复合函数中管道运算符的行为 purrr - dplyr

Question

purrr-dplyr 包中管道运算符的使用（简而言之）定义如下：

y%>%f(x,.,z) is the same as f(x,y,z)

我正在尝试使用管道运算符完成以下任务。首先，我向您展示了不使用管道的任务：

#####for reproducibility
set.seed(50)
z0<-factor(sample(c(letters[1:3],NA),100,replace = T))
###the task
rep(1,length(table(z0)))

现在我想用管道来做这个：

z0%>%table%>%rep(1,length(.))

但是结果不一样。管道运算符似乎无法处理对函数组合的正确分配。即

y%>%f(x,g(.)) should be the same as f(x,g(y))

所以，具体问题是是否可以做到

y%>%f(x,g(.))

提前感谢您的意见。

Answer 1

%>%实现了第一个参数规则，即如果.不是直接参数，它将前面的数据作为第一个参数传递给函数；在你的第二种情况下，rep 的参数是 1 和 length(.)，所以第一个参数规则生效；为避免这种情况，请使用 {} 将表达式括起来；您可以阅读更多相关信息 at Re-using the placeholder for attributes:

Re-using the placeholder for attributes

It is straight-forward to use the placeholder several times in a right-hand side expression. However, when the placeholder only appears in a nested expressions magrittr will still apply the first-argument rule. The reason is that in most cases this results more clean code.
 x %>% f(y = nrow(.), z = ncol(.)) is equivalent to f(x, y = nrow(x), z = ncol(x))
The behavior can be overruled by enclosing the right-hand side in braces:
x %>% {f(y = nrow(.), z = ncol(.))} is equivalent to f(y = nrow(x), z = ncol(x))

rep(1,length(table(z0)))
# [1] 1 1 1

等价于：

z0 %>% table %>% {rep(1,length(.))}
# [1] 1 1 1

复合函数中管道运算符的行为 purrr - dplyr

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