获取经计算得出的表达式，该表达式在`magrittr`调用的函数中为点 [英] Get expression that evaluated to dot in function called by `magrittr`
问题描述
我有一个函数x_expression()，它打印传递给参数x的表达式.
pacman::p_load(magrittr, rlang)
x_expression <- function(x) {
print(enquo(x))
}
y <- 1
x_expression(y)
#> <quosure>
#> expr: ^y
#> env: global
y %>% x_expression()
#> <quosure>
#> expr: ^.
#> env: 0x7ff27c36a610
因此您可以看到它知道已将y传递给它，但是将y与%>%一起传递时，该函数将返回打印内容..如果y已通过管道插入，是否有办法恢复它，或者它永远消失了?简而言之，我想要的是类似x_expression()的函数，但是在上述两种情况下都可以打印y的函数.
这个问题确实类似于获取通过R中的管道传递的数据框的名称，但是它稍微通用一些.这个人只想要数据框的名称，我想要表达式，无论它是什么.但是，相同的答案很可能适用于两者.我不喜欢这个几乎重复的问题的答案，也不喜欢这个答案的作者.
y并非永远消失"，因为管道调用了您的函数，并且它也了解y.有一种恢复y的方法，但是它需要遍历调用堆栈.要了解发生了什么，我们将使用?sys.frames和?sys.calls:
'sys.calls'和'sys.frames'分别给出所有活动调用和帧的配对列表，'sys.parents'返回每个帧的父帧索引的整数向量./p>
如果将它们撒在整个x_expression()中，我们可以看到从全局环境调用y %>% x_expression()时会发生什么:
x_expression <- function(x) {
print( enquo(x) )
# <quosure>
# expr: ^.
# env: 0x55c03f142828 <---
str(sys.frames())
# Dotted pair list of 9
# $ :<environment: 0x55c03f151fa0>
# $ :<environment: 0x55c03f142010>
# ...
# $ :<environment: 0x55c03f142828> <---
# $ :<environment: 0x55c03f142940>
str(sys.calls())
# Dotted pair list of 9
# $ : language y %>% x_expression() <---
# $ : language withVisible(eval(...
# ...
# $ : language function_list[[k]...
# $ : language x_expression(.)
}
我用<---突出显示了重要部分.请注意，enquo捕获的quores位于函数的父环境中(位于堆栈底部的第二个位置)，而了解y的管道调用始终位于堆栈的顶部.
有两种方法可以遍历堆栈. @MrFlick的答案和类似问题以及抽象句法树( AST):
# Recursively constructs Abstract Syntax Tree for a given expression
getAST <- function(ee) purrr::map_if(as.list(ee), is.call, getAST)
# Example: getAST( quote(a %>% b) )
# List of 3
# $ : symbol %>%
# $ : symbol a
# $ : symbol b
我们现在可以系统地将此功能应用于整个sys.calls()堆栈.目的是识别第一个元素为%>%的AST.然后，第二个元素将对应于管道的左侧(在a %>% b示例中为symbol a).如果有多个这样的AST，那么我们处于嵌套的%>%管道方案中.在这种情况下，列表中的最后一个AST在调用堆栈中将是最低的，并且最接近我们的函数.
x_expression2 <- function(x) {
sc <- sys.calls()
ASTs <- purrr::map( as.list(sc), getAST ) %>%
purrr::keep( ~identical(.[[1]], quote(`%>%`)) ) # Match first element to %>%
if( length(ASTs) == 0 ) return( enexpr(x) ) # Not in a pipe
dplyr::last( ASTs )[[2]] # Second element is the left-hand side
}
(次要注意:我使用enexpr()而不是enquo()来确保函数在管道内外的行为一致.由于sys.calls()遍历返回的是表达式，而不是等式，因此我们希望执行相同的操作在默认情况下也是如此.)
新功能非常强大，可以在其他功能中使用，包括嵌套的%>%管道:
x_expression2(y)
# y
y %>% x_expression2()
# y
f <- function() {x_expression2(v)}
f()
# v
g <- function() {u <- 1; u %>% x_expression2()}
g()
# u
y %>% (function(z) {w <- 1; w %>% x_expression2()}) # Note the nested pipes
# w
I have a function x_expression() which prints the expression passed to argument x.
pacman::p_load(magrittr, rlang)
x_expression <- function(x) {
print(enquo(x))
}
y <- 1
x_expression(y)
#> <quosure>
#> expr: ^y
#> env: global
y %>% x_expression()
#> <quosure>
#> expr: ^.
#> env: 0x7ff27c36a610
So you can see that it knows y was passed to it, but when y is piped in with %>%, the function returns prints .. Is there a way to recover the y in the case that it is piped in, or is it gone forever? In brief, what I want is a function like x_expression() but one that would print y in both cases above.
This question is indeed similar to Get name of dataframe passed through pipe in R, however it is slightly more general. This person just wants the name of the data frame, I want the expression, whatever it is. However, the same answer will likely apply to both. I don't like the answer of this near-duplicate question, nor does the author of that answer.
y is not "gone forever", because the pipe calls your function, and it also knows about y. There's a way to recover y, but it requires some traversal of the calling stack. To understand what's happening, we'll use ?sys.frames and ?sys.calls:
‘sys.calls’ and ‘sys.frames’ give a pairlist of all the active calls and frames, respectively, and ‘sys.parents’ returns an integer vector of indices of the parent frames of each of those frames.
If we sprinkle these throughout your x_expression(), we can see what happens when we call y %>% x_expression() from the global environment:
x_expression <- function(x) {
print( enquo(x) )
# <quosure>
# expr: ^.
# env: 0x55c03f142828 <---
str(sys.frames())
# Dotted pair list of 9
# $ :<environment: 0x55c03f151fa0>
# $ :<environment: 0x55c03f142010>
# ...
# $ :<environment: 0x55c03f142828> <---
# $ :<environment: 0x55c03f142940>
str(sys.calls())
# Dotted pair list of 9
# $ : language y %>% x_expression() <---
# $ : language withVisible(eval(...
# ...
# $ : language function_list[[k]...
# $ : language x_expression(.)
}
I highlighted the important parts with <---. Notice that the quosure captured by enquo lives in the parent environment of the function (second from the bottom of the stack), while the pipe call that knows about y is all the way at the top of the stack.
There's a couple of ways to traverse the stack. @MrFlick's answer to a similar question as well as this GitHub issue traverse the frames / environments from sys.frames(). Here, I will show an alternative that traverses sys.calls() and parses the expressions to find %>%.
The first piece of the puzzle is to define a function that converts an expression to its Abstract Sytax Tree(AST):
# Recursively constructs Abstract Syntax Tree for a given expression
getAST <- function(ee) purrr::map_if(as.list(ee), is.call, getAST)
# Example: getAST( quote(a %>% b) )
# List of 3
# $ : symbol %>%
# $ : symbol a
# $ : symbol b
We can now systematically apply this function to the entire sys.calls() stack. The goal is to identify ASTs where the first element is %>%; the second element will then correspond to the left-hand side of the pipe (symbol a in the a %>% b example). If there is more than one such AST, then we're in a nested %>% pipe scenario. In this case, the last AST in the list will be the lowest in the calling stack and closest to our function.
x_expression2 <- function(x) {
sc <- sys.calls()
ASTs <- purrr::map( as.list(sc), getAST ) %>%
purrr::keep( ~identical(.[[1]], quote(`%>%`)) ) # Match first element to %>%
if( length(ASTs) == 0 ) return( enexpr(x) ) # Not in a pipe
dplyr::last( ASTs )[[2]] # Second element is the left-hand side
}
(Minor note: I used enexpr() instead of enquo() to ensure consistent behavior of the function in and out of the pipe. Since sys.calls() traversal returns an expression, not a quosure, we want to do the same in the default case as well.)
The new function is pretty robust and works inside other functions, including nested %>% pipes:
x_expression2(y)
# y
y %>% x_expression2()
# y
f <- function() {x_expression2(v)}
f()
# v
g <- function() {u <- 1; u %>% x_expression2()}
g()
# u
y %>% (function(z) {w <- 1; w %>% x_expression2()}) # Note the nested pipes
# w
这篇关于获取经计算得出的表达式，该表达式在`magrittr`调用的函数中为点的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
