问题描述
似乎是一个基本问题，也许我只是遗漏了一些明显的东西......但是有没有办法pluck一个子列表(使用purrr)?更具体地说，这是一个初始列表:
Seems like a basic question and perhaps I'm just missing something obvious ... but is there any way to pluck a sublist (with purrr)?More specifically, here's an initial list:
l <- list(a = "foo", b = "bar", c = "baz")
我想返回一个只有元素 a 和 b 的新(子)列表.通常，我只会做基础"R 子列表:
And I want to return a new (sub-)list with only elements a and b.Normally, I'd just do 'base' R sub-listing:
l[c("a", "b")]
但这并没有提供 pluck 的良好 .default 处理.我的理解是 pluck 'replaces' [[，但是是否有一个 purrr 等价物来替换 [?
But this doesn't provide the nice .default handling of pluck.My understanding is that pluck 'replaces' [[, but is there a purrr equivalent for replacing [?
推荐答案
类似于 jzadra 的帖子，你也可以这样做:
Similar to jzadra's post, you could also do:
l %>% keep(names(.) %in% c("a","b"))
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