Re: [R-es] Resumen de R-help-es, Vol 176, Envío 1

[Amable Moreno]

Me chas gracias!!
Ya lo pude instalar!!!

El El sáb, 7 de oct. de 2023 a la(s) 07:01, 
escribió:

> Envíe los mensajes para la lista R-help-es a
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> "Re: Contents of R-help-es digest...". Además, por favor, incluya en
> la respuesta sólo aquellas partes del mensaje a las que está
> respondiendo.
>
>
> Asuntos del día:
>
>1. TeachingDemos (Amable Moreno)
>2. Re: TeachingDemos (Javier Marcuzzi)
>
> ------
>
> Message: 1
> Date: Fri, 6 Oct 2023 17:26:43 -0300
> From: Amable Moreno 
> To: r-help-es@r-project.org
> Subject: [R-es] TeachingDemos
> Message-ID:
> <
> cadothbhp6s5_ulc5eitx0otc7e37kx6pp2c+rekfcioj9bf...@mail.gmail.com>
> Content-Type: text/plain; charset="utf-8"
>
> Estimados colegas
>
> Hasta el año pasado he usado el paquete "TeachingDemos"
> Pero ahora, este paque no lo tengo instalado y no lo puedo volver a
> instalar
> me marca error
> Agradecweria me constesten, por favor, si alguien sabe porque ocurre
>
> [[alternative HTML version deleted]]
>
>
>
>
> --
>
> Message: 2
> Date: Fri, 6 Oct 2023 19:54:10 -0300
> From: Javier Marcuzzi 
> To: Amable Moreno 
> Cc: "R-help-es@r-project.org" 
> Subject: Re: [R-es] TeachingDemos
> Message-ID: <32218d4b-5644-4ac8-8489-f7c01ecdd...@gmail.com>
> Content-Type: text/plain; charset="utf-8"
>
> Estimada Amable Moreno
>
> En Mac, R actualizado, anda. Observe mi escritura y la respuesta.
>
> > install.packages("TeachingDemos")
> trying URL '
> https://cran.rstudio.com/bin/macosx/big-sur-arm64/contrib/4.3/TeachingDemos_2.12.tgz
> '
> Content type 'application/x-gzip' length 1237702 bytes (1.2 MB)
> ==
> downloaded 1.2 MB
>
>
> Javier Rubén Marcuzzi
>
> > El 6 oct 2023, a las 17:26, Amable Moreno 
> escribió:
> >
> > TeachingDemos
>
>
> [[alternative HTML version deleted]]
>
>
>
>
> --
>
> Subject: Pié de página del digest
>
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> Fin de Resumen de R-help-es, Vol 176, Envío 1
> *
>

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[R-es] TeachingDemos

[Amable Moreno]

Estimados colegas

Hasta el año pasado he usado el paquete "TeachingDemos"
Pero ahora, este paque no lo tengo instalado y no lo puedo volver a instalar
me marca error
Agradecweria me constesten, por favor, si alguien sabe porque ocurre

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Re: [R-es] error

[Amable Moreno]

Muchas gracias por la ayuda.

El El lun, 10 de jul. de 2023 a la(s) 08:43, Carlos Ortega <
c...@qualityexcellence.es> escribió:

> Hola,
>
> Sí, el error no es estrictamente de "SemPlot" si no de una librería de la
> que depende...
>
>- *namespace ‘vctrs’ 0.4.1 is being loaded, but >= 0.5.0 is required"*
>
> Tienes que actualizar la librería "vctrs".
> De forma general, si haces un "Update" de todas las librerías, seguramente
> se resuelvan este y otros problemas de dependencias.
>
> Gracias,
> Carlos Ortega
> www.qualityexcellence.es
>
>
> El dom, 9 jul 2023 a las 4:08, Amable Moreno ()
> escribió:
>
>> Necesito que me ayuden a resolver el siguiente error al intentar usar el
>> paquete "semPlot": El error es:
>> "package or namespace load failed for ‘semPlot’ in loadNamespace(i,
>> c(lib.loc, .libPaths()), versionCheck = vI[[i]]):
>>  namespace ‘vctrs’ 0.4.1 is being loaded, but >= 0.5.0 is required"
>>
>> [[alternative HTML version deleted]]
>>
>> ___
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>
>
> --
> Saludos,
> Carlos Ortega
> www.qualityexcellence.es
>

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[R-es] error

[Amable Moreno]

Necesito que me ayuden a resolver el siguiente error al intentar usar el
paquete "semPlot": El error es:
"package or namespace load failed for ‘semPlot’ in loadNamespace(i,
c(lib.loc, .libPaths()), versionCheck = vI[[i]]):
 namespace ‘vctrs’ 0.4.1 is being loaded, but >= 0.5.0 is required"

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Re: [R-es] Resumen de R-help-es, Vol 172, Envío 23

[Amable Moreno]

Buen dia
Tengo un problema con el paquete "semPlot", no puedo realizar un digrama de
senderos porque me da este error::
"package or namespace load failed for ‘semPlot’ in loadNamespace(i,
c(lib.loc, .libPaths()), versionCheck = vI[[i]]):
 namespace ‘vctrs’ 0.4.1 is being loaded, but >= 0.5.0 is required"
Agradeceria que alguien me ayude

El vie, 16 jun 2023 a la(s) 17:16, 
escribió:

> Envíe los mensajes para la lista R-help-es a
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> linea del asunto (subject) para que el texto sea mas especifico que:
> "Re: Contents of R-help-es digest...". Además, por favor, incluya en
> la respuesta sólo aquellas partes del mensaje a las que está
> respondiendo.
>
>
> Asuntos del día:
>
>1. Re: Seleccionar valores consecutivos en un dataframe
>   (Juan Carlos Lopez Mesa)
>2. Supuestos de una ANOVA (Yesica Pallavicini Fernandez)
>3. Opciones de guardado - gráfico en png (David Camilo Gomez Medina)
>
> --
>
> Message: 1
> Date: Fri, 16 Jun 2023 05:02:14 -0500
> From: Juan Carlos Lopez Mesa 
> To: jose luis 
> Cc: "r-help-es@r-project.org" , Jaume Tormo
> 
> Subject: Re: [R-es] Seleccionar valores consecutivos en un dataframe
> Message-ID:
> <
> cahuwr6oentz4dpwxyuop0k39txmyfv4nrdmrkxn4kayp1k5...@mail.gmail.com>
> Content-Type: text/plain; charset="utf-8"
>
> Hola,
> Este código le puede servir para lo que necesita
> library(dplyr)
>
> ejemplo$dia <- as.Date(ejemplo$dia)
>
> ejemplo <- arrange(ejemplo, dia)
>
> ejemplo$suma <- NA
>
> ejemplo$suma[1] <- as.numeric(ejemplo$germ)
>
> for(i in 2:nrow(ejemplo)){
>
>   ejemplo$suma[i] <- ifelse(ejemplo$germ[i],
> ejemplo$suma[i-1] + ejemplo$p[i],
> 0)
> }
>
>
> Saludos
>
>
>
> El vie, 16 jun 2023 a las 3:21, jose luis via R-help-es (<
> r-help-es@r-project.org>) escribió:
>
> >  Hola Jaumecomo todos los valores de p son menores de 1he entendido
> > esto: por ejemplo en la fila 5 que sería el segundo TRUE, quieres que la
> > nueva variable sea la suma del valor de la fila 4 mas el de la fila 5 y
> asi
> > sucesivamente. ¿o me desvío mucho?
> > En viernes, 16 de junio de 2023, 09:28:18 CEST, Jaume Tormo via
> > R-help-es  escribió:
> >
> >  Estimados eRReros,
> > Tengo un df como el adjunto (en txt y como objeto de R)Como veréis hay
> una
> > columna T/F que se llama germ y cada fila corresponde a datos de un día.
> > Me gustaría que R fuera siguiendo la columna germ y en cada serie de días
> > con TRUE sumara el valor de la columna p. Se trata de saber si en esa
> serie
> > de días con valor T el total de p es mayor que 1 o no.
> > He hecho algo parecido con rle() que me cuenta la longitud de las series
> > de TRUE, pero este siguiente paso no se como darlo.Si uso apply o subset
> me
> > toma todas las filas del df con T en la columna germ. Lo que no se el
> como
> > decirle a R que empiece por el principio y vaya tomando grupo a grupo.Me
> > imagino que podría llegar a construir un bucle que lo hiciera, pero no
> > quiero pasarme tres horas dándole vueltas si hay una función o
> combinación
> > de funciones que lo hace ¿Alguna sugerencia o me pongo ya con el bucle?
> >
> > Muchas gracias.
> > Jaume.
> >
> >
> > --
> > Dr. Jaume Tormo.
> > Area of Ecology
> > Department of Agrarian and Environmental Sciences
> > Technological College. Agri-food and Environment
> > University of Zaragoza, Spain
> > 0034 974292678
> > https://flipboard.com/@jaumetormo/hallazgos-interesantes-bj8opmboy
> > https://acercad.wordpress.com/
> >
> > ___
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> >
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[R-es] covarianza

[Amable Moreno]

Quisiera saber, si existe algún paquete de R específico para los modelos
lineales, en particular el modelo de covarianza

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[R-es] paquete npsm

[Amable Moreno]

Buen dia a todos
Quisiera saber con qué paquete hacer el test de Fligner Killen
porque lo solia resolver con el paquete npsm pero ahora ese paquete
no se puede bajar

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[R] recovering PLSDA equations for prediction (outside R)

[Diego Ruiz Moreno]

Hi all,
I'm trying to use the result of a PLSDA model outside R, but I'm having a
really hard time finding documentation on how to write the model from the
results in the fit object.
In order to provide a good starting point I believe that this code creates
a good model that is stored in the fit variable. I know that I can use the
predict function to create the predictions but I would like to "write the
predict function outside R".

Can anyone provide me with the equations? Thanks


# load the package
library(caret)
data(iris)
x <- iris[,1:4]
y <- iris[,5]
# fit model
fit <- plsda(x, y, probMethod="Bayes")
# summarize the fit
summary(fit)
# make predictions
predictions <- predict(fit, iris[,1:4])
# summarize accuracy
table(predictions, iris$Species)

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Re: [R] Soft Question: Where to find this reference.

[Oscar Moreno]

See book: Statistical Models in S (ISBN-10: 041283040X; ISBN-13: 
978-0412830402) Chapter 5, p. 145. 

• Analysis of Variance; Designed Experiments, J. M. Chambers, A. E. Freeny and 
R. M. Heiberger, Statistical Models in S, J. M. Chambers and T. J. Hastie 
(editors), Wadsworth & Brooks/Cole, Pacific Grove, California, 1992.

> On Jul 25, 2016, at 06:48, Justin Thong  wrote:
> 
> I notice a lot of r documentation refer to this reference below. I can't
> seem to find it anywhere.
> Does anyone have a link to point to where I can either view it or buy it?
> 
> 
> *Chambers, J. M., Freeny, A and Heiberger, R. M. (1992) Analysis of
> variance; designed experiments*
> 
> -- 
> Yours sincerely,
> Justin
> 
> *I check my email at 9AM and 4PM everyday*
> *If you have an EMERGENCY, contact me at +447938674419(UK) or
> +60125056192(Malaysia)*
> 
>   [[alternative HTML version deleted]]
> 
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[R-es] Problemas de instalación

[Justo de Jorge Moreno]

Buenos días,
He cambiado la versión de R como siempre a través de:
install.packages("installr")
library(installr)
updateR()
En esta ocasión trabajando bajo windows 10. La versión 3.2.2 me ofrece
menos mirror, pero mi principal problema es que no soy capaz de trabajar
con Rcommander (ahora lo hago con RStudio) me da la siguiente salida:

> utils:::menuInstallPkgs()
--- Please select a CRAN mirror for use in this session ---
probando la URL '
https://ftp.cixug.es/CRAN/bin/windows/contrib/3.2/Rcmdr_2.2-3.zip'
Content type 'application/zip' length 5464077 bytes (5.2 MB)
downloaded 5.2 MB

package ‘Rcmdr’ successfully unpacked and MD5 sums checked

The downloaded binary packages are in
C:\Users\JUSMI\AppData\Local\Temp\Rtmpeui6Qa\downloaded_packages

pero no lanza la interfaz, dando el siguiente fallo:

> library(Rcmdr)
Loading required package: car
Error in loadNamespace(i, c(lib.loc, .libPaths()), versionCheck = vI[[i]])
:
  there is no package called ‘lme4’
Error: package ‘car’ could not be loaded

Pensaba que había algún problema con la descarga de archivo, pero no parce.

Alguien puede ayudarme
Gracias por anticipado
Saludos
Justo

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Re: [R] ncdf size error

[Hernan A. Moreno Ramirez]

Sure, here it is. Thanks for any help with this Dr. Pierce:


# Exporting to NETCDF files 
#
# define the netcdf coordinate variables 
library(ncdf)
dim1 = dim.def.ncdf( Nodes,, seq(1,19000))
dim2= dim.def.ncdf( Time,Hours since 2005-01-01 00:00:00,seq(1:2190)) 
#final[1,2,]))

varT2 = var.def.ncdf(T2,celsius, list(dim1,dim2), -9, longname=T2)
varQ2 = var.def.ncdf(Q2,kg kg-1, list(dim1,dim2), -9, longname=Q2)
varQVAPOR = var.def.ncdf(QVAPOR,kg kg-1, list(dim1,dim2), -9, 
longname=QVAPOR)
varQCLOUD = var.def.ncdf(QCLOUD,kg kg-1, list(dim1,dim2), -9, 
longname=QCLOUD)
varPSFC = var.def.ncdf(PSFC,Pa, list(dim1,dim2), -9, longname=PSFC)
varW10 = var.def.ncdf(W10,ms-1, list(dim1,dim2), -9, longname=W10)
varW = var.def.ncdf(W,ms-1, list(dim1,dim2), -9, longname=W)
varVEGFRA = var.def.ncdf(VEGFRA,, list(dim1,dim2), -9, 
longname=VEGFRA)
varMAXVEGFRA = var.def.ncdf(MAXVEGFRA,, list(dim1,dim2), -9, 
longname=MAXVEGFRA)
varTPREC = var.def.ncdf(TPREC,mm, list(dim1,dim2), -9, longname=TPREC)
varFRAC_FROZ_PREC = var.def.ncdf(FRAC_FROZ_PREC,mm, list(dim1,dim2), 
-9, longname=FRAC_FROZ_PREC)
varSWDOWN = var.def.ncdf(SWDOWN,W m-2, list(dim1,dim2), -9, 
longname=SWDOWN)
varGLW = var.def.ncdf(GLW,W m-2, list(dim1,dim2), -9, longname=GLW)
varPBLH = var.def.ncdf(PBLH,m, list(dim1,dim2), -9, longname=PBLH)
varTSLB = var.def.ncdf(TSLB,Celsius, list(dim1,dim2), -9, 
longname=TSLB)

# associate the netcdf variable with a netcdf file   
# put the variable into the file, and
# close

nc.ex = create.ncdf(/media/D/output.nc,list(varT2, varQ2, varQVAPOR, 
varQCLOUD, varPSFC, varW10, varW, varVEGFRA, varMAXVEGFRA, varTPREC, 
varFRAC_FROZ_PREC, varSWDOWN, varGLW, varPBLH, varTSLB))

Error in R_nc_enddef: NetCDF: One or more variable sizes violate format 
constraints
Error in R_nc_sync: NetCDF: Operation not allowed in define mode

---
Hernan A. Moreno, Ph.D.
Postdoctoral Research Associate
Department of Civil  Architectural Engineering
Room 3038, Engineering Building
University of Wyoming
Phone: 480-3990571
http://www.public.asu.edu/~hamoreno/


From: davidwilliampie...@gmail.com davidwilliampie...@gmail.com on behalf of 
David W. Pierce dpie...@ucsd.edu
Sent: Monday, September 15, 2014 7:00 PM
To: Hernan A. Moreno Ramirez
Cc: r-help@r-project.org
Subject: Re: [R] ncdf size error

On Mon, Sep 15, 2014 at 4:20 PM, Hernan A. Moreno Ramirez
hmore...@uwyo.edu wrote:

 Hi I am using both ncdf and ncdf4 libraries and with both I keep getting the
 same error: Error in R_nc_enddef: NetCDF: One or more variable sizes violate
 format constraints. Error in R_nc_sync: NetCDF: Operation not allowed in
 define mode. This happens when I try to create.ncdf() a file with more than
 13 variables. I think is a problem of memory size. What would you recommend?
 Any help will be appreciated

Hi Hernan,

can you supply an example that shows the problem?

Regards,

--Dave

---
David W. Pierce
Division of Climate, Atmospheric Science, and Physical Oceanography
Scripps Institution of Oceanography, La Jolla, California, USA
(858) 534-8276 (voice)  /  (858) 534-8561 (fax)dpie...@ucsd.edu

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Re: [R] ncdf size error

[Hernan A. Moreno Ramirez]

Hi Professor,

Thanks for you valuable help. I did change my code to use ncdf4 package and 
still I get the same error:

# Exporting to NETCDF files 
#
# define the netcdf coordinate variables -- note these have values!
library(ncdf4)
cat(--Exporting final NETCDF file to Outputfolder,fill=TRUE)
dim1 = ncdim_def( Nodes,, seq(1,19000))
dim2= ncdim_def( Time,Hours since 2005-01-01 00:00:00,seq(1:2190)) 
#final[1,2,]))
#dim2= dim.def.ncdf( Time,,as.data.frame(final[1,2,])) 


# define the EMPTY (elevation) netcdf variable
varT2 = ncvar_def(T2,celsius, list(dim1,dim2), -9, longname=T2)
varQ2 = ncvar_def(Q2,kg kg-1, list(dim1,dim2), -9, longname=Q2)
varQVAPOR = ncvar_def(QVAPOR,kg kg-1, list(dim1,dim2), -9, 
longname=QVAPOR)
varQCLOUD = ncvar_def(QCLOUD,kg kg-1, list(dim1,dim2), -9, 
longname=QCLOUD)
varPSFC = ncvar_def(PSFC,Pa, list(dim1,dim2), -9, longname=PSFC)
varW10 = ncvar_def(W10,ms-1, list(dim1,dim2), -9, longname=W10)
varW = ncvar_def(W,ms-1, list(dim1,dim2), -9, longname=W)
varVEGFRA = ncvar_def(VEGFRA,, list(dim1,dim2), -9, longname=VEGFRA)
varMAXVEGFRA = ncvar_def(MAXVEGFRA,, list(dim1,dim2), -9, 
longname=MAXVEGFRA)
varTPREC = ncvar_def(TPREC,mm, list(dim1,dim2), -9, longname=TPREC)
varFRAC_FROZ_PREC = ncvar_def(FRAC_FROZ_PREC,mm, list(dim1,dim2), -9, 
longname=FRAC_FROZ_PREC)
varSWDOWN = ncvar_def(SWDOWN,W m-2, list(dim1,dim2), -9, 
longname=SWDOWN)
varGLW = ncvar_def(GLW,W m-2, list(dim1,dim2), -9, longname=GLW)
varPBLH = ncvar_def(PBLH,m, list(dim1,dim2), -9, longname=PBLH)
varTSLB = ncvar_def(TSLB,Celsius, list(dim1,dim2), -9, longname=TSLB)

nc.ex = nc_create(/media/D/output.nc,list(varT2, varQ2, varQVAPOR, varQCLOUD, 
varPSFC, varW10, varW, varVEGFRA, varMAXVEGFRA, varTPREC, varFRAC_FROZ_PREC, 
varSWDOWN, varGLW, varPBLH, varTSLB))

Error in R_nc4_enddef: NetCDF: One or more variable sizes violate format 
constraints
Error in R_nc4_sync: NetCDF: Operation not allowed in define mode

Thanks a lot for more help

---
Hernan A. Moreno, Ph.D.
Postdoctoral Research Associate
Department of Civil  Architectural Engineering
Room 3038, Engineering Building
University of Wyoming
Phone: 480-3990571
http://www.public.asu.edu/~hamoreno/


From: davidwilliampie...@gmail.com davidwilliampie...@gmail.com on behalf of 
David W. Pierce dpie...@ucsd.edu
Sent: Tuesday, September 16, 2014 2:21 PM
To: Hernan A. Moreno Ramirez
Cc: r-help@r-project.org
Subject: Re: [R] ncdf size error

On Tue, Sep 16, 2014 at 11:36 AM, Hernan A. Moreno Ramirez
hmore...@uwyo.edu wrote:
 Sure, here it is. Thanks for any help with this Dr. Pierce:


 # Exporting to NETCDF files 
 #
 # define the netcdf coordinate variables
 library(ncdf)

... rest of example omitted ...

Hi Hernan,

The file you are trying to make is violating size constraints imposed
by the netcdf library/R interface, which is 2 GB in the R ncdf
package. To get around this switch to the R ncdf4 package, which is
the replacement for the ncdf package (I stopped supporting the ncdf
package in 2010) and set force_v4=TRUE in the nc_create() call. This
requires a modern version of the netcdf library to be installed on
your machine, which it probably is, since the netcdf version 4 library
has been out for many years now.

Regards,

--Dave

 http://www.public.asu.edu/~hamoreno/

 
 From: davidwilliampie...@gmail.com davidwilliampie...@gmail.com on behalf 
 of David W. Pierce dpie...@ucsd.edu
 Sent: Monday, September 15, 2014 7:00 PM
 To: Hernan A. Moreno Ramirez
 Cc: r-help@r-project.org
 Subject: Re: [R] ncdf size error

 On Mon, Sep 15, 2014 at 4:20 PM, Hernan A. Moreno Ramirez
 hmore...@uwyo.edu wrote:

 Hi I am using both ncdf and ncdf4 libraries and with both I keep getting the
 same error: Error in R_nc_enddef: NetCDF: One or more variable sizes violate
 format constraints. Error in R_nc_sync: NetCDF: Operation not allowed in
 define mode. This happens when I try to create.ncdf() a file with more than
 13 variables. I think is a problem of memory size. What would you recommend?
 Any help will be appreciated

 Hi Hernan,

 can you supply an example that shows the problem?

 Regards,

 --Dave

 ---
 David W. Pierce
 Division of Climate, Atmospheric Science, and Physical Oceanography
 Scripps Institution of Oceanography, La Jolla, California, USA
 (858) 534-8276 (voice)  /  (858) 534-8561 (fax)dpie...@ucsd.edu



--
David W. Pierce
Division of Climate, Atmospheric Science, and Physical Oceanography
Scripps Institution of Oceanography, La Jolla, California, USA
(858) 534-8276 (voice)  /  (858) 534-8561 (fax)dpie...@ucsd.edu

__
R-help@r-project.org

[R] ncdf size error

[Hernan A. Moreno Ramirez]

Hi I am using both ncdf and ncdf4 libraries and with both I keep getting the
same error: Error in R_nc_enddef: NetCDF: One or more variable sizes violate
format constraints. Error in R_nc_sync: NetCDF: Operation not allowed in
define mode. This happens when I try to create.ncdf() a file with more than
13 variables. I think is a problem of memory size. What would you recommend?
Any help will be appreciated,.
Sent from my iPhone

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Re: [R] writing spdiags function for R

[moreno]

I am posting here the brilliant solutions, gently provided by  Prof JC Nash
nashjc at uottawa.ca, to me; so that people struggling in the future with
the same issue can find a way through. 

FYI, compared to the original Matlab implementation: 1) it does not handle
the case with more than one input, and  
2) (m  n) matrices give the B matrix columns in a different order, but the
d vector of indices will also be changed
accordingly, so the set of columns is OK, just ordered differently cif. JN.

Copied below the .R version of the spdiags code, a Fortran implementation of
it, and an R wrapper to run the .f

thanks again JN, your help was really invaluable :)

#
 

R version

#
spdiagsj - function(A) { # A is a matrix
   m - dim(A)[[1]]
   n - dim(A)[[2]]
   k - min(m, n) # length of diagonals
   Bdata-NULL # start with nothing in B matrix (as vector)
   jb-0 # column index of last column saved for B
   d-NULL # index vector of diagonals from A 
   # d contains 0 for the principal diagonal, -i for i'th lower
   # diagonal (prefaced with zeros), +j for j'th upper diagonal
   # (suffixed by zeros)
   q-(m-1)+n # There are m-1 subdiagonals and n-1 superdiagonals + main
diagonal

   if (m  n) { # tall matrix
  Adata - as.vector(t(A)) # convert to vector BY ROWS
   } else { # fat or square matrix (m = n)
  Adata - as.vector(A) # convert to vector BY COLUMNS 
   }
   # Augment the data with columns of zeros fore and aft
   Adata-c(rep(0,(k-1)*k), Adata, rep(0,(k-1)*k))
   cat(Augmented Adata with ,length(Adata), elements:\n)
   print(Adata)
   for (i in 1:q) {
   tv - c(Adata[[k*(i-1)+1]], rep(0,k-1)) # top element of augmented
column
  # plus enough zeros to pad it out (some zeros may be replaced
below)
  # i.e., start at 1, then m+1, 2*m+1 etc.
  qx-min((q-i), (k-1)) # 
  cat( qx=,qx,\n)
  if (qx  0) { # qx will be 0 when we are at last
superdiagonal,i.e., i == q
 for (j in 1:qx) { # get the rest of the diagonal elements
 tv[[j+1]] - Adata[[k*(i+j-1)+j+1]]
 }
  }
  if (any(tv != 0)){ # check for non-zeros, if there are, then save
 jb-jb+1 # next column of B
 d-c(d,(i-k)) # record the index
 Bdata-c(Bdata, tv) # save the diagonal as column of B in
vector form
  }
  }
   if (m  n) d - -d # reset index
   cat(Bdata:);  print(Bdata)

   B - matrix(Bdata, nrow=k, byrow=FALSE) # convert to matrix form
   result-list(B=B, d=d)
}

cat(Matlab example 1\n)
dta - c(0, 5, 0, 10, 0, 0, 0, 0, 6, 0, 11, 0, 3, 0, 0, 7, 0, 12, 1, 4, 0,
0, 8, 0, 0, 2, 5, 0, 0, 9)
A1 - matrix(dta, nrow=5, ncol=6, byrow=TRUE)

print(A1)
res1-spdiagsj(A1)
print(res1)
tmp-readline(Next)

cat(Matlab example 2\n)
n-10 # choose 10 for an example
A2-matrix(rep(0, n*n), nrow=n, ncol=n)

for (i in 1:n) {
for (j in 1:n) {
   if (i == j) A2[i, j] - -2
   if ( (i == (j-1)) || (i == (j+1))) A2[i,j] - 1
}
}
print(A2)
res2-spdiagsj(A2)
print(res2)
tmp-readline(Next)

cat(Matlab example 3\n)
dta3 - c(11, 0, 13, 0, 0, 22, 0, 24, 0, 0, 33, 0, 41, 0, 0, 44,
0, 52, 0, 0, 0, 0, 63, 0, 0, 0, 0, 74)
A3 - matrix(dta3, nrow=7, ncol=4, byrow=TRUE)

print(A3)
res3-spdiagsj(A3)
print(res3)
tmp-readline(Next)

cat(try transpose\n)
A3T-t(A3)
print(A3T)
res3T-spdiagsj(A3T)
print(res3T)
tmp-readline(Next)

cat(Example 5B \n)
dta5b1-c(6, 0, 13, 0, 0, 0, 7, 0, 14, 0, 1, 0, 8, 0, 15, 0, 2, 0, 9, 0, 0,
0, 3, 0, 10)
A5b1-matrix(dta5b1, nrow=5, ncol=5, byrow=TRUE)
print(A5b1)
res5b1-spdiagsj(A5b1)
print(res5b1)

#
 

Fortran version

#

  subroutine jspd(m, n, k, Adata, jb, Bdata, d, tv, na, nb, nd)
C   Central part of spdiags for R
C m and n are row and column sizes of A (underlying matrix)
C jb will be number of returned diagonals
C returns jb, Bdata, d
  integer m, n, na, nb, nd, jb, d(nd)
  integer i, j, k, kend, q, mn, kk1, js, je, qx
  double precision Adata(na), Bdata(nb), tv(k)
  LOGICAL not0
C  k = min(m, n)
C ??  check if k=1
C  Bdata-NULL # start with nothing in B matrix (as vector)
  jb = 0 
  kend=0
C column index of last column saved for B
C   d = NULL # index vector of diagonals from A 
C   # d contains 0 for the principal diagonal, -i for i'th lower
C   # diagonal (prefaced with zeros), +j for j'th upper diagonal
C   # (suffixed by zeros)
  q = (m-1)+n 
C  There are m-1 subdiagonals and n-1 superdiagonals + main diagonal
C   assume we have already built Adata for tall or fat matrix
C   # Augment the data with columns of zeros fore and aft
  mn = m*n
C  print *,Original Adata
C  print 

Re: [R] writing spdiags function for R

[moreno]

Dear R-list,

l have been working on a translation of a matlab library into R, it took me
a while but I am almost there to submit it to CRAN...however, for this
library to be computationally competitive I need to solve an issue with the
home-made R version of the spdiags.m function (an old issue coming back).
What I have now working is: 

# A = rbind(c(1,3,1,0), c(1,1,0,2), c(2,1,0,0) )
# A = replicate(100, rnorm(50)) 
A = replicate(1000, rnorm(1000)) 

exdiag - function(mat, off) {mat[row(mat)+off == col(mat)]}
spdiags - function(A){
require(Matrix)
 
indx = which(A != 0, arr.ind = T)
i = indx[,1]; j = indx[,2]
   
d = sort(j-i);
d = d[which(diff(c(-Inf, d) ) != 0)]

m = nrow(A); n = ncol(A);
p = length(d)
empty = vector()

A = as.matrix(A) ## make A logical, otherwise exdiag throws horrible
warnings
   ## might have to condition the above command on the type of input matrix
 
B =  Matrix(0, nrow = min(c(m,n)), ncol = p, sparse = TRUE);   

system.time(
for (k in 1:p){
print(k)

if (m = n){
i = max(c(1, 1 + d[k])):min(c(n, m + d[k]) )
} else {
i = max(c(1, 1 - d[k])):min(c(m, n-d[k]) )
}

if (length(i) != 0){

B[i, k] = exdiag(A, d[k])

 #  print(B)
 #  print(i)
 #  print(k)
 
}
   
}
)
   
   return (list( B = B, d = d) )
   
}

the advantages are: 1) that it works with matrices of any dimension
(previous versions worked only with squared matrices).  2) it does not need
other functions, beside exdiag to work.

However, I run into serious issues with computational efficiency. As I am
extracting non-zero diagonals and filling B in a for loop, the time it takes
to complete each iteration highly depends on the dimensions of the matrix. 
Try the different starting A matrices to test this. 

Unfortunately, I really need to optimize this aspect, as the people who I am
expecting to use this library would have really long time-series, i.e. it
would make .m win over .R

So, as I am wondering if there is a more efficient solution to do this
(perhaps using apply?) that I cannot currently see. Perhaps using sparse
matrices could help, but exdiag throws a warning:  sparse[ logic ] :
.M.sub.i.logical() maybe inefficient (noted also in previous responses to
this post of mine).

Using the function ?band has the drawback of padding the matrix with 0
(except the diagonal considered, e.g., band(A,1,1) ). Often, however, you
want to keep the 0s along the diagonal of interest, distinguished from the
0s of the other diagonals. Thus, doing something like:  which(band(A,1,1) !=
0, arr.ind = TRUE)  to get the non-zero element would also discard the 0s
along the diagonal.

The nice getband,R posted by Ben above, makes a combined use of diag and
band, together with a system of indeces, and tries to work around the issue
of the zero padding.  However, I did not manage to use/adapt this function
to work with matrices that are not squared. And, as I am not really 100% a
programmer, I did not find a solution to it... 
Moreover, I believe the problem of computational efficiency would still be
there, even fixing the dimensionality issue.

I really hope somebody out there can help me to figure this out. 

thanks a lot for your precious time and attention,

Moreno



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[R] choose many files

[RODRIGUEZ MORENO VICTOR MANUEL]

Hello everyone on the list
I want to run a FindU function on as many files as  in a directory. Due of my 
limitations on data management I am calling the function, wait until finish the 
process, call for the next file and so on. I have almost 120 CSV files on the 
same directory. This the example of calling the function

FindU(InSeries=C:/RHTests/data/tmax/00032015tmax.csv, 
MissingValueCode=-99.9, p.lev=0.95, Iadj=1, Mq=10, Ny4a=0, 
output=C:/RHTests/data/tmax/32015tmx)
As you can see my input file is 00032015tmax.csv and my output file is 
32015tmx.
Any suggestions? Thanks a lot in advance


Dr Victor M Rodríguez M
Doctor en Ciencias, en Ciencias de la Tierra / Geociencias Ambientales
INIFAP. CEPAB. (465) 9580167 y 86 Ext 108 y 220
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difusion o extensión a terceros, se requiere del consentimiento expresso del 
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extenso de sus responsabilidades .
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[R] problem on reading many files

[RODRIGUEZ MORENO VICTOR MANUEL]

Hi, I have to run almost 120 stations files of temperatura (mx and min), 
separated, and rain. I am following the instructions on RHtests tutorial as on 
http://www.cmc.org.ve/mediawiki/index.php?title=Preparando_los_datos link. Bit 
I have no success on running on multiple files. I have the .ls file which 
include the names of the file station, which I include on this msg for your 
consideration.
The instruction that I wrote on the console is

listatmn - readLines (“tmaxcopia.ls”, warn=false)
for(ifile in listatmn) 
FindU(paste(./,ifile,sep=),paste(salidas/,ifile,sep=),-999.9)





Of course the tmaxcopia.ls is the list of stations for tmax data. But once 
running on the console, this is the error msg I got

 listatmn - readLines (‘tmaxcopia.ls’, warn=false)
Error: inesperado entrada in listatmn - readLines (‘
 for(ifile in listatmn) 
 FindU(paste(./,ifile,sep=),paste(salidas/,ifile,sep=),-999.9)
 Error: objeto 'ifnames' no encontrado

What I'd like to do is to make easier and fast running the RHtest

Thank's a lot in advance for your help.




Dr Victor M Rodríguez M
Doctor en Ciencias, en Ciencias de la Tierra / Geociencias Ambientales
INIFAP. CEPAB. (465) 9580167 y 86 Ext 108 y 220

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fines descritos en su contenido. Queda expresamente prohibida bajo la ley de 
protección de datos del Gobierno de los Estados Unidos Mexicanos y para su 
difusion o extensión a terceros, se requiere del consentimiento expresso del 
titular de la cuenta de correo electrónico y/o del administrador del sitio en 
extenso de sus responsabilidades .
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[R] changing cex pointwise in lattice

[José M. Blanco Moreno]

, 10, 1, 13, 5, 14, 4, 21, 6, 16, 3, 0, 
1, 0, 0, 8, 3, 0, 0, 0, 0, 1, 0, 3, 0, 0, 0, 14, 2, 5, 3, 3, 4, 14, 6, 9, 4, 4, 
3, 14, 2, 5, 85, 17, 4, 4, 1, 3, 0, 8, 2, 10, 11, 43, 10, 27, 6, 16, 11, 2, 0, 
3, 0, 2, 0, 1, 3, 4, 1, 3, 1, 10, 3, 3, 1, 10, 2, 1, 1, 11, 15, 3, 1, 0, 0, 6, 
2, 3, 0, 0, 1, 3, 0, 3, 2, 2, 0, 1, 2, 2, 2, 3, 0, 3, 0, 2, 3, 6, 1, 7, 0, 1, 
2, 3, 3, 1, 0, 2, 1, 1, 0, 0, 0, 6, 2, 0, 2, 12, 4, 6, 5, 8, 4, 8, 4, 12, 3, 0, 
0, 4, 0, 4, 0, 5, 2, 4, 1, 8, 3, 3, 1, 1, 2, 1, 1, 8, 0, 0, 0, 13, 1, 18, 1, 5, 
0, 14, 2, 2, 1, 2, 0, 4, 1, 1, 1, 4, 3, 3, 2, 3, 4, 8, 2, 2, 2, 8, 4, 4, 2, 6, 
4, 7, 4, 3, 1, 10, 5, 4, 40, 4, 2, 14, 3, 5, 1, 7, 4, 4, 2, 6, 4, 2, 3, 9, 1, 
13, 6, 22, 8)), .Names = c(y, x, group, z), class = data.frame, 
row.names = c(35L, 83L, 131L, 179L, 227L, 275L, 323L, 371L, 419L, 467L, 515L, 
563L, 611L, 659L, 707L, 755L, 803L, 851L, 899L, 947L, 995L, 1043L, 1091L, 
1139L, 1187L, 1235L, 1283L, 1331L, 1379L, 1427L, 1475L, 1523L, 1571L, 1619L, 
1667L, 1715L, 1763L, 1811L, 1859L, 1907L, 1955L, 2003L, 2051L, 2099L, 2147L, 
2195L, 2243L, 2291L, 2339L, 2387L, 2435L, 2483L, 2531L, 2579L, 2627L, 2675L, 
2723L, 2771L, 2819L, 2867L, 2915L, 2963L, 3011L, 3059L, 3107L, 3155L, 3203L, 
3251L, 3299L, 3347L, 3395L, 3443L, 3491L, 3539L, 3587L, 3635L, 3683L, 3731L, 
3779L, 3827L, 3875L, 3923L, 3971L, 4019L, 4067L, 4115L, 4163L, 4211L, 4259L, 
4307L, 4355L, 4403L, 4451L, 4499L, 4547L, 4595L, 4643L, 4691L, 4739L, 4787L, 
4835L, 4883L, 4931L, 4979L, 5027L, 5075L, 5123L, 5171L, 5219L, 5267L, 5315L, 
5363L, 5411L, 5459L, 5507L, L, 5603L, 5651L, 5699L, 5747L, 5795L, 5843L, 
5891L, 5939L, 5987L, 6035L, 6083L, 6131L, 6179L, 6227L, 6275L, 6323L, 6371L, 
6419L, 6467L, 6515L, 6563L, 6611L, 6659L, 6707L, 6755L, 6803L, 6851L, 6899L, 
6947L, 6995L, 7043L, 7091L, 7139L, 7187L, 7235L, 7283L, 7331L, 7379L, 7427L, 
7475L, 7523L, 7571L, 7619L, 7667L, 7715L, 7763L, 7811L, 7859L, 7907L, 7955L, 
8003L, 8051L, 8099L, 8147L, 8195L, 8243L, 8291L, 8339L, 8387L, 8435L, 8483L, 
8531L, 8579L, 8627L, 8675L, 8723L, 8771L, 8819L, 8867L, 8915L, 8963L, 9011L, 
9059L, 9107L, 9155L, 9203L, 9251L, 9299L, 9347L, 9395L, 9443L, 9491L, 9539L, 
9587L, 9635L, 9683L, 9731L, 9779L, 9827L, 9875L, 9923L, 9971L, 10019L, 10067L, 
10115L, 10163L, 10211L, 10259L, 10307L, 10355L, 10403L, 10451L, 10499L, 10547L, 
10595L, 10643L, 10691L, 10739L, 10787L, 10835L, 10883L, 10931L, 10979L, 11027L, 
11075L, 11123L, 11171L, 11219L, 11267L, 11315L, 11363L, 11411L, 11459L, 11507L, 
11555L, 11603L, 11651L, 11699L, 11747L, 11795L, 11843L, 11891L, 11939L, 11987L, 
12035L, 12083L, 12131L, 12179L, 12227L, 12275L))


--
---
José M. Blanco-Moreno
Dept. de Biologia Vegetal (Botànica)
Facultat de Biologia
Universitat de Barcelona
Av. Diagonal 643
08028 Barcelona
SPAIN
---
phone: (+34) 934 039 863
fax: (+34) 934 112 842
---


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Re: [R] changing cex pointwise in lattice

[José M. Blanco Moreno]

Sorry about that, when I asked before I meant that I had tried your
proposed syntax. That has a problem, and it is that it uses the same set
of values for the character expansion *for both groups*, and it is not
what I wanted to do. If you plot the data (now a reduced set! sorry!)
you'll see that both plots are exactly the same: I have data on the same
positions, but the values that I want to represent with cex are
different for each group. The intention is to take advantage of the fact
that (I think that) lattice will use the same scaling on both plots.

I think I have found it...
# This is just to make sure that a group has a value of 1 and the other of 2
dd$z - (1:50)%%2 + 1

xyplot(y~x|group, panel=function(x,y,subscripts){panel.xyplot(x,y,
cex=dd$z[subscripts])}, data=dd)

Cheers,

José M. Blanco

El 20/07/2012 21:34, Peter Ehlers escribió:

Sorry, I had meant to include the following line:

Try: xyplot(y~x|group,cex=dd$z,data=dd)

Peter Ehlers

structure(list(y = c(3L, 3L, 3L, 3L, 9L, 9L, 9L, 9L, 38L, 38L,
38L, 38L, 44L, 44L, 44L, 44L, 38L, 38L, 38L, 38L, 44L, 44L, 44L,
44L, 38L, 38L, 38L, 38L, 44L, 44L, 44L, 44L, 54L, 54L, 54L, 54L,
60L, 60L, 60L, 60L, 54L, 54L, 54L, 54L, 60L, 60L, 60L, 60L, 54L,
54L), x = c(3.25, 3.25, 9.75, 9.75, 3.25, 3.25, 9.75, 9.75, 16.25,
16.25, 22.75, 22.75, 16.25, 16.25, 22.75, 22.75, 29.25, 29.25,
35.75, 35.75, 29.25, 29.25, 35.75, 35.75, 42.25, 42.25, 48.75,
48.75, 42.25, 42.25, 48.75, 48.75, 3.25, 3.25, 9.75, 9.75, 3.25,
3.25, 9.75, 9.75, 16.25, 16.25, 22.75, 22.75, 16.25, 16.25, 22.75,
22.75, 29.25, 29.25), group = structure(c(2L, 1L, 2L, 1L, 2L,
1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L,
1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L,
1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L), .Label = c(B,
A), class = factor), z = c(2, 0, 8, 3, 3, 4, 5, 4, 14, 11,
4, 2, 37, 7, 6, 5, 4, 3, 14, 5, 9, 3, 3, 1, 9, 4, 9, 7, 4, 2,
9, 1, 9, 2, 9, 1, 5, 2, 10, 1, 13, 5, 14, 4, 21, 6, 16, 3, 0,
1)), .Names = c(y, x, group, z), row.names = c(NA, 50L
), class = data.frame)

--
---
José M. Blanco-Moreno
Dept. de Biologia Vegetal (Botànica)
Facultat de Biologia
Universitat de Barcelona
Av. Diagonal 643
08028 Barcelona
SPAIN
---
phone: (+34) 934 039 863
fax: (+34) 934 112 842
---


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Re: [R] writing spdiags function for R

[Moreno I. Coco]

Hi Ben,


  A couple of quick comments:

 the semicolons at the end of each line are unnecessary (they're
harmless, but considered bad style -- most common in code of C and
MATLAB coders


yes, I know, just forgot to clean it up after translating :)


  You don't need to source(getband.R) inside the body of spdiags,
probably better to do it outside (so it doesn't get run every time the
function is called)

if (length(which(bd) == T) != 0)

  can be shortened to

  if (any(bd!=0))

  I haven't looked too carefully but can't all the i/j/d/p stuff be
replaced by

  n - nrow(A)
  p - 2*n-1
  d - seq(-(n-1),(n-1))


yes, all good points, I have acted the changes you suggested,
and it works smooth.

thanks,

Moreno


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Re: [R] writing spdiags function for R

[Moreno I. Coco]

Hi Ben,

thank you soo much:) your code worked, and we
managed to halven computing time.
I am copying below the final version of the function,
if anybody is willing to use/improve it.

thanks again,

Moreno

###

spdiags = function(A){
require(Matrix)
source(getband.R)

# Find all nonzero diagonals
i = rep(seq(1, nrow(A),1),nrow(A));
j = sort(i);
d = sort(j-i);

d = unique(d);
p = length(d); ##the nr. col of the new matrix
m = nrow(A); n = ncol(A);

B =  Matrix(FALSE, nrow = min(c(m,n)), ncol = p, sparse = TRUE);
count = 0

  for (k in 1:p){

  if (m = n){
 i = max(1, 1+d[k]):min(n, m+d[k]);
  } else { i = max(1, 1-d[k]):min(m,n-d[k]); }

  if (length(i) != 0){
bd = getband(A, d[k]);

if (length(which(bd) == T) != 0){
  count = count + 1;
  # print( paste( non-zero diagonal, count) )
  B[i,count] = bd
 } else {
  # print (paste (ncol, ncol(B), diag, k ) )
   B = B[, -ncol(B)] }
  }
  }

return (list( B = B, d = d) )

}





Quoting Ben Bolker bbol...@gmail.com on Fri, 13 Apr 2012 20:09:52 +:


Ben Bolker bbolker at gmail.com writes:



  I'm not quite sure how to do it, but I think you should look
at the ?band function in Matrix.  In combination with diag() of a
suitably truncated matrix, you should be able to extract bands
of sparse matrices efficiently ...




getband - function(A,k) {
n - nrow(A)
if (abs(k)(n-1)) stop(bad band requested)
if (k0) {
v - seq(n-k) ## -seq((n-k+1),n)
w - seq(k+1,n) ## -seq(n-k-1)
} else if (k0) {
v - seq(-k+1,n)
w - seq(n+k)
} else return(diag(A))
diag(band(A,k,k)[v,w,drop=FALSE])
}

PS: I think this should extract the k^th off-diagonal
band in a way that should (?) work reasonably efficiently
with sparse matrices.  I have not tested it carefully,
nor benchmarked it.

  Ben Bolker

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[R] writing spdiags function for R

[Moreno I. Coco]

Dear R-list,

I am in the process of translating a long function written in Matlab
into R (mainly because I am a big of fan of R, and folks will not
have to pay to use it :). In the translation of this function
I got stack because they use spdiags, which, as far as I can tell
it is not available in R. I have explored the Matrix package, from
which I borrowed some of the functions (e.g., sparseMatrix), but
I could not actually find an equivalent to spdiags (please, let
me know if it is there somewhere).

So, I have written my own spdiags function (below); following
also a suggestion in an old, and perhaps unique post, about
this issue.

It works only for square matrices (that's my need), however I
have a couple of issues, mainly related to computational
efficiency:

1) if I use it with a sparseMatrix, it throws a tedious warning
sparse[ logic ] : .M.sub.i.logical() maybe inefficient;
can I suppress this warning somehow, this is slowing the computation
very radically;

2) I can go around this problem by translating a sparseMatrix back
into a logical matrix before I run spdiags on it. However, the loop
gets very slow for large matrices (e.g., 2000x2000), which is the
kind of matrices I have to handle. If you look in the code,
I have placed a system.time() where the code is slowing down, and
it takes about:

user  system elapsed
   0.280.050.33

to complete an iteration...thus, I was wondering whether there is
a more efficient way to do what I am doing...also, if you spot
other places where the function could be optimized I would be
very glad to hear it!

thank you very much in advance for your kind help,

Best,

Moreno

###

## it works only for square matrices
## it could work with sparse matrices but it spits a tedious warning
## it is definitely inefficient compared to the original matlab code

## choose below different matrices to test the function.
# r = c(2,3,5,5); c = c(2,1,4,5)
# A = sparseMatrix(r, c)
# A = replicate(1000, rnorm(1000) )
# A = rbind(c(1,2,3),c(2,3,4),c(3,4,5))

spdiags = function(A){

# Find all nonzero diagonals
i = rep(seq(1, nrow(A),1),nrow(A));
j = sort(i);
d = sort(j-i);

  # d = d(find(diff([-inf; d]))); ## from Matlab ...
  # d = c(d[which(diff(d) == 1)], d[length(d)] ) ## this emulate 
above but needs to stick in last element


d = unique(d); ##this should work just fine and it is simpler
p = length(d); ##the nr. col of the new matrix
m = nrow(A); n = ncol(A);

B = matrix(0, nrow = min(c(m,n)), ncol = p);

  for (k in 1:p){
 # print(k)
 cl = vector();

  if (m = n){
 i = max(1, 1+d[k]):min(n, m+d[k]);
  } else { i = max(1, 1-d[k]):min(m,n-d[k]); }

  system.time(
  if (length(i) != 0){
 B[i,k] = A[ col(A) == row (A) - d[k]]
  } )
}

return (list( B = B, d = d) )

}




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[R] XML package: modify info of internal node

[Moreno Coco]

Dear All,

I have a very simple question to which I did not find any
answer online (apologize if it exists).
I have an XML file (attached) on which I just want
to modify the info of a node.
So, I load my xml file:

current =  xmlTreeParse(file, useInternalNodes=T)

and I want to change the information of:

filename1.jpg/filename

to be, for example: 1-1.jpg (newname)

I do:

xmlValue( xpathApply( current, /annotation/filename)[[1]] ) = newname

now, the crazy thing is that it does the change to the xml
but it also through the following error:

Error in xmlValue(xpathApply(current, /annotation/filename)[[1]]) =  
newname : could not find function xpathApply-


it seems that it works, but for a mysterious reason, it also
seems that it does not work.

Since I have to apply this change many times, and write
the resulting modified .xmls, the loop would stop as a
result of the error...

what am i missing here? is there any simpler way to do it?

thanks a lot for your kind help,

Moreno

P.S. I also try:

xmlSApply(xpathApply( current, /annotation/filename)[[1]], xmlValue)  
= newname


obtaining the same result.






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[R] modulo operation

[José M. Blanco Moreno]

 Dear R-users,
May be there is something that I am not understanding, missed or else...
Why do these operations yield these results?
 25%/%0.2
[1] 124
 25%%0.2
[1] 0.2

I would expect (although I know that what I do expect and what is really 
intended in the code may be different things)

 25/0.2
[1] 125
 25 - floor(25/0.25)*0.25
[1] 0

(At least this second one is what I would expect from the code in 
arithmetic.c, lines 168 to 178)


--
---
José M. Blanco-Moreno

Dept. de Biologia Vegetal (Botànica)
Facultat de Biologia
Universitat de Barcelona
Av. Diagonal 645
08028 Barcelona
SPAIN
---

phone: (+34) 934 039 863
fax: (+34) 934 112 842

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Re: [R] modulo operation

[José M. Blanco Moreno]

Did you read the documentation before you read the code?

‘%%’ and ‘x %/% y’ can be used for non-integer ‘y’, e.g. ‘1 %/%
0.2’, but the results are subject to rounding error and so may be
platform-dependent. Because the IEC 60059 representation of ‘0.2’
is a binary fraction slightly larger than ‘0.2’, the answer to ‘1
%/% 0.2’ should be ‘4’ but most platforms give ‘5’.

I suspect that is relevant to your interests



Yes. I think José is assuming that 25 %/% 0.2 and floor(25/0.2) are 
equal, but they are not, because rounding affects them differently. 
(The first is a single operation with no rounding except in the 
representation of 0.2; the second is two operations and is subject to 
another set of rounding.)


Duncan Murdoch
Thank you (both) very much for the info. Indeed I wasn't aware of that 
piece of documentation and of the implications of rounding. Excuse me 
for my hasty question when facing this behaviour.


--
---
José M. Blanco-Moreno

Dept. de Biologia Vegetal (Botànica)
Facultat de Biologia
Universitat de Barcelona
Av. Diagonal 645
08028 Barcelona
SPAIN
---

phone: (+34) 934 039 863
fax: (+34) 934 112 842

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[R] Neyman-Scott in 1-dimension?

[Moreno I. Coco]

Dear R-Users,

I would like to use the spatstat package in R for simulation and  
analysis of point processes in 1d = 1 dimension, i.e. on the line. I  
tried to use the owin class to obtain a 1-dimensional window, e.g.  
[1,..L], but it doesn't work.


Does anyone know if the spatstat package can deal with linear point  
processes? If yes, how?


Thank you very much for your help,

Moreno

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[R] updating specific arguments of formula

[Moreno Ignazio Coco]

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[R] updating specific arguments of formula

[Moreno Ignazio Coco]

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[R] updating specific arguments of formula

[Moreno Ignazio Coco]

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[R] updating arguments of formulae

[Moreno Ignazio Coco]

Dear R-Community,

I am relatively new with R, so sorry for things which for you might be  
obvious...

I am trying to automatically update lmer formulae.

the variables of the model are:

depM= my dependent measure
Sb2= a random factor
OS = a predictor
VR= another predictor

So, I am building the first model with random intercept only:

model = lmer(depM ~ (1 |Sb2))

then I update the formula adding the first predictor

model1 = update(model, as.formula(paste(. ~ . + , OS)))

the resulting formula will be:

depM ~ (1 |Sb2) + OS

let suppose now I want to update the model to have OS both as a fixed  
effect and in the random term, something like:


depM ~ (1 + OS |Sb2) + OS

I can do something very ugly (please tell me if there is a  more  
elegant way to do it) that looks like:


model2 = update(model1, as.formula(paste(paste(paste(paste(. ~ . + (1  
+ , OS), | ), Sb2), 


the resulting model2 formula will be:

depM ~ (1 |Sb2) + OS + (1 + OS | Sb2)

one first thing I am wondering at this point is whether having
(1 |Sb2) and (1 + OS | Sb2) in the same expression is redundant.
in the output it will obviously tell me that group Sb2 is considered twice:

number of obs: 6514, groups:  Sb2, 23; Sb2, 23

and i am not sure if am doing it correctly...any advice?

So let suppose now I want to add the new predictor VR again both in  
the fixed and in the random part of the formula.

If i just repeat the two steps above:

model3 = update(model2, as.formula(paste(. ~ . + , VR)))

and then:

model4 = update(model3, as.formula(paste(paste(paste(paste(. ~ . + (1  
+ , VR), | ), Sb2), 


the formula I get is:

depM ~ (1 |Sb2) + OS + (1 + OS | Sb2) + VR + (1 + VR | Sb2)

so, basically I am adding new stuff on the right side of the formula...

My first question at this point is whether the above formula is equivalent to:

depM ~ OS + VR + (1 + OS + VR | Sb2)

if is not equivalent, which one of the two is correct?

obviously in the second case, group Sb2, is considered only once.

If the second version of the formula is the correct one, I don't  
understand how I can update arguments inside the formula rather than  
adding things on his right side...


thus, in the ideal case,  how do I go from something like this:

depM ~ OS + (1 + OS | Sb2)

to something like this:

depM ~ OS + VR + (1 + OS + VR | Sb2)

Thanks a lot for your help,
Best,

Moreno


























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Re: [R] updating arguments of formulae

[Moreno Ignazio Coco]

Michael,

Thanks a lot for your reply, I have now understood how to fiddle  
around with the formulae updates...my question (see my previous e-mail  
where I was sketching this problem out) about LME models remains open...

whether:

depM ~ (1 |Sb2) + OS + (1 + OS | Sb2) + VR + (1 + VR | Sb2)

is equivalent to:

depM ~ OS + VR + (1 + OS + VR | Sb2)

and if probably not what is the best approach to it and where I can  
find a kind of guideline/rule of thumb list to build  
semi-automatically linear mixed effect models with fixed effects and  
random intercepts/slopes on it.


I am putting in copy the group  you suggested me...

Thanks again,

Moreno

Quoting Meyners,Michael,LAUSANNE,AppliedMathematics  
michael.meyn...@rdls.nestle.com:



Moreno,

I leave the discussion on the mixed models to others (you might consider
the SIG group on mixed models as well for this), but try a few hints to
make your code more accessible:

* The . in updating a formula is substituted with the respective old
formula (depending on the side), but is not mandatory. You could give
the new formula explicitly, i.e. consider something like
model1 = update(model, . ~ (1 |Sb2) + OS)
if you loose control about your models. See ?update.formula

* I don't see the need for using your construct with
as.formula(paste()), this makes things unnecessarily complicated. See my
above example, which should work as well on your data (and see ?update)

* There is also the - operator available in update.formula to remove
terms (because it uses formula, see ?formula). As to your question on
how to move from

depM ~ OS + (1 + OS | Sb2)
to
depM ~ OS + VR + (1 + OS + VR | Sb2)

try something like
update(model1, .~. - (1 + OS|Sb2) + VR + (1 + OS + VR | Sb2))
while it goes without saying that in this case, it would be easier to
drop the . and use something like
update(model1, .~ OS + VR + (1 + OS + VR | Sb2))
directly.

* paste accepts more than just two arguments to be pasted: Try somthing
like
model2 = update(model1, as.formula(paste(. ~ . + (1 + , OS, | ,
Sb2, )))
instead of your construct with several nested calls to paste, and see
?paste. (Note that I added quotes to OS and Sb2, it didn't work for
me otherwise as I have no object OS, not sure what happens if you have
such an object on our search path, but I would suspect you encounter
problems as well.)

If you work yourself through these and thereby simplify your code, you
are more likely to get responses to your questions on which model to use
(which is actually independent from the use of update). As far as I see
it, it doesn't make sense to use a formula like in your model4, but the
mixed model experts might tell me wrong (and I got a bit lost in your
code as well). Please also try to provide commented, minimal,
self-contained, reproducible code for further enquiries (use e.g. one of
the examples on ?lmer to create appropriate examples for your
questions).

HTH, Michael



-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Moreno Ignazio Coco
Sent: Donnerstag, 10. Dezember 2009 13:35
To: R-help@r-project.org
Subject: [R] updating arguments of formulae

Dear R-Community,

I am relatively new with R, so sorry for things which for you
might be obvious...
I am trying to automatically update lmer formulae.

the variables of the model are:

depM= my dependent measure
Sb2= a random factor
OS = a predictor
VR= another predictor

So, I am building the first model with random intercept only:

model = lmer(depM ~ (1 |Sb2))

then I update the formula adding the first predictor

model1 = update(model, as.formula(paste(. ~ . + , OS)))

the resulting formula will be:

depM ~ (1 |Sb2) + OS

let suppose now I want to update the model to have OS both as
a fixed effect and in the random term, something like:

depM ~ (1 + OS |Sb2) + OS

I can do something very ugly (please tell me if there is a
more elegant way to do it) that looks like:

model2 = update(model1, as.formula(paste(paste(paste(paste(.
~ . + (1
+ , OS), | ), Sb2), 

the resulting model2 formula will be:

depM ~ (1 |Sb2) + OS + (1 + OS | Sb2)

one first thing I am wondering at this point is whether having
(1 |Sb2) and (1 + OS | Sb2) in the same expression is redundant.
in the output it will obviously tell me that group Sb2 is
considered twice:

number of obs: 6514, groups:  Sb2, 23; Sb2, 23

and i am not sure if am doing it correctly...any advice?

So let suppose now I want to add the new predictor VR again
both in the fixed and in the random part of the formula.
If i just repeat the two steps above:

model3 = update(model2, as.formula(paste(. ~ . + , VR)))

and then:

model4 = update(model3, as.formula(paste(paste(paste(paste(.
~ . + (1
+ , VR), | ), Sb2), 

the formula I get is:

depM ~ (1 |Sb2) + OS + (1 + OS | Sb2) + VR + (1 + VR | Sb2)

so, basically I am adding new stuff on the right side of the
formula...

My first question at this point is whether the above formula

Re: [R] Combine vectors in order to form matrixes with combn

[Moreno Mancosu]

 Moreno,

 I don't understand exactly what it is you are trying to 
do.  Can you explain what you want your matrices to look like?  Perhaps 
give an example.

 Bryan


Dear Bryan,

let's try with an example: i 
have 8 column vectors

v1-matrix(c(2,5,6,8))
v2-matrix(c(3,7,9,11))

v3-matrix(c(13,4,2,7))
v4-matrix(c(3.4,6,11,21))
v5-matrix(c
(1,9,45,3))
v6-matrix(c(2,76,4,2))
v7-matrix(c(34,7,8,1))
v8-matrix(c
(2,6,1,9))

I would like to form a combination of theese vectors in 
order to obtain a set of matrixes in the form
k-combinations from a 
set of n elements, when the set is the number of vectors (in this case 
8) 
and k is the number of columns (in this case 4):

in this example, 
i would like to obtain


matrix_1

 [,1] [,2] [,3] [,4]

[1,]23   13  3.4
[2,]574  6.0
[3,]692 
11.0
[4,]8   117 21.0

matrix_2

 [,1] [,2] [,3] [,4]

[1,]23   131
[2,]5749
[3,]692   
45
[4,]8   1173

matrix_3

 [,1] [,2] [,3] [,4]
[1,]
23   132
[2,]574   76
[3,]6924

[4,]8   1172

[...]

matrix_n!/k!(n-k!)

 [,1] [,2] [,
3] [,4]
[1,]12   342
[2,]9   7676
[3,]   45
481
[4,]3219

I hope I was more clear!


However, thanks in advance,

Moreno


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[R] saving large matrices with decimal numbers

[Moreno Ignazio Coco]

Hi everybody,

I am struggling to save txt. or csv. large sparse matrices (1024 X 768  
image size), initialized at 0.1.


 image1[0:10,0:10]
  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
 [2,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
 [3,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
 [4,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
 [5,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
 [6,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
 [7,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
 [8,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
 [9,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
[10,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1

with small distributions located at different points of matrix.
If I try the same matrix but initialized at 0, it can save it.

It seems that R manages to save up around 800 X 600 decimal matrices  
but for bigger sizes the output is an empty file. Does anyone know  
whether there is a limit in saving such matrices and how can I work  
around this problem?


Thanks a lot,

Regards,

Moreno



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Re: [R] saving large matrices with decimal numbers

[Moreno Ignazio Coco]

Please tell us what you did exactly when you got the empty file:
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


well, that's exactly the point, I didn't do anything fancy or special,  
just save the matrix I gave you before (1024 X 768) either in format  
txt.


write.table(image, ~/mydata/blablabla/data.txt, col.names=F, row.names=F)

or csv:

write.csv (image, ~/mydata/blablabla/data.csv)


both cases the output is an empty file. Do you know what can be the problem?

Thanks a lot,

Moreno

 write.csv (image,   
/afs/inf.ed.ac.uk/user/s07/s0788806/R/r_data/ProdEyeTrack/DataFrame/Clustered/Clut/Cluster1/Cl1Verb-tr.csv.1)




Quoting Uwe Ligges lig...@statistik.tu-dortmund.de:


Moreno Ignazio Coco wrote:

Hi everybody,

I am struggling to save txt. or csv. large sparse matrices (1024 X   
768 image size), initialized at 0.1.


image1[0:10,0:10]
 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
[2,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
[3,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
[4,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
[5,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
[6,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
[7,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
[8,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
[9,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1
[10,]  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1  0.1   0.1

with small distributions located at different points of matrix.
If I try the same matrix but initialized at 0, it can save it.

It seems that R manages to save up around 800 X 600 decimal   
matrices but for bigger sizes the output is an empty file. Does   
anyone know whether there is a limit in saving such matrices and   
how can I work around this problem?


Please tell us what you did exactly when you got the empty file:
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Best wishes,
Uwe Ligges





Thanks a lot,

Regards,

Moreno







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Re: [R] Solutions of equation systems

[Moreno Mancosu]

Ravi Varadhan wrote:

[cut]

It seems to work!

Thanks a lot Ravi and Berend.

Moreno.

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Re: [R] Solutions of equation systems

[Moreno Mancosu]

Greg Snow wrote:

?solve

  

Dear Greg,

I tried to use function solve, but I have some problems with it.
Let's try with a simple equation:

x + 3y -2z = 5
3x + 5y + 6z =7

 a-matrix(c(1,3,-2,3,5,6),nrow=2,ncol=3)
 a
[,1] [,2] [,3]
[1,]1   -25
[2,]336
 b-matrix(c(5,7),nrow=2,ncol=1)
 b
[,1]
[1,]5
[2,]7

When I try to solve this system, I find this error:

 solve(a,b)
Error in solve.default(a, b) : 'b' must be compatible with 'a'

Also, when i try to solve a system with n equation and n variables, 
solve works perfectly.


Thanks in advance!

Moreno

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Re: [R] Solutions of equation systems

[Moreno Mancosu]

Moshe Olshansky wrote:

Is your system of equations linear?
  

dear sir,

Yes, the system is linear. It can have this form:

a x + b y + c z = d
e x + f y + g z = h

But it can be more complex. I would like to calculate in R the solutions 
under the form of equation of x,y and z.

Any idea?

Thanks in advance,

Moreno

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[R] Solutions of equation systems

[Moreno Mancosu]

Hello all!

Maybe it's a newbie question(in fact I _am_, a newbie), but I hope 
you'll find the solution.

I have an equation system which has k equation and n variables (kn).
I would like to obtain a description of the solutions (which can be the 
equation of lines or a plane, or everything else) with the lesser degree 
of freedom, obviously using R.
In other words, I would like to obtain a result which is very similar to 
the results of Maxima or similar software.
I tried to search on internet and i found the systemfit package, but I 
think it is not related with my problem.

Any idea?

Thanks in advance,

Moreno

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Re: [R] [R-sig-Geo] error in trmesh (alphahull package)

[José Manuel Blanco Moreno]

Hello,
The problem is that your first three points CANNOT be collinear. This is 
a feature of trmesh in tripack. So: either you unsort your data 
(keeping the indices you can sort them again) or alter (e.g. jitter) the 
coordinates very slightly. Either solution will do the work.

Cheers,
José M. Blanco


Murray Richardson escribió:

Hello R community,

I have cross-posted with r-sig-geo as this issue could fall under 
either interest group I believe.
I just came accross the alphahull package and am very pleased I may 
not need to use CGAL anymore for this purpose.   However, I am having 
a problem computing alpha shapes with my point data, and it seems to 
have to do with the spatial configuration of my points (which form 
long, skinny, contiguous clusters).  If I use rnorm to generate the 
same sample size with the same mean and SD, I can compute the alpha 
shape without trouble (which is why I think it might have to do with 
the spatial arrangement of my points).  See below:


# first here's a quick sample of my data (UTM coordinates)
 xcoords[1:5]
[1] 670080.2 670080.2 670080.2 670080.2 670080.2
 ycoords[1:5]
[1] 5005501 5005499 5005498 5005497 5005495
 xcoords[1:5]
#try the ashape routine with error
 alpha.shape-ashape(xcoords,ycoords,15)
Error in tri.mesh(X) : error in trmesh

#get statistics to generate a similar dataset to test against
 length(xcoords)
[1] 26257
 length(ycoords)
[1] 26257
 mean(xcoords)
[1] 670462.4
 mean(ycoords)
[1] 5005382
 sd(xcoords)
[1] 149.3114
 sd(ycoords)
[1] 181.5950

#generate the test data
 xtest-rnorm(26257,670462.4,149.3)
 ytest-rnorm(26257,5005382,181.60)

# try ashape routine with success
 alpha.shape-ashape(xtest,ytest,15)
 class(alpha.shape)
[1] ashape

Thanks for any insight into this!

Murray

ps I am able to compute the alpha shapes for this same dataset without 
problem using CGAL but I find it a pain to work with


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---
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Dept. de Biologia Vegetal (Botànica)
Universitat de Barcelona
Av. Diagonal 645 
08028 Barcelona SPAIN

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fax: (+34) 934 112 842
e-mail: jmbla...@ub.edu 


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Re: [R] heatmap-changing column or row names

[Moreno Ignazio Coco]

Hi Kevin,

thanks a lot for the advices. The first suggestion worked, overall, I  
have to find a way to scale the names/numbers on the interval  
considered. But it was the good way to look at the problem.
The second suggestion didn't really work. It is not a problem of  
custom fitting, it is more where in the plot, zero is centered.

If I say

abline(v=0)

it produces a vertical line much more right on the x axis than the  
actual 0. It is a problem of where the axis of the plots are set and  
they don't seem to match the layout of the heatmap. Any suggestion on  
this?


Moreno

Quoting Kevin J. Thompson [EMAIL PROTECTED]:




would this work for you?


cl2=c(rep('',4500))
for (i in seq(500,4500,by=500)){

+ cl2[i]=as.character(i)}


heatmap(x, labCol=cl2)


  you might have to custom fit the lines


d=seq(.2,.8,by=.1)
abline(v=d)


hope it helps,
kt

- Original Message -
From: Moreno Ignazio Coco [EMAIL PROTECTED]
Date: Wednesday, May 28, 2008 12:55 pm
Subject: [R] heatmap-changing column or row names


Dear R Community,

I am trying to create an heatmap for the following set of data:

##example of data matrix
o4
   V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18
green  27 28 29 29 28 28 26 25 25  23  23  22  22  21  21  22  22  22
yellow  6  8  8  7  7  7  6  6  6   7   7   7   6   6   6   6   6
5
red15 15 15 15 15 15 14 13 12  11  12  10   9   8   7   6   8
9
pink   11 11 11 11 11 10 12 11 13  14  14  15  15  14  14  17  17  17
blue   17 15 15 16 17 17 17 17 18  18  18  19  20  20  20  21  22  21

the column names are timepoints 450 in total. When I run the
heatmap code:

x  - as.matrix(o4)
ramp - colorRamp(c(yellow,green,blue))
cv-rgb( ramp(seq(0, 1, length = 83)), max = 255)

heatmap(x, col = cv, Colv=NA, Rowv=NA,xaxt=n, yaxt=n,
scale=column, margins=c(5,10), labCol=NA,xlab = Time, ylab=
Colors,main = )

I have quite few problems to change the column names. Being 450
timepoints too many to be readable on the plot, I wanted to
reduce/scale them to, let say 45. So, as you can see from the code

above, I have removed the x and y, row and column information.
Then I have tried to create a new name vector:

##create col vector to add names

Colname-as.character(seq(0,4500,500))

##putting it as arguments of (labCol=Colname, in the heatmap function)

but this solution didn't work. It looks like as it cannot find the

exact point on the axis where to stick the label in and it
clutters
all the datapoint on the same spot without being able to unfold it.

Therefore I have tried to introduce an axis:

axis(1, at=seq(0,450,1),line=1)

and then adding some text on top

mtext(paste(seq(0,4500, 250),sep=),side=1,
at=seq(0,450,25),line=2,cex=0.6,las=1)

also this way didn't work out. Actually, it has created a kind of
axis
but the zero of it was aligned to the middle of the plot. So, it
was
completely shifted.
That was my first problem.

Second small problem.
I am trying to mark on the heatmap some crucial timepoints. I
wanted
to do it by adding a straight abline on the heatmap plot

abline(v=80,col=green)

but also this doesn't seem to have any result
It seems that it is quite difficult to add any information to the
heatmap, but perhaps some of you might know how to do it.

Many thanks in advance

Moreno




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[R] 3dscatterplot -different colors for different factors of the same variable

[Moreno Ignazio Coco]

Dear R users,

I was trying to do a 3d scatterplot for the following set of datas:

obj  time X Y
1  yellow  333 388.7 492.3
2  yellow  567 388.7 492.3
3  green   621 135.5 371.7
4  green   1039 135.5 371.7
5  red 1373 744.1 205.0
6  red 1763 744.1 205.0

The points should be drew in the plot taking information of time,  
X and Y. The colors of these points should be assigned according  
to the information contained in variable obj.

What I did but it didn't work out is:

1) Plotting only one set of points (the green one)

s3d-scatterplot3d(greenpts[,3],greenpts[,2],greenpts[,4],type=p,scale.y=4,  
angle=10, color=green)


and try to overlay the others (as you normally would do with plot):

s3d$points3d(redpts[,3],redpts[,2],redpts[,4],col=red,type=o)
s3d$points3d(yellowpts[,3],yellowpts[,2],yellowpts[,4], col=yellow,type=o)
s3d$points3d(bluepts[,3],bluepts[,2],bluepts[,4],col=blue,type=o)
s3d$points3d(pinkpts[,3],pinkpts[,2],pinkpts[,4],col=pink,type=o)

R has plotted only the first set and wasn't able to overlay other points.

2) I have generated a for loop that should have taken points  
information ,arranged into a list object, plotting them one by one.

- listpoints = list(X,Y,time)
-color-rep(black,26) ##solution found in one of the discussion  
around scatterplot

color[mat1$obj==green]-green
color[mat1$obj==yellow]-yellow
color[mat1$obj==red]-red
color[mat1$obj==pink]-pink
color[mat1$obj==blue]-blue

for (i in 1:26)
points3d(listpoints[i],type=1, col=color[i])

Also this way didn't work out. Do you have any solution?
Many thanks

Moreno

P.S. I have tried also points3d(rgl) trying to follow an explanation  
given in the forum but i haven't understood how to set the function.


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[R] install problem 2.6 mac os x 10.5

[Oscar Moreno]

I am having the very SAME problem.  I removed Version 2.5.1 before  
trying to install 2.6.0. Please help.
TIA, Oscar

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[R] standard errors of regression coefficients

[Alexander Moreno]

Hi,

If I have two vectors x and y and I do lm(y~x) and now I want to define
variables that are the standard errors of the slope and intercept, how do I
do that?

Thanks,
Alex

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