, Jun 2, 2011 at 11:43 AM, Marius Hofert m_hof...@web.de wrote:
 Dear all,
 
 I have a vector of expressions and would like to paste some string to it 
 before using it in a plot:
 
 vars - vector(expression, 2)
 vars[1] - expression(alpha)
 vars[2] - expression(beta)
 plot(0, 0, main=substitute(bold(Foo ~~ VAR), list(VAR=vars[2]) ))
 
 Although I tried hard, I just can't figure out how to solve this. The title 
 should be Foo theta, where theta is the greek letter. I tried some 
 constructions with bquote but that wasn't successful... I also looked in the 
 mailing list but couldn't find anything helpful [I am sure I overlooked 
 something].
 
 Cheers,
 
 Marius
 
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Re: [R] plotmath: paste string and expression [from a vectorof expressions]

2011-06-02 Thread William Dunlap

 -Original Message-
 From: r-help-boun...@r-project.org 
 [mailto:r-help-boun...@r-project.org] On Behalf Of Marius Hofert
 Sent: Thursday, June 02, 2011 1:20 PM
 To: Uwe Ligges
 Cc: Help R
 Subject: Re: [R] plotmath: paste string and expression [from 
 a vectorof expressions]

 Dear Uwe,

 thanks for your help. Actually, I first thought about writing 
 your solution in the email in order to make clear that it is 
 not the solution I'm looking for :-) My goal is to work with 
 the vector vars of expressions. The example is only a 
 minimal example and for that your solution is perfectly fine, 
 but my original problem is more complicated and there it 
 makes sense to work with a vector of expressions. Do you know 
 a solution to that? I tried many things... the obvious 
 plot(0, 0, main=substitute(bold(Foo ~~ VAR), 
 list(VAR=vars[2]) )) did not work...

Use VAR=vars[[2]] (double brackets) there.  You can see the
difference if you look at the output of your call to substitute.
[[ gives you an element of the expression and [ gives you
an expression containing an element:

   substitute(bold(Foo ~~ VAR), list(VAR=vars[[2]]) )
  bold(Foo ~ ~beta)
   substitute(bold(Foo ~~ VAR), list(VAR=vars[2]) )
  bold(Foo ~ ~expression(beta))

The same holds for the bquote() solution that David W. suggested.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com  

 Cheers,

 Marius

 On 2011-06-02, at 22:14 , Uwe Ligges wrote:

  On 02.06.2011 20:43, Marius Hofert wrote:
  Dear all,

  I have a vector of expressions and would like to paste 
 some string to it before using it in a plot:

  vars- vector(expression, 2)
  vars[1]- expression(alpha)
  vars[2]- expression(beta)
  plot(0, 0, main=substitute(bold(Foo ~~ VAR), list(VAR=vars[2]) ))

  Although I tried hard, I just can't figure out how to 
 solve this. The title should be Footheta, wheretheta  
 is the greek letter. I tried some constructions with bquote 
 but that wasn't successful... I also looked in the mailing 
 list but couldn't find anything helpful [I am sure I 
 overlooked something].

  plot(0, 0, main=expression(Foo ~~ theta))

  Uwe Ligges

  Cheers,

  Marius

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  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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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide 
 http://www.R-project.org/posting-guide.html
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Re: [R] Problem with package development

2011-06-02 Thread steven mosher

I hope you're successful because I'm having issues as well building a simple
package on
windows.  maybe when you're finished  you can share back a step by step
guide.

On Thu, Jun 2, 2011 at 12:21 PM, Nipesh Bajaj bajaj141...@gmail.com wrote:

 I have run R CMD check trial1 and saw an error. This says that:

 * checking pdf version of manual without hyperrefs or index  ... ERROR
 Re-running with no redirection of stdout/stderr.
 Hmm... looks like a package
 Error in texi2dvi(Rd2.tex, pdf = (out_ext == pdf), quiet = FALSE, :
 pdflatex is not available
 Error in running tools:: texi2dvi

 Does this information hwlp you to suggest something? Please let me
 know what else I can provide.

 Thanks,

 On Fri, Jun 3, 2011 at 12:00 AM, Nipesh Bajaj bajaj141...@gmail.com
 wrote:
  What else I need to do? In the Read-and-delete-me file following steps
  are asked to perform:
  * Edit the help file skeletons in 'man', possibly combining help files
  for multiple functions.
  * Edit the exports in 'NAMESPACE', and add necessary imports.
  * Put any C/C++/Fortran code in 'src'.
  * If you have compiled code, add a useDynLib() directive to 'NAMESPACE'.
  * Run R CMD build to build the package tarball.
  * Run R CMD check to check the package tarball.
 
  I editied the help page for fn1() function (as I already communicated
  in previous mail) as follows:
  \name{fn1}
  \alias{fn1}
 
  \title{
  A function.
  }
 
  \description{
  A function.
  }
 
  \usage{
  A function.
  }
 
  \arguments{
  A function.
  }
 
  \value{
  A function.
  }
 
  \author{
  \bold{Me}
  \cr
  \email{m...@me.com}
  }
 
  And regarding th Namespace file, this time I put
  package.skeleton(trial1,namespace = FALSE, code_files =
  f:/trial.r)
  
 
  I do not have any C/C++ code so I ignored 3rd step.
 
  then Read-and-delete-me file asking me to build the package, so in
  cmd, I run following:
  cd C:\R_PackageBuild
  Rcmd build âbinary trial1
 
  What I am missing in this entire process? Do you please point me?
 
  Thanks,
 
 
  On Thu, Jun 2, 2011 at 11:40 PM, Duncan Murdoch
  murdoch.dun...@gmail.com wrote:
  On 02/06/2011 2:03 PM, Nipesh Bajaj wrote:
 
  Thanks Prof. Ripley and Duncan for your pointers. Noting down your
  points I have modified my way of building package and have done
  following so far:
 
  1. In my C: drive I create one working folder naming R_PackageBuild
  2. In R console I have written following codes:
setwd(c:/R_packageBuild)
package.skeleton(trial1,namespace = TRUE, code_files =
 f:/trial.r)
  3. then I opened cmd and wrote following:
  cd C:\R_PackageBuild
  Rcmd build âbinary trial1
 
  This process halted with following error:
  Error: unexpected symbol in 
  âtools:::.test_load_package(âtrial1â²,â¦.)â
  Execution halted
  ERROR: loading failed
 
  What I have missed in this process? Can you please help me how to
  solve this issue?
 
  You haven't done the manual changes required between steps 2 and 3.
   package.skeleton() creates the skeleton of a package; you run it once
 as
  you are starting development, the do a lot of manual updates, described
 on
  the ?package.skeleton help page, and in the  âRead-and-delete-meâ file.
   Once those are done, step 3 should succeed.
 
  Duncan Murdoch
 
  Thanks,
 
  PS: I am sorry I missplet 'Program Files'. Thanks Prof. Ripley for this
  pointer.
 
  On Wed, Jun 1, 2011 at 11:41 AM, Prof Brian Ripley
  rip...@stats.ox.ac.uk  wrote:
On Wed, 1 Jun 2011, Nipesh Bajaj wrote:
  
I have been struggling for last one hour but not yet any through.
  
However again I recreate the package.skeleton and run R CMD check
   trial3
  
Here are the errors:
  
warning in dir.create(pkgoutdir, mode = 0755):
cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
reason .
Error in printLog(Log, , text, \n): object 'Log' not found
Execution haulted
  
Why I am getting this error? what is that Log. I will really
appreciate if somebody please help me to figure out.
  
R CMD check writes a (in your case) trial3.Rcheck directory, and in
   there in
file 00check.log a copy of the log.  If it cannot create
 trial3.Rcheck
   it
cannot write the log.
  
I would be surprised that even on Windows Vista the message was
   literally
  
reason .
  
but if it was, blame Microsoft for their error messages.
But
  
cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
  
is clear enough.  You need to run 'R CMD check' in your user area.
In case you did this because that is where you though 'R' was, it is
   not the
correct R.exe.   You may need to add
  
c:\Program Files\R\R-2.13.0\bin\i386
  
(assuming 32-bit R) to your path.
  
However, your use of e.g. 'Program files' suggests you are not
   accurately
transmitting the messages you got.
  
  
  
Thanks,
  
On Wed, Jun 1, 2011 at 1:20 AM, Nipesh

Re: [R] complex search between dataframes

2011-06-02 Thread Marc Schwartz

On Jun 2, 2011, at 1:42 PM, Filippo Beleggia wrote:

 Hi!
 
 I am very new to R, I hope someone can help me.
 
 I have two dataframes:
 
 data1-data.frame(from=c(1,12,16,40,55,81,101),to=c(10,13,23,45,67,99,123))
 data2-data.frame(name=c(1,2,3,4,5,6,7,8,9),position=c(2,14,20,50,150,2000,2001,2002,85))
 
 
 I want to know which of the entries in position of data2 are included 
 between 
 any from and the corresponding to of data1.
 
 So in this case I would need to somehow be able to extract 2,20 and 85, 
 corrisponding to the names 1,3 and 9.
 
 Thank you very much!
 Filippo


See ?findInterval

Coerce data1 into a matrix, so that the interval boundaries are in increasing 
order by columns, which is then actually used by findInterval as a vector (eg. 
c(1, 10, 12, ...)):

 t(data1)
 [,1] [,2] [,3] [,4] [,5] [,6] [,7]
from1   12   16   40   55   81  101
to 10   13   23   45   67   99  123


findInterval() will return the interval indices for each data2$position value 
within the sorted intervals. Since your actual intervals are discontinuous, you 
only want the values that fit in the odd intervals, which is where the use of 
%in% seq(1, 13, 2) comes in. Prior to that, findInterval() returns:

 findInterval(data2$position, t(data1))
[1]  1  4  5  8 14 14 14 14 11

With it:

 findInterval(data2$position, t(data1)) %in% seq(1, 13, 2)
[1]  TRUE FALSE  TRUE FALSE FALSE FALSE FALSE FALSE  TRUE

Now you can use the TRUE values to index data2$name:

 data2$name[findInterval(data2$position, t(data1)) %in% seq(1, 13, 2)]
[1] 1 3 9

or data2$position:

 data2$position[findInterval(data2$position, t(data1)) %in% seq(1, 13, 2)]
[1]  2 20 85



HTH,

Marc Schwartz

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

Re: [R] Problem with package development

Still I am struggling to get some inputs from the experts here :(
Definitely We can shere our experiences once I am done (really, it
seems to me very hard nut to crack!) I strongly feel that related
documentations should be more Statistician-friendly, than some Engg.
guys

Thanks,

On Fri, Jun 3, 2011 at 2:55 AM, steven mosher mosherste...@gmail.com wrote:
 I hope you're successful because I'm having issues as well building a simple
 package on
 windows.  maybe when you're finished  you can share back a step by step
 guide.

 On Thu, Jun 2, 2011 at 12:21 PM, Nipesh Bajaj bajaj141...@gmail.com wrote:

 I have run R CMD check trial1 and saw an error. This says that:

 * checking pdf version of manual without hyperrefs or index  ... ERROR
 Re-running with no redirection of stdout/stderr.
 Hmm... looks like a package
 Error in texi2dvi(Rd2.tex, pdf = (out_ext == pdf), quiet = FALSE, :
 pdflatex is not available
 Error in running tools:: texi2dvi

 Does this information hwlp you to suggest something? Please let me
 know what else I can provide.

 Thanks,

 On Fri, Jun 3, 2011 at 12:00 AM, Nipesh Bajaj bajaj141...@gmail.com
 wrote:
  What else I need to do? In the Read-and-delete-me file following steps
  are asked to perform:
  * Edit the help file skeletons in 'man', possibly combining help files
  for multiple functions.
  * Edit the exports in 'NAMESPACE', and add necessary imports.
  * Put any C/C++/Fortran code in 'src'.
  * If you have compiled code, add a useDynLib() directive to 'NAMESPACE'.
  * Run R CMD build to build the package tarball.
  * Run R CMD check to check the package tarball.
 
  I editied the help page for fn1() function (as I already communicated
  in previous mail) as follows:
  \name{fn1}
  \alias{fn1}
 
  \title{
  A function.
  }
 
  \description{
  A function.
  }
 
  \usage{
  A function.
  }
 
  \arguments{
  A function.
  }
 
  \value{
  A function.
  }
 
  \author{
  \bold{Me}
  \cr
  \email{m...@me.com}
  }
 
  And regarding th Namespace file, this time I put
  package.skeleton(trial1,namespace = FALSE, code_files =
  f:/trial.r)
  
 
  I do not have any C/C++ code so I ignored 3rd step.
 
  then Read-and-delete-me file asking me to build the package, so in
  cmd, I run following:
  cd C:\R_PackageBuild
  Rcmd build –binary trial1
 
  What I am missing in this entire process? Do you please point me?
 
  Thanks,
 
 
  On Thu, Jun 2, 2011 at 11:40 PM, Duncan Murdoch
  murdoch.dun...@gmail.com wrote:
  On 02/06/2011 2:03 PM, Nipesh Bajaj wrote:
 
  Thanks Prof. Ripley and Duncan for your pointers. Noting down your
  points I have modified my way of building package and have done
  following so far:
 
  1. In my C: drive I create one working folder naming R_PackageBuild
  2. In R console I have written following codes:
    setwd(c:/R_packageBuild)
    package.skeleton(trial1,namespace = TRUE, code_files =
   f:/trial.r)
  3. then I opened cmd and wrote following:
  cd C:\R_PackageBuild
  Rcmd build –binary trial1
 
  This process halted with following error:
  Error: unexpected symbol in “tools:::.test_load_package(‘trial1′,….)”
  Execution halted
  ERROR: loading failed
 
  What I have missed in this process? Can you please help me how to
  solve this issue?
 
  You haven't done the manual changes required between steps 2 and 3.
   package.skeleton() creates the skeleton of a package; you run it once
  as
  you are starting development, the do a lot of manual updates, described
  on
  the ?package.skeleton help page, and in the  ‘Read-and-delete-me’ file.
   Once those are done, step 3 should succeed.
 
  Duncan Murdoch
 
  Thanks,
 
  PS: I am sorry I missplet 'Program Files'. Thanks Prof. Ripley for
  this
  pointer.
 
  On Wed, Jun 1, 2011 at 11:41 AM, Prof Brian Ripley
  rip...@stats.ox.ac.uk  wrote:
    On Wed, 1 Jun 2011, Nipesh Bajaj wrote:
  
    I have been struggling for last one hour but not yet any through.
  
    However again I recreate the package.skeleton and run R CMD check
   trial3
  
    Here are the errors:
  
    warning in dir.create(pkgoutdir, mode = 0755):
    cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
    reason .
    Error in printLog(Log, , text, \n): object 'Log' not found
    Execution haulted
  
    Why I am getting this error? what is that Log. I will really
    appreciate if somebody please help me to figure out.
  
    R CMD check writes a (in your case) trial3.Rcheck directory, and in
   there in
    file 00check.log a copy of the log.  If it cannot create
   trial3.Rcheck
   it
    cannot write the log.
  
    I would be surprised that even on Windows Vista the message was
   literally
  
    reason .
  
    but if it was, blame Microsoft for their error messages.
    But
  
    cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
  
    is clear enough.  You need to run 'R CMD check' in your user area.
    In case you did this because that is where you though 'R' was, it

Re: [R] Use line break at scrip but avoid line break on graphics


On 03/06/11 09:03, Walmes Zeviani wrote:

Hello list,

I have plots with long strings in main=, ylab= or xlab=. So, in I my script
I use break long lines to avoid lines hiden on my monitor and in sweave
document pages. I use graphics like this

plot(1, main=
   )

but I would like a plot result like this

plot(1, main= )

I remember once I saw a meta character like \n that avoid this breack line

plot(1, main=\(?)
   )

Does someone know that?


I think that

plot(1,main=paste(,
   ))

might do what you want.

cheers,

Rolf Turner

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

Re: [R] Problem with package development

2011-06-02 Thread Sarah Goslee

I don't use windows, but this error message you report:
 Error in texi2dvi(Rd2.tex, pdf = (out_ext == pdf), quiet = FALSE, :
 pdflatex is not available
 Error in running tools:: texi2dvi

strongly indicates that you need to install pdflatex, don't you think?

If you don't have non-R code (c or fortran) you can probably get away
without having the MinGW compiler tools, but you do need the rest of
the things listed in the admin manual as required for building
packages:
http://cran.r-project.org/doc/manuals/R-admin.html#The-Windows-toolset

In addition, googling for building R packages on Windows turns up
very many detailed guides for going through the process, including
discussions of what additional software you need and how to install
it.

Sarah

On Thu, Jun 2, 2011 at 6:03 PM, Nipesh Bajaj bajaj141...@gmail.com wrote:
 Still I am struggling to get some inputs from the experts here :(
 Definitely We can shere our experiences once I am done (really, it
 seems to me very hard nut to crack!) I strongly feel that related
 documentations should be more Statistician-friendly, than some Engg.
 guys

 Thanks,

 On Fri, Jun 3, 2011 at 2:55 AM, steven mosher mosherste...@gmail.com wrote:
 I hope you're successful because I'm having issues as well building a simple
 package on
 windows.  maybe when you're finished  you can share back a step by step
 guide.

 On Thu, Jun 2, 2011 at 12:21 PM, Nipesh Bajaj bajaj141...@gmail.com wrote:

 I have run R CMD check trial1 and saw an error. This says that:

 * checking pdf version of manual without hyperrefs or index  ... ERROR
 Re-running with no redirection of stdout/stderr.
 Hmm... looks like a package
 Error in texi2dvi(Rd2.tex, pdf = (out_ext == pdf), quiet = FALSE, :
 pdflatex is not available
 Error in running tools:: texi2dvi

 Does this information hwlp you to suggest something? Please let me
 know what else I can provide.

 Thanks,

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

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Re: [R] Use line break at scrip but avoid line break on graphics

2011-06-02 Thread Peter Langfelder

On Thu, Jun 2, 2011 at 3:13 PM, Rolf Turner rolf.tur...@xtra.co.nz wrote:
 On 03/06/11 09:03, Walmes Zeviani wrote:

 Hello list,

 I have plots with long strings in main=, ylab= or xlab=. So, in I my
 script
 I use break long lines to avoid lines hiden on my monitor and in sweave
 document pages. I use graphics like this

 plot(1, main=
               )

 but I would like a plot result like this

 plot(1, main= )

 I remember once I saw a meta character like \n that avoid this breack
 line

 plot(1, main=\(?)
               )

 Does someone know that?

 I think that

    plot(1,main=paste(,
                                   ))

 might do what you want.

I think he needs

plot(1,main=paste(\n,
 , sep = ))

or simply

plot(1, main=\n)

Peter

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Re: [R] plotmath: paste string and expression [from a vector of expressions]



On Jun 2, 2011, at 5:07 PM, Marius Hofert wrote:


Dear Dennis, Dear Uwe, Dear David,

many thanks for helping. Dennis and David, your solutions seemed  
perfectly fine, but when I applied it to my original problem, it did  
not show a title. Below is a (longer) minimal example (the first  
part is from the help page of bbmle). Is this a bug in bbmle? Hmmm...


library(bbmle)

x - 0:10
y - c(26, 17, 13, 12, 20, 5, 9, 8, 5, 4, 8)
d - data.frame(x,y)

## in general it is best practice to use the `data' argument,
##  but variables can also be drawn from the global environment
LL - function(ymax=15, xhalf=6)
   -sum(stats::dpois(y, lambda=ymax/(1+x/xhalf), log=TRUE))

## uses default parameters of LL
(fit - mle2(LL))
ml - mle2(LL, fixed=list(xhalf=6))
mlp - profile(ml)

vars - c(quote(theta), quote(beta))
plot(mlp, main=bquote(bold(Foo~.(vars[[2]]


I do not have experience with that package but my guess is that the  
plot.mle2 (or whatever its name might be... )  function does something  
different. class(mlp) returns profile.mle2. Looking at the  
documentation you see that it is using S4 methods and using Methods()  
one sees that there is a 'plot' method for 'profile.mle2' objects.  
That's about as far as I go.


whine-mode on
 Navigating S4 methods is an arcane art into which I have not  
initiated myself. Unlike S3 methods where you just type the function  
name and it's easy to figure out what the names will be, there are  
several levels of specification and the help pages for Methods'   
are not, ... helpful to one who approaches it without more  
experience than I have. The help page regarding acceptable expressions  
to pass to main arguments is not particularly helpful, either. I  
would advise asking the package author or maintainer.


Before anyone berates me for insufficient effort at self-learning, I  
swear that my copy of Chambers (2008) arrived this week. I do think it  
would be helpful to have a worked example near the top of examples  
on help(Methods) that shows HOW_TO get at the functional machinery for  
a plot method when one knows the class of an object.


After some further experimentation I have a theory that the 'main'  
argument will not accept a language object but that one can coerce to  
an expression object and succeed. (Didn't I go through this once  
before? Maybe this was what Hadley was trying to teach me about a  
month ago.)

whine-mode off

# -Answer

plot(mlp, main=as.expression(bquote(bold(Foo~.(vars[[2]])  )) ) )

# --- back your regularly scheduled programming ---
--
DAvid.




Cheers,

Marius

On 2011-06-02, at 22:23 , Dennis Murphy wrote:


Hi:

This seems to work:

vars2 - c(quote(alpha), quote(beta))   # returns a list of mode call
plot(0, 0, main = bquote(bold('Foo '~.(vars2[[2]]

Expressions are only evaluated once, which means that inner
expressions are not evaluated. You need a call object rather than an
expression inside of bquote().

HTH,
Dennis

On Thu, Jun 2, 2011 at 11:43 AM, Marius Hofert m_hof...@web.de  
wrote:

Dear all,

I have a vector of expressions and would like to paste some  
string to it before using it in a plot:


vars - vector(expression, 2)
vars[1] - expression(alpha)
vars[2] - expression(beta)
plot(0, 0, main=substitute(bold(Foo ~~ VAR), list(VAR=vars[2]) ))

Although I tried hard, I just can't figure out how to solve this.  
The title should be Foo theta, where theta is the greek  
letter. I tried some constructions with bquote but that wasn't  
successful... I also looked in the mailing list but couldn't find  
anything helpful [I am sure I overlooked something].


Cheers,

Marius

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David Winsemius, MD
West Hartford, CT

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Re: [R] Use line break at scrip but avoid line break on graphics



On Jun 2, 2011, at 6:13 PM, Rolf Turner wrote:


On 03/06/11 09:03, Walmes Zeviani wrote:

Hello list,

I have plots with long strings in main=, ylab= or xlab=. So, in I  
my script
I use break long lines to avoid lines hiden on my monitor and in  
sweave

document pages. I use graphics like this

plot(1, main=
  )

but I would like a plot result like this

plot(1, main= )

I remember once I saw a meta character like \n that avoid this  
breack line


plot(1, main=\(?)
  )

Does someone know that?


I think that

   plot(1,main=paste(,
  ))


Look, Ma, no quotes!  Er, well, maybe one quote.

plot(1, main=quote(~
  ) )

--

David Winsemius, MD
West Hartford, CT

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[R] lattice + plotmath: how to get a variable in bold face?

Dear all,

How can I get a bold 1000 in the title? I would like to use a variable (as 
opposed to putting in 1000 directly).

library(lattice)
N - 1000
xyplot(0~0, xlab.top=list(label=as.expression(bquote(bold(foo ~ .(N) ~ 
bar))), font=2, cex=1.2)) 
## = font=2 is ignored (of course)

Cheers,

Marius
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Re: [R] lattice + plotmath: how to get a variable in bold face?

2011-06-02 Thread Peter Ehlers


On 2011-06-02 15:50, Marius Hofert wrote:

Dear all,

How can I get a bold 1000 in the title? I would like to use a variable (as opposed to 
putting in 1000 directly).

library(lattice)
N- 1000
xyplot(0~0, xlab.top=list(label=as.expression(bquote(bold(foo ~ .(N) ~ 
bar))), font=2, cex=1.2))
## =  font=2 is ignored (of course)


You could add

 N - as.character(N)

before your call to xyplot.

Peter Ehlers



Cheers,

Marius
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Re: [R] Problem with package development

2011-06-02 Thread Joshua Wiley

Hi,

On Thu, Jun 2, 2011 at 11:30 AM, Nipesh Bajaj bajaj141...@gmail.com wrote:
 I editied the help page for fn1() function (as I already communicated
 in previous mail) as follows:
 \name{fn1}
 \alias{fn1}

 \title{
 A function.
 }

 \description{
 A function.
 }

 \usage{
 A function.

What makes you think this qualifies as editing?  Please read the
Writing R Extensions manual thoroughly.  If you had, you would see
that what you have written will clearly not work (though I would think
this intuitively obvious if you have ever read documentation for
R...what function is simply documented, A function.?).  Here is a
link to the manual:

http://cran.r-project.org/doc/manuals/R-exts.html

it actually provides a wealth of information and when I was writing my
first package, I kept it open and constantly referred to it as I was
writing documentation, naming files, creating the DESCRIPTION file,
etc.  The other resource that helped me tremendously was Dr. Chambers
book Software for Data Analysis.  You might also consider downloading
the source code for a package (I just used the SoDA package) so you
have an example of how someone who knows what they are doing writes a
package.

I feel pretty strongly that:

I strongly feel that related documentations should be more
Statistician-friendly, than some Engg. guys

is quite an unfair statement given all the time and effort R core has
put not only into writing R, but making manuals and documentation
that, while complex at times, are extremely thorough to the point that
I, as a psychologist, with no training in computer science,
engineering, or statistics, was able to blunder my way along just by
carefully going through their work.

Best of luck to you,

Josh

 }

 \arguments{
 A function.
 }

 \value{
 A function.
 }

 \author{
 \bold{Me}
 \cr
 \email{m...@me.com}
 }

 And regarding th Namespace file, this time I put
 package.skeleton(trial1,namespace = FALSE, code_files =
 f:/trial.r)
 

 I do not have any C/C++ code so I ignored 3rd step.

 then Read-and-delete-me file asking me to build the package, so in
 cmd, I run following:
 cd C:\R_PackageBuild
 Rcmd build –binary trial1

 What I am missing in this entire process? Do you please point me?

 Thanks,


 On Thu, Jun 2, 2011 at 11:40 PM, Duncan Murdoch
 murdoch.dun...@gmail.com wrote:
 On 02/06/2011 2:03 PM, Nipesh Bajaj wrote:

 Thanks Prof. Ripley and Duncan for your pointers. Noting down your
 points I have modified my way of building package and have done
 following so far:

 1. In my C: drive I create one working folder naming R_PackageBuild
 2. In R console I have written following codes:
   setwd(c:/R_packageBuild)
   package.skeleton(trial1,namespace = TRUE, code_files = f:/trial.r)
 3. then I opened cmd and wrote following:
 cd C:\R_PackageBuild
 Rcmd build –binary trial1

 This process halted with following error:
 Error: unexpected symbol in “tools:::.test_load_package(‘trial1′,….)”
 Execution halted
 ERROR: loading failed

 What I have missed in this process? Can you please help me how to
 solve this issue?

 You haven't done the manual changes required between steps 2 and 3.
  package.skeleton() creates the skeleton of a package; you run it once as
 you are starting development, the do a lot of manual updates, described on
 the ?package.skeleton help page, and in the  ‘Read-and-delete-me’ file.
  Once those are done, step 3 should succeed.

 Duncan Murdoch

 Thanks,

 PS: I am sorry I missplet 'Program Files'. Thanks Prof. Ripley for this
 pointer.

 On Wed, Jun 1, 2011 at 11:41 AM, Prof Brian Ripley
 rip...@stats.ox.ac.uk  wrote:
   On Wed, 1 Jun 2011, Nipesh Bajaj wrote:
 
   I have been struggling for last one hour but not yet any through.
 
   However again I recreate the package.skeleton and run R CMD check
  trial3
 
   Here are the errors:
 
   warning in dir.create(pkgoutdir, mode = 0755):
   cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
   reason .
   Error in printLog(Log, , text, \n): object 'Log' not found
   Execution haulted
 
   Why I am getting this error? what is that Log. I will really
   appreciate if somebody please help me to figure out.
 
   R CMD check writes a (in your case) trial3.Rcheck directory, and in
  there in
   file 00check.log a copy of the log.  If it cannot create trial3.Rcheck
  it
   cannot write the log.
 
   I would be surprised that even on Windows Vista the message was
  literally
 
   reason .
 
   but if it was, blame Microsoft for their error messages.
   But
 
   cannot create dir 'c:\Program files\R\R-2.13.0\bin\trial3.Rcheck',
 
   is clear enough.  You need to run 'R CMD check' in your user area.
   In case you did this because that is where you though 'R' was, it is
  not the
   correct R.exe.   You may need to add
 
   c:\Program Files\R\R-2.13.0\bin\i386
 
   (assuming 32-bit R) to your path.
 
   However, your use of e.g. 'Program files' suggests you are not
  accurately
   transmitting the

Re: [R] Use line break at scrip but avoid line break on graphics


On 03/06/11 10:16, Peter Langfelder wrote:

On Thu, Jun 2, 2011 at 3:13 PM, Rolf Turnerrolf.tur...@xtra.co.nz  wrote:

On 03/06/11 09:03, Walmes Zeviani wrote:

Hello list,

I have plots with long strings in main=, ylab= or xlab=. So, in I my
script
I use break long lines to avoid lines hiden on my monitor and in sweave
document pages. I use graphics like this

plot(1, main=
   )

but I would like a plot result like this

plot(1, main= )

I remember once I saw a meta character like \n that avoid this breack
line

plot(1, main=\(?)
   )

Does someone know that?

I think that

plot(1,main=paste(,
   ))

might do what you want.

I think he needs

plot(1,main=paste(\n,
  , sep = ))

or simply

plot(1, main=\n)


Absolutely not!  Did you *read* the OP's question?  He wants to break
the line in the code --- for readable code presumably --- but ***not***
in the output!   Your advice is the antithesis of what is required.

cheers,

Rolf Turner

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Re: [R] Use line break at scrip but avoid line break on graphics

2011-06-02 Thread Peter Langfelder

 Absolutely not!  Did you *read* the OP's question?  He wants to break
 the line in the code --- for readable code presumably --- but ***not***
 in the output!   Your advice is the antithesis of what is required.

    cheers,

        Rolf Turner

I stand corrected, indeed I missed the crucial part of the OP's post.

Peter

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Re: [R] Problem with package development


On 03/06/11 11:33, Joshua Wiley wrote:

Hi,

On Thu, Jun 2, 2011 at 11:30 AM, Nipesh Bajajbajaj141...@gmail.com  wrote:

I editied the help page for fn1() function (as I already communicated
in previous mail) as follows:
\name{fn1}
\alias{fn1}

\title{
A function.
}

\description{
A function.
}

\usage{
A function.

What makes you think this qualifies as editing?  Please read the
Writing R Extensions manual thoroughly.  If you had, you would see
that what you have written will clearly not work (though I would think
this intuitively obvious if you have ever read documentation for
R...what function is simply documented, A function.?).


SNIP

Huh?  What on earth are you on about?  This is just a toy example
to get things working and in such instances a title such as ``A 
function''

is perfectly acceptable.

It  looks to me like the OP's editing was done adequately, and is *not*
the source of the OP's problems.

Sarah Goslee has already pointed out what at least one of the sources
of his problems is.

cheers,

Rolf Turner

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Re: [R] Problem with package development

2011-06-02 Thread Joshua Wiley

On Thu, Jun 2, 2011 at 4:49 PM, Rolf Turner rolf.tur...@xtra.co.nz wrote:
 On 03/06/11 11:33, Joshua Wiley wrote:

 Hi,

 On Thu, Jun 2, 2011 at 11:30 AM, Nipesh Bajajbajaj141...@gmail.com
  wrote:

 I editied the help page for fn1() function (as I already communicated
 in previous mail) as follows:
 \name{fn1}
 \alias{fn1}

 \title{
 A function.
 }

 \description{
 A function.
 }

 \usage{
 A function.

 What makes you think this qualifies as editing?  Please read the
 Writing R Extensions manual thoroughly.  If you had, you would see
 that what you have written will clearly not work (though I would think
 this intuitively obvious if you have ever read documentation for
 R...what function is simply documented, A function.?).

 SNIP

    Huh?  What on earth are you on about?  This is just a toy example
    to get things working and in such instances a title such as ``A
 function''
    is perfectly acceptable.

Really?  R CMD check screams bloody murder at me:

Bad \usage lines found in documentation object 'fn1':
  A function.

Functions with \usage entries need to have the appropriate \alias
entries, and all their arguments documented.
The \usage entries must correspond to syntactically valid R code.
See the chapter 'Writing R documentation files' in the 'Writing R
Extensions' manual.


    It  looks to me like the OP's editing was done adequately, and is *not*
    the source of the OP's problems.

    Sarah Goslee has already pointed out what at least one of the sources
    of his problems is.

        cheers,

            Rolf Turner

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

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Re: [R] Problem with package development

2011-06-02 Thread Joshua Wiley

Okay, that might have been a little strong.  screams bloody murder
is a warning, not technically an error, and does not occur when simply
running R CMD build.  That said, the OP did mention using R CMD check
and pdflatex is not an issue when only building anyway.  Still, it is
not the cause of the earlier problems and I was probably overly harsh
and I give my sincerest apology.

Josh

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Re: [R] Problem with package development