similar to: Setting the names of a data.frame

Hi, I am Ramu from India, My rails application needs to have sms feature I searched for couple of rails plugins like sms-fu and all. But it seems that none of them has a support to india mobiles. Please help me in finding out the free sms plugin for rails in india. Thanks, Ramu.

I would like to incorporate multiple summary plots from the rms package into a single device and to control the titles, and also to open a new device when I reach a specified number of plots. Currently I am only getting a single "plot(summary(" graph in the upper left- hand corner of each successive device. However, in the rms documention I see instances of a loop being used with

Muchas gracias, pero claro en una muestra de 50 datos se ejecuta, en la muestra original de 1000 registros me tira error :( 2013/10/30 daniel <daniel319@gmail.com> > Amalia, > > No obtengo tus resultados. Corrí tus formulas y datos y el resultado es > x <- structure(list(ï..psraid = c(202517L, 202518L, 202520L, 202523L, > + 202527L, 202537L, 202543L, 202544L, 202551L,

Saludo gente, antes que nada gracias por la ayuda que puedan aportarme, soy iniciante en R, estoy usando el paquete Mice para realizar imputaciones múltiples sobre variables en su mayoría categóricas. El problema está que cuando expresó este comando imp <- mice(dataset,method="polr",maxit=1) donde el dataset es un data.frame me tirá este error : iter imp variable 1 1 pial1a

Hi All, I am attempting to work with some data from loggers. I have read in a .csv exported from MS Access that already has my dates and times (in 24 clock format), (with StringsAsFactors=FALSE). > head(tdata) LogData date time 1 77.16 2008/04/24 02:00 2 61.78 2008/04/24 04:00 3 75.44 2008/04/24 06:00 4 89.43 2008/04/24

Dear, When i multiply the y of a regression by 10, I would expect that the coefficient would be multiply by 10 and the z value to stay constant. Here some reproducible code to support the case. *Ex 1* library(mvtnorm) library(VGAM) set.seed(1) x=rmvnorm(1000,sigma=matrix(c(1,0.75,0.75,1),2,2))

Hello R users, I am new to R and am wondering if anyone can help me out with the following issue: I wrote a function to build ts models using different inputs, but when R displays the call for a model, I cannot tell which variables it is using because it shows the arguments instead of the real variables passed to the function. (e.g Call: lm(formula = dyn(dep ~ lag(dep, -1) + indep)) --->

Hello R users, I have been trying to output all my results (text, plots, etc) into the same postscript file as one document, but have been unable to...Can anyone help me improve my code below so that I can accomplish this? Currently I have to output them separately then piece them back together into one document.. Thanks in Advance for any help! options (scipen=999, digits=7)

Hi, I have been trying draw tickmark labels along the y - axis perpendicular to the y axis while labels along the x - axis parallel to x axis while making box plot. Here is my test dataset. TData ID Ratio 1 0 7.075 2 0 7.414 3 0 7.403 4 0 7.168 5 0 6.820 6 0 7.294 7 0 7.238 8 0 7.938 9 1 7.708 10 1 8.691 11 1 8.714 12 1 8.066 13 1 8.949 14 1 8.590 15 1 8.714 16 1

I am attempting to add a calculated column to a data frame. Basically, adding a column called "newcol2" which are the stock closing prices from 1 day to the next. The one little hang up is the name of the column. There seems to be an additional data column name included in the attributes (dimnames?). So when i run HEAD(DATAFRAMENAME) i get the column name = "Open". but

Hi I'm trying to extract complete rows from a dataframe by group based on the maximum in a column within that group. Thus I have a dataframe: cvd_basestudy ... es_time ... _____________ study1 ... 0.3091667 study2 ... 0.3091667 study2 ... 0.2625000 study3 ... 0.3033333 study3 ... 0.2625000 __________ etc I can extract the basestudy and the max(es_time)

Hi, All, This is my program ts1.sim <- arima.sim(list(order = c(1,1,0), ar = c(0.7)), n = 200) ts2.sim <- arima.sim(list(order = c(1,1,0), ar = c(0.5)), n = 200) tdata<-ts(c(ts1.sim[-1],ts2.sim[-1])) tre<-c(rep(0,200),rep(1,200)) gender<-rbinom(400,1,.5) x<-matrix(0,2,400) x[1,]<-tre x[2,]<-gender fit <- arima(tdata, c(1, 1, 0), method = "CSS",xreg=t(x))

Consider the following simple data set and a call to ave: > tdata <- data.frame(f1=c(1,1,1,1,2,2,2,2), f2=c(1,2,1,2,1,1,1,1), y=1:8) > with(tdata, table(f1, f2)) f2 f1 1 2 1 2 2 2 4 0 > with(tdata, ave(y, f1, f2, FUN=max)) [1] 3 4 3 4 8 8 8 8 Warning message: In FUN(X[[i]], ...) : no non-missing arguments to max; returning -Inf There are no missing values in the

A user sent me an example where coxph fails, and the root of the failure is a case where names(mf) is not equal to the term.labels attribute of the formula -- the latter has an extraneous newline. Here is an example that does not use the survival library. # first create a data set with many long names n <- 30? # number of rows for the dummy data set vname <- vector("character",

I don't normally have to go anywhere near this stuff , but it seems to me that this should be a straight-forward process in R. For the purposes of this enquiry I thought I would use something I can work out on my own. So I have my matrix and the right hand results from that matrix tdata <- matrix(c(0,1,0,-1,-1,2,0,0,-5,-6,0,0,3,-5,-6,1,-1,-1,0,0),byrow = T,ncol = 5) sumtd <-

Hi All, My application is using paperclip and its working fine without any problems, I am able to generate the thumb images and big images from actual uploaded images. There are many images already uploaded by the users. Now I want to create small images from the original uploaded image. For the newly uploaded images its working fine but how can I do the same for already uploaded image?? if

The survexp function can fail when called from another function. The "why" of this has me baffled, however. Here is a simple test case, using a very stripped down version of survexp: survexp.test <- function(formula, data, weights, subset, na.action, rmap, times, cohort=TRUE, conditional=FALSE, ratetable=survexp.us, scale=1, npoints, se.fit,

Amalia, No obtengo tus resultados. Corrí tus formulas y datos y el resultado es x <- structure(list(ï..psraid = c(202517L, 202518L, 202520L, 202523L, + 202527L, 202537L, 202543L, 202544L, 202551L, 202566L, 202570L, + 202571L, 202606L, 202619L, 202624L, 202629L, 202631L, 202632L, + 202633L, 202648L, 202657L, 202663L, 202676L, 202683L, 202685L, + 202706L, 202708L, 202709L, 202710L, 202734L,

On Mon, 21 Jan 2019 15:53:47 +0530 venkat ramu <ramut123 at gmail.com> wrote: > > [inherit] > path = /srv/samba/test/inherit > valid users = +"SBX\Inherit-Group", at +"SBX\Inherit-Group" > invalid users = +"SBX\Test-Group" > writeable = yes > > [inherit1] > writeable = yes > comment = inherit1 > valid users =

Dear Group, I have the following data matrix which is a timeseries. > dput(tData) structure(list(A = c(0.2, 0.13, 0.05, 0.1, 0.02, 0.18, 0.09, 0.06, 0.13), B = c(0.15, 0.06, 0.09, 0.02, 0.03, 0.12, 0.01, 0.15, 0.06), C = c(-0.1, 0, -0.07, -0.06, -0.05, -0.05, -0.06, -0.08, -0.07), D = c(-0.15, -0.05, -0.1, -0.03, -0.13, -0.04, -0.1, -0.04, -0.15), E = c(-0.17, -0.16, -0.08, -0.07, -0.09,