magrittr

Take this object as an example: expr <- substitute(mean(exp(sqrt(.)), .)) It is a nested list. I want to find every element that matches quote(.) . For example, magrittr 's solution matches only the first level of the call: dots <- c(FALSE, vapply(expr[-1], identical, quote(.), FUN.VALUE = logical(1))) dots [1] FALSE FALSE TRUE But I wanted to find every "." in an arbitrary nested list. In this particular case this would be these two dots: expr[[3]] expr[[2]][[2]][[2]] And then these dots should be replaced: expr[[3]] <- as.name("replacement") expr[[2]][[2]][[2]] <- as.name("replacement") expr

Is there a way to output the result of a pipeline at each step without doing it manually? (eg. without selecting and running only the selected chunks) I often find myself running a pipeline line-by-line to remember what it was doing or when I am developing some analysis. For example: library(dplyr) mtcars %>% group_by(cyl) %>% sample_frac(0.1) %>% summarise(res = mean(mpg)) # Source: local data frame [3 x 2] # # cyl res # 1 4 33.9 # 2 6 18.1 # 3 8 18.7 I'd to select and run: mtcars %>% group_by(cyl) and then... mtcars %>% group_by(cyl) %>% sample_frac(0.1) and so on... But selecting and CMD

I want to do the opposite of: Convert row names into first column Somewhere down the chain of pipes I would like to add row names to the data frame, for example, I would like to do the following actions using pipes: rownames(mtcars) <- as.character(1:nrow(mtcars)) so that it looks like: library(magrittr) mtcars <- mtcars %>% ...??? Note that @akrun's answer shows that if the pipe is used in conjection with a function that coerces the data frame into a tbl_df , the row names will be lost. You can use row.names<- : mtcars <- mtcars %>% `row.names<-`(as.character(1:nrow(mtcars))) Should work. As

I recently discovered the pipe operator %>% , which can make code more readable. Here is my MWE . library(dplyr) # for the pipe operator library(lsr) # for the cohensD function set.seed(4) # make it reproducible dat <- data.frame( # create data frame subj = c(1:6), pre = sample(1:6, replace = TRUE), post = sample(1:6, replace = TRUE) ) dat %>% select(pre, post) %>% sapply(., mean) # works as expected However, I struggle using the pipe operator in this particular case dat %>% select(pre, post) %>% cohensD(.$pre, .$post) # piping returns an error cohensD(dat$pre, dat$post) # classical way works