This does not sound like a clustering problem to me. Instead, you are looking for a decision tree, on the target variable “grade > 4”. 0 solved Best data mining algorithm for detect the profile of students with excellent grades
I believe the quantity you are looking for is called variance. This describes consistency, if you were to say, use the time each day that a student works. 2 solved Algorithm to check consistency? [closed]
Lets assume that set.seed(44) deaths<- 10:1 + sample.int(3, 10, replace = T) and spent<- seq(100, 550, by = 50 ) The very first thing you want to do when you get your data is literally to look at it. This can be done relatively painlessly with plot(spent, deaths) which yields So it looks like the … Read more
Here you go. library(dplyr) Mean_of_a_group <- dataset %>% group_by(age_group) %>% summarize(happy = mean(happy)) I suggest using ggplot2 to do what you want here. ggplot(data = Mean_of_a_group, aes(age_group, happy))+ geom_point Dummy example library(ggplot2) ggplot(aes(x = disp, y= mpg), data = mtcars)+ geom_smooth(aes(colour = “red”), data = mtcars, stat = “smooth”, formula = y ~ x , … Read more
If you are looking for an R solution, here’s one way to do it: The trick is using [ combined with rowSums FRAMETRUE$Group1 <- rowSums(FRAMETRUE[, 2:8], na.rm = TRUE) solved Sum rows by interval Dataframe
There is no standard deviation function C++, so you’d need to do write all the necessary functions yourself — Generate random numbers and calculate the standard deviation. double stDev(const vector<double>& data) { double mean = std::accumulate(data.begin(), data.end(), 0.0) / data.size(); double sqSum = std::inner_product(data.begin(), data.end(), data.begin(), 0.0); return std::sqrt(sqSum / data.size() – mean * mean); … Read more
Some parts of your question are unclear. It might help to give the context of what you’re trying to achieve, rather than what are the specific steps you’re taking. 1) + 3) In a Normal distribution – fitting the distribution, and estimating the mean and standard deviation – are basically the same thing. The mean … Read more
As the first google match stated, you can try doing as: n<-50 x<-sample(40:70,n,rep=T) y<-.7*x+rnorm(n,sd=5) plot(x,y,xlim=c(20,90),ylim=c(0,80)) mylm<-lm(y~x) abline(mylm,col=”red”) newx<-seq(20,90) prd<-predict(mylm,newdata=data.frame(x=newx),interval = c(“confidence”), level = 0.90,type=”response”) lines(newx,prd[,2],col=”red”,lty=2) lines(newx,prd[,3],col=”red”,lty=2) Of course it is one of possibilities Another example would be: x <- rnorm(15) y <- x + rnorm(15) new <- data.frame(x = seq(-3, 3, 0.5)) pred.w.clim <- predict(lm(y … Read more
Often you can use a named vector to do these sorts of mapping operations simply: students <- data.frame(name=c(“J. Holly”,”H. Kally”, “P. Spertson”, “A. Hikh”, “R. Wizht”), CRSE_GRADE_OFF=c(“A”,”D”,”E”,”A”,”A”)) scores = c(A=1, B=2, C=3, D=4, E=5, F=6) students$grade <- scores[as.character(students$CRSE_GRADE_OFF)] students # name CRSE_GRADE_OFF grade # 1 J. Holly A 1 # 2 H. Kally D 4 … Read more