Displaying 20 results from an estimated 8000 matches similar to: "Help"
2008 Nov 26
1
multiple imputation with fit.mult.impute in Hmisc - how to replace NA with imputed value?
I am doing multiple imputation with Hmisc, and
can't figure out how to replace the NA values with
the imputed values.
Here's a general ourline of the process:
> set.seed(23)
> library("mice")
> library("Hmisc")
> library("Design")
> d <- read.table("DailyDataRaw_01.txt",header=T)
> length(d);length(d[,1])
[1] 43
[1] 2666
2013 Sep 27
1
RV: cronbach
rm(list = ls()) #borra todo lo anterior en memoria
setwd("G:/Public/Documents/R/EPICALC/") #como se llama la data desde su path
sanda<-read.csv("sandavid2.csv",header=TRUE, sep=",", dec=".")
use(sanda)
attach (sanda)
library (MASS)
label.cronbach <- label.var(p01, "¿Consume mucho pan? ") ####¿Hay alguna manera de nombrarlos
2013 Sep 27
2
RV: cronbach
¿existe algún método en el cual no sea necesario hacer este trabajo y que
aparezcan los nombres de las preguntas?
label.cronbach <- label.var(p01, "¿Le agrada el programa que se le ha
mostrado? ")
label.cronbach <- label.var(p02, "¿Cree que ayuda en el aprendizaje?")
label.cronbach <- label.var(p03, "¿Propicia el trabajo en el equipo?")
label.cronbach
2013 Sep 27
0
RV: cronbach
Hola,
Todos estamos muy bien, gracias.
Con respecto de tu consulta, creo que ofreces poca información para poder
reproducir o intentar emular tu inquietud (aunque en este grupo de
voluntarios hay gente muy hábil que seguramente podrá ayudarte mejor). Yo
me pregunto, por ejemplo: ¿cómo son tus datos originalmente?¿los estás
leyendo de otro programa?¿los tienes en un texto plano?¿podrías enviar
2013 Sep 27
0
cronbach
¿existe algún método en el cual no sea necesario hacer este trabajo y que
aparezcan los nombres de las preguntas?
label.cronbach <- label.var(p01, "¿Le agrada el programa que se le ha
mostrado? ")
label.cronbach <- label.var(p02, "¿Cree que ayuda en el aprendizaje?")
label.cronbach <- label.var(p03, "¿Propicia el trabajo en el equipo?")
label.cronbach
2005 Nov 09
2
error in NORM lib
Dear alltogether,
I experience very strange behavior of imputation of NA's with the NORM
library. I use R 2.2.0, win32.
The code is below and the same dataset was also tried with MICE and
aregImpute() from HMISC _without_ any problem.
The problem is as follows:
(1) using the whole dataset results in very strange imputations - values
far beyond the maximum of the respective column, >
2009 Mar 27
1
General help for a function I'm attempting to write
Hello,
I have written a small function ('JostD' based upon a recent molecular
ecology paper) to calculate genetic distance between populations (columns in
my data set). As I have it now I have to tell it which 2 columns to use (X,
Y). I would like it to automatically calculate 'JostD' for all combinations
of columns, perhaps returning a matrix of distances. Thanks for any help
2004 Jul 22
1
[PATCH] add LOGIN authentication mechanism
Hello,
attached patch (1.0-test29) adds LOGIN authentication mechanism.
Tested with KMail and seems working.
Please consider applying.
Best regards.
P.S. I also have NTLM authentication working and plan to
submit it RSN.
--
Andrey Panin | Linux and UNIX system administrator
pazke at donpac.ru | PGP key: wwwkeys.pgp.net
-------------- next part --------------
diff -urpNX /usr/share/dontdiff
2013 Oct 03
0
cronbach
Estimado José,
Es la segunda vez que veo te pregunta en este Foro, la primera vez se te
pidió mas detalles y luego se te dio una respuesta. El hecho que reiteres
tu pregunta, sin detalles y sin mencionar que es lo que no funcionó con la
respuesta anterior, me hace recordar algunas recomendaciones que leí en
una excelente respuesta en http://stackoverflow.com/, que trato de adaptar
a
2004 Jul 22
0
[PATCH] POP3 CAPA command RFC violation
Hello,
looks like SASL capability reported by dovecot in response to POP3 CAPA
command doesn't conforms with rfc2449 which says:
Examples:
C: CAPA
S: +OK Capability list follows
S: TOP
S: USER
S: SASL CRAM-MD5 KERBEROS_V4
S: RESP-CODES
S: LOGIN-DELAY 900
S: PIPELINING
2007 Dec 19
1
strange timings in convolve(x,y,type="open")
Dear R-ophiles,
I've found something very odd when I apply convolve
to ever larger vectors. Here is an example below
with vectors ranging from 2^11 to 2^17. There is
a funny bump up at 2^12. Then it gets very slow at 2^16.
> for( i in 11:20 )print( system.time(convolve(1:2^i,1:2^i,type="o")))
user system elapsed
0.002 0.000 0.002
user system elapsed
0.373
2001 Jun 05
2
a bug? (PR#968)
--T4sUOijqQbZv57TR
Content-Type: text/plain; charset=us-ascii
Content-Disposition: inline
Dear R,
I would like to report what I think is a bug in R. I am running R
within emacs on a Digital AlphaStation. See the version information
at the end of my R session for details. I also attach a copy of the
file that is read in the `read.table' command.
Here's my R session, with a few
2008 Nov 26
2
Chi-Square Test Disagreement
I was asked by my boss to do an analysis on a large data set, and I am
trying to convince him to let me use R rather than SPSS. I think Sweave
could make my life much much easier. To get me a little closer to this
goal, I ran my analysis through R and SPSS and compared the resulting
values. In all but one case, they were the same. Given the matrix
[,1] [,2]
[1,] 110 358
[2,] 71 312
[3,]
2008 Nov 25
1
Efficient passing through big data.frame and modifying select
> -----Original Message-----
> From: William Dunlap
> Sent: Tuesday, November 25, 2008 9:16 AM
> To: 'johannes_graumann at web.de'
> Subject: Re: [R] Efficient passing through big data.frame and
> modifying select fields
>
> > Johannes Graumann johannes_graumann at web.de
> > Tue Nov 25 15:16:01 CET 2008
> >
> > Hi all,
> >
> >
2008 Dec 23
1
quotation problem/dataframe names as function input argument.
Dear R friends:
Can someone help me with the following problem? Many thanks in advance.
# Problem Description:
# I want to write functions which take a (character) vector of dataframe
names as input argument.
# For example, I want to extract the number of observations from a number of
dataframes.
# I tried the following:
nobs.fun <- function (dframe.vec)
{
nobs.vec <-
2012 Mar 30
1
How to use access results of gregexpr in data frames
Hello,
I'm trying to figure out how to find the index of the second occurrence of "/" in a string (which happens to represent a date) within a data frame column.
I've used the following code successfully to find the first instance of "/".
dframe <- data.frame(date=c("5/14/2011", "4/7/2011"))
dframe$x1 <- regexpr("/", dframe[, 1])
2011 Jun 09
3
How to subset based on column name that is a number ?
Hi,
I have a data frame with column names "1", "2", "3", ... and I'd like to extract
a subset based on the values in the first column. None of the methods I tried
worked (below).
x <- subset(dframe, 1 = = "My Text")
x <- subset(dframe, "1" = = "My Text")
x <- subset(dframe, names(dframe)[1] = = "My Text")
Q
2006 Mar 21
3
Rsync 4TB datafiles...?
I need to rsync 4 TB datafiles to remote server and clone to a new oracle
database..I have about 40 drives that contains this 4 TB data. I would like
to do rsync from a directory level by using --files-from=FILE option. But
the problem is what will happen if the network connection fails the whole
rsync will fail right.
rsync -a srchost:/ / --files-from=dbf-list
and dbf-list would contain this:
1998 Jul 29
0
Printing on a Window Client from AIX
Hi all...
First, thank you to Matt Chapman for pointing out why I could not connect to a
non-domain participating machine. I needed to have a local account on that
machine (it wasn't checking back with the domain controllers as I thought it
would). I guess I'm going to have to start venturing out of the UNIX world and
into the NT one. Scary :-)
Either way, I now successfully am
2011 Jul 08
1
Referencing a vector of data labels in ggplot function
Hi,
I really feel I've looked everywhere, although I know this can't be a hard
problem. I'd like to be able to call the graph below as a function, but I
can't get the function to recognize variables beyond 'dframe'. I've read
through many papers on writing functions in R, but I can't get this to work.
data <- data.frame('date' = as.Date(rep(c(15101,