search for: mydt

...ef)/(1+exp(x%*%truecoef)) # prob is the true probability of class 1 y <- rbinom(n+n.test,1,prob1) # separate the data into train and test sets mydata <- data.frame(y=y[1:n],x=x[1:n,]) mydata.test <- data.frame(y=y[n+(1:n.test)],x=x[n+(1: n.test),]) ########################## library(e1071) mydt.nb<-naiveBayes(y~ ., data=mydata) m.pr<-predict(mydt.nb, mydata[,-1], type="class") regards, Leah [[alternative HTML version deleted]]

...") > head(myData) var1 var2 id 1 31.59 33.78 0m4 2 32.21 33.25 0m4 3 31.78 NA 0m4 4 31.34 32.05 0m5 5 31.61 32.59 0m5 6 31.61 NA 0m5 results1 <- aggregate(. ~ id ,data=myData,FUN=mean,na.rm=T) head(results1,1) # id var1 var2 # 1 0m11 30.79 32.27 library(data.table) mydt <- data.table(myData) setkey(mydt,id) results2 <- mydt[,lapply(.SD,mean,na.rm=TRUE),by=id] head(results2,1) # id var1 var2 # [1,] 0m11 30.84 32.27 library(plyr) results3 <- ddply(myData,.(id),colwise(mean),na.rm=TRUE) head(results3,1) # id var1 var2 # 1 0m11 30.84 32.27 &g...

Hola, Bueno, puedes hacer el cálculo de una forma mucho más compacta y rápida. Esta forma es especialmente recomendable cuando tienes muchas columnas y muchas filas. > library(data.table) > myDT <- as.data.table(mtcars) > myDTlong <- melt(myDT, measure.vars=1:ncol(myDT)) > myDTlong[ , list(p_value = shapiro.test(value)$p.value, v_stat = shapiro.test(value)$statistic) , by = .(variable)] variable p_value v_stat 1: mpg 1.228814e-01 0.9475647 2: cyl 6...

...t;- function(x) { myvec <- rep(0,length(x)) lastInd <- length(myvec) myvec[lastInd] = 1 myvec } plyrSolution <- ddply(timeData, "id", transform, censor = makeCensor(mygroup)) # here is a data table solution # use makeCensor function from above library(data.table) mydt <- data.table(myData) setkey(mydt,id,mygroup) timeData <- mydt[,list(mytime=length(gender)),by=list(id,mygroup)] makeCensor <- function(x) { myvec <- rep(0,length(x)) lastInd <- length(myvec) myvec[lastInd] = 1 myvec } mycensor <- timeData[,list(censor=makeCensor(myg...

Dear all, I?m having trouble getting a list of regression variables back into a dataframe. mydf <- data.frame(x1=rnorm(100), x2=rnorm(100), x3=rnorm(100)) mydf$fac<-factor(sample((0:2),replace=T,100)) mydf$y<- mydf$x1+0.01+mydf$x2*3-mydf$x3*19+rnorm(100) dlply(mydf,.(fac),function(df) lm(y~x1+x2+x3,data=df))->dl here I?d like to use ldply(dl,coef(summary)) or something

Buen día estimados Estoy tratando de hacer un tibble con los resultados de un apply que se muestran en la consola que me da R, no estoy seguro si eso se pueda hacer, pero me gustaría organizar los resultados de esa manera. mi código es: data("mtcars") Mtcars_matriz <- as.matrix(mtcars) apply(Mtcars_matriz, MARGIN =2, FUN = shapiro.test) DF2 <- tibble(Variable = NA, W = NA, Pvalue =

...oef)/(1+exp(x%*%truecoef)) # prob is the true probability of class 1 y <- rbinom(n+n.test,1,prob1) # separate the data into train and test sets mydata <- data.frame(y=y[1:n],x=x[1:n,]) mydata.test <- data.frame(y=y[n+(1:n.test)],x=x[n+(1:n.test),]) ########################## library(e1071) mydt.nb<-naiveBayes(y~ ., data=mydata) m.pr<-predict(mydt.nb, mydata[,-1], type="class") regards, Leah [[alternative HTML version deleted]]

I have a data set with about 6 million rows and 50 columns. It is a mixture of dates, factors, and numerics. What I am trying to accomplish can be seen with the following simplified data, which is given as dput output below. > head(myData) mydate gender mygroup id 1 2012-03-25 F A 1 2 2005-05-23 F B 2 3 2005-09-08 F B 2 4 2005-12-07 F B 2

This is no doubt trivial but after searching the help files and the web, I cannot seem to find it. 1) How do I convert 'hgt' into 'HGT' in R? 2) How should I have used the help facilities to find this? At the end of the day, all I want to do is case insensitive string matching... i.e. 'if ("HGT" == 'hgt') print('this should be true')' I tried