r - How does dplyr pass in non-string parameters -

a lot of time in dplyr, like:

mydat %>% select(., mycol1, mycol2, mycol3) 

however, mycol1, mycol2, , mycol3 not strings text in r. how function know convert string.

for instance, if do:

 dat <- data.frame(blue = rnorm(100), red= rnorm(100))  mysum <- function(dat, x, y){   browser()   return (sum(dat$x)+ sum(dat$y))   }   mysum(dat, blue, red) 

your function going deliver 0 because $ infix function uses non-standard evaluation of right-hand side argument. (as point out, non-standard evaluation favorite mechanism in @hadley's functions. me it's barrier, many people seems welcome strategy.) if write function in manner (using $) fail want:

 mysum(dat, blue, red) [1] 0   # wrong answer 

you said earlier that: "however, mycol1, mycol2, , mycol3 not strings text in r." guess trying mycol not enclosed in quotes , not character literal. in r such "text" (a sequence of unquoted characters) called 'symbol' or 'name'. (up point not talking dplyr.) if want write function deliver sum, (avoiding $ operation):

mysum <- function(dat, x, y){   return (sum(dat[[x]])+ sum(dat[[y]]))  }   mysum(dat, 'blue', 'red') [1] 19.16727 

if want retrieve argument name matched parameter need use deparse( substitute(.))-maneuver:

 dat <- data.frame(blue = rnorm(10), red= rnorm(10))   mysum2 <- function(dfrm, arg1, arg2){      a1 <- deparse(substitute(arg1)); a2 <-  deparse(substitute(arg2))       sum(dfrm[[a1]]) +sum(dfrm[[a2]]) }  mysum2(dat, blue, red) #[1] -0.5754979  mysum(dat, "blue", "red") #[1] -0.5754979 

if want see how @hadley does, type:

> dplyr::select function (.data, ...)  {     select_(.data, .dots = lazyeval::lazy_dots(...)) } <environment: namespace:dplyr> 

.... doesn't deliver answer, it? need try this:

 help(pac=lazyeval) 

... has accompanying vignette named "lazyeval::lazyeval" --> "lazyeval: new approach nse". hadley argues lazyeval functions superior traditional substitute because carry forward environments, , suppose agree.