Re: [R] Loading .Rdata within an R function

[Giles Crane]

Thank you for your consideration of this question.
I have tried both your suggestions.
However, the data is not loaded within the function.

When I specify load(mydata.Rdata,.globalEnv),
the data is loaded into the top level environment,
and the function does access the data in the top level environment.

However, I would like to load the data into the function environment,
so that the data goes away when the function terminates.

Cordially,
Giles Crane
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[R] Appropriate tests for logistic regression with a continuous predictor variable and Bernoulli response variable

[Kiyoshi Sasaki]

I have a data with binary response variable, repcnd (pregnant or not) and one 
predictor continuous variable, svl (body size) as shown below. I did 
Hosmer-Lemeshow test as a goodness of fit (as suggested by a kind 
“R-helper” previously). To test whether the predictor (svl, or body size) 
has significant effect on predicting whether or not a female snake is pregnant, 
I used the differences between null deviance and residual deviance using a code 
as following:

 
1-pchisq(mod.fit$null.deviance - mod.fit$deviance, mod.fit$df.null - 
mod.fit$df.residual)
 
Could anyone tell me whether I did the test properly? I did this test because 
I thought Wald test/z score listed in the output from summary(mod.fit) is 
not appropriate for a kind of data I have.  Does R have automated function 
to run appropriate tests? I have pasted my R output below.
 
Thank you in advance for your time and help.
 
Kiyoshi
 
 
            repcnd             svl
1          1                      51.5
2          1                      52.5
edited
294      0                      59.8
298      1                      60.0
300      1                      51.7
301      1                      57.4
302      1                      60.9
303      0                      56.8
304      0                      50.0
---
 mod.fit - glm(formula = gb.no.M$repcnd ~ gb.no.M$svl, family = binomial(link 
 = logit), data = gb.no.M, na.action = na.exclude, control = list(epsilon = 
 0.0001, maxit = 50, trace = F))
 summary(mod.fit)
 
Call:
glm(formula = gb.no.M$repcnd ~ gb.no.M$svl, family = binomial(link = logit), 
    data = gb.no.M, na.action = na.exclude, control = list(epsilon = 1e-04, 
        maxit = 50, trace = F))
 
Deviance Residuals: 
   Min      1Q  Median      3Q     Max  
-1.757  -1.109   0.734   1.113   1.632  
 
Coefficients:
            Estimate Std. Error z value Pr(|z|)    
(Intercept) -7.08565    1.84106  -3.849 0.000119 ***
gb.no.M$svl  0.13529    0.03474   3.894 9.85e-05 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ 
’ 1 
 
(Dispersion parameter for binomial family taken to be 1)
 
    Null deviance: 301.92  on 217  degrees of freedom
Residual deviance: 285.04  on 216  degrees of freedom
  (8 observations deleted due to missingness)
AIC: 289.04
 
Number of Fisher Scoring iterations: 3
---
 Hosmer-Lemeshow test
 
 hosmerlem - function (y, yhat, g = 10) 
+ {
+ cutyhat - cut(yhat, breaks = quantile(yhat, probs = seq(0, 1, 1/g)), 
include.lowest = T)
+ obs - xtabs(cbind(1 - y, y) ~ cutyhat)
+ expect - xtabs(cbind(1 - yhat, yhat) ~ cutyhat)
+  chisq - sum((obs - expect)^2/expect)
+ P - 1 - pchisq(chisq, g - 2)
+ c(X^2 = chisq, Df = g - 2, P(Chi) = P)
+ }
 
 mod.fit - glm(formula = no.NA$repcnd ~  no.NA$svl, family = binomial(link = 
 logit), data =  no.NA, na.action = na.exclude, control = list(epsilon = 
 0.0001, maxit = 50, trace = F))
 
 hosmerlem(no.NA$repcnd, fitted(mod.fit))
      X^2        Df   P(Chi) 
6.8742531 8.000 0.5502587
---
 list(p.value = round(1-pchisq(mod.fit$null.deviance - mod.fit$deviance,
+ mod.fit$df.null- mod.fit$df.residual),6), 
+ df = mod.fit$df.null- mod.fit$df.residual,
+ change = mod.fit$null.deviance - mod.fit$deviance)
 
$p.value
[1] 4e-05
 
$df
[1] 1
 
$change
[1] 16.87895


  
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[R] Aide pour faire ACM

[housseima guiga]

Bonjour,

Je travaille sur R et je veux faire MCA, je veux savoir comment calcules les 
valeurs de Contribution, cosinus carré, coordonnée et distance au centre des 
modalités des différentes variables. Je vous remercie beaucoup d'avance.

Housseima
  
_
Hotmail : une messagerie performante et gratuite avec une sécurité signée 
Microsoft

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[R] Non-parametric regression

[Ralf B]

I have two data sets, each a vector of 1000 numbers, each vector
representing a distribution (i.e. 1000 numbers each of which
representing a frequency at one point on a scale between 1 and 1000).
For similfication, here an short version with only 5 points.


a - c(8,10,8,12,4)
b - c(7,11,8,10,5)

Leaving the obvious discussion about causality aside fro a moment, I
would like to see how well i can predict b from a using a regression.
Since I do not know anything about the distribution type and already
discovered non-normality I cannot use parametric regression or
anything GLM for that matter.

How should I proceed in using non-parametric regression to model
vector a and see how well it predicts b? Perhaps you could extend the
given lines into a short example script to give me an idea? Are there
any other options?

Best,
Ralf

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Re: [R] Non-parametric regression

[Tal Galili]

From reviewing the first google page result for Non-parametric regression
R, I hope this link will prove useful:

http://socserv.mcmaster.ca/jfox/Courses/Oxford-2005/R-nonparametric-regression.html



Contact
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--




On Fri, Jul 9, 2010 at 11:01 AM, Ralf B ralf.bie...@gmail.com wrote:

 I have two data sets, each a vector of 1000 numbers, each vector
 representing a distribution (i.e. 1000 numbers each of which
 representing a frequency at one point on a scale between 1 and 1000).
 For similfication, here an short version with only 5 points.


 a - c(8,10,8,12,4)
 b - c(7,11,8,10,5)

 Leaving the obvious discussion about causality aside fro a moment, I
 would like to see how well i can predict b from a using a regression.
 Since I do not know anything about the distribution type and already
 discovered non-normality I cannot use parametric regression or
 anything GLM for that matter.

 How should I proceed in using non-parametric regression to model
 vector a and see how well it predicts b? Perhaps you could extend the
 given lines into a short example script to give me an idea? Are there
 any other options?

 Best,
 Ralf

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Re: [R] Kite diagrams

[Jim Lemon]

On 07/09/2010 07:23 AM, Graham Smith wrote:

I asked the same question on  R-sig-eco, and Ben Bolker provided this
solution, which as I assume this should show up in a search I copy
here.

However, if someone can come up with a single function, that would be good.


Hi Graham,

library(plotrix)
kiteChart(t(X))

Jim

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[R] R^2 in loess and predict?

[Ralf B]

Parametric regression produces R^2 as a measure of how well the model
predicts the sample and adjusted R^2 as a measure of how well it
models the population. What is the equalvalent for non-parametric
regression (e.g. loess function) ?

Ralf

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Re: [R] R^2 in loess and predict?

[Joris Meys]

I do not agree with your interpretation of the adjusted R^2. The R^2
is no more than the ratio of the explained variance by the total
variance, expressed in sums of squares. The adjusted R^2 is adjusted
for the degrees of freedom, and can only be used for selection
purposes. The interpretation towards the final model is hard, and
definitely not a measure of how well it models the population.

For a loess regression this can be calculated as well. But the loess
is a local regression technique, highly flexible, and highly dependent
on the window you use. In these cases, R^2 (or any other goodness of
fit test) tells you even less. You can get an R^2 of 1 if you chose
the window small enough.

If you want to do inference on nonlinear regression techniques, I
strongly suggest you use Generalized Additive Models, eg from the
package mgcv. There you can use the framework of likelihood ratio
tests for determination of goodness of fit by comparing models.

Cheers
Joris

On Fri, Jul 9, 2010 at 10:42 AM, Ralf B ralf.bie...@gmail.com wrote:
 Parametric regression produces R^2 as a measure of how well the model
 predicts the sample and adjusted R^2 as a measure of how well it
 models the population. What is the equalvalent for non-parametric
 regression (e.g. loess function) ?

 Ralf

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-- 
Joris Meys
Statistical consultant

Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control

tel : +32 9 264 59 87
joris.m...@ugent.be
---
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Re: [R] Kite diagrams

[Graham Smith]

Jim,

This is very good news, not so much for me, as I don't use them, but I
have colleagues who do, and its an expected graphic in student
assignments.

So I have passed on the information.

So many thanks for adding this, I a sure many people will find it useful.

Graham

On 9 July 2010 09:41, Jim Lemon j...@bitwrit.com.au wrote:
 On 07/09/2010 07:23 AM, Graham Smith wrote:

 I asked the same question on  R-sig-eco, and Ben Bolker provided this
 solution, which as I assume this should show up in a search I copy
 here.

 However, if someone can come up with a single function, that would be
 good.

 Hi Graham,

 library(plotrix)
 kiteChart(t(X))

 Jim


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[R] accessing return variables from a function

[Noah Silverman]

Hi,

I am trying to figure out a short way to access two values output from
the sort function.

x - c(3,4,3,6,78,3,1,2)
sort(x, index.return=T)
$x
[1]  1  2  3  3  3  4  6 78

$ix
[1] 7 8 1 3 6 2 4 5


It would be great to do something like this (doesn't work.):

c(y, indexes) - sort(x, index.return=T)

But that doesn't work.

I CAN grab the output of sort in a variable and then access it twice to
get the values, but that seems wasteful.

Any  ideas?

-N

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[R] R crashes with large vectors

[Jeremie Smaga]

Good afternoon,

I have been experiencing a lot of crashes working with large vectors in R.

Specifically, I am using XTS of length of minimum 120k elements.

My problem is that I cannot display the vector (otherwise R crashes), I
cannot plot it either (otherwise R crashes). That could be solved by
reducing the amount of points.

However, I have been performing some statistical opreations on is and even
sd(myXTS) crashes R.

By crashes, I mean shuts down without any warning whatsoever.

I use R 2.11.1 (64).

Has anyone had the same kind of problem?

Can we solve this?

Best,


-- 
Jeremie

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[R] Current script name from R

[Ralf B]

Is there a way for a script to find out about its own name ?

Ralf

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[R] Plotting text in existing plot?

[Ralf B]

I would like to plot some text in a existing plot graph. Is there a
very simple way to do that. It does not need to be pretty at all (just
maybe a way to center it or define a position within the plot). ( ? )

Ralf

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Re: [R] Plotting text in existing plot?

[Tal Galili]

see

?text


Tal

Contact
Details:---
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On Fri, Jul 9, 2010 at 12:52 PM, Ralf B ralf.bie...@gmail.com wrote:

 I would like to plot some text in a existing plot graph. Is there a
 very simple way to do that. It does not need to be pretty at all (just
 maybe a way to center it or define a position within the plot). ( ? )

 Ralf

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Re: [R] Loading .Rdata within an R function

[Duncan Murdoch]

Giles Crane wrote:

Thank you for your consideration of this question.
I have tried both your suggestions.
However, the data is not loaded within the function.

When I specify load(mydata.Rdata,.globalEnv),
the data is loaded into the top level environment,
and the function does access the data in the top level environment.

However, I would like to load the data into the function environment,
so that the data goes away when the function terminates.
  


For the benefit of others:  an offline followup showed that the data 
were being loaded in the right place, but a function in a contributed 
package wasn't looking there and didn't see it.



Duncan Murdoch

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Re: [R] Plotting text in existing plot?

[Jim Lemon]

On 07/09/2010 07:52 PM, Ralf B wrote:

I would like to plot some text in a existing plot graph. Is there a
very simple way to do that. It does not need to be pretty at all (just
maybe a way to center it or define a position within the plot). ( ? )


Hi Ralf,
The text function in the graphics package will place text on your 
plot. It centers the text by default, so that:


text(3,4,my neat\ntwo liner)

will place the two lines of text centered at x=3 and y=4.

Jim

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[R] split with list

[n.via...@libero.it]

Dear List I would like to ask you something concenting a better print of the R 
output:
I have a bit data frame which has the following structure:
CFISCALE  RAGSOCBANNO   VAR1VAR2.
9853312 astra 2005   6  
 45

9853312 astra 2006  78  
45


9853312 astra 2007   55 
 76


9653421  geox 2005   35 
89



9653421 geox 200624 
  33

9653421  geox 2007   54 
 55


The first thing I did is to split my data frame for CFISCALE. The result is 
that R has transformed my data frame into a list. The second step was to 
transpose each element of my list. 
repo=split(rep,rep$CFISCALE)
repor=lapply(repo,function(x){
t(x)})


When I print my list the format is the following 
$9853312
   1   2
3

CFISCALE9853312 9853312 9853312   

RAGSOCBastra   astraastra

ANNO   20052006  2007

VAR1   6 78 
 55

VAR2   4545 
76


There is a way to remove the  first row I mean 1, 2 , 3 and to have just one 
CFISCALE and RAGSOCB???
For the second problem I tried to use unique but it seems that it doesnt work 
for list. So what I would like to get is: 
$9853312


  


CFISCALE9853312 


RAGSOCBastra  
ANNO   20052006  2007

VAR1   6 78 
 55

VAR2   4545 
76


This is because I next run xtable on my list in order to get a table in Latex, 
which I woud like to be in a nice format.
Thanks a lot for your attention!





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Re: [R] C module causing rounding errors?

[Duncan Murdoch]

Joseph N. Paulson wrote:

Hi all!

I am currently writing a C-module for a for loop in which I permute columns
in a matrix (bootstrapping) and I send a couple of variables into C
initially. All of it is working, except the initial values I send to R are
rounded/truncated (I believe rounded).

I am using a 32 bit machine to compile, I am using (I believe) 32 bit R

While debugging I print the values I am sending to C, and then I print the
same values using Rprintf and the number gets rounded to 10^-6, which is
actually causing some errors for me. Is there any way to correct/prevent the
error?


sample output from R

 [1,]  1.000
 [2,]  1.0256242
 [3,]  1.1826277
 [4,] -0.6937246
 [5,]  1.3633604

sample output from C
1.00
1.025624
1.182628
0.693725
1.363360
  


It looks as though you are confusing the display of numbers with their 
internal values.  R is printing 7 decimal places, C is printing 6.  As 
far as we can tell, that's the only difference.


Duncan Murdoch

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Re: [R] accessing return variables from a function

[David Winsemius]

On Jul 9, 2010, at 5:20 AM, Noah Silverman wrote:


Hi,

I am trying to figure out a short way to access two values output  
from

the sort function.


x - c(3,4,3,6,78,3,1,2)
sort(x, index.return=T)

$x
[1]  1  2  3  3  3  4  6 78

$ix
[1] 7 8 1 3 6 2 4 5


It would be great to do something like this (doesn't work.):


 yoursort - function(x) {sx - sort(x, index.return=T); return( c(sx 
$x, sx$ix) )}

 yoursort(x)
 [1]  1  2  3  3  3  4  6 78  7  8  1  3  6  2  4  5


c(y, indexes) - sort(x, index.return=T)

But that doesn't work.

I CAN grab the output of sort in a variable and then access it twice  
to

get the values, but that seems wasteful.

Any  ideas?



David Winsemius, MD
West Hartford, CT

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Re: [R] Non-parametric regression

[David Winsemius]

On Jul 9, 2010, at 4:01 AM, Ralf B wrote:


I have two data sets, each a vector of 1000 numbers, each vector
representing a distribution (i.e. 1000 numbers each of which
representing a frequency at one point on a scale between 1 and 1000).
For similfication, here an short version with only 5 points.


a - c(8,10,8,12,4)
b - c(7,11,8,10,5)

Leaving the obvious discussion about causality aside fro a moment, I
would like to see how well i can predict b from a using a regression.


You can use density estimation,. There was a recent thread that  
included worked examples using MASS::kde2d and locfit::locfit for  
graphical display of joint distributions.




Since I do not know anything about the distribution type and already
discovered non-normality I cannot use parametric regression or
anything GLM for that matter.

How should I proceed in using non-parametric regression to model
vector a and see how well it predicts b? Perhaps you could extend the
given lines into a short example script to give me an idea? Are there
any other options?

Best,
Ralf


David Winsemius, MD
West Hartford, CT

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[R] KLdiv produces NA. Why?

[Ralf B]

I am trying to calculate a Kullback-Leibler divergence from two
vectors with integers but get NA as a result when trying to calulate
the measure. Why?

x - cbind(stuff$X, morestuff$X)

x[1:5,]

 [,1] [,2]
[1,]  293  938
[2,]  293  942
[3,]  297  949
[4,]  290  956
[5,]  294  959

KLdiv(x)


 [,1] [,2]
[1,]0   NA
[2,]   NA0


Best,
Ralf

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Re: [R] accessing return variables from a function

[Barry Rowlingson]

On Fri, Jul 9, 2010 at 10:20 AM, Noah Silverman n...@smartmediacorp.com wrote:
 Hi,

 I am trying to figure out a short way to access two values output from
 the sort function.

x - c(3,4,3,6,78,3,1,2)
sort(x, index.return=T)
 $x
 [1]  1  2  3  3  3  4  6 78

 $ix
 [1] 7 8 1 3 6 2 4 5


 It would be great to do something like this (doesn't work.):

 c(y, indexes) - sort(x, index.return=T)

 But that doesn't work.

 I CAN grab the output of sort in a variable and then access it twice to
 get the values, but that seems wasteful.

 Care to do a little study to see how wasteful? The difference is
going to be between:

 foo = sort(...)
 repeat loads of times{
   f(foo$x)
   g(foo$ix)
 }

and

 foo = sort(...)
 x=foo$x
 ix = foo$ix
 repeat loads of times{
f(x)
g(ix)
  }

  I'd guess that for most applications the difference is less than
something very close to zero.

Barry

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Re: [R] Plotting text in existing plot?

[Barry Rowlingson]

On Fri, Jul 9, 2010 at 11:22 AM, Jim Lemon j...@bitwrit.com.au wrote:

 The text function in the graphics package will place text on your plot. It
 centers the text by default, so that:

 text(3,4,my neat\ntwo liner)

 will place the two lines of text centered at x=3 and y=4.

 BUT it only works on 'base' graphics plots, so if you're using
lattice or ggplot then there's another way to do it. Perhaps.

 Original poster wanted a simple way to do it, but when R has three
graphics systems, four OO systems, and a zillion helpful people
there's never a simple way :)

Barry

-- 
blog: http://geospaced.blogspot.com/
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web: http://www.rowlingson.com/
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Re: [R] metafor and meta-analysis at arm-level

[Viechtbauer Wolfgang (STAT)]

With appropriate design matrix, I mean the X matrix in the mixed-effects 
model y = Xb + u + e, where y is the vector of outcomes, u is a vector of 
(possibly correlated) random effects, and e is a vector of (possibly) random 
errors. The X matrix is specified via the 'mods' argument in the rma() 
function. If y consists of arm-level outcomes, then you need appropriate dummy 
variables in X to code what type of arm the outcome corresponds to.

Have you read, for example:

Salanti, G., Higgins, J. P. T., Ades, A. E.,  Ioannidis, J. P. A. (2008). 
Evaluation of networks of randomized trials. Statistical Methods in Medical 
Research, 17(3), 279-301.

This article may be helpful.

Best,

--
Wolfgang Viechtbauerhttp://www.wvbauer.com/
Department of Methodology and StatisticsTel: +31 (0)43 388-2277
School for Public Health and Primary Care   Office Location:
Maastricht University, P.O. Box 616 Room B2.01 (second floor)
6200 MD Maastricht, The Netherlands Debyeplein 1 (Randwyck)


Original Message
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Angelo Franchini
Sent: Tuesday, July 06, 2010 10:37 To: Wolfgang Viechtbauer
Cc: r-help@r-project.org; Angelo Franchini
Subject: Re: [R] metafor and meta-analysis at arm-level

 Hello Wolfgang,

 Thank you very much for your response.
 When you mentionthe appropriate design matrix, do you mean by that
 the 'n1i, n2i, m1i, m2i, sd1i, sd2i' arguments of the rma function,
 or am I missing something? I read the documentation on metafor
 (introduction), rma/rma.uni and escalc, and that was the only way
 that I could find for the package to use information at the arm-level
 rather than the trial one.

 As for the complexity of possible correlations between effects, that
 is something to be considered for the network analysis case, correct?

 Many thanks.

 Best regards,
 Angelo



 On Sun, July 4, 2010 6:06 am, Wolfgang Viechtbauer wrote:
 Hello Angelo,

 You can either supply the arm-level outcomes and corresponding
 sampling variances directly (via the yi and vi arguments) or supply
 the necessary information so that the escalc() or rma() functions can
 calculate an appropriate arm-level outcome (such as the log odds).
 See the documentation of the escalc() function and in particular the
 part about proportions and tranaformations thereof as possible
 outcome measures.

 This is the easy part. Then you need to set up an appropriate design
 matrix to code what arm each observed outcome corresponds to. And
 finally comes the tricky/problematic part. The rma() function assumes
 independent sampling errors and independent random effects for each
 observed outcome. Independent sampling errors is (usually) ok when
 using arm-level outcomes, but the independent random errors part may
 not be appropriate. This is why I am working on functions that do not
 make this independence assumption. With those functions, you can then
 carry out multivariate and network-type meta-analyses. These
 functions will become part of the metafor package in the future.

 Best,

 --
 Wolfgang Viechtbauer
 http://www.wvbauer.com

 Angelo Franchini angelo.franch...@bristol.ac.uk wrote:

 Hi,

 I have been looking for an R package which allowed to do
 meta-analysis (both pairwise and network/mixed-treatment) at
 arm-level rather than at trial-level, the latter being the common
 way in which meta-analysis is done. By arm-level meta-analysis I
 mean one that accounts for data provided at the level of the
 individual arms of each trial and that does not simply derive the
 difference between arms and do the meta-analysis on that.

 I am not sure metafor can do that, but hopefully someone more
 experienced on it can clarify that to me. I can see that it can take
 data in both forms, arm and trial level, but it looks as if the
 arm-level information would be converted into trial one through
 escalc and the latter then used for the meta-analysis. Is that
 right?

 Many thanks.

 Angelo


 --
 NIHR Research Methods Training Fellow,
 Department of Community Based Medicine
 University of Bristol
 25 Belgrave Road
 Bristol BS8 2AA

 Tel. 0779 265-6552

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Re: [R] accessing return variables from a function

[Eik Vettorazzi]

which is just
unlist(sort(x, index.return=T))

but I think, Noah would like to have something like
srt-sort(x, index.return=T)
names(srt)-c(y,indexes)
attach(srt)

but that is wasteful either.


Am 09.07.2010 12:28, schrieb David Winsemius:

 On Jul 9, 2010, at 5:20 AM, Noah Silverman wrote:

 Hi,

 I am trying to figure out a short way to access two values output from
 the sort function.

 x - c(3,4,3,6,78,3,1,2)
 sort(x, index.return=T)
 $x
 [1]  1  2  3  3  3  4  6 78

 $ix
 [1] 7 8 1 3 6 2 4 5


 It would be great to do something like this (doesn't work.):

  yoursort - function(x) {sx - sort(x, index.return=T); return(
 c(sx$x, sx$ix) )}
  yoursort(x)
  [1]  1  2  3  3  3  4  6 78  7  8  1  3  6  2  4  5

 c(y, indexes) - sort(x, index.return=T)

 But that doesn't work.

 I CAN grab the output of sort in a variable and then access it twice to
 get the values, but that seems wasteful.

 Any  ideas?


 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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-- 
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Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/7410-58243
F ++49/40/7410-57790

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Re: [R] Current script name from R

[Allan Engelhardt]

I'm assuming you are using Rscript (please provide self-contained 
examples when posting) in which case you could look for the element in 
(base|R.utils)::commandArgs() that begin with the string --file= - the 
rest is the file name.  See the asValues= parameter in 
help(commandArgs, package=R.utils) for a nice way to get the parameter.


For an invocation of the form R  foo.R you'd need to inspect your 
system's process table (so don't do that).


Hope this helps.

Allan

On 09/07/10 10:48, Ralf B wrote:

Is there a way for a script to find out about its own name ?

Ralf

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Re: [R] Current script name from R

[Ralf B]

I am using RGUI, the  command line or the StatET Eclipse environment.
Should this not all be the same?

Ralf

On Fri, Jul 9, 2010 at 7:11 AM, Allan Engelhardt all...@cybaea.com wrote:
 I'm assuming you are using Rscript (please provide self-contained examples
 when posting) in which case you could look for the element in
 (base|R.utils)::commandArgs() that begin with the string --file= - the
 rest is the file name.  See the asValues= parameter in help(commandArgs,
 package=R.utils) for a nice way to get the parameter.

 For an invocation of the form R  foo.R you'd need to inspect your system's
 process table (so don't do that).

 Hope this helps.

 Allan

 On 09/07/10 10:48, Ralf B wrote:

 Is there a way for a script to find out about its own name ?

 Ralf

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Re: [R] transformation of data.frame

[Assa Yeroslaviz]

Hello Petr,

sorry for the mixed up. your example works perfectly fine.

The one from Søren has shown the mentioned error.  But even after reading
the columns as character

 go - read.table(go.txt, header= TRUE, colClasses = c(character,
character))
or
 go - read.table(go.txt, header= TRUE, as.is = 1)


it didn't solve the problem.
the command:
gmt - lapplyBy(~GO, data = go, FUN = function(uu) {as.list(uu$GO[1],
paste(uu$gen, collapse =  ))})

tries to convert my first column into integers and thand add 'NA's.

What I don't understand is why.
Does lapplyBy can work only with integers?

THX,

Assa


2010/7/8 Petr PIKAL petr.pi...@precheza.cz

 Hi

 r-help-boun...@r-project.org napsal dne 08.07.2010 12:02:45:

  I don't understand it. When I'm doing this example it wirks fine, but
 when
  I'm adding the GO: to the beginning of the first column (as to see in
 my
  wanted result table:
  GO0042787
  GO0016070
  GO0016070
 
  I'm getting a list of warning:
  Warning messages:
  1: In storage.mode(xi) - a$sm : NAs introduced by coercion
  2: In storage.mode(xi) - a$sm : NAs introduced by coercion
  ...
  9: In storage.mode(xi) - a$sm : NAs introduced by coercion
  10: In storage.mode(xi) - a$sm : NAs introduced by coercion

 Not sure what is wrong, it seems to me that your ID become factor.

 Having your data in dataframe test as character columns

 see ?str

 test.ag-aggregate(test$X.gen, list(test$ID), function(x) paste(x,
 collapse=:))

 I can make aggregated data frame

 paste(GO,test.ag[,1], sep=)
 [1] GO0006417 GO0006511 GO0007409 GO0016070 GO0042787

 and it is strightforward to add GO at the beginning.

 I leave how to add this result to your aggregated data frame as an
 exercise.

 Regards
 Petr


 
  What did I do wrong here?
 
  Assa
 
  On Thu, Jul 8, 2010 at 11:09, Søren Højsgaard
 soren.hojsga...@agrsci.dkwrote:
 
   Like this?
  
library(doBy)
(ddd - read.table(foo.txt,header=T))
   ID  gen
   1 42787 gen2
   2 16070 gen2
   3 16070 gen3
   4  7409 Gen1
   5  7409 gen3
   6  6511 gen2
   7  6417 gen3
   8 16070 gen4
   9  6511 gen4
aa-lapplyBy(~ID, data=ddd,
   +   FUN=function(uu){
   +   list(uu$ID[1], paste(uu$gen, collapse=:))
   + })
   
do.call(rbind,aa)
[,1]  [,2]
   42787 42787 gen2
   16070 16070 gen2:gen3:gen4
   7409  7409  Gen1:gen3
   6511  6511  gen2:gen4
   6417  6417  gen3
  
   Regards
   Søren
  
  
  
  
  
   -Oprindelig meddelelse-
   Fra: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
 ] PĂĄ
   vegne af Assa Yeroslaviz
   Sendt: 8. juli 2010 10:45
   Til: r-h...@stat.math.ethz.ch
   Emne: [R] transformation of data.frame
  
   Hello all R users,
  
   I have a problems transforming (or maybe better regrouping) a
 data.frame.
I have a big data.frame, which I would like to sum up according to a
   specific column.
  
   This is an example of my matrix:
   IDgen
   0042787gen2
   0016070gen2
   0016070gen3
   0007409Gen1
   0007409gen3
   0006511gen2
   0006417gen3
   0016070gen4
   0006511gen4
  
   I want to rearrange the matrix according to column GO, so that it will
 look
   likes that:
  
   GO:0042787 gen2
   GO:0016070gen2  :  gen3  :  gen4
   GO:0007409gen1  :  gen3
   GO:0006511gen2  :  gen4
   GO:0006417gen3
  
   I've tried it with the package doBy (lapplyBy and paste) but it just
   doesn't
   work out.
  
   I will be very happy for any suggestions you might have to help me.
  
   Thanks
  
   Assa
  
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[R] sarima.Sim function

[Jason Overstreet]

Does anyone have some a nice simple example of how this function is used?

Thanks,

Jason

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[R] Mirror axis on hist2d plot - how?

[Ralf B]

The following code produces a heatmap based on normalized data. I
would like to mirror x and y axis for this plot. Any idea how to do
that?


require(gplots)
x - rnorm(500)
y - rnorm(500)
hist2d(x, y, freq=TRUE, nbins=50, col = c(white,heat.colors(256)))

Best,
Ralf

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Re: [R] How not to print '\\' as '\\'

[Gavin Simpson]

On Tue, 2010-07-06 at 19:08 -0700, Joshua Wiley wrote:
 Try cat(), for instance:
 
  cat(\\\n)
 \
 
 
 The extra \n is needed for a proper line break.

Or

writeLines(\\)

Which handles the newline for you. ?writeLines.

HTH

G

 
 HTH,
 
 Josh
 
 On Tue, Jul 6, 2010 at 7:01 PM, thmsfuller...@gmail.com
 thmsfuller...@gmail.com wrote:
  Hello All,
 
  '\\' is printed as '\\', but it is actually only one character.
  Sometimes, I'd rather print it as a single '\'. Is there a function to
  do so in R?
 
  nchar('\\')
  [1] 1
  print('\\')
  [1] \\
 
  --
  Tom
 
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-- 
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 Dr. Gavin Simpson [t] +44 (0)20 7679 0522
 ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
 Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/
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Re: [R] Appropriate tests for logistic regression with a continuous predictor variable and Bernoulli response variable

[Eik Vettorazzi]

Have a look at
?anova
or rather
?anova.glm

hth

Am 09.07.2010 04:46, schrieb Kiyoshi Sasaki:
 I have a data with binary response variable, repcnd (pregnant or not) and one 
 predictor continuous variable, svl (body size) as shown below. I did 
 Hosmer-Lemeshow test as a goodness of fit (as suggested by a kind “R-helper” 
 previously). To test whether the predictor (svl, or body size) has 
 significant effect on predicting whether or not a female snake is pregnant, I 
 used the differences between null deviance and residual deviance using a code 
 as following:

  
 1-pchisq(mod.fit$null.deviance - mod.fit$deviance, mod.fit$df.null - 
 mod.fit$df.residual)
  
 Could anyone tell me whether I did the test properly? I did this test because 
 I thought Wald test/z score listed in the output from summary(mod.fit) is 
 not appropriate for a kind of data I have.  Does R have automated function to 
 run appropriate tests? I have pasted my R output below.
  
 Thank you in advance for your time and help.
  
 Kiyoshi
  
  
 repcnd svl
 1  1  51.5
 2  1  52.5
 edited
 294  0  59.8
 298  1  60.0
 300  1  51.7
 301  1  57.4
 302  1  60.9
 303  0  56.8
 304  0  50.0
 ---
   
 mod.fit - glm(formula = gb.no.M$repcnd ~ gb.no.M$svl, family = 
 binomial(link = logit), data = gb.no.M, na.action = na.exclude, control = 
 list(epsilon = 0.0001, maxit = 50, trace = F))
 summary(mod.fit)
 
  
 Call:
 glm(formula = gb.no.M$repcnd ~ gb.no.M$svl, family = binomial(link = logit), 
 data = gb.no.M, na.action = na.exclude, control = list(epsilon = 1e-04, 
 maxit = 50, trace = F))
  
 Deviance Residuals: 
Min  1Q  Median  3Q Max  
 -1.757  -1.109   0.734   1.113   1.632  
  
 Coefficients:
 Estimate Std. Error z value Pr(|z|)
 (Intercept) -7.085651.84106  -3.849 0.000119 ***
 gb.no.M$svl  0.135290.03474   3.894 9.85e-05 ***
 ---
 Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 
  
 (Dispersion parameter for binomial family taken to be 1)
  
 Null deviance: 301.92  on 217  degrees of freedom
 Residual deviance: 285.04  on 216  degrees of freedom
   (8 observations deleted due to missingness)
 AIC: 289.04
  
 Number of Fisher Scoring iterations: 3
 ---
   
 Hosmer-Lemeshow test

 hosmerlem - function (y, yhat, g = 10) 
 
 + {
 + cutyhat - cut(yhat, breaks = quantile(yhat, probs = seq(0, 1, 1/g)), 
 include.lowest = T)
 + obs - xtabs(cbind(1 - y, y) ~ cutyhat)
 + expect - xtabs(cbind(1 - yhat, yhat) ~ cutyhat)
 +  chisq - sum((obs - expect)^2/expect)
 + P - 1 - pchisq(chisq, g - 2)
 + c(X^2 = chisq, Df = g - 2, P(Chi) = P)
 + }
   
 mod.fit - glm(formula = no.NA$repcnd ~  no.NA$svl, family = binomial(link = 
 logit), data =  no.NA, na.action = na.exclude, control = list(epsilon = 
 0.0001, maxit = 50, trace = F))
 
  
   
 hosmerlem(no.NA$repcnd, fitted(mod.fit))
 
   X^2Df   P(Chi) 
 6.8742531 8.000 0.5502587
 ---
   
 list(p.value = round(1-pchisq(mod.fit$null.deviance - mod.fit$deviance,
 
 + mod.fit$df.null- mod.fit$df.residual),6), 
 + df = mod.fit$df.null- mod.fit$df.residual,
 + change = mod.fit$null.deviance - mod.fit$deviance)
  
 $p.value
 [1] 4e-05
  
 $df
 [1] 1
  
 $change
 [1] 16.87895


   
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-- 
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/7410-58243
F ++49/40/7410-57790

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[R] Data format question for triangle.plot package ade4

[Steve_Friedman]

hello,

I am trying to develop a triangle plot but am having difficultly assigning
the row.names to the 3 columns in the data.frame

Here is what I've done,

attach(SoilVegHydro)

dim(SoilVegHydro)
129239

# now  take 3 variables from main data.frame for plotting

dat - cbind.data.frame(TP, meanAnnualDepthAve, BulkDensity)  #  These are
variables held in the data frame SoilVegHydro

row.names(dat) - paste(row.names(SoilVegHydro$Physiogomy), rep(c(1,2,3),
rep(1292, 3)), sep = )  # following the syntax from the help
triangle.plot page

this is returned when the last line is submitted.

row.names(dat) - paste(row.names(SoilVegHydro$Physiogomy), rep(c(1,2,3),
rep(1292,3)), sep=)
Error in `row.names-.data.frame`(`*tmp*`, value = c(1, 1, 1, 1,  :
  invalid 'row.names' length

I'm not certain how to define the row.names .  If anyone can help I'd
appreciate it.

I'm using R 2.11.1 (2010-5-31) on Windows XP

Thanks
Steve


Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034

steve_fried...@nps.gov
Office (305) 224 - 4282
Fax (305) 224 - 4147

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Re: [R] random sample from arrays

[Assa Yeroslaviz]

Hi Joris,
I guess i did it wrong again.
but your example didn't work either. I still get the error massage.

but replicate function just fine. I can even replicate the whole array
lines.

THX

Assa

On Thu, Jul 8, 2010 at 15:20, Joris Meys jorism...@gmail.com wrote:

 Don't know what exactly you're trying to do, but you make a matrix
 with 11 columns and 50 rows, then treat it as a vector. On top of
 that, you try to fill 50 rows/columns with 50 values. Off course that
 doesn't work. Did you check the warning messages when running the
 code?

 Either do :

  for(i in c(1:11)){
 set[,i] -sample(x,50)
print(c(i,-, set), quote = FALSE)
   }

 or

  for(i in c(1:50)){
 set[i,] -sample(x,11)
 print(c(i,-, set), quote = FALSE)
   }

 Or just forget about the loop altogether and do :

 set - replicate(11,sample(x,50))
 or
 set - t(replicate(50,sample(x,11)))

 cheers

 On Thu, Jul 8, 2010 at 8:04 AM, Assa Yeroslaviz fry...@gmail.com wrote:
  Hello R users,
 
  I'm trying to extract random samples from a big array I have.
 
  I have a data frame of over 40k lines and would like to produce around 50
  random sample of around 200 lines each from this array.
 
  this is the matrix
   ID xxx_1c xxx__2c xxx__3c xxx__4c xxx__5T xxx__6T xxx__7T
 xxx__8T
  yyy_1c yyy_1c _2c
  1 A_512  2.150295  2.681759  2.177138  2.142790  2.115344  2.013047
  2.115634  2.189372  1.643328  1.563523
  2 A_134 12.832488 12.596373 12.882581 12.987091 11.956149 11.994779
  11.650336 11.995504 13.024494 12.776322
  3 A_152  2.063276  2.160961  2.067549  2.059732  2.656416  2.075775
  2.033982  2.111937  1.606340  1.548940
  4 A_163  9.570761 10.448615  9.432859  9.732615 10.354234 10.993279
  9.160038  9.104121 10.079177  9.828757
  5 A_184  3.574271  4.680859  4.517047  4.047096  3.623668  3.021356
  3.559434  3.156093  4.308437  4.045098
  6 A_199  7.593952  7.454087  7.513013  7.449552  7.345718  7.367068
  7.410085  7.022582  7.668616  7.953706
  ...
 
  I tried to do it with a for loop:
 
  genelist - read.delim(/user/R/raw_data.txt)
  rownames(genelist) - genelist[,1]
  genes - rownames(genelist)
 
  x - 1:4
  set - matrix(nrow = 50, ncol = 11)
 
  for(i in c(1:50)){
 set[i] -sample(x,50)
 print(c(i,-, set), quote = FALSE)
 }
 
  which basically do the trick, but I just can't save the results outside
 the
  loop.
  After having the random sets of lines it wasn't a problem to extract the
  line from the arrays using subset.
 
  genSet1 -sample(x,50)
  random1 - genes %in% genSet1
  subsetGenelist - subset(genelist, random1)
 
 
  is there a different way of creating these random vectors or saving the
 loop
  results outside tjhe loop so I cn work with them?
 
  Thanks a lot
 
  Assa
 
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 --
 Joris Meys
 Statistical consultant

 Ghent University
 Faculty of Bioscience Engineering
 Department of Applied mathematics, biometrics and process control

 tel : +32 9 264 59 87
 joris.m...@ugent.be
 ---
 Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php


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Re: [R] C module causing rounding errors?

[Joseph N. Paulson]

Sorry for not including enough information everyone.

I have quite a bit of code, so I will just enter relevant pieces...

This is how I call C from R:

The tstats are tstatistics (difference of mean, divided by sqrt of S1+S2)
from an unpermuted matrix. The c code is below...

dyn.load(testp.so)
obj-.C(testp,ptests=as.array(permuted_ttests),as.integer(B),permuted1=as.array(permuted),matrix=as.array(Imatrix),
as.integer(ncols), as.integer(nrows),as.integer(g)*,as.array(abs(tstats)*
),pvalues=as.array(ps))

to test, I decided not to permute my matrix and just send it the original
matrix, Imatrix. Everything in C is a double, or if I use an integer, I cast
it to a double (to divide and get mean, etc). I then compare the values of
the tstats that I sent into C and the tsats I calculate within C...

The following is my C code:

void testp(double *permuted_ttests,int *B,double *permuted,double
*Imatrix,int *nc,int *nr,int *g,*double *Tinitial*,double *ps) {

after which, using the variable above I take the mean of certain elements
(currently the unpermuted matrix to test) via other functions using the same
double and pointers and I store them in C1 and C2 and solve for an absolute
T statistic - and print it (I erased the Ts earlier).


for (i=0; i*nr; i++){
  xbardiff = C1[i][0]-C2[i][0];
  denom = sqrt(C1[i][2]+C2[i][2]);
*  Ts[i]=fabs(xbardiff/denom);*
*  Rprintf(%f Ts\n,Ts[i]);*
*
  if (fabs(Ts[i])fabs(Tinitial[i])){ //summing of permueted_ttests
counter[i]++;
 Rprintf(ts %f and tinitial %f \n, Ts[i],Tinitial[i]);
  }
*
etc...
The issue here - is that I get a few that when printed it appears that they
were rounded up - causing my counter[i] to ++ in some cases.

Should I send more code?

Sorry and thank you very much,




On Fri, Jul 9, 2010 at 6:18 AM, Duncan Murdoch murdoch.dun...@gmail.comwrote:

 Joseph N. Paulson wrote:

 Hi all!

 I am currently writing a C-module for a for loop in which I permute
 columns
 in a matrix (bootstrapping) and I send a couple of variables into C
 initially. All of it is working, except the initial values I send to R are
 rounded/truncated (I believe rounded).

 I am using a 32 bit machine to compile, I am using (I believe) 32 bit
 R

 While debugging I print the values I am sending to C, and then I print the
 same values using Rprintf and the number gets rounded to 10^-6, which is
 actually causing some errors for me. Is there any way to correct/prevent
 the
 error?


 sample output from R

  [1,]  1.000
  [2,]  1.0256242
  [3,]  1.1826277
  [4,] -0.6937246
  [5,]  1.3633604

 sample output from C
 1.00
 1.025624
 1.182628
 0.693725
 1.363360



 It looks as though you are confusing the display of numbers with their
 internal values.  R is printing 7 decimal places, C is printing 6.  As far
 as we can tell, that's the only difference.

 Duncan Murdoch

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-- 
 -  Joseph N. Paulson

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Re: [R] Data format question for triangle.plot package ade4

[David Winsemius]

On Jul 8, 2010, at 4:41 PM, steve_fried...@nps.gov wrote:




hello,

I am trying to develop a triangle plot but am having difficultly  
assigning

the row.names to the 3 columns in the data.frame

Here is what I've done,

attach(SoilVegHydro)

dim(SoilVegHydro)
129239

# now  take 3 variables from main data.frame for plotting

dat - cbind.data.frame(TP, meanAnnualDepthAve, BulkDensity)  #   
These are

variables held in the data frame SoilVegHydro


Did that dat object have what you wanted? The function call did not  
make any reference to SoilVegHydro. What does str(dat) return? Oh,  
never mind, I now see you use attach.




row.names(dat) - paste(row.names(SoilVegHydro$Physiogomy),


Generally row.names is used on a dataframe rather than on a column  
vector.


 dat - data.frame(1:3, LETTERS[1:3])
 row.names(dat$X1)
 row.names(dat)
[1] 1 2 3
 length(row.names(dat$X1))
[1] 0



rep(c(1,2,3),
rep(1292, 3)), sep = )  # following the syntax from the help
triangle.plot page

this is returned when the last line is submitted.

row.names(dat) - paste(row.names(SoilVegHydro$Physiogomy),  
rep(c(1,2,3),

rep(1292,3)), sep=)
Error in `row.names-.data.frame`(`*tmp*`, value = c(1, 1, 1,  
1,  :

 invalid 'row.names' length

I'm not certain how to define the row.names .  If anyone can help I'd
appreciate it.

I'm using R 2.11.1 (2010-5-31) on Windows XP

Thanks
Steve


Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034

steve_fried...@nps.gov
Office (305) 224 - 4282
Fax (305) 224 - 4147

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David Winsemius, MD
West Hartford, CT

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[R] how to plot two histograms overlapped in the same plane coordinate

[Mao Jianfeng]

Dear R-help listers,

I am new. I just want to get helps on how to plot two histograms
overlapped in the same plane coordinate. What I did is very ugly.
Could you please help me to improve it? I want to got a plot with semi-
transparent overlapping region. And, I want to know how to specify the
filled colors of the different histograms.

I also prefer other solutions other than ggplot2.

Many thanks to you.


What I have done:

library(ggplot2)

age-c(rnorm(100, 1.5, 1), rnorm(100, 5, 1))
sex-c(rep(F,100), rep(M, 100))
mydata-cbind(age, sex)
mydata-as.data.frame(mydata)
head(mydata)


qplot(age, data=mydata, geom=histogram, fill=sex, xlab=age,
ylab=count, alpha=I(0.5))


Best,


Mao J-F

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Re: [R] KLdiv produces NA. Why?

[Peter Ehlers]

On 2010-07-09 4:31, Ralf B wrote:

I am trying to calculate a Kullback-Leibler divergence from two
vectors with integers but get NA as a result when trying to calulate
the measure. Why?

x- cbind(stuff$X, morestuff$X)

x[1:5,]

  [,1] [,2]
[1,]  293  938
[2,]  293  942
[3,]  297  949
[4,]  290  956
[5,]  294  959

KLdiv(x)


  [,1] [,2]
[1,]0   NA
[2,]   NA0




I assume that you're using KLdiv() from pkg flexmix.
(You should always indicate any add-on packages used.)

I have no problem with KLdiv on the above 5-by-2 matrix
(R 2.11.1pat, flexmix 2.2-7).

Perhaps one of the following may be the case:

1. your R and/or flexmix is outdated;
2. dependencies were not installed (modeltools?)
3. there's something weird about your x[,]


  -Peter Ehlers



Best,
Ralf



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[R] distributing a value for a given month across the number of weeks in that month

[Dimitri Liakhovitski]

Hello!

Any hint would be greatly appreciated.
I have a data frame that contains (a) monthly dates and (b) a value
that corresponds to each month - see the data frame monthly below:

monthly-data.frame(month=c(20100301,20100401,20100501),monthly.value=c(100,200,300))
monthly$month-as.character(monthly$month)
monthly$month-as.Date(monthly$month,%Y%m%d)
(monthly)

I need to split each month into weeks, e.g., weeks that start on
Monday (it could as well be Sunday - it does not really matter) and
distribute the monthly value evenly across weeks. So, if a month has 5
Mondays, then the monthly value should be dividied by 5, but if a
month has only 4 weeks, then the monthly value should be divided by 4.

The output I need is like this:

week  weekly.value
2010-03-01   20
2010-03-08   20
2010-03-15   20
2010-03-22   20
2010-03-29   20
2010-04-05   50
2010-04-12   50
2010-04-19   50
2010-04-26   50
2010-05-03   60
2010-05-10   60
2010-05-17   60
2010-05-24   60
2010-05-31   60

Thanks a lot for your advice!

-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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[R] Function on columns of a dataframe

[LogLord]

Hi,

I would like to assign the largest value of a column to a specific category
and repeat this for each column (v1 - v4).

 x=c(1:12)
 cat=c(cat1,cat5,cat2,cat2,cat1,cat5,cat3,cat4,cat5,cat2,cat3,cat6)
 v1=rnorm(12,0.5,0.1)
 v2=rnorm(12,0.3,0.2)
 v3=rnorm(12,0.4,0.1)
 v4=rnorm(12,0.6,0.3)
 bla=data.frame(x,cat,v1,v2,v3,v4)
 bla
x  catv1 v2v3 v4
1   1 cat1 0.4013144 0.54839317 0.3946393  0.8679266
2   2 cat5 0.4595873 0.45788906 0.4030078  0.5919596
3   3 cat2 0.4542865 0.21516928 0.2777649  0.6112099
4   4 cat2 0.4787950 0.06252512 0.5095611  0.6450795
5   5 cat1 0.4910746 0.56591049 0.5151813  0.8465181
6   6 cat5 0.4194397 0.16592579 0.4361643  0.6415192
7   7 cat3 0.6148564 0.32240342 0.2690108  0.7114133
8   8 cat4 0.6174652 0.28076152 0.4577064 -0.2567284
9   9 cat5 0.4775395 0.28611768 0.4660210  0.4634120
10 10 cat2 0.4802962 0.03715569 0.4506361  1.0063235
11 11 cat3 0.6495094 0.33303172 0.3352933  1.4390324
12 12 cat6 0.4891481 0.45355589 0.3880739  0.7831656
 

I can assign this by the sqldf() command for each column but I would like to
automate this as I have many columns.

 select=sqldf(select cat, max(v1) FROM bla GROUP BY cat)
 select
   cat   max(v1)
1 cat1 0.4910746
2 cat2 0.4802962
3 cat3 0.6495094
4 cat4 0.6174652
5 cat5 0.4775395
6 cat6 0.4891481
 

Finally, I would like to have a dataframe where where the cat is followed by
each column maximum.

Thanks for your help!
-- 
View this message in context: 
http://r.789695.n4.nabble.com/Function-on-columns-of-a-dataframe-tp2283217p2283217.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Function on columns of a dataframe

[Eik Vettorazzi]

Hi Nils,
have a look at
?tapply
hth.

Am 09.07.2010 15:37, schrieb LogLord:
 Hi,

 I would like to assign the largest value of a column to a specific category
 and repeat this for each column (v1 - v4).

   
 x=c(1:12)
 cat=c(cat1,cat5,cat2,cat2,cat1,cat5,cat3,cat4,cat5,cat2,cat3,cat6)
 v1=rnorm(12,0.5,0.1)
 v2=rnorm(12,0.3,0.2)
 v3=rnorm(12,0.4,0.1)
 v4=rnorm(12,0.6,0.3)
 bla=data.frame(x,cat,v1,v2,v3,v4)
 bla
 
 x  catv1 v2v3 v4
 1   1 cat1 0.4013144 0.54839317 0.3946393  0.8679266
 2   2 cat5 0.4595873 0.45788906 0.4030078  0.5919596
 3   3 cat2 0.4542865 0.21516928 0.2777649  0.6112099
 4   4 cat2 0.4787950 0.06252512 0.5095611  0.6450795
 5   5 cat1 0.4910746 0.56591049 0.5151813  0.8465181
 6   6 cat5 0.4194397 0.16592579 0.4361643  0.6415192
 7   7 cat3 0.6148564 0.32240342 0.2690108  0.7114133
 8   8 cat4 0.6174652 0.28076152 0.4577064 -0.2567284
 9   9 cat5 0.4775395 0.28611768 0.4660210  0.4634120
 10 10 cat2 0.4802962 0.03715569 0.4506361  1.0063235
 11 11 cat3 0.6495094 0.33303172 0.3352933  1.4390324
 12 12 cat6 0.4891481 0.45355589 0.3880739  0.7831656
   
 
 I can assign this by the sqldf() command for each column but I would like to
 automate this as I have many columns.

   
 select=sqldf(select cat, max(v1) FROM bla GROUP BY cat)
 select
 
cat   max(v1)
 1 cat1 0.4910746
 2 cat2 0.4802962
 3 cat3 0.6495094
 4 cat4 0.6174652
 5 cat5 0.4775395
 6 cat6 0.4891481
   
 
 Finally, I would like to have a dataframe where where the cat is followed by
 each column maximum.

 Thanks for your help!
   

-- 
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/7410-58243
F ++49/40/7410-57790

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[R] strange floor rounding

[Trafim Vanishek]

Dear all,

might seem and easy question but I cannot figure it out.

floor(100*(.58))
[1] 57

where is the trick here? And how can I end up with the right answer?

Thanks a lot everybody for your help.
Trafim

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Re: [R] strange floor rounding

[Sarah Goslee]

You too have fallen victim to floating point error (see FAQ 7.31).


 (100*.58) == 58
[1] FALSE
 (100*.58)  58
[1] TRUE


On Fri, Jul 9, 2010 at 9:46 AM, Trafim Vanishek rdapam...@gmail.com wrote:
 Dear all,

 might seem and easy question but I cannot figure it out.

 floor(100*(.58))
 [1] 57

 where is the trick here? And how can I end up with the right answer?

 Thanks a lot everybody for your help.
 Trafim

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

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Re: [R] strange floor rounding

[Marc Schwartz]

On Jul 9, 2010, at 8:46 AM, Trafim Vanishek wrote:

 Dear all,
 
 might seem and easy question but I cannot figure it out.
 
 floor(100*(.58))
 [1] 57
 
 where is the trick here? And how can I end up with the right answer?
 
 Thanks a lot everybody for your help.
 Trafim


 sprintf(%.20f, 100 * .58)
[1] 57.99289457


See R FAQ 7.31

HTH,

Marc Schwartz

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Re: [R] Function on columns of a dataframe

[David Winsemius]

On Jul 9, 2010, at 9:46 AM, Eik Vettorazzi wrote:


Hi Nils,
have a look at
?tapply
hth.


Perhaps this will be part way there (I couldn't really figure out the  
desired structure of the final object):

 lapply( bla[, -(1:2)], function(x) tapply(x, bla$cat, max) )
$v1
 cat1  cat2  cat3  cat4  cat5  cat6
0.4634519 0.4062700 0.4816403 0.6354560 0.6663811 0.5260832

$v2
 cat1  cat2  cat3  cat4  cat5  cat6
0.5274645 0.4282639 0.4996033 0.3558259 0.2154201 0.3934063

$v3
 cat1  cat2  cat3  cat4  cat5  cat6
0.6051479 0.4443707 0.3538144 0.3646292 0.5059900 0.3545962

$v4
 cat1  cat2  cat3  cat4  cat5  cat6
0.7586322 0.8419526 0.9456385 0.1907295 0.7573575 0.6412563




Am 09.07.2010 15:37, schrieb LogLord:

Hi,

I would like to assign the largest value of a column to a specific  
category

and repeat this for each column (v1 - v4).



x=c(1:12)
cat 
= 
c 
(cat1 
,cat5 
,cat2 
,cat2,cat1,cat5,cat3,cat4,cat5,cat2,cat3,cat6)

v1=rnorm(12,0.5,0.1)
v2=rnorm(12,0.3,0.2)
v3=rnorm(12,0.4,0.1)
v4=rnorm(12,0.6,0.3)
bla=data.frame(x,cat,v1,v2,v3,v4)
bla


   x  catv1 v2v3 v4
1   1 cat1 0.4013144 0.54839317 0.3946393  0.8679266
2   2 cat5 0.4595873 0.45788906 0.4030078  0.5919596
3   3 cat2 0.4542865 0.21516928 0.2777649  0.6112099
4   4 cat2 0.4787950 0.06252512 0.5095611  0.6450795
5   5 cat1 0.4910746 0.56591049 0.5151813  0.8465181
6   6 cat5 0.4194397 0.16592579 0.4361643  0.6415192
7   7 cat3 0.6148564 0.32240342 0.2690108  0.7114133
8   8 cat4 0.6174652 0.28076152 0.4577064 -0.2567284
9   9 cat5 0.4775395 0.28611768 0.4660210  0.4634120
10 10 cat2 0.4802962 0.03715569 0.4506361  1.0063235
11 11 cat3 0.6495094 0.33303172 0.3352933  1.4390324
12 12 cat6 0.4891481 0.45355589 0.3880739  0.7831656



I can assign this by the sqldf() command for each column but I  
would like to

automate this as I have many columns.



select=sqldf(select cat, max(v1) FROM bla GROUP BY cat)
select


  cat   max(v1)
1 cat1 0.4910746
2 cat2 0.4802962
3 cat3 0.6495094
4 cat4 0.6174652
5 cat5 0.4775395
6 cat6 0.4891481



Finally, I would like to have a dataframe where where the cat is  
followed by

each column maximum.

Thanks for your help!



--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/7410-58243
F ++49/40/7410-57790

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David Winsemius, MD
West Hartford, CT

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Re: [R] strange floor rounding

[David Winsemius]

On Jul 9, 2010, at 9:46 AM, Trafim Vanishek wrote:


Dear all,

might seem and easy question but I cannot figure it out.

floor(100*(.58))
[1] 57

where is the trick here?


FAQ 7.31


And how can I end up with the right answer?


Define right, please. (There have been several questions in the last  
week for which FAQ 7.31 was the answer and some of the responses had  
useful links.)


--
David Winsemius, MD
West Hartford, CT

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[R] Help for mca

[housseima guiga]

Good eveninig,

I work on R and i want to do an MCA but, i find difficulty in calculating the 
contribution (CTR), cosine(CO2), coord(CORD) and distancefrom the center of the 

modalities of variables. Please help me how to calculate these values.

In fact, when i enter the variables it writes variable not found

Thank you 

Housseima
  
_
Hotmail : un service de messagerie gratuit, fiable et complet

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Re: [R] Non-parametric regression

[Joris Meys]

Just to be correct : gam is mentioned on the page Tal linked to, but
is a semi-parametric approach using maximum likelihood. It stays valid
though.

Another thing : you detect non-normality. But can you use a Poisson
distribution for example? The framework of generalized linear models
and generalized additive models allows you to deal with non-normality
of your data.

In any case, I suggest you contact a statistician nearby for guidance.

Cheers
Joris

On Fri, Jul 9, 2010 at 10:26 AM, Tal Galili tal.gal...@gmail.com wrote:
 From reviewing the first google page result for Non-parametric regression
 R, I hope this link will prove useful:

 http://socserv.mcmaster.ca/jfox/Courses/Oxford-2005/R-nonparametric-regression.html



 Contact
 Details:---
 Contact me: tal.gal...@gmail.com |  972-52-7275845
 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
 www.r-statistics.com (English)
 --




 On Fri, Jul 9, 2010 at 11:01 AM, Ralf B ralf.bie...@gmail.com wrote:

 I have two data sets, each a vector of 1000 numbers, each vector
 representing a distribution (i.e. 1000 numbers each of which
 representing a frequency at one point on a scale between 1 and 1000).
 For similfication, here an short version with only 5 points.


 a - c(8,10,8,12,4)
 b - c(7,11,8,10,5)

 Leaving the obvious discussion about causality aside fro a moment, I
 would like to see how well i can predict b from a using a regression.
 Since I do not know anything about the distribution type and already
 discovered non-normality I cannot use parametric regression or
 anything GLM for that matter.

 How should I proceed in using non-parametric regression to model
 vector a and see how well it predicts b? Perhaps you could extend the
 given lines into a short example script to give me an idea? Are there
 any other options?

 Best,
 Ralf

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-- 
Joris Meys
Statistical consultant

Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control

tel : +32 9 264 59 87
joris.m...@ugent.be
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Re: [R] split with list

[Joris Meys]

One  solution is  to put these unwanted entries to 

repor$9853312 [1:2,2:3] - 

Cheers
Joris

On Fri, Jul 9, 2010 at 12:18 PM, n.via...@libero.it n.via...@libero.it wrote:

 Dear List I would like to ask you something concenting a better print of the 
 R output:
 I have a bit data frame which has the following structure:
 CFISCALE              RAGSOCB            ANNO       VAR1        VAR2.
 9853312                     astra                 2005           6            
    45

 9853312                     astra                 2006          78            
   45


 9853312                     astra                 2007           55           
    76


 9653421                      geox                 2005           35           
   89



 9653421                     geox                 2006            24           
     33

 9653421                      geox                 2007           54           
    55


 The first thing I did is to split my data frame for CFISCALE. The result is 
 that R has transformed my data frame into a list. The second step was to 
 transpose each element of my list.
 repo=split(rep,rep$CFISCALE)
 repor=lapply(repo,function(x){
 t(x)})


 When I print my list the format is the following
 $9853312
                                   1                           2               
          3

 CFISCALE            9853312         9853312             9853312

 RAGSOCB            astra               astra                    astra

 ANNO                   2005                2006                      
 2007

 VAR1                       6                         78                       
        55

 VAR2                       45                        45                       
       76


 There is a way to remove the  first row I mean 1, 2 , 3 and to have just one 
 CFISCALE and RAGSOCB???
 For the second problem I tried to use unique but it seems that it doesnt work 
 for list. So what I would like to get is:
 $9853312




 CFISCALE            9853312


 RAGSOCB            astra
 ANNO                   2005                2006                      
 2007

 VAR1                       6                         78                       
        55

 VAR2                       45                        45                       
       76


 This is because I next run xtable on my list in order to get a table in 
 Latex, which I woud like to be in a nice format.
 Thanks a lot for your attention!





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-- 
Joris Meys
Statistical consultant

Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control

tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

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[R] how can achive step by step execution of the script

[vijaysheegi]

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Re: [R] How can i draw a graph with high and low data points

[Nathaniel Saxe]

Hi Tal, Thanks for your help.


I've had a look at the site, and what i wanted to do was to plot X and Y
where X is a characters and Y is numeric. The problem I'm having now is that
the X axis isn't characters but just numbers from 1 onwards and when i plot
it, the data i have is in descending order which isn't shown on the graph.

I have this at the moment:

plot(1:nrow(dat),dat$Mean,type=b,xaxt=n,
ylim=c(min(dat$lci),max(dat$uci)),
xlab=,ylab=HR,)

It gives me sort of what I want. It has the Y values in descending order,
but it doesn't give me the text on the x axis and I was also thinking of
plotting the upper and lower confidence intervals with a line connecting the
two. I can add in the upper and lower CI values separately, but I don't know
how to join the two together.



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[R] installing packages over ssh without X forwarding

[petr]

Hi,
Is it possible to install packages without the testing if installed
package can be loaded?
I need to install bunch of packages on multiple computers over ssh. Some
packages witch interact with X11 display cannot be installed in this way.
for example after:
 install.packages('cairoDevice',dep=T)
I get
(...)
*** installing help indices
** building package indices ...
** testing if installed package can be loaded
Error : .onLoad failed in loadNamespace() for 'cairoDevice', details:
  call: fun(...)
  error: GDK display not found - please make sure X11 is running
ERROR: loading failed
* removing #8216;/usr/local/lib64/R/library/cairoDevice#8217;
* restoring previous #8216;/usr/local/lib64/R/library/cairoDevice#8217;

The downloaded packages are in
#8216;/tmp/Rtmpk1XxTl/downloaded_packages#8217;
Updating HTML index of packages in '.Library'
Warning message:
In install.packages(cairoDevice, dep = T) :
  installation of package 'cairoDevice' had non-zero exit status

When I connect with remote computer using ssh -X r...@nod1 to enable X11
forwarding then installation works without problems. This would however
require manually connect with each administred computer a do the
installation. cssh which I use now to install packages on multiple
computers does not enable X11 forwarding. I have also tested installation
using R CMD INSTALL with --no-test-load options but the packages are
loaded unsuccesfully anyway. So is it possible to turn of package testing?
Petr

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Re: [R] distributing a value for a given month across the number of weeks in that month

[Gabor Grothendieck]

On Fri, Jul 9, 2010 at 9:35 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
 Hello!

 Any hint would be greatly appreciated.
 I have a data frame that contains (a) monthly dates and (b) a value
 that corresponds to each month - see the data frame monthly below:

 monthly-data.frame(month=c(20100301,20100401,20100501),monthly.value=c(100,200,300))
 monthly$month-as.character(monthly$month)
 monthly$month-as.Date(monthly$month,%Y%m%d)
 (monthly)

 I need to split each month into weeks, e.g., weeks that start on
 Monday (it could as well be Sunday - it does not really matter) and
 distribute the monthly value evenly across weeks. So, if a month has 5
 Mondays, then the monthly value should be dividied by 5, but if a
 month has only 4 weeks, then the monthly value should be divided by 4.

 The output I need is like this:

 week          weekly.value
 2010-03-01   20
 2010-03-08   20
 2010-03-15   20
 2010-03-22   20
 2010-03-29   20
 2010-04-05   50
 2010-04-12   50
 2010-04-19   50
 2010-04-26   50
 2010-05-03   60
 2010-05-10   60
 2010-05-17   60
 2010-05-24   60
 2010-05-31   60



There is new functionality in na.locf in the development version
of zoo that makes it particularly convenient to do this.

First create a zoo object z from monthly and get a vector of all
the mondays.  Then use na.locf to place the monthly value in each
monday and ave to distribute them out.


library(zoo)

# pull in development version of na.locf.zoo
source(http://r-forge.r-project.org/scm/viewvc.php/*checkout*/pkg/zoo/R/na.locf.R?revision=725root=zoo;)

# convert to zoo
z - with(monthly, zoo(monthly.value, month))

# get sequence of all dates and from that get mondays
all.dates - seq(start(z), as.Date(as.yearmon(end(z)), frac = 1), by = day)
mondays - all.dates[weekdays(all.dates) == Monday]

# use na.locf to fill in mondays and ave to distribute them
weeks - na.locf(z, xout = mondays)
weeks[] - ave(weeks, as.yearmon(mondays), FUN = function(x) x[1]/length(x))

# show output in a few different formats
weeks
as.data.frame(weeks)
data.frame(Monday = as.Date(time(weeks)), value = weeks)
data.frame(Monday = as.Date(time(weeks)), value = weeks, row.names = NULL)
plot(weeks)

The output looks like this:

 weeks
2010-03-01 2010-03-08 2010-03-15 2010-03-22 2010-03-29 2010-04-05 2010-04-12
20 20 20 20 20 50 50
2010-04-19 2010-04-26 2010-05-03 2010-05-10 2010-05-17 2010-05-24 2010-05-31
50 50 60 60 60 60 60
 as.data.frame(weeks)
   weeks
2010-03-0120
2010-03-0820
2010-03-1520
2010-03-2220
2010-03-2920
2010-04-0550
2010-04-1250
2010-04-1950
2010-04-2650
2010-05-0360
2010-05-1060
2010-05-1760
2010-05-2460
2010-05-3160


 data.frame(Monday = as.Date(time(weeks)), value = weeks, row.names = NULL)
   Monday value
1  2010-03-0120
2  2010-03-0820
3  2010-03-1520
4  2010-03-2220
5  2010-03-2920
6  2010-04-0550
7  2010-04-1250
8  2010-04-1950
9  2010-04-2650
10 2010-05-0360
11 2010-05-1060
12 2010-05-1760
13 2010-05-2460
14 2010-05-3160

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[R] ttrTests issue in cReturns

[Raghu]

cr-cReturns(spData,ttr=MACD)
Error in ind[t - k] - pos[t - k + 1] - pos[t - k] :
  replacement has length zero
 cr-cReturns(spData,ttr=macd4)


Why is the above error coming? macd4 works alright ( which is a custom
function built by the ttrTests author) while MACD doesnt work.

Thanks
-- 
'Raghu'

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[R] how can achive step by step execution of the script

[vijaysheegi]

Hi R Experts,
I have certain code ,i want to achive interactive execution .
For ex:
1. as part of input ,it should ask file name  or table name as input.
2.in script so many graphs i need to draw,it should wait till certain key is
pressed .
3:i am using windows R,rscript scriptname is not working.



Please some one help me.

Thanks Experts in advance


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Re: [R] Function on columns of a dataframe

[Eik Vettorazzi]

you are right. But maybe aggregate is close to the desired result?

aggregate(bla, list(bla$cat), max)

Am 09.07.2010 16:01, schrieb David Winsemius:

 On Jul 9, 2010, at 9:46 AM, Eik Vettorazzi wrote:

 Hi Nils,
 have a look at
 ?tapply
 hth.

 Perhaps this will be part way there (I couldn't really figure out the
 desired structure of the final object):
  lapply( bla[, -(1:2)], function(x) tapply(x, bla$cat, max) )
 $v1
  cat1  cat2  cat3  cat4  cat5  cat6
 0.4634519 0.4062700 0.4816403 0.6354560 0.6663811 0.5260832

 $v2
  cat1  cat2  cat3  cat4  cat5  cat6
 0.5274645 0.4282639 0.4996033 0.3558259 0.2154201 0.3934063

 $v3
  cat1  cat2  cat3  cat4  cat5  cat6
 0.6051479 0.4443707 0.3538144 0.3646292 0.5059900 0.3545962

 $v4
  cat1  cat2  cat3  cat4  cat5  cat6
 0.7586322 0.8419526 0.9456385 0.1907295 0.7573575 0.6412563



 Am 09.07.2010 15:37, schrieb LogLord:
 Hi,

 I would like to assign the largest value of a column to a specific
 category
 and repeat this for each column (v1 - v4).


 x=c(1:12)
 cat=c(cat1,cat5,cat2,cat2,cat1,cat5,cat3,cat4,cat5,cat2,cat3,cat6)

 v1=rnorm(12,0.5,0.1)
 v2=rnorm(12,0.3,0.2)
 v3=rnorm(12,0.4,0.1)
 v4=rnorm(12,0.6,0.3)
 bla=data.frame(x,cat,v1,v2,v3,v4)
 bla

x  catv1 v2v3 v4
 1   1 cat1 0.4013144 0.54839317 0.3946393  0.8679266
 2   2 cat5 0.4595873 0.45788906 0.4030078  0.5919596
 3   3 cat2 0.4542865 0.21516928 0.2777649  0.6112099
 4   4 cat2 0.4787950 0.06252512 0.5095611  0.6450795
 5   5 cat1 0.4910746 0.56591049 0.5151813  0.8465181
 6   6 cat5 0.4194397 0.16592579 0.4361643  0.6415192
 7   7 cat3 0.6148564 0.32240342 0.2690108  0.7114133
 8   8 cat4 0.6174652 0.28076152 0.4577064 -0.2567284
 9   9 cat5 0.4775395 0.28611768 0.4660210  0.4634120
 10 10 cat2 0.4802962 0.03715569 0.4506361  1.0063235
 11 11 cat3 0.6495094 0.33303172 0.3352933  1.4390324
 12 12 cat6 0.4891481 0.45355589 0.3880739  0.7831656


 I can assign this by the sqldf() command for each column but I would
 like to
 automate this as I have many columns.


 select=sqldf(select cat, max(v1) FROM bla GROUP BY cat)
 select

   cat   max(v1)
 1 cat1 0.4910746
 2 cat2 0.4802962
 3 cat3 0.6495094
 4 cat4 0.6174652
 5 cat5 0.4775395
 6 cat6 0.4891481


 Finally, I would like to have a dataframe where where the cat is
 followed by
 each column maximum.

 Thanks for your help!


 -- 
 Eik Vettorazzi
 Institut für Medizinische Biometrie und Epidemiologie
 Universitätsklinikum Hamburg-Eppendorf

 Martinistr. 52
 20246 Hamburg

 T ++49/40/7410-58243
 F ++49/40/7410-57790

 __
 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.

 David Winsemius, MD
 West Hartford, CT


-- 
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/7410-58243
F ++49/40/7410-57790

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Re: [R] strange floor rounding

[Trafim Vanishek]

Thanks everybody for referring me to FAQ 7.31 but I don't see how to solve
it.
I am giving a concrete number and I need to get 58 not 57. Seems, there is
no way?

On Fri, Jul 9, 2010 at 4:05 PM, David Winsemius dwinsem...@comcast.netwrote:


 On Jul 9, 2010, at 9:46 AM, Trafim Vanishek wrote:

  Dear all,

 might seem and easy question but I cannot figure it out.

 floor(100*(.58))
 [1] 57

 where is the trick here?


 FAQ 7.31


 And how can I end up with the right answer?


 Define right, please. (There have been several questions in the last week
 for which FAQ 7.31 was the answer and some of the responses had useful
 links.)

 --
 David Winsemius, MD
 West Hartford, CT



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[R] print.trellis draw.in

[Mark Connolly]

I am attempting to plot a trellis object on a grid.

vplayout = viewport(layout.pos.row=x, layout.pos.col=y)

grid.newpage()
pushViewport(viewport(layout=grid.layout(2,2)))

g1 = ggplot() ...
g2 = ggplot() ...
g3 = ggplot() ...
p = xyplot() ...

# works as expected
print(g1, vp=vplayout(1,1))
print(g2, vp=vplayout(1,2))
print(g3, vp=vplayout(2,1))

# does not work
print(  p,
 newpage=FALSE,
 draw.in=vplayout(2,2)$name)

Error in grid.Call.graphics(L_downviewport, name$name, strict) :
  Viewport 'GRID.VP.112' was not found


What am I doing wrong?

Thanks!

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Re: [R] installing packages over ssh without X forwarding

[Duncan Murdoch]

On 09/07/2010 7:37 AM, p...@orbit.umbr.cas.cz wrote:

Hi,
Is it possible to install packages without the testing if installed
package can be loaded?
I need to install bunch of packages on multiple computers over ssh. Some
packages witch interact with X11 display cannot be installed in this way.
for example after:
 install.packages('cairoDevice',dep=T)
I get
(...)
*** installing help indices
** building package indices ...
** testing if installed package can be loaded
Error : .onLoad failed in loadNamespace() for 'cairoDevice', details:
  call: fun(...)
  error: GDK display not found - please make sure X11 is running
ERROR: loading failed
* removing #8216;/usr/local/lib64/R/library/cairoDevice#8217;
* restoring previous #8216;/usr/local/lib64/R/library/cairoDevice#8217;

The downloaded packages are in
#8216;/tmp/Rtmpk1XxTl/downloaded_packages#8217;
Updating HTML index of packages in '.Library'
Warning message:
In install.packages(cairoDevice, dep = T) :
  installation of package 'cairoDevice' had non-zero exit status

When I connect with remote computer using ssh -X r...@nod1 to enable X11
forwarding then installation works without problems. This would however
require manually connect with each administred computer a do the
installation. cssh which I use now to install packages on multiple
computers does not enable X11 forwarding. I have also tested installation
using R CMD INSTALL with --no-test-load options but the packages are
loaded unsuccesfully anyway. So is it possible to turn of package testing?
Petr


Yes, use the --no-test-load option to R CMD INSTALL, which you can 
pass through the INSTALL_opts argument to install.packages().


Duncan Murdoch

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Re: [R] strange floor rounding

[David Winsemius]

On Jul 9, 2010, at 10:27 AM, Trafim Vanishek wrote:

Thanks everybody for referring me to FAQ 7.31 but I don't see how to  
solve

it.
I am giving a concrete number and I need to get 58 not 57. Seems,  
there is

no way?

Building on an example on the help page for as.integer,  this seems to  
be working:


 trnc2 - function(x) trunc(x) + (trunc(x)  x)*sign(x)*(abs(x -  
trunc(x))  .Machine$double.eps^0.5)


 trnc2(0.01*100)
[1] 1
 trnc2(seq(0, 1, by=0.01)*100)
  [1]   0   1   2   3   4   5   6   8   8   9  10  11  12  13  15   
15  16  17  18  19
 [21]  20  21  22  23  24  25  26  27  29  28  30  31  32  33  34   
35  36  37  38  39
 [41]  40  41  42  43  44  45  46  47  48  49  50  51  52  53  54   
56  57  58  57  59
 [61]  60  61  62  63  64  65  66  67  68  69  70  71  72  73  74   
75  76  77  78  79
 [81]  80  81  82  83  84  85  86  87  88  89  90  91  92  93  94   
95  96  97  98  99

[101] 100
 trnc2(-seq(0, 1, by=0.01)*100)
  [1]0   -1   -2   -3   -4   -5   -6   -7   -8   -9  -10  -11   
-12  -13  -14  -15
 [17]  -16  -17  -18  -19  -20  -21  -22  -23  -24  -25  -26  -27   
-28  -28  -30  -31
 [33]  -32  -33  -34  -35  -36  -37  -38  -39  -40  -41  -42  -43   
-44  -45  -46  -47
 [49]  -48  -49  -50  -51  -52  -53  -54  -55  -56  -57  -57  -59   
-60  -61  -62  -63
 [65]  -64  -65  -66  -67  -68  -69  -70  -71  -72  -73  -74  -75   
-76  -77  -78  -79
 [81]  -80  -81  -82  -83  -84  -85  -86  -87  -88  -89  -90  -91   
-92  -93  -94  -95

 [97]  -96  -97  -98  -99 -100

And after looking at that a bit I came up with a shorter alternate  
that seems to work as well:


trnc3 - function(x) trunc(x+sign(x)* .Machine$double.eps^0.5)

See also  ?all.equal

--
David.
On Fri, Jul 9, 2010 at 4:05 PM, David Winsemius dwinsem...@comcast.net 
wrote:




On Jul 9, 2010, at 9:46 AM, Trafim Vanishek wrote:

Dear all,


might seem and easy question but I cannot figure it out.

floor(100*(.58))
[1] 57

where is the trick here?



FAQ 7.31


And how can I end up with the right answer?




Define right, please. (There have been several questions in the  
last week
for which FAQ 7.31 was the answer and some of the responses had  
useful

links.)

--
David Winsemius, MD
West Hartford, CT




[[alternative HTML version deleted]]

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David Winsemius, MD
West Hartford, CT

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Re: [R] Current script name from R

[Allan Engelhardt]

On 09/07/10 12:18, Ralf B wrote:

I am using RGUI, the  command line or the StatET Eclipse environment.
Should this not all be the same?
   


No, there is no particular reason why they should.

Allan


Ralf

On Fri, Jul 9, 2010 at 7:11 AM, Allan Engelhardtall...@cybaea.com  wrote:
   

I'm assuming you are using Rscript (please provide self-contained examples
when posting) in which case you could look for the element in
(base|R.utils)::commandArgs() that begin with the string --file= - the
rest is the file name.  See the asValues= parameter in help(commandArgs,
package=R.utils) for a nice way to get the parameter.

For an invocation of the form R  foo.R you'd need to inspect your system's
process table (so don't do that).

Hope this helps.

Allan

On 09/07/10 10:48, Ralf B wrote:
 

Is there a way for a script to find out about its own name ?

Ralf

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Re: [R] strange floor rounding

[Peter Ehlers]

On 2010-07-09 8:27, Trafim Vanishek wrote:

Thanks everybody for referring me to FAQ 7.31 but I don't see how to solve
it.
I am giving a concrete number and I need to get 58 not 57. Seems, there is
no way?


Sure there is: round(100*0.58).

  -Peter Ehlers



On Fri, Jul 9, 2010 at 4:05 PM, David Winsemiusdwinsem...@comcast.netwrote:



On Jul 9, 2010, at 9:46 AM, Trafim Vanishek wrote:

  Dear all,


might seem and easy question but I cannot figure it out.

floor(100*(.58))
[1] 57

where is the trick here?



FAQ 7.31


And how can I end up with the right answer?




Define right, please. (There have been several questions in the last week
for which FAQ 7.31 was the answer and some of the responses had useful
links.)

--
David Winsemius, MD
West Hartford, CT




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

Re: [R] strange floor rounding

[Duncan Murdoch]

On 09/07/2010 10:27 AM, Trafim Vanishek wrote:

Thanks everybody for referring me to FAQ 7.31 but I don't see how to solve
it.
I am giving a concrete number and I need to get 58 not 57. Seems, there is
no way?
  
You aren't giving a concrete number.  You're giving a string of three 
characters, which R interprets as a number that's a little bit less than 
0.58, because
there is no way to represent the number 0.58 in the floating point 
format that R uses.  So it is calculating the right answer.


If you want to work with exact 2 decimal place values, you should not 
use floating point storage.  One alternative is to work with integers 
equal to 100 times the value you want, and then divide by 100 at the 
very end of the operation for printing.


Duncan Murdoch

On Fri, Jul 9, 2010 at 4:05 PM, David Winsemius dwinsem...@comcast.netwrote:


 On Jul 9, 2010, at 9:46 AM, Trafim Vanishek wrote:

  Dear all,

 might seem and easy question but I cannot figure it out.

 floor(100*(.58))
 [1] 57

 where is the trick here?


 FAQ 7.31


 And how can I end up with the right answer?


 Define right, please. (There have been several questions in the last week
 for which FAQ 7.31 was the answer and some of the responses had useful
 links.)

 --
 David Winsemius, MD
 West Hartford, CT



[[alternative HTML version deleted]]

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Re: [R] Function on columns of a dataframe

[David Winsemius]

On Jul 9, 2010, at 10:26 AM, Eik Vettorazzi wrote:


you are right. But maybe aggregate is close to the desired result?

aggregate(bla, list(bla$cat), max)


Right. I couldn't get it to work until I removed the first two columns:

aggregate(bla[,-(1:2)], list(bla$cat), max)

Then I got pretty much the same dataframe as I would have with :

as.data.frame(lapply( bla[, -(1:2)], function(x) tapply(x, bla$cat,  
max) ))

v1v2v3v4
cat1 0.4634519 0.5274645 0.6051479 0.7586322
cat2 0.4062700 0.4282639 0.4443707 0.8419526
cat3 0.4816403 0.4996033 0.3538144 0.9456385
cat4 0.6354560 0.3558259 0.3646292 0.1907295
cat5 0.6663811 0.2154201 0.5059900 0.7573575
cat6 0.5260832 0.3934063 0.3545962 0.6412563

Except that aggregate version returns it with a Group.1 column of  
cats while the other version returned it with the cat names in the  
rownames. A matter of taste?


--
David.


Am 09.07.2010 16:01, schrieb David Winsemius:


On Jul 9, 2010, at 9:46 AM, Eik Vettorazzi wrote:


Hi Nils,
have a look at
?tapply
hth.


Perhaps this will be part way there (I couldn't really figure out the
desired structure of the final object):

lapply( bla[, -(1:2)], function(x) tapply(x, bla$cat, max) )

$v1
cat1  cat2  cat3  cat4  cat5  cat6
0.4634519 0.4062700 0.4816403 0.6354560 0.6663811 0.5260832

$v2
cat1  cat2  cat3  cat4  cat5  cat6
0.5274645 0.4282639 0.4996033 0.3558259 0.2154201 0.3934063

$v3
cat1  cat2  cat3  cat4  cat5  cat6
0.6051479 0.4443707 0.3538144 0.3646292 0.5059900 0.3545962

$v4
cat1  cat2  cat3  cat4  cat5  cat6
0.7586322 0.8419526 0.9456385 0.1907295 0.7573575 0.6412563




Am 09.07.2010 15:37, schrieb LogLord:

Hi,

I would like to assign the largest value of a column to a specific
category
and repeat this for each column (v1 - v4).



x=c(1:12)
cat 
= 
c 
(cat1 
,cat5 
,cat2 
,cat2,cat1,cat5,cat3,cat4,cat5,cat2,cat3,cat6)


v1=rnorm(12,0.5,0.1)
v2=rnorm(12,0.3,0.2)
v3=rnorm(12,0.4,0.1)
v4=rnorm(12,0.6,0.3)
bla=data.frame(x,cat,v1,v2,v3,v4)
bla


  x  catv1 v2v3 v4
1   1 cat1 0.4013144 0.54839317 0.3946393  0.8679266
2   2 cat5 0.4595873 0.45788906 0.4030078  0.5919596
3   3 cat2 0.4542865 0.21516928 0.2777649  0.6112099
4   4 cat2 0.4787950 0.06252512 0.5095611  0.6450795
5   5 cat1 0.4910746 0.56591049 0.5151813  0.8465181
6   6 cat5 0.4194397 0.16592579 0.4361643  0.6415192
7   7 cat3 0.6148564 0.32240342 0.2690108  0.7114133
8   8 cat4 0.6174652 0.28076152 0.4577064 -0.2567284
9   9 cat5 0.4775395 0.28611768 0.4660210  0.4634120
10 10 cat2 0.4802962 0.03715569 0.4506361  1.0063235
11 11 cat3 0.6495094 0.33303172 0.3352933  1.4390324
12 12 cat6 0.4891481 0.45355589 0.3880739  0.7831656



I can assign this by the sqldf() command for each column but I  
would

like to
automate this as I have many columns.



select=sqldf(select cat, max(v1) FROM bla GROUP BY cat)
select


 cat   max(v1)
1 cat1 0.4910746
2 cat2 0.4802962
3 cat3 0.6495094
4 cat4 0.6174652
5 cat5 0.4775395
6 cat6 0.4891481




Finally, I would like to have a dataframe where where the cat is
followed by
each column maximum.

Thanks for your help!



--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/7410-58243
F ++49/40/7410-57790

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David Winsemius, MD
West Hartford, CT



--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/7410-58243
F ++49/40/7410-57790

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David Winsemius, MD
West Hartford, CT

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[R] print.trellis draw.in - plaintext (gmail mishap)

[Mark Connolly]

I am attempting to plot a trellis object on a grid.

vplayout = viewport(layout.pos.row=x, layout.pos.col=y)

grid.newpage()
pushViewport(viewport(layout=grid.layout(2,2)))

g1 = ggplot() ...
g2 = ggplot() ...
g3 = ggplot() ...
p = xyplot() ...

# works as expected
print(g1, vp=vplayout(1,1))
print(g2, vp=vplayout(1,2))
print(g3, vp=vplayout(2,1))

# does not work
print(  p,
 newpage=FALSE,
 draw.in=vplayout(2,2)$name)

Error in grid.Call.graphics(L_downviewport, name$name, strict) :
  Viewport 'GRID.VP.112' was not found


What am I doing wrong?

Thanks!

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Re: [R] distributing a value for a given month across the number of weeks in that month

[Dimitri Liakhovitski]

Wow, Gabor - that's amazing - thank you so much!
Dimitri

On Fri, Jul 9, 2010 at 10:22 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
 On Fri, Jul 9, 2010 at 9:35 AM, Dimitri Liakhovitski
 dimitri.liakhovit...@gmail.com wrote:
 Hello!

 Any hint would be greatly appreciated.
 I have a data frame that contains (a) monthly dates and (b) a value
 that corresponds to each month - see the data frame monthly below:

 monthly-data.frame(month=c(20100301,20100401,20100501),monthly.value=c(100,200,300))
 monthly$month-as.character(monthly$month)
 monthly$month-as.Date(monthly$month,%Y%m%d)
 (monthly)

 I need to split each month into weeks, e.g., weeks that start on
 Monday (it could as well be Sunday - it does not really matter) and
 distribute the monthly value evenly across weeks. So, if a month has 5
 Mondays, then the monthly value should be dividied by 5, but if a
 month has only 4 weeks, then the monthly value should be divided by 4.

 The output I need is like this:

 week          weekly.value
 2010-03-01   20
 2010-03-08   20
 2010-03-15   20
 2010-03-22   20
 2010-03-29   20
 2010-04-05   50
 2010-04-12   50
 2010-04-19   50
 2010-04-26   50
 2010-05-03   60
 2010-05-10   60
 2010-05-17   60
 2010-05-24   60
 2010-05-31   60



 There is new functionality in na.locf in the development version
 of zoo that makes it particularly convenient to do this.

 First create a zoo object z from monthly and get a vector of all
 the mondays.  Then use na.locf to place the monthly value in each
 monday and ave to distribute them out.


 library(zoo)

 # pull in development version of na.locf.zoo
 source(http://r-forge.r-project.org/scm/viewvc.php/*checkout*/pkg/zoo/R/na.locf.R?revision=725root=zoo;)

 # convert to zoo
 z - with(monthly, zoo(monthly.value, month))

 # get sequence of all dates and from that get mondays
 all.dates - seq(start(z), as.Date(as.yearmon(end(z)), frac = 1), by = day)
 mondays - all.dates[weekdays(all.dates) == Monday]

 # use na.locf to fill in mondays and ave to distribute them
 weeks - na.locf(z, xout = mondays)
 weeks[] - ave(weeks, as.yearmon(mondays), FUN = function(x) x[1]/length(x))

 # show output in a few different formats
 weeks
 as.data.frame(weeks)
 data.frame(Monday = as.Date(time(weeks)), value = weeks)
 data.frame(Monday = as.Date(time(weeks)), value = weeks, row.names = NULL)
 plot(weeks)

 The output looks like this:

 weeks
 2010-03-01 2010-03-08 2010-03-15 2010-03-22 2010-03-29 2010-04-05 2010-04-12
        20         20         20         20         20         50         50
 2010-04-19 2010-04-26 2010-05-03 2010-05-10 2010-05-17 2010-05-24 2010-05-31
        50         50         60         60         60         60         60
 as.data.frame(weeks)
           weeks
 2010-03-01    20
 2010-03-08    20
 2010-03-15    20
 2010-03-22    20
 2010-03-29    20
 2010-04-05    50
 2010-04-12    50
 2010-04-19    50
 2010-04-26    50
 2010-05-03    60
 2010-05-10    60
 2010-05-17    60
 2010-05-24    60
 2010-05-31    60


  data.frame(Monday = as.Date(time(weeks)), value = weeks, row.names = NULL)
       Monday value
 1  2010-03-01    20
 2  2010-03-08    20
 3  2010-03-15    20
 4  2010-03-22    20
 5  2010-03-29    20
 6  2010-04-05    50
 7  2010-04-12    50
 8  2010-04-19    50
 9  2010-04-26    50
 10 2010-05-03    60
 11 2010-05-10    60
 12 2010-05-17    60
 13 2010-05-24    60
 14 2010-05-31    60




-- 
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com

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Re: [R] Data Frame Manipulation using function

[harsh yadav]

Hi,

Thanks a lot.
The Vectorize method worked and its much faster than looping through the
data frame.

Regards,
Harsh Yadav

On Thu, Jul 8, 2010 at 11:06 PM, David Winsemius dwinsem...@comcast.netwrote:


 On Jul 8, 2010, at 10:33 PM, Erik Iverson wrote:


  I have a data frame:
 id url
 urlType
 1 1  www.yahoo.com http://www.yahoo.com
  1
 2 2  www.google.com/?search= http://www.google.com/?search=
   2
 3 3  www.google.com http://www.google.com
   1
 4 4  www.yahoo.com/?query= http://www.yahoo.com/?query=
 2
 5 5  www.gmail.com http://www.gmail.com
   1


 This is not output from ?dput, which means more work to read it in.


 Yeah it was kind of pain, but ...

 dta - read.table(textConnection(' id url
   urlType

 1 1  www.yahoo.com http://www.yahoo.com  1
 2 2  www.google.com/?search= http://www.google.com/?search= 2
 3 3  www.google.com http://www.google.com 1
 4 4  www.yahoo.com/?query= http://www.yahoo.com/?query=   2
 5 5  www.gmail.com http://www.gmail.com 1') )




  Here is the definition for WHITELIST:-
 WHITELIST = [?]query=, [?]search=
 WHITELIST - unlist(trim(strsplit(trim(WHITELIST), ,)))


 What is the 'trim' function?  I do not have that defined.

 Perhaps David's answer will work for you...


 Seems to ... after I fixed my incorrect cmd-V paste of the function name
 and guessing that trim was the one in gdata:

  require(gdata)

  checkBaseLine - function(s){
 + for (listItem in WHITELIST){
 + if(regexpr(as.character(listItem), s)[1]  -1){
 + return(TRUE)
 + }
 + }
 + return(FALSE)
 + }
 
  #Here is the definition for WHITELIST:-

 
  WHITELIST = [?]query=, [?]search=
  WHITELIST - unlist(trim(strsplit(trim(WHITELIST), ,)))
  vcheck - Vectorize(checkBaseLine)
 
  vcheck - Vectorize(checkBaseLine)
 
  dta[ dta$urlType != 1  vcheck(dta$url) , url ]
 [1] www.google.com/?search= http://www.google.com/?search=
 www.yahoo.com/?query= http://www.yahoo.com/?query=
 5 Levels: www.gmail.com http://www.gmail.com www.google.com 
 http://www.google.com ... www.yahoo.com/?query= 
 http://www.yahoo.com/?query=

 --
 David.


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Re: [R] random sample from arrays

[Joris Meys]

Could you elaborate?

Both

 x - 1:4
 set - matrix(nrow = 50, ncol = 11)
  for(i in c(1:11)){
set[,i] -sample(x,50)
print(c(i,-, set), quote = FALSE)
   }

and

 x - 1:4
 set - matrix(nrow = 50, ncol = 11)
 for(i in c(1:50)){
   set[i,] -sample(x,11)
   print(c(i,-, set), quote = FALSE)
  }

run perfectly fine on my computer.
Cheers


On Fri, Jul 9, 2010 at 3:10 PM, Assa Yeroslaviz fry...@gmail.com wrote:
 Hi Joris,
 I guess i did it wrong again.
 but your example didn't work either. I still get the error massage.

 but replicate function just fine. I can even replicate the whole array
 lines.

 THX

 Assa

 On Thu, Jul 8, 2010 at 15:20, Joris Meys jorism...@gmail.com wrote:

 Don't know what exactly you're trying to do, but you make a matrix
 with 11 columns and 50 rows, then treat it as a vector. On top of
 that, you try to fill 50 rows/columns with 50 values. Off course that
 doesn't work. Did you check the warning messages when running the
 code?

 Either do :

  for(i in c(1:11)){
    set[,i] -sample(x,50)
    print(c(i,-, set), quote = FALSE)
   }

 or

  for(i in c(1:50)){
    set[i,] -sample(x,11)
    print(c(i,-, set), quote = FALSE)
   }

 Or just forget about the loop altogether and do :

 set - replicate(11,sample(x,50))
 or
 set - t(replicate(50,sample(x,11)))

 cheers

 On Thu, Jul 8, 2010 at 8:04 AM, Assa Yeroslaviz fry...@gmail.com wrote:
  Hello R users,
 
  I'm trying to extract random samples from a big array I have.
 
  I have a data frame of over 40k lines and would like to produce around
  50
  random sample of around 200 lines each from this array.
 
  this is the matrix
           ID xxx_1c xxx__2c xxx__3c xxx__4c xxx__5T xxx__6T xxx__7T
  xxx__8T
  yyy_1c yyy_1c _2c
  1 A_512  2.150295  2.681759  2.177138  2.142790  2.115344  2.013047
  2.115634  2.189372  1.643328  1.563523
  2 A_134 12.832488 12.596373 12.882581 12.987091 11.956149 11.994779
  11.650336 11.995504 13.024494 12.776322
  3 A_152  2.063276  2.160961  2.067549  2.059732  2.656416  2.075775
  2.033982  2.111937  1.606340  1.548940
  4 A_163  9.570761 10.448615  9.432859  9.732615 10.354234 10.993279
  9.160038  9.104121 10.079177  9.828757
  5 A_184  3.574271  4.680859  4.517047  4.047096  3.623668  3.021356
  3.559434  3.156093  4.308437  4.045098
  6 A_199  7.593952  7.454087  7.513013  7.449552  7.345718  7.367068
  7.410085  7.022582  7.668616  7.953706
  ...
 
  I tried to do it with a for loop:
 
  genelist - read.delim(/user/R/raw_data.txt)
  rownames(genelist) - genelist[,1]
  genes - rownames(genelist)
 
  x - 1:4
  set - matrix(nrow = 50, ncol = 11)
 
  for(i in c(1:50)){
     set[i] -sample(x,50)
     print(c(i,-, set), quote = FALSE)
     }
 
  which basically do the trick, but I just can't save the results outside
  the
  loop.
  After having the random sets of lines it wasn't a problem to extract the
  line from the arrays using subset.
 
  genSet1 -sample(x,50)
  random1 - genes %in% genSet1
  subsetGenelist - subset(genelist, random1)
 
 
  is there a different way of creating these random vectors or saving the
  loop
  results outside tjhe loop so I cn work with them?
 
  Thanks a lot
 
  Assa
 
         [[alternative HTML version deleted]]
 
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 Department of Applied mathematics, biometrics and process control

 tel : +32 9 264 59 87
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-- 
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Statistical consultant

Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control

tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php

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Re: [R] Function on columns of a dataframe

[Eik Vettorazzi]

just to satisfy my curiousity,
aggregate(bla, list(bla$cat), max)

works for me and resulted in

  Group.1  x  catv1v2v3v4
1cat1  5 cat1 0.6337076 0.2887081 0.3629962 0.5328683
2cat2 10 cat2 0.5519426 0.6076447 0.4593770 0.9632341
3cat3 11 cat3 0.6094089 0.6152059 0.5670835 0.9084917
4cat4  8 cat4 0.4772603 0.2149017 0.4534723 0.7824375
5cat5  9 cat5 0.6582466 0.3150096 0.5512863 1.3524582
6cat6 12 cat6 0.4632893 0.4498425 0.3926193 0.8023014

so, what didn't work for you, except for the extra columns? (taking
for granted, that random numbers aren't the same)

Am 09.07.2010 16:47, schrieb David Winsemius:

 On Jul 9, 2010, at 10:26 AM, Eik Vettorazzi wrote:

 you are right. But maybe aggregate is close to the desired result?

 aggregate(bla, list(bla$cat), max)

 Right. I couldn't get it to work until I removed the first two columns:

 aggregate(bla[,-(1:2)], list(bla$cat), max)

 Then I got pretty much the same dataframe as I would have with :

 as.data.frame(lapply( bla[, -(1:2)], function(x) tapply(x, bla$cat,
 max) ))
 v1v2v3v4
 cat1 0.4634519 0.5274645 0.6051479 0.7586322
 cat2 0.4062700 0.4282639 0.4443707 0.8419526
 cat3 0.4816403 0.4996033 0.3538144 0.9456385
 cat4 0.6354560 0.3558259 0.3646292 0.1907295
 cat5 0.6663811 0.2154201 0.5059900 0.7573575
 cat6 0.5260832 0.3934063 0.3545962 0.6412563

 Except that aggregate version returns it with a Group.1 column of
 cats while the other version returned it with the cat names in the
 rownames. A matter of taste?


-- 
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf

Martinistr. 52
20246 Hamburg

T ++49/40/7410-58243
F ++49/40/7410-57790

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[R] print.xtable suppress my row.names

[n.via...@libero.it]

Dear list,
someone knows why the print.xtable doesnt print row.names? I dident do 
anything with the options.may depends on the size of my table???
This is my code:

\documentclass[a4paper]{article}
\title{SCHEMA DI BILANCIO PER SINGOLE AZIENDE}
\begin{document}
\maketitle
\hline
echo=F=
library(xtable)
library(plyr)
rep=Bilanci
rep$SPA-as.numeric(NA)
rep$SPP-as.numeric(NA)
rep$CE-as.numeric(NA)
rep$IN-as.numeric(NA)
rep$VA-as.numeric(NA)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,PROD=EC01+EC02+EC03)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,VP=rev(rev(PROD)*c(rev(PROD ^ 
(-1)),0)[-1]))
rep$CTOT-Bilanci$AA01+Bilanci$AA03+Bilanci$AA04
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,CIRCOL=c(NA,rev(rev(AA03)*0.
5+c(rev(AA03)*0.5,NA)[-1])[-1])+c(NA,rev(rev(AA04)*0.5+c(rev(AA04)*0.5,NA)[-1])
[-1])+c(NA,rev(rev(AL04)*0.5+c(rev(AL04)*0.5,NA)[-1])[-1]))
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IE01=(EC04-EC05)/PROD*100)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IE02=EC06/PROD*100)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IE03=EC07/PROD*100)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IE07=(EC11A+EC11C)/PROD*100)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,MOL=100+IE07-IE01-IE02-IE03)
rep$IR-as.numeric(NA)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,ROI=((PROD-EC04+EC05-EC06-
EC07+EC11C+EC11A-EC08)*c(NA,(rev(rev(CTOT)*0.5+c(rev(CTOT)*0.5,NA)[-1])^(-1))
[-1])*100))
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,RNC=((EC12)*c(NA,(rev(rev
(CTOT)*0.5+c(rev(CTOT)*0.5,NA)[-1])^(-1))[-1])*100))
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,EBIT=ROI+RNC)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,RF=((EC10-EC09)*c(NA,(rev(rev
(CTOT)*0.5+c(rev(CTOT)*0.5,NA)[-1])^(-1))[-1])*100))
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,ROE=((EC14)*c(NA,(rev(rev
(AL01)*0.5+c(rev(AL01)*0.5,NA)[-1])^(-1))[-1])*100))
rep$CC-as.numeric(NA)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IC01=(CIRCOL/PROD)*100)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IC02=c(NA,rev(rev(AA03)*0.5+c
(rev(AA03)*0.5,NA)[-1])[-1])/PROD*365)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IC03=c(NA,rev(rev(AA04)*0.5+c
(rev(AA04)*0.5,NA)[-1])[-1])/EC01*365)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IC04=c(NA,rev(rev(AL04)*0.5+c
(rev(AL04)*0.5,NA)[-1])[-1])/(EC04+EC06)*365)
rep$ES-as.numeric(NA)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IF04=(EC14+EC08)/PROD*100)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IN03=c(NA,rev(rev(AA01)*0.5+c
(rev(AA01)*0.5,NA)[-1])[-1])/PROD*100)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IN04=c(NA,rev(rev(CTOT)*0.5+c
(rev(CTOT)*0.5,NA)[-1])[-1])/PROD*100)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IN05=AA02/AA07*100)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IN07=AL01/AL06*100)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IN08=AL05/AL06*100)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IN09=IN08/IN07)
rep-ddply(rep,c(CFISCALE,RAGSOCB),transform,IN10=((EC09)*c(NA,(rev(rev
(AL05)*0.5+c(rev(AL05)*0.5,NA)[-1])^(-1))[-1])*100))
rep1=subset(rep,select=c(RAGSOCB,CFISCALE,ANNO,SPA,AA01,AA01I,AA01M,AA02,AA02B,
AA02L,AA03,AA04,AA05,AA06,AA07,SPP,AL01,AL02,AL03,AL04,AL04A,AL04B,AL05,AL05B,
AL05L,AL99,AL06,CE,EC01,EC02,EC03,EC04,EC05,EC06,EC07,EC08,EC08A,EC08B,EC09,
EC10,EC11,EC11A,EC11C,EC12,EC13,EC14,EC15,EC16,IN,VA,PROD,CTOT,CIRCOL,MOL,IR,
ROI,RNC,EBIT,RF,ROE,CC,IC01,IC02,IC03,IC04,ES,IF04,IN03,IN04,IN05,IN07,IN08,
IN09,IN10))
mynames-names(rep1)
mynames[mynames==SPA]-STATO_PATRIMONIALE_ATTIVO
mynames[mynames==AA01]-Immobilizzazioni_tecniche_nette
mynames[mynames==AA01I]-Immobilizzazioni_imm_nette
mynames[mynames==AA01M]-Immobilizzazioni_mat_nette
mynames[mynames==AA02]-Partecipazioni e crediti fin
mynames[mynames==AA02B]-Attivita fin a breve
mynames[mynames==AA02L]-Immobilizzazioni finan
mynames[mynames==AA03]-Magazzino
mynames[mynames==AA04]-Crediti commerciali
mynames[mynames==AA05]-Liquidita
mynames[mynames==AA06]-Altre attivita
mynames[mynames==AA07]-Tot attivita
mynames[mynames==SPP]-STATO PATRIMONIALE PASSIVO
mynames[mynames==AL01]-Capitale netto
mynames[mynames==AL02]-Fondo tfr
mynames[mynames==AL03]-Altri fondi
mynames[mynames==AL04]-Debiti commerciali
mynames[mynames==AL04A]-Anticipi di clienti
mynames[mynames==AL04B]-Debiti vs fornitori
mynames[mynames==AL05]-Debiti fin tot
mynames[mynames==AL05B]-Debiti fin a breve
mynames[mynames==AL05L]-Debiti fin a medio lungo
mynames[mynames==AL99]-Altre passivita
mynames[mynames==AL06]-Tot passivita
mynames[mynames==CE]-CONTO ECONOMICO
mynames[mynames==EC01]-Ricavi netti
mynames[mynames==EC02]-Produzione int capitalizzate
mynames[mynames==EC03]-Variazione scorte prod finiti
mynames[mynames==EC04]-Acquisti
mynames[mynames==EC05]-Variazioni scorte mat prime
mynames[mynames==EC06]-Costi per servizi god beni terzi
mynames[mynames==EC07]-Costo del lavoro tot
mynames[mynames==EC08]-Ammortamenti e accantonamenti
mynames[mynames==EC08A]-Ammortamenti
mynames[mynames==EC08B]-Accantonamenti e utilizzi di riserve
mynames[mynames==EC09]-Oneri fin
mynames[mynames==EC10]-Proventi fin

Re: [R] Plotting text in existing plot?

[Bert Gunter]

 Original poster wanted a simple way to do it, but when R has three
graphics systems, four OO systems, and a zillion helpful people
there's never a simple way :)

-- Rather, I'd say it has a zillion simple ways. :)
   Bert



Barry

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Re: [R] Data Frame Manipulation using function

[David Winsemius]

Really? I don't usually think of Vectorize as a performance  
enhancement, probably because my use of with a complex function then  
gets applied to 4.5 million records. I need to go out, get a cup of  
coffee, and leave it alone for about half an hour. I tried  recently  
to figure out how I can do the matrix look-up and function application  
without the Vectorize route but gave up after a couple of hours after  
realizing that I had a method that worked and I had spent way more  
time on it than just doing it would have.

Glad it helped.
David.

On Jul 9, 2010, at 11:01 AM, harsh yadav wrote:

 Hi,

 Thanks a lot.
 The Vectorize method worked and its much faster than looping through  
 the data frame.

 Regards,
 Harsh Yadav

 On Thu, Jul 8, 2010 at 11:06 PM, David Winsemius dwinsem...@comcast.net 
  wrote:

 On Jul 8, 2010, at 10:33 PM, Erik Iverson wrote:


 I have a data frame:
 id  
 url urlType
 1 1  www.yahoo.com http:// 
 www.yahoo.com1
 2 2  www.google.com/?search= http://www.google.com/? 
 search= 2
 3 3  www.google.com http:// 
 www.google.com   1
 4 4  www.yahoo.com/?query= http://www.yahoo.com/? 
 query=   2
 5 5  www.gmail.com http:// 
 www.gmail.com 1

 This is not output from ?dput, which means more work to read it in.


 Yeah it was kind of pain, but ...

 dta - read.table(textConnection(' id  
 url urlType

 1 1  www.yahoo.com http://www.yahoo.com  1
 2 2  www.google.com/?search= http://www.google.com/? 
 search= 2
 3 3  www.google.com http://www.google.com 1
 4 4  www.yahoo.com/?query= http://www.yahoo.com/? 
 query=   2
 5 5  www.gmail.com http://www.gmail.com 1') )




 Here is the definition for WHITELIST:-
 WHITELIST = [?]query=, [?]search=
 WHITELIST - unlist(trim(strsplit(trim(WHITELIST), ,)))

 What is the 'trim' function?  I do not have that defined.

 Perhaps David's answer will work for you...

 Seems to ... after I fixed my incorrect cmd-V paste of the function  
 name and guessing that trim was the one in gdata:

  require(gdata)

  checkBaseLine - function(s){
 + for (listItem in WHITELIST){
 + if(regexpr(as.character(listItem), s)[1]  -1){
 + return(TRUE)
 + }
 + }
 + return(FALSE)
 + }
 
  #Here is the definition for WHITELIST:-

 
  WHITELIST = [?]query=, [?]search=
  WHITELIST - unlist(trim(strsplit(trim(WHITELIST), ,)))
  vcheck - Vectorize(checkBaseLine)
 
  vcheck - Vectorize(checkBaseLine)
 
  dta[ dta$urlType != 1  vcheck(dta$url) , url ]
 [1] www.google.com/?search= http://www.google.com/?search= 
 www.yahoo.com/?query= 
  http://www.yahoo.com/?query=
 5 Levels: www.gmail.com http://www.gmail.com www.google.com 
 http://www.google.com 
  ... www.yahoo.com/?query= http://www.yahoo.com/?query=

 -- 
 David.


David Winsemius, MD
West Hartford, CT


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Re: [R] how can achive step by step execution of the script

[Allan Engelhardt]

Sounds like you want devAskNewPage(TRUE) or the related 
options(device.ask.default).  See help(devAskNewPage, 
package=grDevices).


Hope this helps.

Allan

On 09/07/10 13:54, vijaysheegi wrote:

Hi R Experts,
I have certain code ,i want to achive interactive execution .
For ex:
1. as part of input ,it should ask file name  or table name as input.
2.in script so many graphs i need to draw,it should wait till certain key is
pressed .
3:i am using windows R,rscriptscriptname  is not working.



Please some one help me.

Thanks Experts in advance





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Re: [R] Plotting text in existing plot?

[Tal Galili]

Possible fortune.

:)



Contact
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
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On Fri, Jul 9, 2010 at 6:22 PM, Bert Gunter gunter.ber...@gene.com wrote:


  Original poster wanted a simple way to do it, but when R has three
 graphics systems, four OO systems, and a zillion helpful people
 there's never a simple way :)

 -- Rather, I'd say it has a zillion simple ways. :)
Bert



 Barry

 --
 blog: http://geospaced.blogspot.com/
 web: http://www.maths.lancs.ac.uk/~rowlings
 web: http://www.rowlingson.com/
 twitter: http://twitter.com/geospacedman
 pics: http://www.flickr.com/photos/spacedman

 __
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[[alternative HTML version deleted]]

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Re: [R] How can i draw a graph with high and low data points

[Tal Galili]

Hi Nathaniel ,

Could you give us a simple example of your data using the
?dput
Function?

Basically you might want to draw the axis yourself, and connect the lines is
possible through using points(..., type = l)
But I'd rather try and answer this with simple example data to be sure I
understand what you mean.



Tal


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On Fri, Jul 9, 2010 at 1:39 PM, Nathaniel Saxe nathaniels...@hotmail.comwrote:


 Hi Tal, Thanks for your help.


 I've had a look at the site, and what i wanted to do was to plot X and Y
 where X is a characters and Y is numeric. The problem I'm having now is
 that
 the X axis isn't characters but just numbers from 1 onwards and when i plot
 it, the data i have is in descending order which isn't shown on the graph.

 I have this at the moment:

 plot(1:nrow(dat),dat$Mean,type=b,xaxt=n,
ylim=c(min(dat$lci),max(dat$uci)),
xlab=,ylab=HR,)

 It gives me sort of what I want. It has the Y values in descending order,
 but it doesn't give me the text on the x axis and I was also thinking of
 plotting the upper and lower confidence intervals with a line connecting
 the
 two. I can add in the upper and lower CI values separately, but I don't
 know
 how to join the two together.



 --
 View this message in context:
 http://r.789695.n4.nabble.com/How-can-i-draw-a-graph-with-high-and-low-data-points-tp2282524p2283194.html
 Sent from the R help mailing list archive at Nabble.com.

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[R] interpretation of svm models with the e1071 package

[manuel.martin]

Dear all,

after having calibrated a svm model through the svm() command of the 
e1071 package, is there a way to
i) represent the modeled relationships between the y and X variables 
(response variable vs. predictors)?

ii) rank the influence of the predictors used in the model?

Right now I am more interested in regression models, but I guess this 
would be useful for classification too.


Thank you in advance,  manuel


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Re: [R] how can achive step by step execution of the script

[Bert Gunter]

No - devAskNewPage is not what the OP asked for.


1. The following are, but probably only work when R is in interactive mode
(?interactive), e.g. in the GUI:

2. ?winDialog   ?file.choose  ?choose.files  ?select.list  ?readline
and friends
 and for keyboard: ?getGraphicsEvent


Bert Gunter
Genentech Nonclinical Biostatistics
 
 
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Allan Engelhardt
Sent: Friday, July 09, 2010 8:43 AM
To: vijaysheegi
Cc: r-help@r-project.org
Subject: Re: [R] how can achive step by step execution of the script

Sounds like you want devAskNewPage(TRUE) or the related 
options(device.ask.default).  See help(devAskNewPage, 
package=grDevices).

Hope this helps.

Allan

On 09/07/10 13:54, vijaysheegi wrote:
 Hi R Experts,
 I have certain code ,i want to achive interactive execution .
 For ex:
 1. as part of input ,it should ask file name  or table name as input.
 2.in script so many graphs i need to draw,it should wait till certain key
is
 pressed .
 3:i am using windows R,rscriptscriptname  is not working.



 Please some one help me.

 Thanks Experts in advance




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Re: [R] how to plot two histograms overlapped in the same plane coordinate

[Andrew Miles]

I'm not sure what you are trying to do.  Do you want one histogram for  
males and one for females on the same graph? If so, the simplest way  
to put two histograms together is to simply use the add parameter:


age.males=age[which(sex==M)]
age.females=age[which(sex==F)]

hist(age.males, col=blue)
hist(age.females, add=T)

The only problem is that the hist() function does not do semi- 
transparency.  I am not sure if other packages do.  The code above  
will give you a blue histogram for males, and clear histogram for  
females on top of it.  You'll probably have to manually alter the axes  
of the histogram to give the histograms for males and females the same  
break points (i.e. where one bar stops and another begins).  See ?hist  
for more information about that.


Andrew Miles
Department of Sociology
Duke University

On Jul 9, 2010, at 9:29 AM, Mao Jianfeng wrote:


Dear R-help listers,

I am new. I just want to get helps on how to plot two histograms
overlapped in the same plane coordinate. What I did is very ugly.
Could you please help me to improve it? I want to got a plot with  
semi-

transparent overlapping region. And, I want to know how to specify the
filled colors of the different histograms.

I also prefer other solutions other than ggplot2.

Many thanks to you.


What I have done:

library(ggplot2)

age-c(rnorm(100, 1.5, 1), rnorm(100, 5, 1))
sex-c(rep(F,100), rep(M, 100))
mydata-cbind(age, sex)
mydata-as.data.frame(mydata)
head(mydata)


qplot(age, data=mydata, geom=histogram, fill=sex, xlab=age,
ylab=count, alpha=I(0.5))


Best,


Mao J-F

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Re: [R] how to plot two histograms overlapped in the same plane coordinate

[Frank E Harrell Jr]

Empirical CDFs are much better for this purpose, and allow 
superpositioning (see e.g. the Ecdf function in the Hmisc package). 
Otherwise look at histbackback in Hmisc.


Frank

On 07/09/2010 11:40 AM, Andrew Miles wrote:



I'm not sure what you are trying to do. Do you want one histogram for
males and one for females on the same graph? If so, the simplest way to
put two histograms together is to simply use the add parameter:

age.males=age[which(sex==M)]
age.females=age[which(sex==F)]

hist(age.males, col=blue)
hist(age.females, add=T)

The only problem is that the hist() function does not do
semi-transparency. I am not sure if other packages do. The code above
will give you a blue histogram for males, and clear histogram for
females on top of it. You'll probably have to manually alter the axes of
the histogram to give the histograms for males and females the same
break points (i.e. where one bar stops and another begins). See ?hist
for more information about that.

Andrew Miles
Department of Sociology
Duke University

On Jul 9, 2010, at 9:29 AM, Mao Jianfeng wrote:


Dear R-help listers,

I am new. I just want to get helps on how to plot two histograms
overlapped in the same plane coordinate. What I did is very ugly.
Could you please help me to improve it? I want to got a plot with semi-
transparent overlapping region. And, I want to know how to specify the
filled colors of the different histograms.

I also prefer other solutions other than ggplot2.

Many thanks to you.


What I have done:

library(ggplot2)

age-c(rnorm(100, 1.5, 1), rnorm(100, 5, 1))
sex-c(rep(F,100), rep(M, 100))
mydata-cbind(age, sex)
mydata-as.data.frame(mydata)
head(mydata)


qplot(age, data=mydata, geom=histogram, fill=sex, xlab=age,
ylab=count, alpha=I(0.5))


Best,


Mao J-F

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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.


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--
Frank E Harrell Jr   Professor and ChairmanSchool of Medicine
 Department of Biostatistics   Vanderbilt University

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Re: [R] how to plot two histograms overlapped in the same plane coordinate

[Barry Rowlingson]

On Fri, Jul 9, 2010 at 2:29 PM, Mao Jianfeng jianfeng@gmail.com wrote:
 Dear R-help listers,

 I am new. I just want to get helps on how to plot two histograms
 overlapped in the same plane coordinate. What I did is very ugly.
 Could you please help me to improve it? I want to got a plot with semi-
 transparent overlapping region. And, I want to know how to specify the
 filled colors of the different histograms.

 I also prefer other solutions other than ggplot2.

 Many thanks to you.


 What I have done:

 library(ggplot2)

 age-c(rnorm(100, 1.5, 1), rnorm(100, 5, 1))
 sex-c(rep(F,100), rep(M, 100))
 mydata-cbind(age, sex)
 mydata-as.data.frame(mydata)
 head(mydata)


 Tried setting xlim with hist?

 par(mfrow=c(2,1))
 hist(age[sex==M],xlim=range(age))
 hist(age[sex==F],xlim=range(age))

Barry

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Re: [R] Query about using timestamps returned by SQL as 'factor' forsplit

[Matthew Dowle]

Hi Ted,

Well since you mentioned data.table (!) ...

If risk_input is a data.table consisting of 3 columns (m_id, sale_date, 
return_date) where the dates
are of class IDate (recently added to data.table by Tom) then try :

   risk_input[, fitdistr(return_date-sale_date,normal), by=list(m_id, 
year(sale_date), week(sale_date))]

Notice that the 'by' can contain expressions of columns, and lets you group 
by more than one expression.
You don't have to repeat the 'group by' expressions in the select, as you 
would do in SQL. data.table returns
those group columns automatically in the result, alongside the result of the 
j expression applied to each group.

If you need to aggregate by m_id, year and month rather than week another 
way is :

   risk_input[, fitdistr(return_date-sale_date,normal), by=list(m_id, 
round(sale_date,month))]

plyr and sqldf can do this task too by the way, and I'd highly recommend you 
take a look at those packages.

There are also many excellent datetime classes around which you could also 
consider.

The reason we need IDate in data.table is because data.table uses radix 
sorting, see ?sort.list. That is ultra fast for
integers. Again radix is something Tom added to data.table. The radix 
algorithm (see wikipedia) is specifically
designed to sort integers only. We would use Date, but that is stored as 
numeric. IDate is the same as Date
but stored as integer.

HTH,
Matthew


Ted Byers r.ted.by...@gmail.com wrote in message 
news:aanlktinchf3tfzkndcwolrwsxekgpfpjes3f8m5tq...@mail.gmail.com...
I have a simple query as follows:

 SELECT
 m_id,sale_date,YEAR(sale_date),WEEK(sale_date),return_type,DATEDIFF(return_date,sale_date)
 AS elapsed_time FROM risk_input

 I can get, and view, all the data that that query returns.  The question 
 is,
 sale_date is a timestamp, and I need to call split to group this data by
 m_id and the week in which the sale occurred.  Obviously, I would normally
 need both YEAR and WEEK so that data from April this year is not combined
 with that from last year (the system is non-autonomous).  And then I need 
 to
 use lapply to apply fitdistr to each subsample.

 Obviously, I can handle all this data in either a data.frame or in a
 data.table.

 There are two aspects of the question.

 1) Is there a function (or package) that will let me group (or regroup) 
 time
 series data into the week in which the data apply, properly taking into
 account the year that applies, in a single call passing sale_date as the
 argument?  If I can, then I can reduce the amount of data I draw from my
 MySQL server and the computational load it bears.

 2) The example provided for split splits only according to a single 
 variable
 (*g - airquality$Month;l - split(airquality, g)*).  How would that 
 example
 be changed if there were two or more columns in the data.frame that are
 needed to define the groups?  I.E. in my example, I'd need to group by 
 m_id,
 and the year and week values that can be computed from sale_date.

 Thanks

 Ted

 [[alternative HTML version deleted]]


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Re: [R] Calling Gnuplot from R

[Greg Snow]

There is a basic interface between R and gnuplot in the TeachingDemos package, 
see ?gp.open

Not much interest has been shown in this, so it is still pretty alpha level, 
but you can send your R data to gnuplot and have it create a basic plot.

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of Christopher Desjardins
 Sent: Thursday, July 08, 2010 9:22 AM
 To: r-help@r-project.org
 Subject: [R] Calling Gnuplot from R
 
 Hi,
 I am wondering if there is a way to call Gnuplot from R and/or if
 anyone can
 recommend a package on CRAN capable of doing this?
 Thanks,
 Chris
 PS - Please cc me on the response.
 
   [[alternative HTML version deleted]]
 
 __
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 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
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[R] Ctree Question

[Steve_Friedman]

Hello,

I've been using ctree and have developed a 55 node - 28 terminal solution.
As can be imagined, the plot is difficult to travel down each of the major
branches.

I've read the help files for ctree I saw where terminal nodes can be color
coded.

 plot(airct, type = simple)
 plot(airct, terminal_panel = node_boxplot(airct, col = blue,  + fill =
hsv(2/3, 0.5, 1)))

Here is my question:

Since my model has 55 nodes and 28 terminal nodes,(ie many branches)  is it
feasible to color code the each of the major branches and track the paths
down the decision tree?

 R 2.11.1, Windox XP


Thanks
Steve



Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034

steve_fried...@nps.gov
Office (305) 224 - 4282
Fax (305) 224 - 4147

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

Re: [R] Plotting text in existing plot?

[Greg Snow]

Others pointed you to the text function for the base graphics system.  But if 
what you want to do is use text, but have a simple way of specifying the center 
of the plot without computing the user coordinates by hand of the center, then 
look at the grconvertX and grconvertY functions.

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
 project.org] On Behalf Of Ralf B
 Sent: Friday, July 09, 2010 3:52 AM
 To: r-help@r-project.org
 Subject: [R] Plotting text in existing plot?
 
 I would like to plot some text in a existing plot graph. Is there a
 very simple way to do that. It does not need to be pretty at all (just
 maybe a way to center it or define a position within the plot). ( ? )
 
 Ralf
 
 __
 R-help@r-project.org mailing list
 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
 and provide commented, minimal, self-contained, reproducible code.

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Re: [R] how to plot two histograms overlapped in the same planecoordinate

[Bert Gunter]

Don't do this. The overlapping will confuse.

Plot them in a lattice display with one group above the other on the same
horizontal scale. See ?histogram.

Bert Gunter
Genentech Nonclinical Biostatistics
 
 

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Barry Rowlingson
Sent: Friday, July 09, 2010 9:47 AM
To: Mao Jianfeng
Cc: r-help@r-project.org
Subject: Re: [R] how to plot two histograms overlapped in the same
planecoordinate

On Fri, Jul 9, 2010 at 2:29 PM, Mao Jianfeng jianfeng@gmail.com wrote:
 Dear R-help listers,

 I am new. I just want to get helps on how to plot two histograms
 overlapped in the same plane coordinate. What I did is very ugly.
 Could you please help me to improve it? I want to got a plot with semi-
 transparent overlapping region. And, I want to know how to specify the
 filled colors of the different histograms.

 I also prefer other solutions other than ggplot2.

 Many thanks to you.


 What I have done:

 library(ggplot2)

 age-c(rnorm(100, 1.5, 1), rnorm(100, 5, 1))
 sex-c(rep(F,100), rep(M, 100))
 mydata-cbind(age, sex)
 mydata-as.data.frame(mydata)
 head(mydata)


 Tried setting xlim with hist?

 par(mfrow=c(2,1))
 hist(age[sex==M],xlim=range(age))
 hist(age[sex==F],xlim=range(age))

Barry

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

[R] nls error regarding numerics vs logicals

[Jim Bouldin]

I am trying to perform an nls for a valid negative exponential function:

zz=nls(y~constant+a.est*2.7183^(b.est*x),start=list(constant=4.0,a.est=-4,b.est
= -.005),trace=T)

and am getting a number of different error messages, the most problematic
of  which is Error in nls(ring.area ~ constant + a.est * 2.7183^(b.est *
ba.beg), start = list(constant = 4,  : 
  REAL() can only be applied to a 'numeric', not a 'logical'

I can't see where there are any logicals in this equation to cause this
problem.  Any help appreciated. Thank you.

Jim Bouldin

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Re: [R] R crashes with large vectors

[Tengfei Yin]

Hi Jeremie,

Maybe you can take a look at the bigmemory  package.
 If you have multi core or have access to clusters, you may want to use any
parallel computing strategy.

For plotting of large data, if you are using basic R graphics, first try to
use line instead of using 'point',

if this still doesn't working, you may want to try an alternative way by
using qtinterface (R-forge project), install qtbase and qtpaint, they are
still under development, but painting in QT interface is a lot faster for
large data set.

Best

Tengfei

On Fri, Jul 9, 2010 at 2:42 AM, Jeremie Smaga jere...@4ecap.com wrote:

 Good afternoon,

 I have been experiencing a lot of crashes working with large vectors in R.

 Specifically, I am using XTS of length of minimum 120k elements.

 My problem is that I cannot display the vector (otherwise R crashes), I
 cannot plot it either (otherwise R crashes). That could be solved by
 reducing the amount of points.

 However, I have been performing some statistical opreations on is and even
 sd(myXTS) crashes R.

 By crashes, I mean shuts down without any warning whatsoever.

 I use R 2.11.1 (64).

 Has anyone had the same kind of problem?

 Can we solve this?

 Best,


 --
 Jeremie

[[alternative HTML version deleted]]

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




-- 
Tengfei Yin
MCDB PhD student
1620 Howe Hall, 2274,
Iowa State University
Ames, IA,50011-2274
Homepage: www.tengfei.name

[[alternative HTML version deleted]]

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Re: [R] nls error regarding numerics vs logicals

[Duncan Murdoch]

On 09/07/2010 1:51 PM, Jim Bouldin wrote:

I am trying to perform an nls for a valid negative exponential function:

zz=nls(y~constant+a.est*2.7183^(b.est*x),start=list(constant=4.0,a.est=-4,b.est
= -.005),trace=T)

and am getting a number of different error messages, the most problematic
of  which is Error in nls(ring.area ~ constant + a.est * 2.7183^(b.est *
ba.beg), start = list(constant = 4,  : 
  REAL() can only be applied to a 'numeric', not a 'logical'


I can't see where there are any logicals in this equation to cause this
problem.  Any help appreciated. Thank you.


1.  The expression you gave us is clearly not the one that produced the 
error:  it involved ring.area and ba.beg.


2.  You don't tell us what x and y are, so we can't reproduce anything.

We can't help you if you don't tell us what the problem is.

Duncan Murdoch

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[R] select columns from vector of column names

[Jonathan Flowers]

Hi

I want to extract columns from a data frame using a vector with the desired
column names.

This short example uses the select argument in the subset function to
accomplish what I am trying to do.  Is there a better solution?

#names of desired columns
colnames - c(col1,col3)

#my data
data - data.frame(col1=c(1,2,3),col2=c(A,B,C),col3=c(4,5,6))

fun - function(colname,dframe){
nframe - subset(dframe,select=colname)
vec - nframe[,1]
return(vec)
}

fun(colnames[1],data)
fun(colnames[2],data)

[[alternative HTML version deleted]]

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Re: [R] Calling Gnuplot from R

[Tal Galili]

Dear Greg,

I keep on being amazed at the abundance of functions you have packed into
the TeachingDemos package - thank you!

Tal



Contact
Details:---
Contact me: tal.gal...@gmail.com |  972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--




On Fri, Jul 9, 2010 at 8:18 PM, Greg Snow greg.s...@imail.org wrote:

 There is a basic interface between R and gnuplot in the TeachingDemos
 package, see ?gp.open

 Not much interest has been shown in this, so it is still pretty alpha
 level, but you can send your R data to gnuplot and have it create a basic
 plot.

 --
 Gregory (Greg) L. Snow Ph.D.
 Statistical Data Center
 Intermountain Healthcare
 greg.s...@imail.org
 801.408.8111


  -Original Message-
  From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
  project.org] On Behalf Of Christopher Desjardins
  Sent: Thursday, July 08, 2010 9:22 AM
  To: r-help@r-project.org
  Subject: [R] Calling Gnuplot from R
 
  Hi,
  I am wondering if there is a way to call Gnuplot from R and/or if
  anyone can
  recommend a package on CRAN capable of doing this?
  Thanks,
  Chris
  PS - Please cc me on the response.
 
[[alternative HTML version deleted]]
 
  __
  R-help@r-project.org mailing list
  https://stat.ethz.ch/mailman/listinfo/r-help
  PLEASE do read the posting guide http://www.R-project.org/posting-
  guide.html
  and provide commented, minimal, self-contained, reproducible code.

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


[[alternative HTML version deleted]]

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

[R] select columns from dataframe

[Jonathan Flowers]

Hi,

I would like to extract columns from a dataframe using a vector of desired
column names.

The following working example uses the select argument in the subset
function to accomplish what I am trying to do.  Is there a better solution?

Thanks.

#my data
data - data.frame(col1=c(1,2,3),col2=c(A,B,C),col3=c(4,5,6))

#names of desired columns
colnames - c(col1,col3)

fun - function(colname,dframe){
nframe - subset(dframe,select=colname)
vec - nframe[,1]
return(vec)
}

fun(colnames[1],data)
fun(colnames[2],data)

[[alternative HTML version deleted]]

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Re: [R] Ctree Question

[Horace Tso]

Steve, I'm not sure if your task could be accomplished with a ready-made 
function in party. But, if you could manage to convert your tree structure to a 
dendrogram, then it's straightforward using dendrapply. In fact, there is an 
example in dendrapply help page showing how leaves are colored. 

?dendrapply

HTH.

H

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of steve_fried...@nps.gov
Sent: Friday, July 09, 2010 10:22 AM
To: r-help@r-project.org
Subject: [R] Ctree Question


Hello,

I've been using ctree and have developed a 55 node - 28 terminal solution.
As can be imagined, the plot is difficult to travel down each of the major
branches.

I've read the help files for ctree I saw where terminal nodes can be color
coded.

 plot(airct, type = simple)
 plot(airct, terminal_panel = node_boxplot(airct, col = blue,  + fill =
hsv(2/3, 0.5, 1)))

Here is my question:

Since my model has 55 nodes and 28 terminal nodes,(ie many branches)  is it
feasible to color code the each of the major branches and track the paths
down the decision tree?

 R 2.11.1, Windox XP


Thanks
Steve



Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034

steve_fried...@nps.gov
Office (305) 224 - 4282
Fax (305) 224 - 4147

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Re: [R] select columns from vector of column names

[Horace Tso]

How about

data[,colnames(data)%in%colnames] 


-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of Jonathan Flowers
Sent: Friday, July 09, 2010 11:27 AM
To: r-help@r-project.org
Subject: [R] select columns from vector of column names

Hi

I want to extract columns from a data frame using a vector with the desired
column names.

This short example uses the select argument in the subset function to
accomplish what I am trying to do.  Is there a better solution?

#names of desired columns
colnames - c(col1,col3)

#my data
data - data.frame(col1=c(1,2,3),col2=c(A,B,C),col3=c(4,5,6))

fun - function(colname,dframe){
nframe - subset(dframe,select=colname)
vec - nframe[,1]
return(vec)
}

fun(colnames[1],data)
fun(colnames[2],data)

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[R] Not nice behaviour of nlminb (windows 32 bit, version 2.11.1)

[Matthew Killeya]

 nlminb( obj = function(x) x, start=1, lower=-Inf, upper=Inf )
$par
[1] 0

$objective
[1] 0

$convergence
[1] 0

$message
[1] absolute function convergence (6)

$iterations
[1] 1

$evaluations
function gradient
   22

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Re: [R] split with list

[Ista Zahn]

n.via...@libero.it n.via...@libero.it writes:
Hi,

 Dear List I would like to ask you something concenting a better print of the 
 R output:
 I have a bit data frame which has the following structure:
 CFISCALE  RAGSOCBANNO   VAR1VAR2.
 9853312 astra 2005   6
45

 9853312 astra 2006  78
   45


 9853312 astra 2007   55   
76


 9653421  geox 2005   35   
   89



 9653421 geox 200624   
 33

 9653421  geox 2007   54   
55


 The first thing I did is to split my data frame for CFISCALE. The result is 
 that R has transformed my data frame into a list. The second step was to 
 transpose each element of my list. 
 repo=split(rep,rep$CFISCALE)
 repor=lapply(repo,function(x){
 t(x)})


 When I print my list the format is the following 
 $9853312
1   2  
   3

 CFISCALE9853312 9853312 9853312   

 RAGSOCBastra   astraastra

 ANNO   20052006  
 2007

 VAR1   6 78   
55

 VAR2   4545   
   76


So far so good.


 There is a way to remove the  first row I mean 1, 2 , 3 and to have just one 
 CFISCALE and RAGSOCB???
 For the second problem I tried to use unique but it seems that it
 doesnt work for list. So what I would like to get is: 

Well I'm not sure what the first problem is. But if I understand the
second problem, you can do something like this:

dup.null - function(mat) {
  mat[CFISCALE, duplicated(mat[CFISCALE,])] - NA
  mat[RAGSOCB, duplicated(mat[RAGSOCB,])] - NA
  return(mat)
  }

repor - lapply(repor, dup.null)


Best,
Ista

 $9853312

   
 


 CFISCALE9853312 


 RAGSOCBastra  
 ANNO   20052006  
 2007

 VAR1   6 78   
55

 VAR2   4545   
   76


 This is because I next run xtable on my list in order to get a table in 
 Latex, which I woud like to be in a nice format.
 Thanks a lot for your attention!





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[R] eval and assign in loop problem

[S.Nicholas]

deaR useRs,

I am trying to assign different values to different objects in a for loop.
The following is a toy example of the part that has been giving me a hard
time.

The first for loop generates four objects, b0, b1, b2, b3 with random
numbers.
And, the second for loop is equivalent to
b1 = b0
b2 = b1
b3 = b2
b4 = b3

But, when I run this code, the result is equivalent to
b1 = b0
b2 = b0
b3 = b0
b4 = b0

So, the increment does not seem to be properly working for the second part
of the assign function.
Why would this be?

for (i in 0:3)
   {
   r = runif(1)
   assign(paste('b',i,sep=''),r)
   }


for (i in 1:4)
   {

assign(paste('b',i,sep=''),eval(parse(text=paste('b',i-1,sep=''
   }

Thank you.

Nic

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[R] July BaselR Meeting

[Sarah Lewis]

BaselR meeting July 2010

Thank you to everyone who supported our inaugural BaselR meeting on April 
28th; we were fortunate to have three excellent presentations which prompted 
lively discussion - our thanks go to Andreas Krause, Yann Abraham and Charles 
Roosen for their presentations, details of which are available here

The next BaselR meeting is Wednesday 28th July 2010

Venue:
transBARent
Viaduktstrasse 3
CH-4051 Basel
Tel. 061 222 21 31
Fax 061 222 21 32
i...@transbarent.ch

http://transbarent.sv-group.ch/de.html  

Agenda:

* Introduction - Charles Roosen, Mango Solutions AG
* Desktop Publishing with Sweave - Andrew Ellis, ETH Zurich
* Professional Reporting with RExcel - Dominik Locher, THETA AG
* R Generator Tool for Google Motion Charts - Sebastian Pérez Saaibi, ETH Zurich

If you would like to join the BaselR mailing list and receive details of all 
BaselR meetings please email us at bas...@mango-solutions.com 

What is BaselR?

Similar to the well-known LondonR, this informal meeting is intended to serve 
as a platform for all local (and regional) R users to present and exchange 
their experiences and ideas around the usage of R. 

Mango Solutions aims to host such meetings about every quarter. A typical 
BaselR meeting will consist of 3-4 talks of about 20-25 min to give plenty of 
room for sharing your R experiences, discussions and exchange of ideas.

How to contribute? 

We are always looking for volunteers to present at subsequent meetings. If you 
think you have something interesting to present or know of someone who has, 
please contact us.

For ideas on presentations, take a look at previous presentations given at 
LondonR (here).

For more information about Mango Solutions please contact us or visit our 
website www.mango-solutions.ch

Sarah Lewis

Hadley Wickham, Creator of ggplot2 - first time teaching in the UK. 1st - 2nd  
November 2010. 
To book your seat please go to http://mango-solutions.com/news.html 

T: +44 (0)1249 767700 Ext: 200
F: +44 (0)1249 767707
M: +44 (0)7746 224226
www.mango-solutions.com
Unit 2 Greenways Business Park 
Bellinger Close
Chippenham
Wilts
SN15 1BN
UK 

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[R] syntax for highlighting table rows and columns using R plugin for spss

[Anderson, Chris]

I am using  the R plug-in for spss 18, and would like to know if there is any R 
code that will highlight and bold rows (particularly the Totals row) within 
an spss table. The current option that spss has is to use a python plug-in 
which doesn't seem to work on my windows 7 machine.

Chris Anderson
Data Analyst
Medical Affairs
wk: 925-677-4870
cell: 707-315-8486
Fax:925-677-4670


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[R] Compress string memCompress/Decompress

[Erik Wright]

Hello,

I would like to compress a long string (character vector), store the compressed 
string in the text field of a SQLite database (using RSQLite), and then load 
the text back into memory and decompress it back into the the original string.  
My character vector can be compressed considerably using standard gzip/bzip2 
compression.  In theory it should be much faster for me to compress/decompress 
a long string than to write the whole string to the hard drive and then read it 
back (not to mention the saved hard drive space).

I have tried accomplishing this task using memCompress() and memDecompress() 
without success.  It seems memCompress can only convert a character vector to 
raw type which cannot be treated as a string.  Does anyone have ideas on how I 
can go about doing this, especially using the standard base packages?

Thanks!,
Erik


 sessionInfo()
R version 2.11.0 (2010-04-22) 
x86_64-apple-darwin9.8.0 

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

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

loaded via a namespace (and not attached):
[1] tools_2.11.0

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[R] Help finding valid permutations

[Michael D]

So I have an array A of length n with multiple attributes and for a
selected attribute y I need to list all valid permutations where a
valid permutation is of the form:
A[1,y] != A[n,y]
A[i,y] != A[i+1,y]

I've tried using the 'combinat' package, but with the vector lengths
I'm using the permn function fails with the vector(list, gamma(n +
1)) assignment.


For example:
13 cards from a standard 52 card deck. attributes facevalue and suit.
A = (9C, 9H, 9D, 8C, 7C, 6C, 6S, 5C, 4C, 4H, 3C, 2C, 2D)

And I want to find all possible arrangements where the facevalue of
each consecutive card is different but the orders are unique because
of the suit.

Also, I need to consider the case where the suit of each consecutive
case is different but the orders are unique because of the facevalue.

For A above, note that there are 8 clubs out of 13 cards so 3 of the
clubs must be dropped from the draw an all valid of ten of the cards
must be found (5 same suit, 5 off suit)


I hope this is clear. Thanks.

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