emmeans() 与 fct_na_value_to_level() 一起使用时的错误报告

Question

这是一个 MRE，显示 emmeans() 与 fct_na_value_to_level() 的使用不一致。 从我最初的代码中找出错误的原因并不容易;)

我更喜欢将其放在此处而不是 GitHub 上，因为我不知道哪个包可以修复该行为。

你觉得怎么样？ 谢谢你

# I was dealing with na values in some factors and I decided to use the fct_na_value_to_level() function
warpbreaks2 <- warpbreaks |> 
  dplyr::mutate(
    tension_na_value = forcats::fct_na_level_to_value(tension, "L"), # just to simulate some missing values in my datatset
    tension_na_level = forcats::fct_na_value_to_level(tension_na_value, "missing"), # I want to make the missing values explicit for my later model
    tension_na_level2 = forcats::fct_na_value_to_level(tension_na_value) # this is another try, without renaming the new explicit level
  )

# For your information
levels(warpbreaks2$tension_na_level)
#> [1] "M"       "H"       "missing"
levels(warpbreaks2$tension_na_level2)
#> [1] "M" "H" NA

# First try : everything is ok
lm(breaks ~ wool * tension_na_level, data = warpbreaks2) |> 
  emmeans::emmeans (~ wool | tension_na_level)
#> tension_na_level = M:
#>  wool emmean   SE df lower.CL upper.CL
#>  A      24.0 3.65 48     16.7     31.3
#>  B      28.8 3.65 48     21.4     36.1
#> 
#> tension_na_level = H:
#>  wool emmean   SE df lower.CL upper.CL
#>  A      24.6 3.65 48     17.2     31.9
#>  B      18.8 3.65 48     11.4     26.1
#> 
#> tension_na_level = missing:
#>  wool emmean   SE df lower.CL upper.CL
#>  A      44.6 3.65 48     37.2     51.9
#>  B      28.2 3.65 48     20.9     35.6
#> 
#> Confidence level used: 0.95

# Second try throws an error. the behaviour of emmeans() is not consistant, but maybe it's an issue with how fct_na_value_to_level() is written?
lm(breaks ~ wool * tension_na_level2, data = warpbreaks2) |> 
  emmeans::emmeans (~ wool | tension_na_level2)
#> Error in X[, nm, drop = FALSE]: indice hors limites

^{创建于 2024-07-08，使用 reprex v2.1.1}

会议信息

sessioninfo::session_info()
#> ─ Session info ───────────────────────────────────────────────────────────────
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#>  language (EN)
#>  collate  C
#>  ctype    fr_FR.UTF-8
#>  tz       Indian/Reunion
#>  date     2024-07-08
#>  pandoc   3.1.11 @ /usr/lib/rstudio/resources/app/bin/quarto/bin/tools/x86_64/ (via rmarkdown)
#> 
#> ─ Packages ───────────────────────────────────────────────────────────────────
#>  package      * version  date (UTC) lib source
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#> ──────────────────────────────────────────────────────────────────────────────

Answer 1

我想说这是 emmeans 包（第 81 行）中的一个错误，特别是

ref_grid

函数。

80号线：

if (is.factor(x) && !(nm %in% coerced$covariates)) 
      xlev[[nm]] = levels(factor(x))

上面的第二行删除了

x

级别中的 NA，因为

factor(x)

删除了 x 中的级别

NA

。然后，对象

xlev

不再包含级别

NA

，并且该函数会在轨道上进一步生成错误（具体来说，在第 11 行

emm_basis

调用的函数中）。我认为上面第81行应该很简单：

xlev[[nm]] = levels(x)

或者也许

xlev[[nm]] = levels(factor(x, exclude=NULL))

但这由软件包作者决定。