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为什么是匹配。调用有用吗?

call r

Usobi · 技术社区 · 9 年前

例如,在一些R函数的主体中 lm 我看到有人打电话给 match.call 作用正如其帮助页面所述, 在函数内部使用时 匹配.call 返回指定参数名称的调用 ; 这对于向另一个函数传递大量参数是有用的。

例如,在 流明 函数我们看到对该函数的调用 model.frame ...

function (formula, data, subset, weights, na.action, method = "qr", 
model = TRUE, x = FALSE, y = FALSE, qr = TRUE, singular.ok = TRUE, 
contrasts = NULL, offset, ...) 
{
  cl <- match.call()
  mf <- match.call(expand.dots = FALSE)
  m <- match(c("formula", "data", "subset", "weights", "na.action", 
      "offset"), names(mf), 0L)
  mf <- mf[c(1L, m)]

  mf$drop.unused.levels <- TRUE
  mf[[1L]] <- quote(stats::model.frame)
  mf <- eval(mf, parent.frame())
  ...

... 为什么这比直接打电话给 型号.框架 指定参数名称 我下一步做什么?

function (formula, data, subset, weights, na.action, method = "qr", 
model = TRUE, x = FALSE, y = FALSE, qr = TRUE, singular.ok = TRUE, 
contrasts = NULL, offset, ...) 
{
  mf <- model.frame(formula = formula, data = data,
                    subset = subset, weights = weights, subset = subset)
  ...

(注意 匹配.call 有另一个我没有讨论的用途,将调用存储在结果对象中。)

2 回复 | 直到 5 年前

BrodieG 8 年前

与此相关的一个原因是 match.call 捕获呼叫的语言而不进行评估,在这种情况下,它允许 lm 将某些“缺失”变量视为“可选”。考虑:

lm(x ~ y, data.frame(x=1:10, y=runif(10)))

Vs公司:

lm2 <- function (
  formula, data, subset, weights, na.action, method = "qr", 
  model = TRUE, x = FALSE, y = FALSE, qr = TRUE, singular.ok = TRUE, 
  contrasts = NULL, offset, ...
) {
  mf <- model.frame(
    formula = formula, data = data, subset = subset, weights = weights
  ) 
}
lm2(x ~ y, data.frame(x=1:10, y=runif(10)))
## Error in model.frame.default(formula = formula, data = data, subset = subset,  :
##   invalid type (closure) for variable '(weights)'

在里面 lm2 自从 weights “缺失”,但您仍在 weights=weights ,R尝试使用 stats::weights 显然不是预期的功能。你可以通过在打电话之前测试是否丢失来解决这个问题 model.frame ,但此时 匹配.call 看起来很不错。看看如果我们 debug 呼叫:

debug(lm2)
lm2(x ~ y, data.frame(x=1:10, y=runif(10)))
## debugging in: lm2(x ~ y, data.frame(x = 1:10, y = runif(10)))
## debug at #5: {
##     mf <- model.frame(formula = formula, data = data, subset = subset,
##         weights = weights)
## }
Browse[2]> match.call()
## lm2(formula = x ~ y, data = data.frame(x = 1:10, y = runif(10)))

匹配.call 根本不涉及缺少的参数。

您可以争辩说,可选参数应该通过默认值显式地设置为可选的,但这里并不是这样。

-1

Phil van Kleur 4 年前

这里有一个例子。在其中,calc_1是一个函数,它包含大量的数值参数,希望对它们进行加法和乘法运算。它将这项工作委托给calc_2,calc_2是一个辅助函数,它接受大部分这些参数。但calc_2还需要一些额外的参数(q到t),calc_1无法从其自身的实际参数提供这些参数。相反,它把它们当作额外的东西。

如果将对calc_2的调用编写为显示calc_1传递给它的所有内容,那么它将非常可怕。因此,我们假设如果calc_1和calc_2共享一个形式参数,那么它们会给它取相同的名称。这使得可以编写一个调用者来计算calc_1可以传递给calc_2的参数,构造一个调用,并输入额外的值来完成它。下面代码中的注释应该清楚说明这一点。

顺便说一句,只有%>%才需要库“tidyverse”和str_c(我用它定义了calc_2),以及一个断言的库“assertthat”。(虽然在一个现实的程序中,我会提出断言来检查论点。)

以下是输出:

> calc_1( a=1, b=11, c=2, d=22, e=3, f=33, g=4, h=44, i=5, j=55, k=6
+       , l=66, m=7, n=77, o=8, p=88 
+       )
[1] "87654321QRST"

下面是代码:

library( tidyverse )
library( rlang )
library( assertthat )


`%(%` <- call_with_extras
#
# This is the operator for calling
# a function with arguments passed
# from its parent, supplemented 
# with extras. See call_with_extras()
# below.


# A function with a very long
# argument list. It wants to call
# a related function which takes
# most of these arguments and
# so has a long argument list too.
# The second function takes some
# extra arguments.
#
calc_1 <- function( a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p )
{
  calc_2 %(% list( t = "T", q = "Q", s = "S", r = "R" )
  #
  # Call it with those extras, passing 
  # all the others that calc_2() needs
  # as well. %(% is my function for
  # doing so: see below.
}


# The function that we call above. It
# uses its own arguments q to t , as
# well as those from calc_1() .
#
calc_2 <- function( a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t )
{
  ( a + c * 10 + e * 100 + g * 1000 + i * 10000 + k * 100000 +
  m * 1000000 + o * 10000000 ) %>%
  str_c( q, r, s, t )
} 


# Calls function f2 . Passes f2 whichever
# arguments it needs from its caller. 
# Corresponding formals should have the
# same name in both. Also passes f2 extra
# arguments from the named list extra. 
# The names should have the same names as
# corresponding formals of f2 .
#
call_with_extras <- function( f2, extras )
{   
  f1_call <- match.call( sys.function(1), sys.call(1) )  
  # A call object.

  f1_actuals <- as.list( f1_call %>% tail(-1) ) 
  # Named list of f1's actuals.

  f1_formals <- names( f1_actuals )
  # Names of f1's formals.

  f2_formals <- names( formals( f2 ) )
  # Names of f2's formals.

  f2_formals_from_f1 <- intersect( f2_formals, f1_formals )
  # Names of f2's formals which f1 can supply.

  f2_formals_not_from_f1 <- setdiff( f2_formals, f1_formals )
  # Names of f2's formals which f1 can't supply.

  extra_formals <- names( extras ) 
  # Names of f2's formals supplied as extras.

  assert_that( setequal( extra_formals, f2_formals_not_from_f1 ) )
  # The last two should be equal.

  f2_actuals_from_f1 <- f1_actuals[ f2_formals_from_f1 ]
  # List of actuals which f1 can supply to f2.

  f2_actuals <- append( f2_actuals_from_f1, extras )
  # All f2's actuals.

  f2_call <- call2( f2, !!! f2_actuals )
  # Call to f2.

  eval( f2_call )
  # Run it.
}


# Test it.
#
calc_1( a=1, b=11, c=2, d=22, e=3, f=33, g=4, h=44, i=5, j=55, k=6
      , l=66, m=7, n=77, o=8, p=88 
      )