Re: [R] Writing data onto xlsx file without cell formatting

[Christofer Bogaso]

openxlsx is not solving my problem either. It is corrupting my xlsx file.

I have a large data.frame, which I want to export to an existing xlsx
file, without formatting that existing file. With XLconnect there is
an option "setStyleAction(wb,XLC$"STYLE_ACTION.NONE")" which does it
so. I am looking for a similar codeline for xlsx package which will
enable me to save my data.frame in my existing file without formatting
my xlsx file.

Thanks,

On Tue, Sep 27, 2016 at 3:39 AM, jim holtman  wrote:
> I use the "openxlsx" package to handle spreadsheets.
>
>
> Jim Holtman
> Data Munger Guru
>
> What is the problem that you are trying to solve?
> Tell me what you want to do, not how you want to do it.
>
> On Mon, Sep 26, 2016 at 5:56 PM, Christofer Bogaso
>  wrote:
>>
>> Hi again,
>>
>> I have been following above suggestion to export data from R to xlsx
>> file using XLconnect. However recently I am facing Java memory
>> allocation problem with large dataset (looks like a known issue with
>> this package) and therefore decided to move to using "xlsx" package.
>>
>> Now I started facing that same problem of losing my existing formating
>> when I use xlsx package for data export. Can someone help me with some
>> pointer on how can I preserve the cell formating after exporting
>> data.frame to some existing xlsx file using "xlsx" package.
>>
>> Thanks for your time.
>>
>> On Mon, Jul 11, 2016 at 10:43 AM, Ismail SEZEN 
>> wrote:
>> > I think, this is what you are looking for:
>> >
>> >
>> > http://stackoverflow.com/questions/11228942/write-from-r-into-template-in-excel-while-preserving-formatting
>> >
>> > On 11 Jul 2016, at 03:43, Christofer Bogaso
>> > 
>> > wrote:
>> >
>> > Hi again,
>> >
>> > I am trying to write a data frame to an existing Excel file (xlsx)
>> > from row 5 and column 6 of the 1st Sheet. I was going through a
>> > previous instruction which is available here :
>> >
>> >
>> > http://stackoverflow.com/questions/32632137/using-write-xlsx-in-r-how-to-write-in-a-specific-row-or-column-in-excel-file
>> >
>> > However trouble is that it is modifying/removing formatting of all the
>> > affected cells. I have predefined formatting of those cells where data
>> > to be pasted, and I dont want to modify or remove that formatting.
>> >
>> > Any idea if I need to pass some additional argument.
>> >
>> > Appreciate your valuable feedback.
>> >
>> > Thanks,
>> >
>> > __
>> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> > 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 -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>

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[R-es] Asignacion de colclasses

[Cesar Lopez]

Buenas Noches,


Les escribo para solicitarles una ayuda con lo siguiente:

"Primero leo las primeras 100 lineas de mi archivo"
inicial <- read.csv("cartera1.csv", header=TRUE, sep=";", 
fileEncoding="latin1", nrows=100)

"Segundo determino las clases o tipos de objetos"

clases <- sapply(inicial, class)


"Tercero reviso clases"

clases
USUARIO  NOMBRE
  "integer""factor"
  DIRECCION ESTRATO
   "factor"   "integer"
  CICLO DEUDA_PENDIENTE
  "integer""factor"
   INTERESES_PENDIENTES_MES INTERESES_PENDIENTES_ACUMULADOS
  "numeric"   "numeric"
 MESES_MORA   CUOTA_INICIAL
   "factor"   "integer"
CUOTA_SALDO CUOTA_VALOR
  "numeric"   "numeric"
CUOTA_INTERESESCUOTA_PENDIENTES
  "numeric"   "integer"
PAGADAS   FACTURADO
  "integer"   "numeric"
 ABONOS
  "integer"


"Cuarto intento importar el archivo con el colclasses"

datos <- read.csv("cartera1.csv", header=TRUE, sep=";", fileEncoding= "latin1", 
colClasses=clases)


"Me aparece el siguiente error"

Error in scan(file = file, what = what, sep = sep, quote = quote, dec = dec,  :
  scan() expected 'a real', got '13.060.496'


Que es lo que estoy haciendo mal o porque me sale dicho error ?


Agradeciendo todo su apoyo,


CESAR AUGUSTO LOPEZ

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[R] ggplot grouped barchart based on marginal proportions

[Paul Sanfilippo]

I am trying to create a grouped barplot that uses marginal (row) proportions 
rather than cell proportions and can't figure out how to change:

y = (..count..)/sum(..count..)
in ggplot to do this.

Using the mtcars dataset as an example and considering two categorical 
variables (cyl and am - purely for the sake of the example taking cyl as the 
response and am as the explanatory variable). Can anyone help me to do this:

data(mtcars)
# Get Proportions
mtcars_xtab <- table(mtcars$cyl,mtcars$am)
mtcars_xtab
margin.table(mtcars_xtab, 1) # A frequencies (summed over B) 
margin.table(mtcars_xtab, 2) # B frequencies (summed over A)
prop.table(mtcars_xtab) # cell percentages - THIS IS WHAT'S USED IN THE PLOT
prop.table(mtcars_xtab, 1) # row percentages - THESE ARE WHAT I WANT TO USE IN 
THE PLOT

# Make Plot
mtcars$cyl <- as.factor(mtcars$cyl) 
mtcars$am <- as.factor(mtcars$am) 
ggplot(mtcars, aes(x=am, fill=cyl)) +
geom_bar(aes(y = (..count..)/sum(..count..)), position = "dodge") +
  scale_fill_brewer(palette="Set2")
Thank you.




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Re: [R] src/Makevars ignored ?

[Jeff Newmiller]

You failed to read the Posting Guide, which would have told you which mailing 
list to post this question to. (Hint: not this one.)
-- 
Sent from my phone. Please excuse my brevity.

On September 26, 2016 4:46:06 AM PDT, Eric Deveaud  wrote:
>
>
>   Hello,
>
>as far as I understood the R library generic compilation mechanism, 
>compilation of C//C++ sources is controlde
>
>1) at system level by the ocntentos RHOME/etc/Makeconf
>2) at user level by the content of ~/.R/Makevars
>3) at package level by the content of src/Makevars
>
>Problem I have is that src/Makevars is ignored
>
>
>see following example:
>
>R is compiled and use the following CC and CFLAGS definition
>
>bigmess:epactsR/src > R CMD config CC
>gcc -std=gnu99
>bigmess:epactsR/src > R CMD config CFLAGS
>-Wall -g
>
>so building C sources lead to the following
>
>bigmess:epactsR/src > R CMD SHLIB index.c
>gcc -std=gnu99 -I/local/gensoft2/adm/lib64/R/include -DNDEBUG 
>-I/usr/local/include-fpic  -Wall -g  -c index.c -o index.o
>
>normal, it uses defintion from RHOME/etc/Makeconf
>
>
>when I set upp a ~/.R/Makevars that overwrite CC and CFLAGS definition.
>
>bigmess:epactsR/src > cat ~/.R/Makevars
>CC=gcc
>CFLAGS=-O3
>bigmess:epactsR/src > R CMD SHLIB index.c
>gcc -I/local/gensoft2/adm/lib64/R/include -DNDEBUG 
>-I/usr/local/include 
>-fpic  -O3 -c index.c -o index.o
>gcc -std=gnu99 -shared -L/usr/local/lib64 -o index.so index.o
>
>
>OK CC and CFLAGS are honored and set accordingly to ~/.R/Makevars
>
>
>but when I try to use src/Makevars, it is ignored
>
>bigmess:epactsR/src > cat ~/.R/Makevars
>cat: /home/edeveaud/.R/Makevars: No such file or directory
>bigmess:epactsR/src > cat ./Makevars
>CC = gcc
>CFLAGS=-O3
>bigmess:epactsR/src > R CMD SHLIB index.c
>gcc -std=gnu99 -I/local/gensoft2/adm/lib64/R/include -DNDEBUG 
>-I/usr/local/include-fpic  -Wall -g  -c index.c -o index.o
>
>
>what I have missed or is there something wrong ?
>
>
>PS I tested the ssame behaviour with various version of R from R/2.15
>to 
>R/3.3
>
>   best regards
>
>   Eric
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>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] Writing data onto xlsx file without cell formatting

[jim holtman]

I use the "openxlsx" package to handle spreadsheets.


Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.

On Mon, Sep 26, 2016 at 5:56 PM, Christofer Bogaso <
bogaso.christo...@gmail.com> wrote:

> Hi again,
>
> I have been following above suggestion to export data from R to xlsx
> file using XLconnect. However recently I am facing Java memory
> allocation problem with large dataset (looks like a known issue with
> this package) and therefore decided to move to using "xlsx" package.
>
> Now I started facing that same problem of losing my existing formating
> when I use xlsx package for data export. Can someone help me with some
> pointer on how can I preserve the cell formating after exporting
> data.frame to some existing xlsx file using "xlsx" package.
>
> Thanks for your time.
>
> On Mon, Jul 11, 2016 at 10:43 AM, Ismail SEZEN 
> wrote:
> > I think, this is what you are looking for:
> >
> > http://stackoverflow.com/questions/11228942/write-from-
> r-into-template-in-excel-while-preserving-formatting
> >
> > On 11 Jul 2016, at 03:43, Christofer Bogaso  >
> > wrote:
> >
> > Hi again,
> >
> > I am trying to write a data frame to an existing Excel file (xlsx)
> > from row 5 and column 6 of the 1st Sheet. I was going through a
> > previous instruction which is available here :
> >
> > http://stackoverflow.com/questions/32632137/using-
> write-xlsx-in-r-how-to-write-in-a-specific-row-or-column-in-excel-file
> >
> > However trouble is that it is modifying/removing formatting of all the
> > affected cells. I have predefined formatting of those cells where data
> > to be pasted, and I dont want to modify or remove that formatting.
> >
> > Any idea if I need to pass some additional argument.
> >
> > Appreciate your valuable feedback.
> >
> > Thanks,
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > 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 -- To UNSUBSCRIBE and more, see
> 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] Writing data onto xlsx file without cell formatting

[Christofer Bogaso]

Hi again,

I have been following above suggestion to export data from R to xlsx
file using XLconnect. However recently I am facing Java memory
allocation problem with large dataset (looks like a known issue with
this package) and therefore decided to move to using "xlsx" package.

Now I started facing that same problem of losing my existing formating
when I use xlsx package for data export. Can someone help me with some
pointer on how can I preserve the cell formating after exporting
data.frame to some existing xlsx file using "xlsx" package.

Thanks for your time.

On Mon, Jul 11, 2016 at 10:43 AM, Ismail SEZEN  wrote:
> I think, this is what you are looking for:
>
> http://stackoverflow.com/questions/11228942/write-from-r-into-template-in-excel-while-preserving-formatting
>
> On 11 Jul 2016, at 03:43, Christofer Bogaso 
> wrote:
>
> Hi again,
>
> I am trying to write a data frame to an existing Excel file (xlsx)
> from row 5 and column 6 of the 1st Sheet. I was going through a
> previous instruction which is available here :
>
> http://stackoverflow.com/questions/32632137/using-write-xlsx-in-r-how-to-write-in-a-specific-row-or-column-in-excel-file
>
> However trouble is that it is modifying/removing formatting of all the
> affected cells. I have predefined formatting of those cells where data
> to be pasted, and I dont want to modify or remove that formatting.
>
> Any idea if I need to pass some additional argument.
>
> Appreciate your valuable feedback.
>
> Thanks,
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Using lapply in R data table

[Bert Gunter]

... and just for fun, here's an alternative in which mapply() is used
to vectorize switch(); again, whether you like it may be just a matter
of taste, although I suspect it might be less efficient than ifelse(),
which is already vectorized:

DT <- within(DT,
exposure <- {
  mapply(function(x,fac)switch(as.character(fac),
  a = 1,
  b = difftime(as.Date("2007-01-01"), x,
units="days")/365.25,
  c = .5
),
  x = fini,
  fac =
cut(fini,as.Date(c("2000-01-01","2006-01-01","2006-06-30","2006-12-21")),
labels= letters[1:3])
  )}
  )


> DT
  id   fini group  exposure
1  2 2005-04-20 A 1.000
2  2 2005-04-20 A 1.000
3  2 2005-04-20 A 1.000
4  5 2006-02-19 B 0.8651608
5  5 2006-06-29 B 0.5092402
6  7 2006-10-08 A 0.500
7  7 2006-10-08 A 0.500


Cheers,
Bert


Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Mon, Sep 26, 2016 at 1:27 PM, Bert Gunter  wrote:
> Ista:
>
> Aha -- now I see the point. My bad. You are right. I was careless.
>
> However, cut() with ifelse() might simplify the code a bit and/or make
> it more readable. To be clear, this is just a matter of taste; e.g.
> using your data and a data frame instead of a data table:
>
>> DT <- within(DT,
> exposure <- {
>   f 
> <-cut(fini,as.Date(c("2000-01-01","2006-01-01","2006-06-30","2006-12-21")),
>   labels= letters[1:3])
>   ifelse(f == "a", 1,
>  ifelse( f == "c", .5,
> difftime(as.Date("2007-01-01"), fini, 
> units="days")/365.25))
> }
> )
>
>
>> DT
>   id   fini group  exposure f
> 1  2 2005-04-20 A 1.000 a
> 2  2 2005-04-20 A 1.000 a
> 3  2 2005-04-20 A 1.000 a
> 4  5 2006-02-19 B 0.8651608 b
> 5  5 2006-06-29 B 0.5092402 b
> 6  7 2006-10-08 A 0.500 c
> 7  7 2006-10-08 A 0.500 c
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Mon, Sep 26, 2016 at 12:07 PM, Ista Zahn  wrote:
>> On Mon, Sep 26, 2016 at 2:48 PM, Bert Gunter  wrote:
>>> I thought that that was a typo from the OP, as it disagrees with his
>>> example. But the labels are arbitrary, so in fact cut() will do it
>>> whichever way he meant.
>>
>> I don't see how cut will do it, at least not conveniently. Consider
>> this slightly altered example:
>>
>> library(data.table)
>> DT <- data.table(
>>   id = rep(c(2, 5, 7), c(3, 2, 2)),
>>   fini = rep(as.Date(c('2005-04-20',
>>'2006-02-19',
>>'2006-06-29',
>>'2006-10-08')),
>>  c(3, 1, 1, 2)),
>>   group = rep(c("A", "B", "A"), c(3, 2, 2))  )
>>
>> DT[, exposure := vector(mode = "numeric", length = .N)]
>> DT[fini < as.Date("2006-01-01"), exposure := 1]
>> DT[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30"),
>>exposure := difftime(as.Date("2007-01-01"), fini, units="days")/365.25]
>> DT[fini >= as.Date("2006-07-01"), exposure := 0.5]
>>
>> DT
>>
>> ##id   fini group  exposure
>> ## 1:  2 2005-04-20 A 1.000
>> ## 2:  2 2005-04-20 A 1.000
>> ## 3:  2 2005-04-20 A 1.000
>> ## 4:  5 2006-02-19 B 0.8651608
>> ## 5:  5 2006-06-29 B 0.5092402
>> ## 6:  7 2006-10-08 A 0.500
>> ## 7:  7 2006-10-08 A 0.500
>>
>> Best,
>> Ista
>>
>>>
>>> -- Bert
>>> Bert Gunter
>>>
>>> "The trouble with having an open mind is that people keep coming along
>>> and sticking things into it."
>>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>>
>>>
>>> On Mon, Sep 26, 2016 at 11:37 AM, Ista Zahn  wrote:
 On Mon, Sep 26, 2016 at 1:59 PM, Bert Gunter  
 wrote:
> This seems like a job for cut() .

 I thought that at first two, but the middle group shouldn't be .87 but 
 rather

 exposure" = "2007-01-01" - "fini"

 so, I think cut alone won't do it.

 Best,
 Ista
>
> (I made DT a data frame to avoid loading the data table package. But I
> assume it would work with a data table too, Check this, though!)
>
>> DT <- within(DT, exposure <- 
>> cut(fini,as.Date(c("2000-01-01","2006-01-01","2006-06-30","2006-12-21")),
>>  labels= c(1,.87,.5)))
>
>> DT
>   id   fini group exposure
> 1  2 2005-04-20 A1
> 2  2 2005-04-20 A1
> 3  2 2005-04-20 A1
> 4  5 2006-02-19 B 0.87
> 5  5 2006-02-19 B 0.87
> 6  7 

Re: [R] Using lapply in R data table

[Bert Gunter]

Ista:

Aha -- now I see the point. My bad. You are right. I was careless.

However, cut() with ifelse() might simplify the code a bit and/or make
it more readable. To be clear, this is just a matter of taste; e.g.
using your data and a data frame instead of a data table:

> DT <- within(DT,
exposure <- {
  f 
<-cut(fini,as.Date(c("2000-01-01","2006-01-01","2006-06-30","2006-12-21")),
  labels= letters[1:3])
  ifelse(f == "a", 1,
 ifelse( f == "c", .5,
difftime(as.Date("2007-01-01"), fini, units="days")/365.25))
}
)


> DT
  id   fini group  exposure f
1  2 2005-04-20 A 1.000 a
2  2 2005-04-20 A 1.000 a
3  2 2005-04-20 A 1.000 a
4  5 2006-02-19 B 0.8651608 b
5  5 2006-06-29 B 0.5092402 b
6  7 2006-10-08 A 0.500 c
7  7 2006-10-08 A 0.500 c
Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Mon, Sep 26, 2016 at 12:07 PM, Ista Zahn  wrote:
> On Mon, Sep 26, 2016 at 2:48 PM, Bert Gunter  wrote:
>> I thought that that was a typo from the OP, as it disagrees with his
>> example. But the labels are arbitrary, so in fact cut() will do it
>> whichever way he meant.
>
> I don't see how cut will do it, at least not conveniently. Consider
> this slightly altered example:
>
> library(data.table)
> DT <- data.table(
>   id = rep(c(2, 5, 7), c(3, 2, 2)),
>   fini = rep(as.Date(c('2005-04-20',
>'2006-02-19',
>'2006-06-29',
>'2006-10-08')),
>  c(3, 1, 1, 2)),
>   group = rep(c("A", "B", "A"), c(3, 2, 2))  )
>
> DT[, exposure := vector(mode = "numeric", length = .N)]
> DT[fini < as.Date("2006-01-01"), exposure := 1]
> DT[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30"),
>exposure := difftime(as.Date("2007-01-01"), fini, units="days")/365.25]
> DT[fini >= as.Date("2006-07-01"), exposure := 0.5]
>
> DT
>
> ##id   fini group  exposure
> ## 1:  2 2005-04-20 A 1.000
> ## 2:  2 2005-04-20 A 1.000
> ## 3:  2 2005-04-20 A 1.000
> ## 4:  5 2006-02-19 B 0.8651608
> ## 5:  5 2006-06-29 B 0.5092402
> ## 6:  7 2006-10-08 A 0.500
> ## 7:  7 2006-10-08 A 0.500
>
> Best,
> Ista
>
>>
>> -- Bert
>> Bert Gunter
>>
>> "The trouble with having an open mind is that people keep coming along
>> and sticking things into it."
>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>
>>
>> On Mon, Sep 26, 2016 at 11:37 AM, Ista Zahn  wrote:
>>> On Mon, Sep 26, 2016 at 1:59 PM, Bert Gunter  wrote:
 This seems like a job for cut() .
>>>
>>> I thought that at first two, but the middle group shouldn't be .87 but 
>>> rather
>>>
>>> exposure" = "2007-01-01" - "fini"
>>>
>>> so, I think cut alone won't do it.
>>>
>>> Best,
>>> Ista

 (I made DT a data frame to avoid loading the data table package. But I
 assume it would work with a data table too, Check this, though!)

> DT <- within(DT, exposure <- 
> cut(fini,as.Date(c("2000-01-01","2006-01-01","2006-06-30","2006-12-21")), 
> labels= c(1,.87,.5)))

> DT
   id   fini group exposure
 1  2 2005-04-20 A1
 2  2 2005-04-20 A1
 3  2 2005-04-20 A1
 4  5 2006-02-19 B 0.87
 5  5 2006-02-19 B 0.87
 6  7 2006-10-08 A  0.5
 7  7 2006-10-08 A  0.5


 (but note that exposure is a factor, not numeric)


 Cheers,
 Bert

 Bert Gunter

 "The trouble with having an open mind is that people keep coming along
 and sticking things into it."
 -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


 On Mon, Sep 26, 2016 at 10:05 AM, Ista Zahn  wrote:
> Hi Frank,
>
> lapply(DT) iterates over each column. That doesn't seem to be what you 
> want.
>
> There are probably better ways, but here is one approach.
>
> DT[, exposure := vector(mode = "numeric", length = .N)]
> DT[fini < as.Date("2006-01-01"), exposure := 1]
> DT[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30"),
>   exposure := difftime(as.Date("2007-01-01"), fini, 
> units="days")/365.25]
> DT[fini >= as.Date("2006-07-01"), exposure := 0.5]
>
> Best,
> Ista
>
> On Mon, Sep 26, 2016 at 11:28 AM, Frank S.  wrote:
>> Dear all,
>>
>> I have a R data table like this:
>>
>> DT <- data.table(
>>   id = rep(c(2, 5, 7), c(3, 2, 2)),
>>   fini = rep(as.Date(c('2005-04-20', '2006-02-19', '2006-10-08')), c(3, 
>> 2, 2)),
>>   group = rep(c("A", "B", "A"), c(3, 2, 2))  )
>>
>>

Re: [R-es] Concatenación de tablas

[Carlos J. Gil Bellosta]

?merge

El 26 de septiembre de 2016, 22:09, Cesar Lopez 
escribió:

> Buenas Tardes,
>
>
> Les escribo para solicitarles una ayuda dado que tengo 2 tablas, una con
> los campos:
>
>
> cedula | nombre | direccion
>
>
> y la otra con la tabla:
>
>
> cedula | barrio | municipio
>
>
> Lo que necesito es hacer una comparación del campo cedula de las dos
> tablas y si son iguales, agregarle los campos barrio y municipio de la
> segunda tabla a la fila correspondiente de esa cedula de la primera tabla
> al lado de direccion.
>
>
> Agradeciendo todo su apoyo,
>
>
> CESAR AUGUSTO LOPEZ
>
> [[alternative HTML version deleted]]
>
>
> ___
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> https://stat.ethz.ch/mailman/listinfo/r-help-es
>

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Re: [R] 32 and 64 bit R

[Duncan Murdoch]

On 26/09/2016 6:29 AM, Mike meyer wrote:

Hello,
  
I have both 32 and 64 bit verions of R installed. What happens if I open a workspace saved from 64 bit R

in the 32 bit version or conversely?
I am fairly careless but never noticed any problems.


No problems will arise because of the different word size.

You will possibly see problems if you have the two versions set up to 
use different libraries; occasionally a workspace will fail to load if 
it needs a package that is not installed.  Normally on WIndows the same 
library can be used for both 32 and 64 bit R, but you could always 
choose to break that.


Duncan Murdoch

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Re: [R] Using lapply in R data table

[Ista Zahn]

On Mon, Sep 26, 2016 at 2:48 PM, Bert Gunter  wrote:
> I thought that that was a typo from the OP, as it disagrees with his
> example. But the labels are arbitrary, so in fact cut() will do it
> whichever way he meant.

I don't see how cut will do it, at least not conveniently. Consider
this slightly altered example:

library(data.table)
DT <- data.table(
  id = rep(c(2, 5, 7), c(3, 2, 2)),
  fini = rep(as.Date(c('2005-04-20',
   '2006-02-19',
   '2006-06-29',
   '2006-10-08')),
 c(3, 1, 1, 2)),
  group = rep(c("A", "B", "A"), c(3, 2, 2))  )

DT[, exposure := vector(mode = "numeric", length = .N)]
DT[fini < as.Date("2006-01-01"), exposure := 1]
DT[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30"),
   exposure := difftime(as.Date("2007-01-01"), fini, units="days")/365.25]
DT[fini >= as.Date("2006-07-01"), exposure := 0.5]

DT

##id   fini group  exposure
## 1:  2 2005-04-20 A 1.000
## 2:  2 2005-04-20 A 1.000
## 3:  2 2005-04-20 A 1.000
## 4:  5 2006-02-19 B 0.8651608
## 5:  5 2006-06-29 B 0.5092402
## 6:  7 2006-10-08 A 0.500
## 7:  7 2006-10-08 A 0.500

Best,
Ista

>
> -- Bert
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Mon, Sep 26, 2016 at 11:37 AM, Ista Zahn  wrote:
>> On Mon, Sep 26, 2016 at 1:59 PM, Bert Gunter  wrote:
>>> This seems like a job for cut() .
>>
>> I thought that at first two, but the middle group shouldn't be .87 but rather
>>
>> exposure" = "2007-01-01" - "fini"
>>
>> so, I think cut alone won't do it.
>>
>> Best,
>> Ista
>>>
>>> (I made DT a data frame to avoid loading the data table package. But I
>>> assume it would work with a data table too, Check this, though!)
>>>
 DT <- within(DT, exposure <- 
 cut(fini,as.Date(c("2000-01-01","2006-01-01","2006-06-30","2006-12-21")), 
 labels= c(1,.87,.5)))
>>>
 DT
>>>   id   fini group exposure
>>> 1  2 2005-04-20 A1
>>> 2  2 2005-04-20 A1
>>> 3  2 2005-04-20 A1
>>> 4  5 2006-02-19 B 0.87
>>> 5  5 2006-02-19 B 0.87
>>> 6  7 2006-10-08 A  0.5
>>> 7  7 2006-10-08 A  0.5
>>>
>>>
>>> (but note that exposure is a factor, not numeric)
>>>
>>>
>>> Cheers,
>>> Bert
>>>
>>> Bert Gunter
>>>
>>> "The trouble with having an open mind is that people keep coming along
>>> and sticking things into it."
>>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>>
>>>
>>> On Mon, Sep 26, 2016 at 10:05 AM, Ista Zahn  wrote:
 Hi Frank,

 lapply(DT) iterates over each column. That doesn't seem to be what you 
 want.

 There are probably better ways, but here is one approach.

 DT[, exposure := vector(mode = "numeric", length = .N)]
 DT[fini < as.Date("2006-01-01"), exposure := 1]
 DT[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30"),
   exposure := difftime(as.Date("2007-01-01"), fini, 
 units="days")/365.25]
 DT[fini >= as.Date("2006-07-01"), exposure := 0.5]

 Best,
 Ista

 On Mon, Sep 26, 2016 at 11:28 AM, Frank S.  wrote:
> Dear all,
>
> I have a R data table like this:
>
> DT <- data.table(
>   id = rep(c(2, 5, 7), c(3, 2, 2)),
>   fini = rep(as.Date(c('2005-04-20', '2006-02-19', '2006-10-08')), c(3, 
> 2, 2)),
>   group = rep(c("A", "B", "A"), c(3, 2, 2))  )
>
>
> I want to construct a new variable "exposure" defined as follows:
>
> 1) If "fini" earlier than 2006-01-01 --> "exposure" = 1
> 2) If "fini" in [2006-01-01, 2006-06-30] --> "exposure" = "2007-01-01" - 
> "fini"
> 3) If "fini" in [2006-07-01, 2006-12-31] --> "exposure" = 0.5
>
>
> So the desired output would be the following data table:
>
>idfini exposure group
> 1:  2 2005-04-201.00A
> 2:  2 2005-04-201.00A
> 3:  2 2005-04-201.00A
> 4:  5 2006-02-190.87B
> 5:  5 2006-02-190.87B
> 6:  7 2006-10-080.50A
> 7:  7 2006-10-080.50A
>
>
> I have tried:
>
> DT <- DT[ , list(id, fini, exposure = 0, group)]
> DT.new <- lapply(DT, function(exposure){
>   exposure[fini < as.Date("2006-01-01")] <- 1   # 1st case
>   exposure[fini >= as.Date("2006-01-01") & fini <= 
> as.Date("2006-06-30")] <- difftime(as.Date("2007-01-01"), fini, 
> units="days")/365.25 # 2nd case
> exposure[fini >= as.Date("2006-07-01") & fini <= 
> as.Date("2006-12-31")] <- 0.5   # 3rd case
>   exposure  # return value
>   })
>
>

Re: [R] curve() doesn't seem to use the whole range of x? And Error: longer object length is not a multiple of shorter object length

[Jeff Newmiller]

I think you are going to have to be more specific than "having some trouble". 
Your plot used lka as the x-axis.

FWIW note that

lm(ruotsi.pist ~ mies + koulu + clka + koulu*clka, data=dta)

is the same as

lm(ruotsi.pist ~ mies + koulu*clka, data=dta)
-- 
Sent from my phone. Please excuse my brevity.

On September 26, 2016 9:41:57 AM PDT, Matti Viljamaa  wrote:
>
>> On 26 Sep 2016, at 19:41, Matti Viljamaa  wrote:
>> 
>> Thank you.
>> 
>> However, I’m having some trouble converting your code to use clka,
>because the model I was using was:
>> 
>> fit2 <- lm(ruotsi.pist ~ mies + koulu + clka + koulu*clka, data=dta)
>
>I mean, not to use clka to replace lka. But to use the above fit2,
>rather than your fit2.
>
>>> On 25 Sep 2016, at 21:23, Jeff Newmiller 
>wrote:
>>> 
>>> This illustrates why you need to post a reproducible example. You
>have a number of confounding factors in your code.
>>> 
>>> First, "data" is a commonly-used function... avoid using it for
>variable names.
>>> 
>>> Second, using the attach function this way leads to confusion...
>best to forget this function until you start building packages.
>>> 
>>> Third, clka is linearly dependent on lka, so having them both in the
>regression is not possible. In this case lm has chosen to ignore clka
>so that bs("clka") is NA.
>>> 
>>> Fourth, curve expects you to give it a function, and instead you
>have given it a vector.
>>> 
>>> Fifth, you are plotting versus lka, but attempting to vary clka in
>the curve call.
>>> 
>>> There are a number of directions you could go with this to get a
>working output... below is my version.
>>> 
>>> dta <- read.table(
>"http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt;,
>header=TRUE )
>>> fit2 <- lm( ruotsi.pist ~ mies + koulu*lka, data=dta )
>>> bs <- coef( fit2 )
>>> rpBylka <- function( lka ) {
>>> kouluB <- factor( "B", levels = levels( dta$koulu ) )
>>> newdta <- expand.grid( mies=0, koulu=kouluB, lka=lka )
>>> predict( fit2, newdata = newdta )
>>> }
>>> dtaKouluB <- subset( dta, koulu == "B" )
>>> varitB <- dtaKouluB$mies
>>> varitB[ varitB == 0 ] <- 2
>>> plot( dtaKouluB$lka
>>>   , dtaKouluB$ruotsi.pist
>>>   , col=varitB
>>>   , pch=16
>>>   , xlab='lka'
>>>   , ylab='ruotsi.pist'
>>>   , main='Lukio B'
>>>   )
>>> curve( rpBylka, from = min( dta$lka ), max( dta$lka ), add=TRUE,
>col="red" )
>>> 
>>> On Sun, 25 Sep 2016, Matti Viljamaa wrote:
>>> 
 
> On 25 Sep 2016, at 19:37, Matti Viljamaa 
>wrote:
> 
> Okay here?s a pretty short code to reproduce it:
> 
> data <-
>read.table("http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt;,
>header=TRUE)
 
 data$clka <- I(data$lka - mean(data$lka))
 
> attach(data)
> 
> fit2 <- lm(ruotsi.pist ~ mies + koulu + lka + koulu*clka)
> 
> bs <- coef(fit2)
> 
> varitB <- c(data[koulu == 'B',]$mies)
> varitB[varitB == 0] = 2
> plot(data[data$koulu == 'B',]$lka, data[koulu ==
>'B',]$ruotsi.pist, col=varitB, pch=16, xlab='', ylab='', main='Lukio
>B?)
> 
>
>curve(bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*x+bs["kouluB:clka"]*clka,
>from=min(lka), to=max(lka), add=TRUE, col='red')
> 
> 
>> On 25 Sep 2016, at 19:24, Jeff Newmiller
> wrote:
>> 
>> Go directly to C. Do not pass go, do not collect $200.
>> 
>> You think curve does something, but you are missing what it
>actually does. Since you don't seem to be learning from reading ?curve
>or from our responses, you need to give us an example you can learn
>from.
>> --
>> Sent from my phone. Please excuse my brevity.
>> 
>> On September 25, 2016 9:04:09 AM PDT, mviljamaa
> wrote:
>>> On 2016-09-25 18:52, Jeff Newmiller wrote:
 You seem to be confused about what curve is doing vs. what you
>are
 doing.
>>> 
>>> But my x-range in curve()'s parameters from and to should be the
>entire
>>> 
>>> lka vector, since they are from=min(lka) and to=max(lka). Then
>why does
>>> 
>>> this not span the entire of lka? Because of duplicate entries or
>what?
>>> 
>>> It seems like I cannot use curve(), since my x-axis must be
>exactly lka
>>> 
>>> for the function to plot the y value for every lka value.
>>> 
 A) Compute the points you want to plot and put them into 2
>vectors.
 Then figure out how to plot those vectors. Then (perhaps)
>consider
 putting that all into one line of code again.
 
 B) The predict function is the preferred way to compute points.
>It
>>> may
 be educational for you to do the computations by hand at first,
>but
>>> in
 the long run using predict will help you avoid problems getting
>the
 equations right in multiple places in your script.
 
 C) Learn what makes an example reproducible 

Re: [R] curve() doesn't seem to use the whole range of x? And Error: longer object length is not a multiple of shorter object length

[Greg Snow]

If your goal is to visualize the predicted curve from an lm fit (or
other model fit) then you may want to look at the Predict.Plot and
TkPredict functions from the TeachingDemos package.




On Sun, Sep 25, 2016 at 7:01 AM, Matti Viljamaa  wrote:
> I’m trying to plot regression lines using curve()
>
> The way I do it is:
>
> bs <- coef(fit2)
>
> and then for example:
>
> curve(bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*x+bs["kouluB:clka"]*clka,
>  from=min(lka), to=max(lka), add=TRUE, col='red')
>
> This above code runs into error:
>
> Error in curve(bs["(Intercept)"] + bs["mies"] * 0 + bs["kouluB"] + bs["lka"] 
> *  :
>   'expr' did not evaluate to an object of length 'n'
> In addition: Warning message:
> In bs["(Intercept)"] + bs["mies"] * 0 + bs["kouluB"] + bs["lka"] *  :
>   longer object length is not a multiple of shorter object length
>
> Which I’ve investigated might be related to the lengths of the different 
> objects being multiplied or summed.
> Taking length(g$x) or length(g$y) of
>
> g <- curve(bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*x, 
> from=min(lka), to=max(lka), add=TRUE, col='red')
>
> returns 101.
>
> However length(lka) is 375. But perhaps these being different is not the 
> problem?
>
> I however do see that the whole range of lka is not plotted, for some reason. 
> So how can I be sure
> that it passes through all x-values in lka? And i.e. that the lengths of 
> objects inside curve() are correct?
>
> What can I do?
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.



-- 
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com

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Re: [R] Using lapply in R data table

[Bert Gunter]

I thought that that was a typo from the OP, as it disagrees with his
example. But the labels are arbitrary, so in fact cut() will do it
whichever way he meant.

-- Bert
Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Mon, Sep 26, 2016 at 11:37 AM, Ista Zahn  wrote:
> On Mon, Sep 26, 2016 at 1:59 PM, Bert Gunter  wrote:
>> This seems like a job for cut() .
>
> I thought that at first two, but the middle group shouldn't be .87 but rather
>
> exposure" = "2007-01-01" - "fini"
>
> so, I think cut alone won't do it.
>
> Best,
> Ista
>>
>> (I made DT a data frame to avoid loading the data table package. But I
>> assume it would work with a data table too, Check this, though!)
>>
>>> DT <- within(DT, exposure <- 
>>> cut(fini,as.Date(c("2000-01-01","2006-01-01","2006-06-30","2006-12-21")), 
>>> labels= c(1,.87,.5)))
>>
>>> DT
>>   id   fini group exposure
>> 1  2 2005-04-20 A1
>> 2  2 2005-04-20 A1
>> 3  2 2005-04-20 A1
>> 4  5 2006-02-19 B 0.87
>> 5  5 2006-02-19 B 0.87
>> 6  7 2006-10-08 A  0.5
>> 7  7 2006-10-08 A  0.5
>>
>>
>> (but note that exposure is a factor, not numeric)
>>
>>
>> Cheers,
>> Bert
>>
>> Bert Gunter
>>
>> "The trouble with having an open mind is that people keep coming along
>> and sticking things into it."
>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>
>>
>> On Mon, Sep 26, 2016 at 10:05 AM, Ista Zahn  wrote:
>>> Hi Frank,
>>>
>>> lapply(DT) iterates over each column. That doesn't seem to be what you want.
>>>
>>> There are probably better ways, but here is one approach.
>>>
>>> DT[, exposure := vector(mode = "numeric", length = .N)]
>>> DT[fini < as.Date("2006-01-01"), exposure := 1]
>>> DT[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30"),
>>>   exposure := difftime(as.Date("2007-01-01"), fini, 
>>> units="days")/365.25]
>>> DT[fini >= as.Date("2006-07-01"), exposure := 0.5]
>>>
>>> Best,
>>> Ista
>>>
>>> On Mon, Sep 26, 2016 at 11:28 AM, Frank S.  wrote:
 Dear all,

 I have a R data table like this:

 DT <- data.table(
   id = rep(c(2, 5, 7), c(3, 2, 2)),
   fini = rep(as.Date(c('2005-04-20', '2006-02-19', '2006-10-08')), c(3, 2, 
 2)),
   group = rep(c("A", "B", "A"), c(3, 2, 2))  )


 I want to construct a new variable "exposure" defined as follows:

 1) If "fini" earlier than 2006-01-01 --> "exposure" = 1
 2) If "fini" in [2006-01-01, 2006-06-30] --> "exposure" = "2007-01-01" - 
 "fini"
 3) If "fini" in [2006-07-01, 2006-12-31] --> "exposure" = 0.5


 So the desired output would be the following data table:

idfini exposure group
 1:  2 2005-04-201.00A
 2:  2 2005-04-201.00A
 3:  2 2005-04-201.00A
 4:  5 2006-02-190.87B
 5:  5 2006-02-190.87B
 6:  7 2006-10-080.50A
 7:  7 2006-10-080.50A


 I have tried:

 DT <- DT[ , list(id, fini, exposure = 0, group)]
 DT.new <- lapply(DT, function(exposure){
   exposure[fini < as.Date("2006-01-01")] <- 1   # 1st case
   exposure[fini >= as.Date("2006-01-01") & fini <= 
 as.Date("2006-06-30")] <- difftime(as.Date("2007-01-01"), fini, 
 units="days")/365.25 # 2nd case
 exposure[fini >= as.Date("2006-07-01") & fini <= 
 as.Date("2006-12-31")] <- 0.5   # 3rd case
   exposure  # return value
   })


 But I get an error message.

 Thanks for any help!!


 Frank S.


 [[alternative HTML version deleted]]

 __
 R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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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.
>>>
>>> __
>>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> 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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Using lapply in R data table

[Ista Zahn]

On Mon, Sep 26, 2016 at 1:59 PM, Bert Gunter  wrote:
> This seems like a job for cut() .

I thought that at first two, but the middle group shouldn't be .87 but rather

exposure" = "2007-01-01" - "fini"

so, I think cut alone won't do it.

Best,
Ista
>
> (I made DT a data frame to avoid loading the data table package. But I
> assume it would work with a data table too, Check this, though!)
>
>> DT <- within(DT, exposure <- 
>> cut(fini,as.Date(c("2000-01-01","2006-01-01","2006-06-30","2006-12-21")), 
>> labels= c(1,.87,.5)))
>
>> DT
>   id   fini group exposure
> 1  2 2005-04-20 A1
> 2  2 2005-04-20 A1
> 3  2 2005-04-20 A1
> 4  5 2006-02-19 B 0.87
> 5  5 2006-02-19 B 0.87
> 6  7 2006-10-08 A  0.5
> 7  7 2006-10-08 A  0.5
>
>
> (but note that exposure is a factor, not numeric)
>
>
> Cheers,
> Bert
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Mon, Sep 26, 2016 at 10:05 AM, Ista Zahn  wrote:
>> Hi Frank,
>>
>> lapply(DT) iterates over each column. That doesn't seem to be what you want.
>>
>> There are probably better ways, but here is one approach.
>>
>> DT[, exposure := vector(mode = "numeric", length = .N)]
>> DT[fini < as.Date("2006-01-01"), exposure := 1]
>> DT[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30"),
>>   exposure := difftime(as.Date("2007-01-01"), fini, units="days")/365.25]
>> DT[fini >= as.Date("2006-07-01"), exposure := 0.5]
>>
>> Best,
>> Ista
>>
>> On Mon, Sep 26, 2016 at 11:28 AM, Frank S.  wrote:
>>> Dear all,
>>>
>>> I have a R data table like this:
>>>
>>> DT <- data.table(
>>>   id = rep(c(2, 5, 7), c(3, 2, 2)),
>>>   fini = rep(as.Date(c('2005-04-20', '2006-02-19', '2006-10-08')), c(3, 2, 
>>> 2)),
>>>   group = rep(c("A", "B", "A"), c(3, 2, 2))  )
>>>
>>>
>>> I want to construct a new variable "exposure" defined as follows:
>>>
>>> 1) If "fini" earlier than 2006-01-01 --> "exposure" = 1
>>> 2) If "fini" in [2006-01-01, 2006-06-30] --> "exposure" = "2007-01-01" - 
>>> "fini"
>>> 3) If "fini" in [2006-07-01, 2006-12-31] --> "exposure" = 0.5
>>>
>>>
>>> So the desired output would be the following data table:
>>>
>>>idfini exposure group
>>> 1:  2 2005-04-201.00A
>>> 2:  2 2005-04-201.00A
>>> 3:  2 2005-04-201.00A
>>> 4:  5 2006-02-190.87B
>>> 5:  5 2006-02-190.87B
>>> 6:  7 2006-10-080.50A
>>> 7:  7 2006-10-080.50A
>>>
>>>
>>> I have tried:
>>>
>>> DT <- DT[ , list(id, fini, exposure = 0, group)]
>>> DT.new <- lapply(DT, function(exposure){
>>>   exposure[fini < as.Date("2006-01-01")] <- 1   # 1st case
>>>   exposure[fini >= as.Date("2006-01-01") & fini <= 
>>> as.Date("2006-06-30")] <- difftime(as.Date("2007-01-01"), fini, 
>>> units="days")/365.25 # 2nd case
>>> exposure[fini >= as.Date("2006-07-01") & fini <= as.Date("2006-12-31")] 
>>> <- 0.5   # 3rd case
>>>   exposure  # return value
>>>   })
>>>
>>>
>>> But I get an error message.
>>>
>>> Thanks for any help!!
>>>
>>>
>>> Frank S.
>>>
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> __
>>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> 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 -- To UNSUBSCRIBE and more, see
>> 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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Using lapply in R data table

[Bert Gunter]

This seems like a job for cut() .

(I made DT a data frame to avoid loading the data table package. But I
assume it would work with a data table too, Check this, though!)

> DT <- within(DT, exposure <- 
> cut(fini,as.Date(c("2000-01-01","2006-01-01","2006-06-30","2006-12-21")), 
> labels= c(1,.87,.5)))

> DT
  id   fini group exposure
1  2 2005-04-20 A1
2  2 2005-04-20 A1
3  2 2005-04-20 A1
4  5 2006-02-19 B 0.87
5  5 2006-02-19 B 0.87
6  7 2006-10-08 A  0.5
7  7 2006-10-08 A  0.5


(but note that exposure is a factor, not numeric)


Cheers,
Bert

Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Mon, Sep 26, 2016 at 10:05 AM, Ista Zahn  wrote:
> Hi Frank,
>
> lapply(DT) iterates over each column. That doesn't seem to be what you want.
>
> There are probably better ways, but here is one approach.
>
> DT[, exposure := vector(mode = "numeric", length = .N)]
> DT[fini < as.Date("2006-01-01"), exposure := 1]
> DT[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30"),
>   exposure := difftime(as.Date("2007-01-01"), fini, units="days")/365.25]
> DT[fini >= as.Date("2006-07-01"), exposure := 0.5]
>
> Best,
> Ista
>
> On Mon, Sep 26, 2016 at 11:28 AM, Frank S.  wrote:
>> Dear all,
>>
>> I have a R data table like this:
>>
>> DT <- data.table(
>>   id = rep(c(2, 5, 7), c(3, 2, 2)),
>>   fini = rep(as.Date(c('2005-04-20', '2006-02-19', '2006-10-08')), c(3, 2, 
>> 2)),
>>   group = rep(c("A", "B", "A"), c(3, 2, 2))  )
>>
>>
>> I want to construct a new variable "exposure" defined as follows:
>>
>> 1) If "fini" earlier than 2006-01-01 --> "exposure" = 1
>> 2) If "fini" in [2006-01-01, 2006-06-30] --> "exposure" = "2007-01-01" - 
>> "fini"
>> 3) If "fini" in [2006-07-01, 2006-12-31] --> "exposure" = 0.5
>>
>>
>> So the desired output would be the following data table:
>>
>>idfini exposure group
>> 1:  2 2005-04-201.00A
>> 2:  2 2005-04-201.00A
>> 3:  2 2005-04-201.00A
>> 4:  5 2006-02-190.87B
>> 5:  5 2006-02-190.87B
>> 6:  7 2006-10-080.50A
>> 7:  7 2006-10-080.50A
>>
>>
>> I have tried:
>>
>> DT <- DT[ , list(id, fini, exposure = 0, group)]
>> DT.new <- lapply(DT, function(exposure){
>>   exposure[fini < as.Date("2006-01-01")] <- 1   # 1st case
>>   exposure[fini >= as.Date("2006-01-01") & fini <= 
>> as.Date("2006-06-30")] <- difftime(as.Date("2007-01-01"), fini, 
>> units="days")/365.25 # 2nd case
>> exposure[fini >= as.Date("2006-07-01") & fini <= as.Date("2006-12-31")] 
>> <- 0.5   # 3rd case
>>   exposure  # return value
>>   })
>>
>>
>> But I get an error message.
>>
>> Thanks for any help!!
>>
>>
>> Frank S.
>>
>>
>> [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> 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 -- To UNSUBSCRIBE and more, see
> 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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R-es] Variable Progresiva

[Carlos J. Gil Bellosta]

Hola, ¿qué tal?

He leído en este hilo la respuesta correcta a la pregunta equivocada.
Realmente, no quieres crear variables con nombres generados con un
subíndice; quieres una lista.

Un saludo,

Carlos J. Gil Bellosta
http://www.datanalytics.com

El 26 de septiembre de 2016, 16:40, Rafael Saturno <
rafael_satu...@hotmail.com> escribió:

> Una duda, si fuese a leer 3 archivos csv que se llamar archivo 1, archivo
> 2 y archivo 3 también se podría usar? gracias
>
>
> 
> De: Isidro Hidalgo Arellano 
> Enviado: lunes, 26 de septiembre de 2016 07:48 a.m.
> Para: 'Rafael Saturno'; 'R'
> Asunto: RE: [R-es] Variable Progresiva
>
> Lo que quieres hacer se hace con la función "assign()". En tu ejemplo:
> assign(paste("k", i, sep = ""), sum(1:i))
> Un saludo
>
> Isidro Hidalgo Arellano
> Observatorio del Mercado de Trabajo
> Consejería de Economía, Empresas y Empleo
> http://www.castillalamancha.es/
> Inicio | Gobierno de Castilla-La Mancha
> www.castillalamancha.es
> Web oficial del gobierno autonómico de Castilla-La Mancha con información
> sobre actividad administrativa, economía, educación, sanidad, servicios
> sociales, sede ...
>
>
>
>
>
>
> -Mensaje original-
> De: R-help-es [mailto:r-help-es-boun...@r-project.org] En nombre de Rafael
> Saturno
> Enviado el: lunes, 26 de septiembre de 2016 1:43
> Para: R 
> Asunto: [R-es] Variable Progresiva
>
> Hola Comunidad,
>
>
> Tengo una duda,
>
>
> Queria que en un For si fuese ejecutando un proceso desde 1 hasta 5 por
> poner un ejemplo , y que el resultado se fuese guardando en variables que
> se
> llamar Ki, es decir k1, k2, k3...
>
>
> Un ejemplo de como crei que funcionaria y no lo hizo xD
>
>
> for (i in 1:3) {
>
> paste("k", i, sep = "") <- sum(1:i)
>
> }
>
> Esperaba se crearan las variables k1 = 1, k2= 3 y k3 = 6
>
>
> Gracias
>
> [[alternative HTML version deleted]]
>
> ___
> R-help-es mailing list
> R-help-es@r-project.org
> https://stat.ethz.ch/mailman/listinfo/r-help-es
>
>
> [[alternative HTML version deleted]]
>
>
> ___
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> https://stat.ethz.ch/mailman/listinfo/r-help-es
>

[[alternative HTML version deleted]]

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Re: [R] Using lapply in R data table

[Ista Zahn]

Hi Frank,

lapply(DT) iterates over each column. That doesn't seem to be what you want.

There are probably better ways, but here is one approach.

DT[, exposure := vector(mode = "numeric", length = .N)]
DT[fini < as.Date("2006-01-01"), exposure := 1]
DT[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30"),
  exposure := difftime(as.Date("2007-01-01"), fini, units="days")/365.25]
DT[fini >= as.Date("2006-07-01"), exposure := 0.5]

Best,
Ista

On Mon, Sep 26, 2016 at 11:28 AM, Frank S.  wrote:
> Dear all,
>
> I have a R data table like this:
>
> DT <- data.table(
>   id = rep(c(2, 5, 7), c(3, 2, 2)),
>   fini = rep(as.Date(c('2005-04-20', '2006-02-19', '2006-10-08')), c(3, 2, 
> 2)),
>   group = rep(c("A", "B", "A"), c(3, 2, 2))  )
>
>
> I want to construct a new variable "exposure" defined as follows:
>
> 1) If "fini" earlier than 2006-01-01 --> "exposure" = 1
> 2) If "fini" in [2006-01-01, 2006-06-30] --> "exposure" = "2007-01-01" - 
> "fini"
> 3) If "fini" in [2006-07-01, 2006-12-31] --> "exposure" = 0.5
>
>
> So the desired output would be the following data table:
>
>idfini exposure group
> 1:  2 2005-04-201.00A
> 2:  2 2005-04-201.00A
> 3:  2 2005-04-201.00A
> 4:  5 2006-02-190.87B
> 5:  5 2006-02-190.87B
> 6:  7 2006-10-080.50A
> 7:  7 2006-10-080.50A
>
>
> I have tried:
>
> DT <- DT[ , list(id, fini, exposure = 0, group)]
> DT.new <- lapply(DT, function(exposure){
>   exposure[fini < as.Date("2006-01-01")] <- 1   # 1st case
>   exposure[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30")] 
> <- difftime(as.Date("2007-01-01"), fini, units="days")/365.25 # 2nd case
> exposure[fini >= as.Date("2006-07-01") & fini <= as.Date("2006-12-31")] 
> <- 0.5   # 3rd case
>   exposure  # return value
>   })
>
>
> But I get an error message.
>
> Thanks for any help!!
>
>
> Frank S.
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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 -- To UNSUBSCRIBE and more, see
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] curve() doesn't seem to use the whole range of x? And Error: longer object length is not a multiple of shorter object length

[Matti Viljamaa]

> On 26 Sep 2016, at 19:41, Matti Viljamaa  wrote:
> 
> Thank you.
> 
> However, I’m having some trouble converting your code to use clka, because 
> the model I was using was:
> 
> fit2 <- lm(ruotsi.pist ~ mies + koulu + clka + koulu*clka, data=dta)

I mean, not to use clka to replace lka. But to use the above fit2, rather than 
your fit2.

>> On 25 Sep 2016, at 21:23, Jeff Newmiller  wrote:
>> 
>> This illustrates why you need to post a reproducible example. You have a 
>> number of confounding factors in your code.
>> 
>> First, "data" is a commonly-used function... avoid using it for variable 
>> names.
>> 
>> Second, using the attach function this way leads to confusion... best to 
>> forget this function until you start building packages.
>> 
>> Third, clka is linearly dependent on lka, so having them both in the 
>> regression is not possible. In this case lm has chosen to ignore clka so 
>> that bs("clka") is NA.
>> 
>> Fourth, curve expects you to give it a function, and instead you have given 
>> it a vector.
>> 
>> Fifth, you are plotting versus lka, but attempting to vary clka in the curve 
>> call.
>> 
>> There are a number of directions you could go with this to get a working 
>> output... below is my version.
>> 
>> dta <- read.table( 
>> "http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt;, header=TRUE )
>> fit2 <- lm( ruotsi.pist ~ mies + koulu*lka, data=dta )
>> bs <- coef( fit2 )
>> rpBylka <- function( lka ) {
>> kouluB <- factor( "B", levels = levels( dta$koulu ) )
>> newdta <- expand.grid( mies=0, koulu=kouluB, lka=lka )
>> predict( fit2, newdata = newdta )
>> }
>> dtaKouluB <- subset( dta, koulu == "B" )
>> varitB <- dtaKouluB$mies
>> varitB[ varitB == 0 ] <- 2
>> plot( dtaKouluB$lka
>>   , dtaKouluB$ruotsi.pist
>>   , col=varitB
>>   , pch=16
>>   , xlab='lka'
>>   , ylab='ruotsi.pist'
>>   , main='Lukio B'
>>   )
>> curve( rpBylka, from = min( dta$lka ), max( dta$lka ), add=TRUE, col="red" )
>> 
>> On Sun, 25 Sep 2016, Matti Viljamaa wrote:
>> 
>>> 
 On 25 Sep 2016, at 19:37, Matti Viljamaa  wrote:
 
 Okay here?s a pretty short code to reproduce it:
 
 data <- 
 read.table("http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt;, 
 header=TRUE)
>>> 
>>> data$clka <- I(data$lka - mean(data$lka))
>>> 
 attach(data)
 
 fit2 <- lm(ruotsi.pist ~ mies + koulu + lka + koulu*clka)
 
 bs <- coef(fit2)
 
 varitB <- c(data[koulu == 'B',]$mies)
 varitB[varitB == 0] = 2
 plot(data[data$koulu == 'B',]$lka, data[koulu == 'B',]$ruotsi.pist, 
 col=varitB, pch=16, xlab='', ylab='', main='Lukio B?)
 
 curve(bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*x+bs["kouluB:clka"]*clka,
  from=min(lka), to=max(lka), add=TRUE, col='red')
 
 
> On 25 Sep 2016, at 19:24, Jeff Newmiller  wrote:
> 
> Go directly to C. Do not pass go, do not collect $200.
> 
> You think curve does something, but you are missing what it actually 
> does. Since you don't seem to be learning from reading ?curve or from our 
> responses, you need to give us an example you can learn from.
> --
> Sent from my phone. Please excuse my brevity.
> 
> On September 25, 2016 9:04:09 AM PDT, mviljamaa  
> wrote:
>> On 2016-09-25 18:52, Jeff Newmiller wrote:
>>> You seem to be confused about what curve is doing vs. what you are
>>> doing.
>> 
>> But my x-range in curve()'s parameters from and to should be the entire
>> 
>> lka vector, since they are from=min(lka) and to=max(lka). Then why does
>> 
>> this not span the entire of lka? Because of duplicate entries or what?
>> 
>> It seems like I cannot use curve(), since my x-axis must be exactly lka
>> 
>> for the function to plot the y value for every lka value.
>> 
>>> A) Compute the points you want to plot and put them into 2 vectors.
>>> Then figure out how to plot those vectors. Then (perhaps) consider
>>> putting that all into one line of code again.
>>> 
>>> B) The predict function is the preferred way to compute points. It
>> may
>>> be educational for you to do the computations by hand at first, but
>> in
>>> the long run using predict will help you avoid problems getting the
>>> equations right in multiple places in your script.
>>> 
>>> C) Learn what makes an example reproducible (e.g. [1] or [2]), and
>> ask
>>> your questions with reproducible code and data so we can give you
>>> concrete responses.
>>> 
>>> [1] http://adv-r.had.co.nz/Reproducibility.html
>>> [2]
>>> 
>> http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
>>> --
>>> Sent from my phone. Please excuse my brevity.
>>> 
>>> On September 25, 2016 8:36:49 AM PDT, mviljamaa 

Re: [R] curve() doesn't seem to use the whole range of x? And Error: longer object length is not a multiple of shorter object length

[Matti Viljamaa]

Thank you.

However, I’m having some trouble converting your code to use clka, because the 
model I was using was:

fit2 <- lm(ruotsi.pist ~ mies + koulu + clka + koulu*clka, data=dta)


> On 25 Sep 2016, at 21:23, Jeff Newmiller  wrote:
> 
> This illustrates why you need to post a reproducible example. You have a 
> number of confounding factors in your code.
> 
> First, "data" is a commonly-used function... avoid using it for variable 
> names.
> 
> Second, using the attach function this way leads to confusion... best to 
> forget this function until you start building packages.
> 
> Third, clka is linearly dependent on lka, so having them both in the 
> regression is not possible. In this case lm has chosen to ignore clka so that 
> bs("clka") is NA.
> 
> Fourth, curve expects you to give it a function, and instead you have given 
> it a vector.
> 
> Fifth, you are plotting versus lka, but attempting to vary clka in the curve 
> call.
> 
> There are a number of directions you could go with this to get a working 
> output... below is my version.
> 
> dta <- read.table( 
> "http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt;, header=TRUE )
> fit2 <- lm( ruotsi.pist ~ mies + koulu*lka, data=dta )
> bs <- coef( fit2 )
> rpBylka <- function( lka ) {
>  kouluB <- factor( "B", levels = levels( dta$koulu ) )
>  newdta <- expand.grid( mies=0, koulu=kouluB, lka=lka )
>  predict( fit2, newdata = newdta )
> }
> dtaKouluB <- subset( dta, koulu == "B" )
> varitB <- dtaKouluB$mies
> varitB[ varitB == 0 ] <- 2
> plot( dtaKouluB$lka
>, dtaKouluB$ruotsi.pist
>, col=varitB
>, pch=16
>, xlab='lka'
>, ylab='ruotsi.pist'
>, main='Lukio B'
>)
> curve( rpBylka, from = min( dta$lka ), max( dta$lka ), add=TRUE, col="red" )
> 
> On Sun, 25 Sep 2016, Matti Viljamaa wrote:
> 
>> 
>>> On 25 Sep 2016, at 19:37, Matti Viljamaa  wrote:
>>> 
>>> Okay here?s a pretty short code to reproduce it:
>>> 
>>> data <- 
>>> read.table("http://users.jyu.fi/~slahola/files/glm1_datoja/yoruotsi.txt;, 
>>> header=TRUE)
>> 
>> data$clka <- I(data$lka - mean(data$lka))
>> 
>>> attach(data)
>>> 
>>> fit2 <- lm(ruotsi.pist ~ mies + koulu + lka + koulu*clka)
>>> 
>>> bs <- coef(fit2)
>>> 
>>> varitB <- c(data[koulu == 'B',]$mies)
>>> varitB[varitB == 0] = 2
>>> plot(data[data$koulu == 'B',]$lka, data[koulu == 'B',]$ruotsi.pist, 
>>> col=varitB, pch=16, xlab='', ylab='', main='Lukio B?)
>>> 
>>> curve(bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*x+bs["kouluB:clka"]*clka,
>>>  from=min(lka), to=max(lka), add=TRUE, col='red')
>>> 
>>> 
 On 25 Sep 2016, at 19:24, Jeff Newmiller  wrote:
 
 Go directly to C. Do not pass go, do not collect $200.
 
 You think curve does something, but you are missing what it actually does. 
 Since you don't seem to be learning from reading ?curve or from our 
 responses, you need to give us an example you can learn from.
 --
 Sent from my phone. Please excuse my brevity.
 
 On September 25, 2016 9:04:09 AM PDT, mviljamaa  wrote:
> On 2016-09-25 18:52, Jeff Newmiller wrote:
>> You seem to be confused about what curve is doing vs. what you are
>> doing.
> 
> But my x-range in curve()'s parameters from and to should be the entire
> 
> lka vector, since they are from=min(lka) and to=max(lka). Then why does
> 
> this not span the entire of lka? Because of duplicate entries or what?
> 
> It seems like I cannot use curve(), since my x-axis must be exactly lka
> 
> for the function to plot the y value for every lka value.
> 
>> A) Compute the points you want to plot and put them into 2 vectors.
>> Then figure out how to plot those vectors. Then (perhaps) consider
>> putting that all into one line of code again.
>> 
>> B) The predict function is the preferred way to compute points. It
> may
>> be educational for you to do the computations by hand at first, but
> in
>> the long run using predict will help you avoid problems getting the
>> equations right in multiple places in your script.
>> 
>> C) Learn what makes an example reproducible (e.g. [1] or [2]), and
> ask
>> your questions with reproducible code and data so we can give you
>> concrete responses.
>> 
>> [1] http://adv-r.had.co.nz/Reproducibility.html
>> [2]
>> 
> http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
>> --
>> Sent from my phone. Please excuse my brevity.
>> 
>> On September 25, 2016 8:36:49 AM PDT, mviljamaa 
>> wrote:
>>> On 2016-09-25 18:30, Duncan Murdoch wrote:
 On 25/09/2016 9:10 AM, Matti Viljamaa wrote:
> Writing:
> 
> 
>>> 
> bs["(Intercept)"]+bs["mies"]*0+bs["kouluB"]+bs["lka"]*lka+bs["kouluB:clka"]*clka
> 
> 

[R] Problem in "cannot allocate vector of size"

[GwanSeon Kim]

Hi R-Users,
I am running raster to point code in R, but I have an error message that
"cannot allocate vector of size 1.7 Gb". One of my friends run the same
code I used, and it is working with his computer. I am using Window 7
64-bit with 16 GB ram. When I check memory size and limit in RStudio, I
have memory.size() : [1] 11205.57 and memory.limit() : [1] 16341. I already
searched on google to solve this problem, but I could not fix it. Could
anyone possibly help me to solve this problem?
Thanks,

Sun

[[alternative HTML version deleted]]

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

[R] [R-pkgs] package Rdice released

[Gennaro Tedesco]

The package "Rdice" has just been released on CRAN. It contains a
collection of functions to simulate dice rolls and the like. In particular,
experiments and exercises can be performed looking at combinations and
permutations of values in dice rolls and coin flips, together with the
corresponding frequencies of occurrences. When applying each function, the
user has to input the number of times (rolls, flips) to toss the dice.
Moreover, the package provides functions to generate non-transitive sets of
dice (like Efron's) and to check whether a given set of dice is
non-transitive with given probability.

A vignette with example and use cases is provided.

Best regards,
Gennaro

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[R] src/Makevars ignored ?

[Eric Deveaud]

Hello,

as far as I understood the R library generic compilation mechanism, 
compilation of C//C++ sources is controlde


1) at system level by the ocntentos RHOME/etc/Makeconf
2) at user level by the content of ~/.R/Makevars
3) at package level by the content of src/Makevars

Problem I have is that src/Makevars is ignored


see following example:

R is compiled and use the following CC and CFLAGS definition

bigmess:epactsR/src > R CMD config CC
gcc -std=gnu99
bigmess:epactsR/src > R CMD config CFLAGS
-Wall -g

so building C sources lead to the following

bigmess:epactsR/src > R CMD SHLIB index.c
gcc -std=gnu99 -I/local/gensoft2/adm/lib64/R/include -DNDEBUG 
-I/usr/local/include-fpic  -Wall -g  -c index.c -o index.o


normal, it uses defintion from RHOME/etc/Makeconf


when I set upp a ~/.R/Makevars that overwrite CC and CFLAGS definition.

bigmess:epactsR/src > cat ~/.R/Makevars
CC=gcc
CFLAGS=-O3
bigmess:epactsR/src > R CMD SHLIB index.c
gcc -I/local/gensoft2/adm/lib64/R/include -DNDEBUG  -I/usr/local/include 
   -fpic  -O3 -c index.c -o index.o

gcc -std=gnu99 -shared -L/usr/local/lib64 -o index.so index.o


OK CC and CFLAGS are honored and set accordingly to ~/.R/Makevars


but when I try to use src/Makevars, it is ignored

bigmess:epactsR/src > cat ~/.R/Makevars
cat: /home/edeveaud/.R/Makevars: No such file or directory
bigmess:epactsR/src > cat ./Makevars
CC = gcc
CFLAGS=-O3
bigmess:epactsR/src > R CMD SHLIB index.c
gcc -std=gnu99 -I/local/gensoft2/adm/lib64/R/include -DNDEBUG 
-I/usr/local/include-fpic  -Wall -g  -c index.c -o index.o



what I have missed or is there something wrong ?


PS I tested the ssame behaviour with various version of R from R/2.15 to 
R/3.3


best regards

Eric

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[R] 32 and 64 bit R

[Mike meyer]

Hello,
 
I have both 32 and 64 bit verions of R installed. What happens if I open a 
workspace saved from 64 bit R
in the 32 bit version or conversely?
I am fairly careless but never noticed any problems. 
 

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[R] Using lapply in R data table

[Frank S.]

Dear all,

I have a R data table like this:

DT <- data.table(
  id = rep(c(2, 5, 7), c(3, 2, 2)),
  fini = rep(as.Date(c('2005-04-20', '2006-02-19', '2006-10-08')), c(3, 2, 2)),
  group = rep(c("A", "B", "A"), c(3, 2, 2))  )


I want to construct a new variable "exposure" defined as follows:

1) If "fini" earlier than 2006-01-01 --> "exposure" = 1
2) If "fini" in [2006-01-01, 2006-06-30] --> "exposure" = "2007-01-01" - "fini"
3) If "fini" in [2006-07-01, 2006-12-31] --> "exposure" = 0.5


So the desired output would be the following data table:

   idfini exposure group
1:  2 2005-04-201.00A
2:  2 2005-04-201.00A
3:  2 2005-04-201.00A
4:  5 2006-02-190.87B
5:  5 2006-02-190.87B
6:  7 2006-10-080.50A
7:  7 2006-10-080.50A


I have tried:

DT <- DT[ , list(id, fini, exposure = 0, group)]
DT.new <- lapply(DT, function(exposure){
  exposure[fini < as.Date("2006-01-01")] <- 1   # 1st case
  exposure[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30")] 
<- difftime(as.Date("2007-01-01"), fini, units="days")/365.25 # 2nd case
exposure[fini >= as.Date("2006-07-01") & fini <= as.Date("2006-12-31")] <- 
0.5   # 3rd case
  exposure  # return value
  })


But I get an error message.

Thanks for any help!!


Frank S.


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Re: [R-es] Listado de Valores

[Carlos Ortega]

Hola,

Puedes hacerlo con "unique(tu_data_frame$columa_fechas)" o con
"table(tu_data_frame$columa_fechas)"

Saludos,
Carlos Ortega
www.qualityexcellence.es

El 26 de septiembre de 2016, 17:14, Rafael Saturno <
rafael_satu...@hotmail.com> escribió:

> Hola Comunidad,
>
>
> Una duda
>
>
> Como hago para saber los valores que hay en un campo de un Frame?
>
> Tengo un csv con mas de un millon de registros y en un campo de fecha
> quiero ver cuales fechas salen
>
>
> Muchas Gracias
>
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-- 
Saludos,
Carlos Ortega
www.qualityexcellence.es

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Re: [R-es] Variable Progresiva

[Rafael Saturno]

Una duda, si fuese a leer 3 archivos csv que se llamar archivo 1, archivo 2 y 
archivo 3 tambi�n se podr�a usar? gracias



De: Isidro Hidalgo Arellano 
Enviado: lunes, 26 de septiembre de 2016 07:48 a.m.
Para: 'Rafael Saturno'; 'R'
Asunto: RE: [R-es] Variable Progresiva

Lo que quieres hacer se hace con la funci�n "assign()". En tu ejemplo:
assign(paste("k", i, sep = ""), sum(1:i))
Un saludo

Isidro Hidalgo Arellano
Observatorio del Mercado de Trabajo
Consejer�a de Econom�a, Empresas y Empleo
http://www.castillalamancha.es/
Inicio | Gobierno de Castilla-La Mancha
www.castillalamancha.es
Web oficial del gobierno auton�mico de Castilla-La Mancha con informaci�n sobre 
actividad administrativa, econom�a, educaci�n, sanidad, servicios sociales, 
sede ...






-Mensaje original-
De: R-help-es [mailto:r-help-es-boun...@r-project.org] En nombre de Rafael
Saturno
Enviado el: lunes, 26 de septiembre de 2016 1:43
Para: R 
Asunto: [R-es] Variable Progresiva

Hola Comunidad,


Tengo una duda,


Queria que en un For si fuese ejecutando un proceso desde 1 hasta 5 por
poner un ejemplo , y que el resultado se fuese guardando en variables que se
llamar Ki, es decir k1, k2, k3...


Un ejemplo de como crei que funcionaria y no lo hizo xD


for (i in 1:3) {

paste("k", i, sep = "") <- sum(1:i)

}

Esperaba se crearan las variables k1 = 1, k2= 3 y k3 = 6


Gracias

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