Re: [R] Basic misunderstanding, or problem with my installation?

[David Parkhurst]

Thanks to the three people who saw what I missed.  I typed my
code in Libre Office as "<" followed by "-", and that program
converted those two characters into a single left arrow symbol.
I copied the commands from Libre into R without noticing that
that had happened.  Wierd.

On 12/31/2013 7:54 PM, Sarah Goslee wrote:

Hi David,

Your code is showing up here with an arrow symbols. If it's an actual
cut and paste, that's your problem: assignment in R is the two-character
<- and not an arrow symbol.Â

Otherwise your code looks fine.

Sarah

On Tuesday, December 31, 2013, David Parkhurst wrote:

I've just uninstalled and then reinstalled R on my windows 7 machine.
To test my understanding of data frames, I'm trying the following code.
(I plan to do other things with it, if it would only work.)
Here's the code, which seems pretty basic to me:
ls()
nums � c(1,2,3,4,5)
ltrs � c(“a�,�b�,�c�,�d�,�e�)
df1 � data.frame(nums,ltrs)

Here's what happens when I try to run it:
 > ls()
character(0)
 > nums � c(1,2,3,4,5)
Error: unexpected input in "nums \"
 > ltrs � c(“a�,�b�,�c�,�d�,�e�)
Error: unexpected input in "ltrs \"
 > df1 � data.frame(nums,ltrs)
Error: unexpected input in "df1 \"
 >

Am I really misunderstanding the basics, or is there something
wrong with my installation?
David


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.



--
Sarah Goslee
http://www.stringpage.com
http://www.sarahgoslee.com
http://www.functionaldiversity.org


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

Re: [R] Basic misunderstanding, or problem with my installation?

[Barry Rowlingson]

On Wed, Jan 1, 2014 at 4:36 AM, David Parkhurst  wrote:
> Thanks to the three people who saw what I missed.  I typed my
> code in Libre Office as "<" followed by "-", and that program
> converted those two characters into a single left arrow symbol.
> I copied the commands from Libre into R without noticing that
> that had happened.  Wierd.

 You should probably install "R Studio" and use that! You'll get
syntax highlighting, bracket matching, and no magic conversion of
arrows!

www.rstudio.com

Barry

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

Re: [R] howto join matrices produced by rcorr()

[Frank Harrell]

This is an unstable process.  I suggest using the bootstrap to get a
confidence interval for the rank of each correlation coefficient among all
non-diagonal correlations.



-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context: 
http://r.789695.n4.nabble.com/howto-join-matrices-produced-by-rcorr-tp4682867p4682932.html
Sent from the R help mailing list archive at Nabble.com.

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

[R] expression in xlab for a plot

[Andras Farkas]

Dear All,

Happy new year!

wonder if you could help with the following:

we have:

hist(runif(1000,0,100),xlab=expression(AUC[0 - 24]~ (xyz)),ylab="Frequency")


the plan is to have part of the xlab expression change dynamically, 
specifically the values of 0 and 24 should be able to update 'automatically" , 
based on vectors low and high, so if we had:
low <-24
high <-48
then we should have this:

hist(runif(1000,0,100),xlab=expression(AUC[24 - 48]~ (xyz)),ylab="Frequency")


appreciate your insights,

Andras

[[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.

Re: [R] expression in xlab for a plot

[arun]

Hi,May be this helps:


  hist(runif(1000,0,100),xlab=bquote(AUC[.(low) - .(high)]~ 
(xyz)),ylab="Frequency")
A.K.


On Wednesday, January 1, 2014 10:30 AM, Andras Farkas  
wrote:
Dear All,

Happy new year!

wonder if you could help with the following:

we have:

hist(runif(1000,0,100),xlab=expression(AUC[0 - 24]~ (xyz)),ylab="Frequency")


the plan is to have part of the xlab expression change dynamically, 
specifically the values of 0 and 24 should be able to update 'automatically" , 
based on vectors low and high, so if we had:
low <-24
high <-48
then we should have this:

hist(runif(1000,0,100),xlab=expression(AUC[24 - 48]~ (xyz)),ylab="Frequency")


appreciate your insights,

Andras

    [[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.

Re: [R] expression in xlab for a plot

[arun]

HI,
Another way would be to use ?substitute()
hist(runif(1000,0,100),xlab=substitute(expression(AUC[low - high]~ 
(xyz)),list(low=low,high=high)),ylab="Frequency")
A.K.




On Wednesday, January 1, 2014 10:36 AM, arun  wrote:
Hi,May be this helps:


  hist(runif(1000,0,100),xlab=bquote(AUC[.(low) - .(high)]~ 
(xyz)),ylab="Frequency")
A.K.



On Wednesday, January 1, 2014 10:30 AM, Andras Farkas  
wrote:
Dear All,

Happy new year!

wonder if you could help with the following:

we have:

hist(runif(1000,0,100),xlab=expression(AUC[0 - 24]~ (xyz)),ylab="Frequency")


the plan is to have part of the xlab expression change dynamically, 
specifically the values of 0 and 24 should be able to update 'automatically" , 
based on vectors low and high, so if we had:
low <-24
high <-48
then we should have this:

hist(runif(1000,0,100),xlab=expression(AUC[24 - 48]~ (xyz)),ylab="Frequency")


appreciate your insights,

Andras

    [[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.

[R] mgcv - markeov random field option

[Helena Baptista]

Dear R-users,

Happy new year to all!

I have been using the mgcv package, and I have run some models using the
option mrf, for saptial data. But I have found quite hard to interpret the
results. I could not find a lot of documentation on that, examples and so
on, so I was wondering if anyone can help me found those.

Thanks a lot.
Cheers
-- 
Helena Baptista

[[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] to modify a dataframe

[Arnaud Michel]

Dear All,

From the dataframe df1

df1 <-
structure(list(Nom = structure(1:9, .Label = c("A1", "A2", "A3",
"B1", "B2", "C1", "C2", "C3", "C4"), class = "factor"), Pays1 = c(1,
1, 0, 0, 1, 0, 0, 0, 0), Pays2 = c(0, 0, 0, 1, 1, 0, 1, 0, 1),
Pays3 = c(0, 0, 0, 0, 1, 0, 0, 0, 0), Pays4 = c(1, 0, 0,
0, 0, 0, 1, 0, 1), Pays5 = c(1, 1, 0, 0, 0, 0, 0, 0, 0)), .Names = c("Nom",
"Pays1", "Pays2", "Pays3", "Pays4", "Pays5"), row.names = c(1L,
3L, 4L, 2L, 5L, 6L, 7L, 8L, 9L), class = "data.frame")


I look for a way to build the new dataframe df2
 
df2 <-

structure(list(Nom = structure(1:9, .Label = c("A1", "A2", "A3",
"B1", "B2", "C1", "C2", "C3", "C4"), class = "factor"), Pays1 = c(1,
1, 1, 1, 1, 0, 0, 0, 0), Pays2 = c(0, 0, 0, 1, 1, 1, 1, 1, 1),
Pays3 = c(0, 0, 0, 1, 1, 0, 0, 0, 0), Pays4 = c(1, 1, 1,
0, 0, 1, 1, 1, 1), Pays5 = c(1, 1, 1, 0, 0, 0, 0, 0, 0)), .Names = c("Nom",
"Pays1", "Pays2", "Pays3", "Pays4", "Pays5"), row.names = c(NA,
-9L), class = "data.frame")

The purpose is to transform df1 it df2 by giving for every group of lines A, B 
and C the value 1 if there is at least a value equal to 1 or a value 0 if there 
is no value equal to 1

Thanks for your helps

--
Michel ARNAUD
Chargé de mission auprès du DRH
DGDRD-Drh - TA 174/04
Av Agropolis 34398 Montpellier cedex 5
tel : 04.67.61.75.38
fax : 04.67.61.57.87
port: 06.47.43.55.31

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

Re: [R] to modify a dataframe

[Bert Gunter]

1. Thank you for the clear reproducible example. This made it easy to
see what you wanted and provide an answer. Hopefully a correct one!

2. Many ways to do this. Here's one, but others may be better.

Step1: First greate a grouping factor for Nom to group the separate
row labels into the logical groups you have specified:

grp <- factor(substring(df1$Nom,1,1))

Note that you may need to use regular expressions or some other method
to do this if your naming system is more complex than you have shown.

Step 2: Create your new structure:

df2 <- df1
df2[,-1]<-lapply(df1[-1],function(x)ave(x,grp,FUN=max))

HTH.

Cheers,
Bert

Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374

"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
H. Gilbert Welch




On Wed, Jan 1, 2014 at 8:55 AM, Arnaud Michel  wrote:
> Dear All,
>
> From the dataframe df1
>
> df1 <-
> structure(list(Nom = structure(1:9, .Label = c("A1", "A2", "A3",
> "B1", "B2", "C1", "C2", "C3", "C4"), class = "factor"), Pays1 = c(1,
> 1, 0, 0, 1, 0, 0, 0, 0), Pays2 = c(0, 0, 0, 1, 1, 0, 1, 0, 1),
> Pays3 = c(0, 0, 0, 0, 1, 0, 0, 0, 0), Pays4 = c(1, 0, 0,
> 0, 0, 0, 1, 0, 1), Pays5 = c(1, 1, 0, 0, 0, 0, 0, 0, 0)), .Names =
> c("Nom",
> "Pays1", "Pays2", "Pays3", "Pays4", "Pays5"), row.names = c(1L,
> 3L, 4L, 2L, 5L, 6L, 7L, 8L, 9L), class = "data.frame")
>
>
> I look for a way to build the new dataframe df2
>  df2 <-
> structure(list(Nom = structure(1:9, .Label = c("A1", "A2", "A3",
> "B1", "B2", "C1", "C2", "C3", "C4"), class = "factor"), Pays1 = c(1,
> 1, 1, 1, 1, 0, 0, 0, 0), Pays2 = c(0, 0, 0, 1, 1, 1, 1, 1, 1),
> Pays3 = c(0, 0, 0, 1, 1, 0, 0, 0, 0), Pays4 = c(1, 1, 1,
> 0, 0, 1, 1, 1, 1), Pays5 = c(1, 1, 1, 0, 0, 0, 0, 0, 0)), .Names =
> c("Nom",
> "Pays1", "Pays2", "Pays3", "Pays4", "Pays5"), row.names = c(NA,
> -9L), class = "data.frame")
>
> The purpose is to transform df1 it df2 by giving for every group of lines A,
> B and C the value 1 if there is at least a value equal to 1 or a value 0 if
> there is no value equal to 1
>
> Thanks for your helps
>
> --
> Michel ARNAUD
> Chargé de mission auprès du DRH
> DGDRD-Drh - TA 174/04
> Av Agropolis 34398 Montpellier cedex 5
> tel : 04.67.61.75.38
> fax : 04.67.61.57.87
> port: 06.47.43.55.31
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

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

Re: [R] to modify a dataframe

[Rui Barradas]

Hello,

Here's one way.



lst1 <- lapply(split(df1, gsub("[0-9]", "", df1$Nom)), function(x){
x[, -1] <- lapply(x[, -1], function(y){
z <- if(any(y == 1)) 1 else 0
rep(z, length(y))
})
x
})
df3 <- do.call(rbind, lst1)
rownames(df3) <- NULL

identical(df2, df3)  # TRUE


Hope this helps,

Rui Barradas

Em 01-01-2014 16:55, Arnaud Michel escreveu:

Dear All,

 From the dataframe df1

df1 <-
structure(list(Nom = structure(1:9, .Label = c("A1", "A2", "A3",
"B1", "B2", "C1", "C2", "C3", "C4"), class = "factor"), Pays1 = c(1,
1, 0, 0, 1, 0, 0, 0, 0), Pays2 = c(0, 0, 0, 1, 1, 0, 1, 0, 1),
 Pays3 = c(0, 0, 0, 0, 1, 0, 0, 0, 0), Pays4 = c(1, 0, 0,
 0, 0, 0, 1, 0, 1), Pays5 = c(1, 1, 0, 0, 0, 0, 0, 0, 0)), .Names =
c("Nom",
"Pays1", "Pays2", "Pays3", "Pays4", "Pays5"), row.names = c(1L,
3L, 4L, 2L, 5L, 6L, 7L, 8L, 9L), class = "data.frame")


I look for a way to build the new dataframe df2

df2 <-
structure(list(Nom = structure(1:9, .Label = c("A1", "A2", "A3",
"B1", "B2", "C1", "C2", "C3", "C4"), class = "factor"), Pays1 = c(1,
1, 1, 1, 1, 0, 0, 0, 0), Pays2 = c(0, 0, 0, 1, 1, 1, 1, 1, 1),
 Pays3 = c(0, 0, 0, 1, 1, 0, 0, 0, 0), Pays4 = c(1, 1, 1,
 0, 0, 1, 1, 1, 1), Pays5 = c(1, 1, 1, 0, 0, 0, 0, 0, 0)), .Names =
c("Nom",
"Pays1", "Pays2", "Pays3", "Pays4", "Pays5"), row.names = c(NA,
-9L), class = "data.frame")

The purpose is to transform df1 it df2 by giving for every group of
lines A, B and C the value 1 if there is at least a value equal to 1 or
a value 0 if there is no value equal to 1

Thanks for your helps



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

Re: [R] to modify a dataframe

[arun]

Hi,
You could try:
 df3 <- df1
library(plyr) 
df3[,-1] <- ddply(df1,.(Nom1=gsub("\\d+","",Nom)),colwise(function(x) 
rep(max(x),length(x[,-1]
attr(df3,"row.names") <- attr(df2,"row.names")
 identical(df2,df3)
#[1] TRUE

A.K.




On Wednesday, January 1, 2014 11:56 AM, Arnaud Michel  
wrote:
Dear All,

>From the dataframe df1

df1 <-
structure(list(Nom = structure(1:9, .Label = c("A1", "A2", "A3",
"B1", "B2", "C1", "C2", "C3", "C4"), class = "factor"), Pays1 = c(1,
1, 0, 0, 1, 0, 0, 0, 0), Pays2 = c(0, 0, 0, 1, 1, 0, 1, 0, 1),
     Pays3 = c(0, 0, 0, 0, 1, 0, 0, 0, 0), Pays4 = c(1, 0, 0,
     0, 0, 0, 1, 0, 1), Pays5 = c(1, 1, 0, 0, 0, 0, 0, 0, 0)), .Names = c("Nom",
"Pays1", "Pays2", "Pays3", "Pays4", "Pays5"), row.names = c(1L,
3L, 4L, 2L, 5L, 6L, 7L, 8L, 9L), class = "data.frame")


I look for a way to build the new dataframe df2
  
df2 <-
structure(list(Nom = structure(1:9, .Label = c("A1", "A2", "A3",
"B1", "B2", "C1", "C2", "C3", "C4"), class = "factor"), Pays1 = c(1,
1, 1, 1, 1, 0, 0, 0, 0), Pays2 = c(0, 0, 0, 1, 1, 1, 1, 1, 1),
     Pays3 = c(0, 0, 0, 1, 1, 0, 0, 0, 0), Pays4 = c(1, 1, 1,
     0, 0, 1, 1, 1, 1), Pays5 = c(1, 1, 1, 0, 0, 0, 0, 0, 0)), .Names = c("Nom",
"Pays1", "Pays2", "Pays3", "Pays4", "Pays5"), row.names = c(NA,
-9L), class = "data.frame")

The purpose is to transform df1 it df2 by giving for every group of lines A, B 
and C the value 1 if there is at least a value equal to 1 or a value 0 if there 
is no value equal to 1

Thanks for your helps

-- 
Michel ARNAUD
Chargé de mission auprès du DRH
DGDRD-Drh - TA 174/04
Av Agropolis 34398 Montpellier cedex 5
tel : 04.67.61.75.38
fax : 04.67.61.57.87
port: 06.47.43.55.31


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

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

Re: [R] seq_len and loops

[William Dunlap]

> 2. However, Bill (and Henrik)  raised the question of replacing '1' with
> '1L'; I understand the meaning of that, but does it matter (in practice)?
> On 12/22/2013 06:57 PM, William Dunlap wrote:
> >>> for (i in seq_len(x - 1) + 1)
> >>>
> >>> should be efficient and safe.
> >>
> >> Oops, not safe when x is 0.
> >
> > Also, the '+ 1' should be '+ 1L' to get the same answer as
> > seq_len(x)[-1].

It depends what your practice involves.

seq_len(n)[-1], 2:n, and seq_len(n-1)+1L all  produce an integer vector (if 
0 if (x > 1){
> for (x in 2:x){
>...
> 
> is the easiest, most effective,  and most easy-to-understand.

The dangerous part of that idiom is what you do in the 'else' part of the 'if' 
statement.
Do both clauses make objects with the same names and types?  I mildly prefer 
avoiding
if statements because it makes reasoning about the results of the code more 
complicated.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
> Behalf
> Of Göran Broström
> Sent: Tuesday, December 31, 2013 4:10 PM
> To: r-help@R-project.org
> Subject: Re: [R] seq_len and loops
> 
> Thanks for the answers from Duncan, Bill, Gabor, and Henrik. You
> convinced me that
> 
> 1. The solution
> 
> if (x > 1){
> for (x in 2:x){
>...
> 
> is the easiest, most effective,  and most easy-to-understand.
> 
> 2. However, Bill (and Henrik)  raised the question of replacing '1' with
> '1L'; I understand the meaning of that, but does it matter (in practice)?
> 
> 3. Noone commented upon
> 
> i <- 1
> while (i < x){
> i <- i + 1
> ...
> }
> 
> I suppose that it means that it is the best solution.
> 
> Thanks, and Happy New Year 2014!
> 
> Göran
> 
> On 12/22/2013 06:57 PM, William Dunlap wrote:
> >>> for (i in seq_len(x - 1) + 1)
> >>>
> >>> should be efficient and safe.
> >>
> >> Oops, not safe when x is 0.
> >
> > Also, the '+ 1' should be '+ 1L' to get the same answer as
> > seq_len(x)[-1].
> >
> > Bill Dunlap
> > Spotfire, TIBCO Software
> > wdunlap tibco.com
> >
> >
> >> -Original Message-
> >> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> >> Of Duncan Murdoch
> >> Sent: Saturday, December 21, 2013 3:52 PM
> >> To: Göran Broström; R-help@r-project.org
> >> Subject: Re: [R] seq_len and loops
> >>
> >> On 13-12-21 6:50 PM, Duncan Murdoch wrote:
> >>> On 13-12-21 5:57 PM, Göran Broström wrote:
>  I was recently reminded on this list that
> 
>  "Using 1:ncol() is bad practice (seq_len is designed for that purpose)"
>  (Ripley)
> 
>  This triggers the following question: What is "good practice" for
>  2:ncol(x)? (This is not a joke; in a recursive situation it often makes
>  sense to perform the calculation for the start value i = 1, then
>  continue with a loop over the rest, "the Fortran way";)
> 
>  I usually use
> 
>  if (ncol(x) > 1)
>  for (i in 2:ncol(x)){
> 
> 
>  but I can think of
> 
>  for (i in seq_len(x - 1)){
>  I <- i + 1
> 
> 
>  and
> 
>  i <- 1
>  while (i < ncol(x)){
>  i <- i + 1
>  
> 
>  What is "good practice" (efficient and safe)?
> >>>
> >>> for (i in seq_len(x - 1) + 1)
> >>>
> >>> should be efficient and safe.
> >>
> >> Oops, not safe when x is 0.
> >>
> >>   >
> >> A little less efficient, but clearer would be
> >>>
> >>> for (i in seq_len(x)[-1])
> >>>
> >>> Duncan Murdoch
> >>>
> >>
> >> __
> >> 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.

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

Re: [R] looping through columns in a matrix or data frame

[arun]

Hi,
May be this helps:

Using your function:
mapply(less,test,4)

#or
 invisible(mapply(less,test,4))
#[1] 2 3
#[1] 3

#or

 for(i in 1:ncol(test)){
 less(test[,i],4)}
#[1] 2 3
#[1] 3


A.K.



Hi, I'm trying to figure out how to loop through columns in a matrix or 
data frame, but what I've been finding online has not been very clear. 
I've written the following simple function that I can use on a column to 
extract all values that are less than a specified number. Consider the 
following example using that function to extract all values less than 4 
from column1 of the table "test" 

> less <- function(x,y){print(x[which(x < y)])} 

> test 
  column1 column2 
1       2       3 
2       3       4 
3       4       5 

> less(test[,1],4) 
[1] 2 3 


What I want to do is loop that function over all the columns
 in the table. Note: I realize that this is a silly example and there 
are better ways to do this particular function in R, so please don't 
respond with better ways to extract values less than a given number. The
 question that I am interested in is merely how do I loop over the 
columns. If you could respond by modifying my silly function so that it 
will loop, that would be the most helpful response. Thanks for the 
advice!

__
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] Question: reproducibility of random sampling with replacement

[Chee Chen]

Dear All,

I would like to ask for your help on "reproducibility of random sampling with 
replacement". For example, one re-samples the rows with replacement of a 
residual matrix and uses the new residual matrix thus obtained to produce a 
statistic ; repeat this for a certain number of times.

My questions:  will the above produce ever be reproducible by setting a seed? 
Namely,  Given the same residual matrix, Ted applies the above process and so 
does Jack, will they get the same results by setting a seed? 

My attempt: setting seed does not freeze the command "sample" from  getting 
different samples, as from the codes:

x= 1:20
S = matrix(0,5,20)
for (i in 1:5) {
  S[i,] = sample(x, replace=FALSE)
}

set.seed(123)

T = matrix(0,5,20)
for (i in 1:5) {
  T[i,] = sample(x, replace=FALSE)
}
sum(S==T)
===

I would appreciate any comments and/or suggestions on this. 
Regards,
Chee

[[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.

Re: [R] Question: reproducibility of random sampling with replacement

[Rui Barradas]

Hello,

Inline.

Em 01-01-2014 22:12, Chee Chen escreveu:

Dear All,

I would like to ask for your help on "reproducibility of random sampling with 
replacement". For example, one re-samples the rows with replacement of a residual 
matrix and uses the new residual matrix thus obtained to produce a statistic ; repeat 
this for a certain number of times.

My questions:  will the above produce ever be reproducible by setting a seed?


Yes.

 Namely,  Given the same residual matrix, Ted applies the above process 
and so does Jack, will they get the same results by setting a seed?


My attempt: setting seed does not freeze the command "sample" from  getting 
different samples,


Yes it does, you are simply not setting the seed before the first for loop.

Rui Barradas
 as from the codes:


x= 1:20
S = matrix(0,5,20)
for (i in 1:5) {
   S[i,] = sample(x, replace=FALSE)
}

set.seed(123)

T = matrix(0,5,20)
for (i in 1:5) {
   T[i,] = sample(x, replace=FALSE)
}
sum(S==T)
===

I would appreciate any comments and/or suggestions on this.
Regards,
Chee

[[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.

Re: [R] Question: reproducibility of random sampling with replacement

[Rolf Turner]

Why on earth would you expect S and T to be the same given
what you have done.  "I am unable to rightly apprehend the
confusion of ideas that could provoke such a question",
(Charles Babbage).

You have to set the *same* seed before each construction.
I.e. do set.seed(123) before creating S; then do set.seed(123)
again before creating T.  If you do so, S and T will be
identical.

cheers,

Rolf Turner

P. S. "T" is not a good name for an object; too easy to confuse
with "TRUE".  Not an egregious sin, but to be avoided.

R. T.

On 02/01/14 11:12, Chee Chen wrote:

Dear All,

I would like to ask for your help on "reproducibility of random sampling with 
replacement". For example, one re-samples the rows with replacement of a residual 
matrix and uses the new residual matrix thus obtained to produce a statistic ; repeat 
this for a certain number of times.

My questions:  will the above produce ever be reproducible by setting a seed? 
Namely,  Given the same residual matrix, Ted applies the above process and so 
does Jack, will they get the same results by setting a seed?

My attempt: setting seed does not freeze the command "sample" from  getting 
different samples, as from the codes:

x= 1:20
S = matrix(0,5,20)
for (i in 1:5) {
   S[i,] = sample(x, replace=FALSE)
}

set.seed(123)

T = matrix(0,5,20)
for (i in 1:5) {
   T[i,] = sample(x, replace=FALSE)
}
sum(S==T)
===

I would appreciate any comments and/or suggestions on this.


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

Re: [R] Question: reproducibility of random sampling with replacement

[Jeff Newmiller]

If you want to reproduce the same sequence twice, then you need to set the seed 
at the beginning of each calculation. You are only doing it for the second 
calculation below.
---
Jeff NewmillerThe .   .  Go Live...
DCN:Basics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

Chee Chen  wrote:
>Dear All,
>
>I would like to ask for your help on "reproducibility of random
>sampling with replacement". For example, one re-samples the rows with
>replacement of a residual matrix and uses the new residual matrix thus
>obtained to produce a statistic ; repeat this for a certain number of
>times.
>
>My questions:  will the above produce ever be reproducible by setting a
>seed? Namely,  Given the same residual matrix, Ted applies the above
>process and so does Jack, will they get the same results by setting a
>seed? 
>
>My attempt: setting seed does not freeze the command "sample" from 
>getting different samples, as from the codes:
>
>x= 1:20
>S = matrix(0,5,20)
>for (i in 1:5) {
>  S[i,] = sample(x, replace=FALSE)
>}
>
>set.seed(123)
>
>T = matrix(0,5,20)
>for (i in 1:5) {
>  T[i,] = sample(x, replace=FALSE)
>}
>sum(S==T)
>===
>
>I would appreciate any comments and/or suggestions on this. 
>Regards,
>Chee
>
>   [[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.

Re: [R] Question: reproducibility of random sampling with replacement

[Bert Gunter]

You have to set the same seed before each random number generation!
You did not do this.



Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374

"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
H. Gilbert Welch




On Wed, Jan 1, 2014 at 2:12 PM, Chee Chen  wrote:
> Dear All,
>
> I would like to ask for your help on "reproducibility of random sampling with 
> replacement". For example, one re-samples the rows with replacement of a 
> residual matrix and uses the new residual matrix thus obtained to produce a 
> statistic ; repeat this for a certain number of times.
>
> My questions:  will the above produce ever be reproducible by setting a seed? 
> Namely,  Given the same residual matrix, Ted applies the above process and so 
> does Jack, will they get the same results by setting a seed?
>
> My attempt: setting seed does not freeze the command "sample" from  getting 
> different samples, as from the codes:
> 
> x= 1:20
> S = matrix(0,5,20)
> for (i in 1:5) {
>   S[i,] = sample(x, replace=FALSE)
> }
>
> set.seed(123)
>
> T = matrix(0,5,20)
> for (i in 1:5) {
>   T[i,] = sample(x, replace=FALSE)
> }
> sum(S==T)
> ===
>
> I would appreciate any comments and/or suggestions on this.
> Regards,
> Chee
>
> [[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.

[R] How to verify char variables contain at least one value

[Luca Meyer]

Happy new year fellows,

I am trying to do something I believe should be fairly straightforward but
I cannot find my way out.

My dataset d2 is 26 rows by 245 columns, exclusively char variables. I
would like to check whether at least one column from V13 till V239 (they
are in numerical sequence) has been filled in, so I try

d2$check <- c(d2$V13:d2$V239)

and/or

d2$check <- paste(d2$V13:d2$V239,sep="")

but I get (translated from Italian):

Error in d2$V13:d2$V239 : argument NA/NaN

I have tried nchar but the same error occurs. I have also tried to run the
above functions on a smaller variable subset (V13, V14, V15, see below for
details) just to double check in case some variable would erroneously be in
another format, but the same occur.

> d2$V13
 [1] """"""
"""""""da -5.1% a -10%"
""
 [9] """"""
""""""""
""
[17] """"""
""""""""
""
[25] """"
> d2$V14
 [1] "" "" ""
"" "" "" "da -10.1% a -15%"
""
 [9] "" "" ""
"" "" "" ""
""
[17] "" "" ""
"" "" "" ""
""
[25] "" ""
> d2$V15
 [1] "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""
"" "" ""

Can anyone suggest an alternative function for me to create a variable that
checks whether there is at least one value for each of the 26 records I
need to analyze?

Thank you in advance,

Luca

[[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] R crashes with memory errors on a 256GB machine (and system shoes only 60GB usage)

[Xebar Saram]

Hi All,

I have a terrible issue i cant seem to debug which is halting my work
completely. I have R 3.02 installed on a linux machine (arch linux-latest)
which I built specifically for running high memory use models. the system
is a 16 core, 256 GB RAM machine. it worked well at the start but in the
recent days i keep getting errors and crashes regarding memory use, such as
"cannot create vector size of XXX, not enough memory" etc

when looking at top (linux system monitor) i see i barley scrape the 60 GB
of ram (out of 256GB)

i really don't know how to debug this and my whole work is halted due to
this so any help would be greatly appreciated

Best wishes

Z

[[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.

Re: [R] How to verify char variables contain at least one value

[Jim Lemon]

On 01/02/2014 05:17 PM, Luca Meyer wrote:

Happy new year fellows,

I am trying to do something I believe should be fairly straightforward but
I cannot find my way out.

My dataset d2 is 26 rows by 245 columns, exclusively char variables. I
would like to check whether at least one column from V13 till V239 (they
are in numerical sequence) has been filled in, so I try

d2$check<- c(d2$V13:d2$V239)

and/or

d2$check<- paste(d2$V13:d2$V239,sep="")

but I get (translated from Italian):

Error in d2$V13:d2$V239 : argument NA/NaN

I have tried nchar but the same error occurs. I have also tried to run the
above functions on a smaller variable subset (V13, V14, V15, see below for
details) just to double check in case some variable would erroneously be in
another format, but the same occur.


d2$V13

  [1] """"""
"""""""da -5.1% a -10%"
""
  [9] """"""
""""""""
""
[17] """"""
""""""""
""
[25] """"

d2$V14

  [1] "" "" ""
"" "" "" "da -10.1% a -15%"
""
  [9] "" "" ""
"" "" "" ""
""
[17] "" "" ""
"" "" "" ""
""
[25] "" ""

d2$V15

  [1] "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""
"" "" ""

Can anyone suggest an alternative function for me to create a variable that
checks whether there is at least one value for each of the 26 records I
need to analyze?


Hi Luca,
Perhaps you are looking for something like this:

d2check<-unlist(apply(as.matrix(d2[,paste("V",13:239,sep="")]),1,nchar))
# to test for any non empty rows
any(d2check)

Jim

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

Re: [R] How to verify char variables contain at least one value

[Gerrit Eichner]

Hello, Luca,

also a happy new year!

It's not quite clear to me what you want to do, but note first that the 
":"-operator is a short-cut for seq() with by = 1 (look at ?seq), and that 
it usually (!) does not work on columns of data frames. Exception: when 
used for the argument subset of function subset().


Second, you seem to want to check in each row of d2 if there is any entry 
different from "", right?


So, does


apply( subset( d2, subset = V13:V239), 1, function( x) any( x != ""))


what you want?

 Hth  --  Gerrit

On Thu, 2 Jan 2014, Luca Meyer wrote:


Happy new year fellows,

I am trying to do something I believe should be fairly straightforward but
I cannot find my way out.

My dataset d2 is 26 rows by 245 columns, exclusively char variables. I
would like to check whether at least one column from V13 till V239 (they
are in numerical sequence) has been filled in, so I try

d2$check <- c(d2$V13:d2$V239)

and/or

d2$check <- paste(d2$V13:d2$V239,sep="")

but I get (translated from Italian):

Error in d2$V13:d2$V239 : argument NA/NaN

I have tried nchar but the same error occurs. I have also tried to run the
above functions on a smaller variable subset (V13, V14, V15, see below for
details) just to double check in case some variable would erroneously be in
another format, but the same occur.


d2$V13

[1] """"""
"""""""da -5.1% a -10%"
""
[9] """"""
""""""""
""
[17] """"""
""""""""
""
[25] """"

d2$V14

[1] "" "" ""
"" "" "" "da -10.1% a -15%"
""
[9] "" "" ""
"" "" "" ""
""
[17] "" "" ""
"" "" "" ""
""
[25] "" ""

d2$V15

[1] "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" "" ""
"" "" ""

Can anyone suggest an alternative function for me to create a variable that
checks whether there is at least one value for each of the 26 records I
need to analyze?

Thank you in advance,

Luca

[[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.