问题
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 fine
Why is it not possible to subset columns using the placeholder .in combination with $? Is it worthwhile to write this line using a pipe operator %>%, or does it complicate syntax? The classical way of writing this seems more concise.
回答1:
Since you're going from a bunch of data into one (row of) value(s), you're summarizing. in a dplyr pipeline you can then use the summarize function, within the summarize function you don't need to subset and can just call pre and post
Like so:
dat %>% select(pre, post) %>% summarize(CD = cohensD(pre, post))
(The select statement isn't actually necessary in this case, but I left it in to show how this works in a pipeline)
回答2:
This would work:
dat %>% select(pre, post) %>% {cohensD(.$pre, .$post)}
Wrapping the last call into curly braces makes it be treated like an expression and not a function call. When you pipe something into an expression, the . gets replaced as expected. I often use this trick to call a function which does not interface well with piping.
What is inside the braces happens to be a function call but could really be any expression of . .
回答3:
It doesn't work because the . operator has to be used directly as an argument, and not inside a nested function (like $...) in your call.
If you really want to use piping, you can do it with the formula interface, but with a little reshaping before (melt is from reshape2 package):
dat %>% select(pre, post) %>% melt %>% cohensD(value~variable, .)
#### [1] 0.8115027
来源：https://stackoverflow.com/questions/38474882/r-further-subset-a-selection-using-the-pipe-and-placeholder