r - 为什么我在按 'with' 选择时得到 NA

Question

作为@allan-cameron 在 Most concise way to get a value from within a tibble 中出色回答的后续

如果我将一些 NA 添加到 band-instruments

bandy <- tibble(name = c("John", "Paul", "Keith", "Eric", NA), 
               plays = c("guitar", "bass", "guitar", NA, "kazoo")) 

with(bandy, plays[name == "John"])
# [1] "guitar" NA  

with(bandy, name[plays == "guitar"])
# [1] "John"  "Keith" NA  

有人可以解释为什么我要退回那些 NA 吗？避免返回的最优雅方式是什么？

Answer 1

我认为那些 NA 与 with 函数或 tibble 格式无关。参见示例:

x <- c(1, 2, NA)
x[x == 1]
#> [1]  1 NA

^{由 reprex package 创建于 2020-05-10 (v0.3.0)}

对于==函数的帮助页面我们可以阅读:

Missing values (NA) and NaN values are regarded as non-comparable even to themselves, so comparisons involving them will always result in NA.

从[函数的帮助页面我们可以读到:

When extracting, a numerical, logical or character NA index picks an unknown element and so returns NA in the corresponding element of a logical, integer, numeric, complex or character result, and NULL for a list. (It returns 00 for a raw result.)

所以这就是代码返回 NA 的原因。

IMO 避免这些 NA 的最简单方法如下:

library(tibble)
bandy <- tibble(name = c("John", "Paul", "Keith", "Eric", NA), 
                plays = c("guitar", "bass", "guitar", NA, "kazoo")) 

with(bandy, plays[name == "John" & !is.na(name)])
#> [1] "guitar"
with(bandy, name[plays == "guitar" & !is.na(plays)])
#> [1] "John"  "Keith"

^{由 reprex package 创建于 2020-05-10 (v0.3.0)}