r - data.table 在 rbind, R 后丢失因子排序

Question

当rbind ing 二 data.table对于有序因素，排序似乎丢失了:

dtb1 = data.table(id = factor(c("a", "b"), levels = c("a", "c", "b"), ordered=T), key="id")
dtb2 = data.table(id = factor(c("c"), levels = c("a", "c", "b"), ordered=T), key="id") 
test = rbind(dtb1, dtb2)
is.ordered(test$id)
#[1] FALSE

有什么想法或想法吗？

Answer 1

data.table做了一些花哨的步法，这意味着 data.table:::.rbind.data.table在 rbind 时调用在对象上调用，包括 data.tables . .rbind.data.table利用与 rbindlist 相关的加速，通过一些额外的检查来匹配名称等。
.rbind.data.table使用 c 处理因子列将它们组合起来(因此保留了 levels 属性)

# the relevant code is
l = lapply(seq_along(allargs[[1L]]), function(i) do.call("c", 
    lapply(allargs, "[[", i)))

在 base R使用 c以这种方式不保留“有序”属性，它甚至不返回一个因素!

例如(在 base R 中)

f <- factor(1:2, levels = 2:1, ordered=TRUE)
g <- factor(1:2, levels = 2:1, ordered=TRUE)
# it isn't ordered!
is.ordered(c(f,g))
# [1] FALSE
# no suprise as it isn't even a factor!
is.factor(c(f,g))
# [1] FALSE

然而 data.table有一个 S3 方法 c.factor ，用于确保返回因子并保留水平。不幸的是，此方法不保留ordered 属性。

getAnywhere('c.factor')
# A single object matching ‘c.factor’ was found
# It was found in the following places
#   namespace:data.table
# with value
# 
# function (...) 
# {
#     args <- list(...)
#     for (i in seq_along(args)) if (!is.factor(args[[i]])) 
#         args[[i]] = as.factor(args[[i]])
#     newlevels = unique(unlist(lapply(args, levels), recursive = TRUE, 
#         use.names = TRUE))
#     ind <- fastorder(list(newlevels))
#     newlevels <- newlevels[ind]
#     nm <- names(unlist(args, recursive = TRUE, use.names = TRUE))
#     ans = unlist(lapply(args, function(x) {
#         m = match(levels(x), newlevels)
#         m[as.integer(x)]
#     }))
    structure(ans, levels = newlevels, names = nm, class = "factor")
}
<bytecode: 0x073f7f70>
<environment: namespace:data.table

所以是的，这是一个错误。现在报告为 #5019 .