为什么我有时必须在`data.frame()`中附加`.`来查找`do`中的命名参数？

Question

为什么这不起作用？

data.frame(x = rnorm(100)) %>% do(df = .)

错误信息:

Error in do_(.data, .dots = lazyeval::lazy_dots(...)) :
  argument ".data" is missing, with no default

相反,我必须封闭.在data.frame():

data.frame(x = rnorm(100)) %>% do(df = data.frame(.))

或者,这也有效:

data.frame(x = rnorm(100)) %>% do(., df = .)

当然,这个例子没有意义.但是在使用时将其保存data.frame为列表变量会很有帮助group_by.

这是一个似乎有关的更复杂的问题:

library("MatchIt")
n <- 5000
DF <- data.frame(
    x1 = rnorm(n),
    x2 = rbinom(n, 1, 0.5),
    group = rbinom(n, 1, 0.5),
    D = rbinom(n, 1, 0.5)) 

现在这会产生错误:

DF %>%
    group_by(group) %>%
    do(m = matchit(D ~ x1, data = ., exact = "x2"))

但是,它的工作原理,当我附上.的data.frame():

DF %>%
    group_by(group) %>%
    do(m = matchit(D ~ x1, data = data.frame(.), exact = "x2"))

我不确定第二个例子matchit是否相关,但在这两种情况下我都要将其括.起来data.frame().

sessionInfo()

> sessionInfo()
R version 3.1.1 (2014-07-10)
Platform: x86_64-apple-darwin13.4.0 (64-bit)

locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base

other attached packages:
[1] MatchIt_2.4-21 MASS_7.3-33    dplyr_0.4.1    Defaults_1.1-1

loaded via a namespace (and not attached):
[1] assertthat_0.1  DBI_0.3.1       lazyeval_0.1.10 magrittr_1.5    parallel_3.1.1  Rcpp_0.11.4     tools_3.1.1

Answer 1

差异来自于magrittr分裂链条的方式。

expr1 <- substitute(data.frame(x = rnorm(100)) %>% do(df = .))
expr2 <- substitute(data.frame(x = rnorm(100)) %>% do(df = (.)))

magrittr:::split_chain(expr1)
magrittr:::split_chain(expr2)