驯服 R 中函数调用的导出

Question

我正在打电话给 goseq::goseq。在那之后，命名空间（如果这是正确的词）都被打乱了。

我想这可以追溯到在包或其依赖项中自由使用 library(...)。

如何防止调用屏蔽我的命名空间中的任何对象？

...
Attaching package: ‘BiocGenerics’

The following objects are masked from ‘package:parallel’:

    clusterApply, clusterApplyLB, clusterCall, clusterEvalQ, clusterExport,
    clusterMap, parApply, parCapply, parLapply, parLapplyLB, parRapply, parSapply,
    parSapplyLB

The following objects are masked from ‘package:dplyr’:

    combine, intersect, setdiff, union

[... ETC, ETC ...]

Attaching package: ‘AnnotationDbi’

The following object is masked from ‘package:dplyr’:

    select

P.S.

如果有人可以用适当的技术术语来表述问题，我会考虑作为答案。例如，为什么 rlang::env_names(rlang::current_env()) 只显示我设置的变量，而不是浮动的对象，我在哪里可以找到它们？由于误报，我找不到好的建议。

Answer 1

作为概念验证，这似乎可行。感谢@KonradRudolph 的评论。

# install.packages("crunch")

# Wrap functions that do `library(...)`
# and force-detach all attached packages
unattach_packages <- function(f) {
  requireNamespace("crunch")

  # wrapper for f
  function(...) {
    base::with(
      crunch::ContextManager(
        enter = base::search,
        exit = function() {
          for (pkg in setdiff(base::search(), enter)) {
            base::detach(pkg, character.only = TRUE, force = TRUE)
          }
        },
        # this overwrites the variable `enter`, if existent
        as = "enter"
      ),
      {
        f(...)
      }
    )
  }
}

# Usage:
unattach_packages(function(x, y) { library(goseq); x + y })(1, 2)

驯服 R 中函数调用的导出

Taming exports of function call in R

scope

r