[ad_1] The first line does the cluster analysis, and stores the cluster labels in a component called cluster_iris$cluster which is just a vector of numbers. The second line puts that cluster number as a categorical label onto the rows of the original data set. So now your iris data has all the petal and sepal … Read more
[ad_1] You do not need more functions. The solution can be simpler from my understanding in 6 lines: pollutantmean <- function(directory, pollutant, id = 1:10) { filenames <- sprintf(“%03d.csv”, id) filenames <- paste(directory, filenames, sep=”https://stackoverflow.com/”) ldf <- lapply(filenames, read.csv) df=ldply(ldf) # df is your list of data.frames mean(df[, pollutant], na.rm = TRUE) } 2 [ad_2] … Read more
[ad_1] You just want to group on-the-fly, which means passing an ad hoc variable to the grouper. In data.table, this would be done by using rep in by: library(data.table) set.seed(01394) DT <- data.table(VAR = rnorm(2520)) DT[ , mean(VAR), by = rep(1:18, each = 140)] # rep V1 # 1: 1 0.029683200 # 2: 2 0.150411987 … Read more
[ad_1] If I understand you correctly, you want to loop while a condition is true and exit the loop when the condition become false. So you should be using the while control-flow operator, not for. i <- 1 while(i<=length(blocks) & length(blocks[[i]])!=0){ i <- i+1 } # check that we exited the loop because there was … Read more
[ad_1] You could do this: df <- data.frame(date=c(“18MAY2013:00:00:00″,”19MAY2013:00:00:00″)) df$year <- format(as.Date(df$date,”%d%B%Y”),”%Y”) df$month <- format(as.Date(df$date,”%d%B%Y”),”%m”) date year month 1 18MAY2013:00:00:00 2013 05 2 19MAY2013:00:00:00 2013 05 1 [ad_2] solved How can I create a new column for month and year from datetime and logically compare date time in R [closed]
[ad_1] You can use sub to get only the number from DF1$A and use %in%to test if the number is present in DF2$A. DF1$C <- “” i <- !sub(“\\D*”, “”, DF1$A) %in% DF2$A DF1$C[i] <-DF1$A[i] DF1 # A B C #1 ABC 1 AA ABC 1 #2 ABC 2 AB #3 ABC 3 AC #4 … Read more
[ad_1] You have to evaluate the Hessian to estimate standard error for a parameter fit = optim(start.theta, fn = function, #start.theta initial value of paramter hessian = TRUE, method = “L-BFGS-B”, #hessian True to calculate Hessian matrix lower = c(), upper = c(), control = list(trace = 1, fnscale = -1)) #fnscale= -1 to maximize … Read more
[ad_1] We can try sub sub(“(V).*(neck)”, “\\1 \\2”, words_1) #[1] “V neck” “V neck” “V neck” Or a general approach would be sub(“([A-Z]+)[^A-Za-z]*([a-z]+)”, “\\1 \\2”, words_1) 2 [ad_2] solved Matching and replacement with gsub [closed]
[ad_1] I really recommend taking a look at the documentation for geom_bar() (as well as the basic ggplot2 documentation), but this might help get you started. You can add an outline to your bars by using colour = “black”, and you can change the axis labels by using labs(). We use the arguments within theme() … Read more
[ad_1] We can try data.table. Convert the ‘data.frame’ to ‘data.table’ (setDT(df2), grouped by ‘RouteId’, if any of the run-length-type id of the logical vector (StopType==’Load’) is greater than 2, we get the Subset of Data.table (.SD). This will give the rows with ‘RouteId’ 103. library(data.table) setDT(df2)[,if(any(rleid(StopType==’Load’) >2)) .SD ,.(RouteId)] # RouteId StopOrder StopType #1: 103 … Read more
[ad_1] Within R studio, you can choose R markdown . Navigate to File Option -> New File -> R Markdown There are no changes in coding. All codes that you write in R would be as-is. Just with a flexibility that you can add comments and explain your code. Make sure in .Rmd file, there … Read more
[ad_1] 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 … Read more
[ad_1] The second column is binary. So one way would be to change it to logical vector, subset the values that correspond to ‘ON’ and ‘OFF’ (i.e. 1 and 0), get the difference of the sum of both and multiply with 40. 40*(sum(ccmix[,1][!!ccmix[,2]]) – sum(ccmix[,1][!ccmix[,2]])) The negate function (!) converts 0 to TRUE and 1 … Read more
[ad_1] In my comment I incorrectly remembered the name of the argument for changing the direction of an order result. The fix is simply to use the correct name: > df[ order(rowSums(is.na(df)), decreasing=TRUE), ] x y z v 4 NA NA NA NA 3 6 NA NA NA 6 NA 18 130 NA 10 NA … Read more
[ad_1] 1. What does the cluster sizes mean? You provided 16 records and told kmeans to find 3 clusters. It clustered those 16 records into 3 groups of A: 3 records, B: 10 records and C: 3 records. 2. What are the cluster means? These numbers signify the location in N-Dimensional space of the centroid … Read more