from r cookbook, where bp is your ggplot: Remove legend for a particular aesthetic (fill): bp + guides(fill=”none”) It can also be done when specifying the scale: bp + scale_fill_discrete(guide=”none”) This removes all legends: bp + theme(legend.position=”none”)
If it’s a list of string, just use the c() function : R> LL <- list(a=”tom”, b=”dick”) R> c(LL, c=”harry”) $a [1] “tom” $b [1] “dick” $c [1] “harry” R> class(LL) [1] “list” R> That works on vectors too, so do I get the bonus points? Edit (2015-Feb-01): This post is coming up on its … Read more
You’re having troubles because strsplit() returns a list which we then need to apply as.data.frame.list() to each element to get it into the proper format that dplyr requires. Even then it would still require a bit more work to get usable results. Long story short, it doesn’t seem like a suitable operation for do(). I … Read more
I haven’t figured out how to put the coalesce_by_column function inside the dplyr pipeline, but this works: coalesce_by_column <- function(df) { return(coalesce(df[1], df[2])) } df %>% group_by(A) %>% summarise_all(coalesce_by_column) ## A B C D E ## <dbl> <dbl> <dbl> <dbl> <dbl> ## 1 1 2 3 2 5 ## 2 2 4 5 3 4 … Read more
Good question, but in my opinion, this shouldn’t have caused a major debugging headache because it is documented quite clearly in multiple places in the manual page for aggregate. First, in the usage section: ## S3 method for class ‘formula’ aggregate(formula, data, FUN, …, subset, na.action = na.omit) Later, in the description: na.action: a function … Read more
This would be faster than an apply based solution (despite it’s cryptic construction): as.numeric(rowSums(`dim<-`(as.matrix(df) %in% vec, dim(df))) >= 1) [1] 1 0 1 0 Update — Some benchmarks Here, we can make up some bigger data to test on…. These benchmarks are on 100k rows. set.seed(1) nrow <- 100000 ncol <- 10 vec <- c(“a”, … Read more
What you have is a vector, not an array. You can use rollapply function from zoo package to get what you need. > x <- c(1, 2, 3, 10, 20, 30) > #library(zoo) > rollapply(x, 3, sum) [1] 6 15 33 60 Take a look at ?rollapply for further details on what rollapply does and … Read more
Try mixedsort from the “gtools” package: > # install.packages(“gtools”) ## Uncomment if not already installed > library(gtools) > mixedsort(cf) [1] “V51” “V108” “V116” “V120” “V155” “V217” “V327” “V440” “V446” “V457” “V477” If you don’t want to use mixedsort (not sure why one wouldn’t), and if your vector has a pretty consistent pattern (eg letters followed … Read more
Some statistical result / background (Link in the picture: Function to calculate R2 (R-squared) in R) Computational details Computations involved here is basically the computation of the variance-covariance matrix. Once we have it, results for all pairwise regression is just element-wise matrix arithmetic. The variance-covariance matrix can be obtained by R function cov, but functions … Read more
Here is a function that allows you to do this: library(lubridate) x <- mdy(c(“1/2/54″,”1/2/68″,”1/2/69″,”1/2/99″,”1/2/04”)) foo <- function(x, year=1968){ m <- year(x) %% 100 year(x) <- ifelse(m > year %% 100, 1900+m, 2000+m) x } Try it out: x [1] “2054-01-02 UTC” “2068-01-02 UTC” “1969-01-02 UTC” “1999-01-02 UTC” [5] “2004-01-02 UTC” foo(x) [1] “2054-01-02 UTC” “2068-01-02 … Read more