search for: mtcars

...ut the points on the graph should be different symbols and colors. It should represent what is in the legend. I tried using the points command but this does not work. Is there another command in R that would allow me to use different symbols and colors for the points? Thank you kindly. data(mtcars) plot(mtcars$wt,mtcars$mpg,xlab= "Weight(lbs/1000)", ylab="Miles per Gallon") mtcars$wt.cyl<-mtcars$wt[mtcars$cyl] mtcars$mpg.cyl<-mtcars$mpg[mtcars$cyl] points(mtcars$wt.cyl, mtcars$mpg.cyl, pch = 17, col = "red") title("Motor Trend Data") text(2...

I am looking for ways to add points to three different plots in parallel. I generate three scatter plots and name them as s3d1, s3d2 and s3d3 s3d1<-scatterplot3d(mtcars[,3],mtcars[,4],mtcars[,5],main="common",pch=20) s3d2<-scatterplot3d(mtcars[,3],mtcars[,4],mtcars[,5],main="common",pch=20) s3d3<-scatterplot3d(mtcars[,3],mtcars[,4],mtcars[,5],main="common",pch=20) If I try to add points to s3d1, s3d1$points3d(mtcars[,3],mtcars...

Dear list, I want the 1st, 2nd, 5th, and 6th columns of mtcars. After copying them, the columns become numeric class rather than data frame. But, when I copy rows, they data frame retains its class. Why is this? I don't see why copying rows vs columns is so different. > class(mtcars) [1] "data.frame" > head(mtcars) m...

...family of functions [lsv]apply(). The idea is to allow one-liner expressions without typing "function(item) {...}" to surround them. The argument to the anonymous function is simply referred as ".". Let take an example. With this new feature, the following call sapply(split(mtcars, mtcars$cyl), function(d) summary(lm(mpg ~ wt, d))$r.squared) #??????? 4???????? 6???????? 8 #0.5086326 0.4645102 0.4229655 could be rewritten as sapply(split(mtcars, mtcars$cyl), summary(lm(mpg ~ wt, .))$r.squared) "Not a big saving in typing" you can say but multiplied by the numbe...

...ysis I am working on (the boot function became quite challenging, for me, for a few reasons). I cannot figure out where I'm going wrong but the estimates from my attempt at the BCP CI are different enough from other methods that I assume I'm doing something wrong. require(boot) data("mtcars") # 1) Bootstrap 95% CI for R-Squared via boot::boot # statmethods.net/advstats/bootstrapping.html # Function for boot rsq <- function(formula, data, indices) { d <- data[indices,] fit <- lm(formula, data=d) return(summary(fit)$r.square) } # bootstrapping with 1000 replicatio...

Hello, When plotting a barchart with ggplot it drops the levels of the factor for which no counts are available. For example: library(ggplot) mtcars$cyl<-factor(mtcars$cyl) ggplot(mtcars[!mtcars$cyl==4,], aes(cyl))+geom_bar() levels(mtcars[!mtcars$cyl==4,]) This shows my problem. Because no counts are available for factorlevel '4', the label 4 dissapears from the plot. However, I would still like it to show up, but without a bar (ze...

I'm sure this exists elsewhere, but, as a trade-off, could you achieve what you want with a separate helper function F(expr) that constructs the function you want to pass to [lsv]apply()? Something that would allow you to write: sapply(split(mtcars, mtcars$cyl), F(summary(lm(mpg ~ wt,.))$r.squared)) Such an F() function would apply elsewhere too. /Henrik On Thu, Apr 16, 2020 at 9:30 AM Michael Mahoney <mike.mahoney.218 at gmail.com> wrote: > > This syntax is already implemented in the {purrr} package, more or > less -- you n...

Dear all Is there an easier way to retrieve the name of an object? For example, > tmp <- 1:10 > as.character(quote(tmp)) [1] "tmp" > as.character(quote(mtcars$cyl)) [1] "$" "mtcars" "cyl" > as.character(quote(mtcars$cyl))[3] [1] "cyl" The last call more than anything seems a hack. Is there a better way? Thank you Liviu

Dear R People: I have a data frame, xm1, which has 12 rows and 4 columns. If I put is xm1[,-4], I get all rows, and columns 1 - 3, which is as it should be. Now, is there a way to use the names of the columns to omit them, please? Thanks so much in advance! Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown

data(mtcars) mtcars[rownames(mtcars)!="Valiant",] # fails mtcars[list(rownames(mtcars))!="Valiant",] # runs but I am not getting the expected result With the latter statement, I expected all rows except the one where the name is "Valiant". I must have got something simple wron...

...sfully using RODBC for a long time (years) to connect to MS SQL Server from R. This week I wanted to try using odbc but I am seeing some problems which may be related to how I set up my driver and/or connection. The dbWriteTable manual page gives as an example command: dbWriteTable( pDB$con, "mtcars", mtcars[1:5,]) When I try this I get the following error Error: <SQL> 'CREATE TABLE "mtcars" ( "row_names" varchar(255), "mpg" FLOAT, "cyl" FLOAT, "disp" FLOAT, "hp" FLOAT, "drat" FLOAT, "wt&q...

Dear all I'm unable to find an example of extracting the rotated scores of a principal components analysis. I can do this easily for the un-rotated version. data(mtcars) .PC <- princomp(~am+carb+cyl+disp+drat+gear+hp+mpg, cor=TRUE, data=mtcars) unclass(loadings(.PC)) # component loadings summary(.PC) # proportions of variance mtcars$PC1 <- .PC$scores[,1] # extract un-rotated scores of 1st principal component mtcars$PC2 <- .PC$scores[,2] # extract un-rota...

Hello. This is to get you started with data.frames, next time please - read the posting guide - see the documentation, especially the builtin "R data import/export" manual form the help menu ## begin R examples, paste into console ## data(mtcars) # builtin database class(mtcars) # what it is mtcars # print it out... head(mtcars) # ...better: see first rows fm <- mpg~hp+wt summary(lm(fm, data=mtcars)) # my favourite regression: see ?mtcars save(mtcars, file="mtcars.rda") # save as R binary in current working directory load(file...

Full_Name: Liming Liang Version: 1.8.1 OS: windows2000 professional Submission from: (NULL) (67.172.81.139) I was looking at the variable 'mpg' of the data file 'mtcars' and make a stem plot, the following is the commend I entered. The stem plot shows the largest observation is 32.9 but actually in the data the largest observation is 33.9, here might be a problem. > data(mtcars) > stem(mtcars$mpg) The decimal point is at the | 10 | 44 12 | 3 1...

...lt;-colnames(x) to.keep<-all.cols[1:2] Kept<-subset(x,select=to.keep) Kept However, if I want to select some columns based on a selection of the most important variables from a random forest then I find myself stuck. The example below demonstrates the problem. library(randomForest) data(mtcars) mtcars.rf <- randomForest(mpg ~ ., data=mtcars,importance=TRUE) Importance<-data.frame(mtcars.rf$importance) Importance MSEImportance<-head(Importance[order(Importance$X.IncMSE, decreasing=TRUE),],3) MSEVars<-row.names(MSEImportance) MSEVars<-data.frame(MSEVars,stringsAsFactors =...

Before I go and do this another way - can I check if anyone has a way of looping through data in odfWeave (or possibly sweave) to do a repeating analysis on subsets of data? For simplicity lets use mtcars dataset in R to explain. Dataset looks like this: > mtcars mpg cyl disp hp drat wt ... Mazda RX4 21.0 6 160 110 3.90 2.62 ... Mazda RX4 Wag 21.0 6 160 110 3.90 2.88 ... Datsun 710 22.8 4 108 93 3.85 2.32 ... ............ Say I wanted to have a...

...allow one-liner expressions without >>>> typing "function(item) {...}" to surround them. The argument to the >>>> anonymous function is simply referred as ".". Let take an example. >>>> With this new feature, the following call sapply(split(mtcars, >>>> mtcars$cyl), function(d) summary(lm(mpg ~ wt, d))$r.squared) # 4 6 >>>> 8 #0.5086326 0.4645102 0.4229655 could be rewritten as >>>> sapply(split(mtcars, mtcars$cyl), summary(lm(mpg ~ wt, >>>> .))$r.squared) "Not a big saving in typing&q...

Hi, I was playing around with something else and I noticed this matrix code for residuals in a linear model doesn't say what lm() says. Please tell me if I am completely misguided here. data(mtcars) Y <- as.matrix(mtcars[,1]) X <- as.matrix(mtcars[,c(2:11)]) # shouldnt this: H <- X %*% solve(t(X) %*% X) %*% t(X) (diag(dim(H)[1]) - H) %*% Y # be equal to this: residuals(lm(Y~X)) # ??? # thanks -- View this message in context: http://r.789695.n4.nabble.com/residuals-from-lm...

...>> The idea is to allow one-liner expressions without typing >> "function(item) {...}" to surround them. The argument to the >> anonymous function is simply referred as ".". Let take an example. >> With this new feature, the following call sapply(split(mtcars, >> mtcars$cyl), function(d) summary(lm(mpg ~ wt, d))$r.squared) # 4 6 8 >> #0.5086326 0.4645102 0.4229655 could be rewritten as >> sapply(split(mtcars, mtcars$cyl), summary(lm(mpg ~ wt, .))$r.squared) >> "Not a big saving in typing" you can say but multiplied...

...a Gandhi Krishi Vishwavidyalaya, Raipur, Chhattisgarh, India. While taking classes, I found the *by() *function producing following error when I use FUN=mean or median and some other functions, however, FUN=summary works. Given below is the output of the example I used on a built-in dataset "mtcars", along with error message reproduced herewith: > by(data=mtcars, INDICES=list(mtcars$am), FUN=mean) : 0 [1] NA ------------------------------------------------------------ : 1 [1] NA Warning messages: 1: In mean.default(data[x, , drop = FALSE], ...) : argument is not numeric or logical:...