This is a generalized way in which you do not have to remember the exact location of the variable: # df = dataframe # old.var.name = The name you don’t like anymore # new.var.name = The name you want to get names(df)[names(df) == ‘old.var.name’] <- ‘new.var.name’ This code pretty much does the following: names(df) looks … Read more
This question was answered in well in the comments by @James, pointing to an excellent explanation by Hadley Wickham of the dangers of subset (and functions like it) [here]. Go read it! It’s a somewhat long read, so it may be helpful to record here the example that Hadley uses that most directly addresses the … Read more
Use bind_rows() from the dplyr package: bind_rows(list_of_dataframes, .id = “column_label”)
The short version: use formatC or sprintf. The longer version: There are several functions available for formatting numbers, including adding leading zeroes. Which one is best depends upon what other formatting you want to do. The example from the question is quite easy since all the values have the same number of digits to begin … Read more
If you have the file locally, then use install.packages() and set the repos=NULL: install.packages(path_to_file, repos = NULL, type=”source”) Where path_to_file would represent the full path and file name: On Windows it will look something like this: “C:\\RJSONIO_0.2-3.tar.gz”. On UNIX it will look like this: “/home/blah/RJSONIO_0.2-3.tar.gz”.
This should work: p <- ggplot(df, aes(x=rating, fill=cond)) + geom_density(alpha=.3) + xlab(“NEW RATING TITLE”) + ylab(“NEW DENSITY TITLE”) p <- p + guides(fill=guide_legend(title=”New Legend Title”)) (or alternatively) p + scale_fill_discrete(name = “New Legend Title”)
paste() is the way to go. As the previous posters pointed out, paste can do two things: concatenate values into one “string”, e.g. > paste(“Hello”, “world”, sep=” “) [1] “Hello world” where the argument sep specifies the character(s) to be used between the arguments to concatenate, or collapse character vectors > x <- c(“Hello”, “World”) … Read more
If your question is “how can I determine how many clusters are appropriate for a kmeans analysis of my data?“, then here are some options. The wikipedia article on determining numbers of clusters has a good review of some of these methods. First, some reproducible data (the data in the Q are… unclear to me): … Read more
Use the colnames() function: R> X <- data.frame(bad=1:3, worse=rnorm(3)) R> X bad worse 1 1 -2.440467 2 2 1.320113 3 3 -0.306639 R> colnames(X) <- c(“good”, “better”) R> X good better 1 1 -2.440467 2 2 1.320113 3 3 -0.306639 You can also subset: R> colnames(X)[2] <- “superduper”
You can just use table(): > a <- table(numbers) > a numbers 4 5 23 34 43 54 56 65 67 324 435 453 456 567 657 2 1 2 2 1 1 2 1 2 1 3 1 1 1 1 Then you can subset it: > a[names(a)==435] 435 3 Or convert it into … Read more